[{"Name":"Introduction to Capacitors","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Calculating Capacitance First Method","Duration":"16m 16s","ChapterTopicVideoID":21494,"CourseChapterTopicPlaylistID":99472,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/21494.jpeg","UploadDate":"2020-04-21T13:40:52.3030000","DurationForVideoObject":"PT16M16S","Description":null,"MetaTitle":"Calculating Capacitance First Method: Video + Workbook | Proprep","MetaDescription":"Capacitors - Introduction to Capacitors. Watch the video made by an expert in the field. Download the workbook and maximize your learning.","Canonical":"https://www.proprep.uk/general-modules/all/physics-2-electricity-and-magnetism/capacitors/introduction-to-capacitors/vid22275","VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.935","Text":"Hello. In this lesson,"},{"Start":"00:01.935 ","End":"00:04.350","Text":"we\u0027ll be speaking about capacitance."},{"Start":"00:04.350 ","End":"00:08.175","Text":"Now, capacitance is denoted by the letter C,"},{"Start":"00:08.175 ","End":"00:13.110","Text":"and it\u0027s defined as q divided by v,"},{"Start":"00:13.110 ","End":"00:16.680","Text":"where q is the charge stored on the capacitor,"},{"Start":"00:16.680 ","End":"00:19.904","Text":"and v is the voltage across the capacitor."},{"Start":"00:19.904 ","End":"00:23.775","Text":"Now the funny thing about capacitance is that even though"},{"Start":"00:23.775 ","End":"00:27.480","Text":"it\u0027s defined by q divided by v,"},{"Start":"00:27.480 ","End":"00:30.540","Text":"a lot of the time our value for capacitance"},{"Start":"00:30.540 ","End":"00:34.405","Text":"is independent of both the charge stored on it,"},{"Start":"00:34.405 ","End":"00:36.575","Text":"and the voltage across it."},{"Start":"00:36.575 ","End":"00:39.620","Text":"The capacitance, in actual fact,"},{"Start":"00:39.620 ","End":"00:44.555","Text":"is due to the geometric shape of the capacitor."},{"Start":"00:44.555 ","End":"00:48.245","Text":"The capacitance is a function of geometry,"},{"Start":"00:48.245 ","End":"00:50.584","Text":"so that means the size of the capacitor,"},{"Start":"00:50.584 ","End":"00:54.145","Text":"the distance between the two capacitors plates."},{"Start":"00:54.145 ","End":"00:57.680","Text":"That means that I can know the capacitance of"},{"Start":"00:57.680 ","End":"01:02.480","Text":"my capacitor before I\u0027ve plugged it in to my circuit,"},{"Start":"01:02.480 ","End":"01:04.663","Text":"and seeing how much charge is stored on it,"},{"Start":"01:04.663 ","End":"01:07.210","Text":"and how much voltage is across it."},{"Start":"01:07.210 ","End":"01:11.539","Text":"In most questions that you\u0027ll see a capacitor is involved,"},{"Start":"01:11.539 ","End":"01:16.729","Text":"they\u0027ll ask you somewhere to work out the capacitance of that capacitor."},{"Start":"01:16.729 ","End":"01:19.955","Text":"There are two ways in order to solve this type of question."},{"Start":"01:19.955 ","End":"01:23.420","Text":"In this lesson, we\u0027re going to speak about the first way."},{"Start":"01:23.420 ","End":"01:25.000","Text":"Method Number 1,"},{"Start":"01:25.000 ","End":"01:31.235","Text":"is calculating the capacitance by the definition of what capacitance is,"},{"Start":"01:31.235 ","End":"01:39.575","Text":"where the definition is that capacitance is equal to q divided by v. Step Number 1 for"},{"Start":"01:39.575 ","End":"01:43.670","Text":"finding the capacitance is to assume"},{"Start":"01:43.670 ","End":"01:48.635","Text":"that there is a charge q on the plates of the capacitor."},{"Start":"01:48.635 ","End":"01:52.790","Text":"Now, usually we are speaking about a capacitor that has two plates,"},{"Start":"01:52.790 ","End":"01:58.705","Text":"but this is correct also if we\u0027re just speaking about one plates."},{"Start":"01:58.705 ","End":"02:05.970","Text":"Step Number 2 is to calculate the electric field between the two plates."},{"Start":"02:06.800 ","End":"02:14.100","Text":"We\u0027ll use either Gauss or Coulomb\u0027s law in order to find the E field between the plates."},{"Start":"02:14.210 ","End":"02:18.980","Text":"That is to calculate the voltage between the two plates."},{"Start":"02:18.980 ","End":"02:20.435","Text":"What is our voltage?"},{"Start":"02:20.435 ","End":"02:25.445","Text":"As we remember, our voltage is equal to the negative integral of"},{"Start":"02:25.445 ","End":"02:28.070","Text":"our electric field dot"},{"Start":"02:28.070 ","End":"02:34.240","Text":"dr. What we really wanted to get to was this value of the voltage between the plates,"},{"Start":"02:34.240 ","End":"02:35.720","Text":"but in order to calculate that,"},{"Start":"02:35.720 ","End":"02:37.685","Text":"we needed to calculate the electric fields."},{"Start":"02:37.685 ","End":"02:39.580","Text":"That\u0027s why we have step Number 2."},{"Start":"02:39.580 ","End":"02:43.640","Text":"Now this minus over here doesn\u0027t really matter because anyway,"},{"Start":"02:43.640 ","End":"02:47.974","Text":"we can see that our capacitance is the absolute value of the voltage."},{"Start":"02:47.974 ","End":"02:51.145","Text":"You can omit the minus, it doesn\u0027t really matter."},{"Start":"02:51.145 ","End":"02:56.705","Text":"Step 4, the final step is to substitute into the equation."},{"Start":"02:56.705 ","End":"02:57.949","Text":"We have a voltage,"},{"Start":"02:57.949 ","End":"03:00.110","Text":"we take the absolute value of a voltage,"},{"Start":"03:00.110 ","End":"03:02.555","Text":"and substitute it into the denominator,"},{"Start":"03:02.555 ","End":"03:04.130","Text":"and into the numerator,"},{"Start":"03:04.130 ","End":"03:05.990","Text":"we just sub in our q,"},{"Start":"03:05.990 ","End":"03:07.610","Text":"which we just guessed."},{"Start":"03:07.610 ","End":"03:10.340","Text":"Then we\u0027ll get our value for the capacitance."},{"Start":"03:10.340 ","End":"03:16.555","Text":"Now, usually our equation for voltage will have some q in the numerator."},{"Start":"03:16.555 ","End":"03:20.495","Text":"Once we substitute in our voltage into our capacitance,"},{"Start":"03:20.495 ","End":"03:22.955","Text":"our Qs, which are unknown,"},{"Start":"03:22.955 ","End":"03:24.290","Text":"will cancel out,"},{"Start":"03:24.290 ","End":"03:26.480","Text":"and that\u0027s great, that\u0027s what we want."},{"Start":"03:26.480 ","End":"03:30.090","Text":"Then we get our equation for the capacitance."},{"Start":"03:30.530 ","End":"03:36.540","Text":"Let\u0027s take now an example and see how to use these four steps."},{"Start":"03:36.950 ","End":"03:40.204","Text":"Here we have a capacitor,"},{"Start":"03:40.204 ","End":"03:43.175","Text":"which is called a parallel plate capacitor."},{"Start":"03:43.175 ","End":"03:46.470","Text":"This is a very common type of capacitor."},{"Start":"03:46.470 ","End":"03:51.635","Text":"We have two parallel plates of some conducting material,"},{"Start":"03:51.635 ","End":"03:57.295","Text":"which are located a distance d one from another."},{"Start":"03:57.295 ","End":"04:04.010","Text":"The surface area of each parallel plates is equal to S or A,"},{"Start":"04:04.010 ","End":"04:05.660","Text":"whatever you want to call it."},{"Start":"04:05.660 ","End":"04:09.540","Text":"Each one has the same surface area."},{"Start":"04:10.400 ","End":"04:14.215","Text":"I\u0027m going to change this a little bit."},{"Start":"04:14.215 ","End":"04:21.943","Text":"What I\u0027m going to say is that the surface area of each parallel plates is A,"},{"Start":"04:21.943 ","End":"04:26.710","Text":"and each side is of length S or whatever."},{"Start":"04:26.710 ","End":"04:29.225","Text":"In order to find the capacitance,"},{"Start":"04:29.225 ","End":"04:31.430","Text":"we have to say that d,"},{"Start":"04:31.430 ","End":"04:38.345","Text":"the distance between these two plates is much smaller than the side,"},{"Start":"04:38.345 ","End":"04:41.915","Text":"let\u0027s say that the side is of length s,"},{"Start":"04:41.915 ","End":"04:46.275","Text":"than the length of the side s. Or,"},{"Start":"04:46.275 ","End":"04:48.600","Text":"in other words, instead of s,"},{"Start":"04:48.600 ","End":"04:50.645","Text":"to make this a little bit easier,"},{"Start":"04:50.645 ","End":"04:53.480","Text":"we can say that d is much smaller than"},{"Start":"04:53.480 ","End":"04:58.410","Text":"the square root of a because that\u0027s the length of one side."},{"Start":"04:58.760 ","End":"05:01.545","Text":"This is our capacitor."},{"Start":"05:01.545 ","End":"05:05.705","Text":"Now let\u0027s go to Step Number 1 of our first method."},{"Start":"05:05.705 ","End":"05:10.385","Text":"We\u0027re going to assume that there is a charge q in the plates of the capacitor."},{"Start":"05:10.385 ","End":"05:12.980","Text":"Generally when we have two plates,"},{"Start":"05:12.980 ","End":"05:17.150","Text":"we\u0027ll assume that the charge on one of the plates is q,"},{"Start":"05:17.150 ","End":"05:21.730","Text":"and on the other plate will have a charge of minus q."},{"Start":"05:21.730 ","End":"05:26.750","Text":"There are a number of reasons why we choose one of our plates to"},{"Start":"05:26.750 ","End":"05:31.805","Text":"be of charge q and the opposite one to be of charge negative q."},{"Start":"05:31.805 ","End":"05:35.615","Text":"The main reason is that this way,"},{"Start":"05:35.615 ","End":"05:41.225","Text":"the electric field outside of the capacitor will therefore be equal to 0,"},{"Start":"05:41.225 ","End":"05:44.780","Text":"which means that we also have 0 capacitance outside."},{"Start":"05:44.780 ","End":"05:50.600","Text":"We only will have a voltage and an electric field between the two parallel plates."},{"Start":"05:50.600 ","End":"05:54.590","Text":"If we didn\u0027t have charges q and minus q,"},{"Start":"05:54.590 ","End":"05:58.730","Text":"then that would mean that there would be an electric field in space,"},{"Start":"05:58.730 ","End":"06:02.685","Text":"around it and the region outside of the parallel plates."},{"Start":"06:02.685 ","End":"06:07.650","Text":"That means that they\u0027ll also be some capacitance there and then the capacitance of"},{"Start":"06:07.650 ","End":"06:14.535","Text":"our capacitor will be dependent or so in what\u0027s going on on the outside."},{"Start":"06:14.535 ","End":"06:16.430","Text":"To avoid all of that,"},{"Start":"06:16.430 ","End":"06:20.285","Text":"we assume that there\u0027s a charge q on one plate,"},{"Start":"06:20.285 ","End":"06:23.200","Text":"a negative q on the other plate."},{"Start":"06:23.200 ","End":"06:26.115","Text":"Now let\u0027s go on to Step Number 2."},{"Start":"06:26.115 ","End":"06:28.489","Text":"What we\u0027re going to do is we\u0027re going to calculate"},{"Start":"06:28.489 ","End":"06:31.505","Text":"the electric field between the two plates."},{"Start":"06:31.505 ","End":"06:36.110","Text":"Now because I know that the distance between the two plates is significantly"},{"Start":"06:36.110 ","End":"06:42.050","Text":"smaller than the length of the sides of the plates,"},{"Start":"06:42.050 ","End":"06:48.270","Text":"that means that we can consider each plate to be an infinite plane."},{"Start":"06:48.410 ","End":"06:51.440","Text":"Now the electric field,"},{"Start":"06:51.440 ","End":"06:54.815","Text":"if we remember from our chapter on Gauss\u0027s Law,"},{"Start":"06:54.815 ","End":"07:00.200","Text":"the electric field of an infinite plane is equal to Sigma"},{"Start":"07:00.200 ","End":"07:06.270","Text":"divided by 2 Epsilon naught in the Z direction."},{"Start":"07:06.270 ","End":"07:09.335","Text":"We can see that the electric field for"},{"Start":"07:09.335 ","End":"07:15.490","Text":"each infinite plane is independent of the distance away from the plane."},{"Start":"07:15.490 ","End":"07:18.980","Text":"Also it\u0027s taking into account that our Sigma,"},{"Start":"07:18.980 ","End":"07:23.700","Text":"so our charged density per unit area is constant."},{"Start":"07:23.910 ","End":"07:28.015","Text":"We don\u0027t have a value for Sigma,"},{"Start":"07:28.015 ","End":"07:32.245","Text":"but we know that our plane has a charge of Q,"},{"Start":"07:32.245 ","End":"07:39.760","Text":"and we\u0027re assuming that our charge Q is distributed uniformly across our plane."},{"Start":"07:39.760 ","End":"07:43.180","Text":"In that case, we can say that Sigma,"},{"Start":"07:43.180 ","End":"07:46.015","Text":"which is the charge density per unit area,"},{"Start":"07:46.015 ","End":"07:50.780","Text":"is equal to the charge per the area."},{"Start":"07:50.910 ","End":"08:00.295","Text":"Now let\u0027s see what\u0027s going on in every region of space in our diagram over here,"},{"Start":"08:00.295 ","End":"08:03.565","Text":"where we have 1 infinite plane of positive charge"},{"Start":"08:03.565 ","End":"08:07.060","Text":"and another infinite plane of negative charge."},{"Start":"08:07.060 ","End":"08:15.310","Text":"The plane of positive charge is going to exert an electric field in this direction."},{"Start":"08:15.310 ","End":"08:17.830","Text":"However, the plane below,"},{"Start":"08:17.830 ","End":"08:26.650","Text":"which has a negative charge is going to exert an equal and oppositely electric field,"},{"Start":"08:26.650 ","End":"08:29.575","Text":"so the electric field is going to be equal,"},{"Start":"08:29.575 ","End":"08:31.970","Text":"but in the opposite direction."},{"Start":"08:31.970 ","End":"08:35.130","Text":"Because we can see that our electric field is"},{"Start":"08:35.130 ","End":"08:38.850","Text":"independent of the distance away from the plane,"},{"Start":"08:38.850 ","End":"08:43.810","Text":"we can see that these E fields really are equal and opposite,"},{"Start":"08:43.810 ","End":"08:47.305","Text":"which means that the total electric field"},{"Start":"08:47.305 ","End":"08:52.280","Text":"outside of the 2 planes is going to be equal to 0."},{"Start":"08:52.440 ","End":"08:56.470","Text":"Similarly, below the bottom,"},{"Start":"08:56.470 ","End":"09:01.060","Text":"infinite plane due to the negative Q we\u0027ll"},{"Start":"09:01.060 ","End":"09:07.585","Text":"have an electric field pointing upwards,"},{"Start":"09:07.585 ","End":"09:11.170","Text":"and due to the positively charged plane,"},{"Start":"09:11.170 ","End":"09:15.745","Text":"we\u0027ll have an equal and opposite electric field pointing downwards."},{"Start":"09:15.745 ","End":"09:20.540","Text":"Again, the E total here will be equal to 0."},{"Start":"09:21.300 ","End":"09:27.715","Text":"Now let\u0027s see what is going on between the 2 plates."},{"Start":"09:27.715 ","End":"09:36.010","Text":"Our negatively charged plane is exerting an electric field in the direction to"},{"Start":"09:36.010 ","End":"09:40.870","Text":"the plane and our positively charged plane is also"},{"Start":"09:40.870 ","End":"09:47.270","Text":"exerting electric field away from itself and to the negatively charged plane."},{"Start":"09:47.790 ","End":"09:54.175","Text":"We can see that we have 2 arrows of electric field"},{"Start":"09:54.175 ","End":"10:00.940","Text":"pointing towards this downwards direction towards the negatively charged plates,"},{"Start":"10:00.940 ","End":"10:06.175","Text":"where of course, 1 of these arrows is red,"},{"Start":"10:06.175 ","End":"10:15.175","Text":"due to the negatively charged infinite plane."},{"Start":"10:15.175 ","End":"10:22.000","Text":"That means that our E field between the 2 infinite planes is"},{"Start":"10:22.000 ","End":"10:28.780","Text":"going to be equal to 2 times the E field due to 1 infinite plane,"},{"Start":"10:28.780 ","End":"10:31.765","Text":"which is Sigma divided by 2 Epsilon,"},{"Start":"10:31.765 ","End":"10:36.130","Text":"so we can see that the E field between the 2 planes"},{"Start":"10:36.130 ","End":"10:41.095","Text":"is equal to Sigma divided by Epsilon naught and of course,"},{"Start":"10:41.095 ","End":"10:44.755","Text":"in the negative z direction this time,"},{"Start":"10:44.755 ","End":"10:47.350","Text":"just because of how we defined"},{"Start":"10:47.350 ","End":"10:52.670","Text":"which plane was positively charged and which one was negatively charged."},{"Start":"10:52.980 ","End":"10:58.255","Text":"If I flip around my Q and my negative Q,"},{"Start":"10:58.255 ","End":"11:00.895","Text":"then my 2 arrows will point upwards,"},{"Start":"11:00.895 ","End":"11:05.140","Text":"and then this will be in the positive z direction."},{"Start":"11:05.140 ","End":"11:07.345","Text":"It doesn\u0027t make a difference, in fact,"},{"Start":"11:07.345 ","End":"11:09.760","Text":"just to make this easier,"},{"Start":"11:09.760 ","End":"11:16.194","Text":"let\u0027s just do that and now we can erase these minus signs."},{"Start":"11:16.194 ","End":"11:23.140","Text":"Perfect. This is the E field between 2"},{"Start":"11:23.140 ","End":"11:30.880","Text":"negatively charged infinite planes when the bottom plane has a charge of positive Q,"},{"Start":"11:30.880 ","End":"11:34.585","Text":"and the top plane has a charge of negative Q,"},{"Start":"11:34.585 ","End":"11:37.615","Text":"so that was step Number 2."},{"Start":"11:37.615 ","End":"11:40.675","Text":"Now we\u0027re going to go on to step Number 3,"},{"Start":"11:40.675 ","End":"11:44.770","Text":"which is to calculate the voltage between the two plates."},{"Start":"11:44.770 ","End":"11:50.170","Text":"The way we\u0027re going to do that is we\u0027re going to integrate along"},{"Start":"11:50.170 ","End":"11:54.850","Text":"our electric field and of course the minus we said here specifically,"},{"Start":"11:54.850 ","End":"11:57.700","Text":"doesn\u0027t really matter because any way we just want"},{"Start":"11:57.700 ","End":"12:01.375","Text":"the absolute value of our voltage and also the minus"},{"Start":"12:01.375 ","End":"12:04.860","Text":"just dictates if we\u0027re integrating from"},{"Start":"12:04.860 ","End":"12:09.280","Text":"the top plane to the bottom or the bottom to the top."},{"Start":"12:09.470 ","End":"12:15.810","Text":"Because we can see that our electric field is in the z direction only,"},{"Start":"12:15.810 ","End":"12:20.290","Text":"so our dr is simply our roots,"},{"Start":"12:20.290 ","End":"12:28.220","Text":"our trajectory is going to be also along the z directory from one plane to the other."},{"Start":"12:29.460 ","End":"12:32.140","Text":"Let\u0027s do this."},{"Start":"12:32.140 ","End":"12:36.010","Text":"We\u0027re integrating along this green dotted line."},{"Start":"12:36.010 ","End":"12:39.505","Text":"The absolute value of our voltage,"},{"Start":"12:39.505 ","End":"12:43.510","Text":"so we don\u0027t need to incorporate the minus is going to be equal to the"},{"Start":"12:43.510 ","End":"12:47.740","Text":"integral of our electric field between the 2 planes,"},{"Start":"12:47.740 ","End":"12:51.640","Text":"which is equal to Sigma divided by Epsilon naughts in"},{"Start":"12:51.640 ","End":"12:56.905","Text":"the z direction and then dot dr, so our dr,"},{"Start":"12:56.905 ","End":"13:00.670","Text":"because our trajectory is only in the z direction,"},{"Start":"13:00.670 ","End":"13:08.095","Text":"so it\u0027s going to be equal to dz in the z direction."},{"Start":"13:08.095 ","End":"13:12.160","Text":"Our z hat.z hat is equal to 1."},{"Start":"13:12.160 ","End":"13:19.990","Text":"We\u0027re simply going to be integrating along Sigma divided by Epsilon naught dz."},{"Start":"13:19.990 ","End":"13:24.205","Text":"Now we have to plug in our bounds."},{"Start":"13:24.205 ","End":"13:27.700","Text":"Let\u0027s say that our bottom infinite plane is located at"},{"Start":"13:27.700 ","End":"13:33.550","Text":"z=0 and our top infinite plane is located at distance d away,"},{"Start":"13:33.550 ","End":"13:40.840","Text":"so at a height d. Now we can"},{"Start":"13:40.840 ","End":"13:44.020","Text":"integrate this and we\u0027ll get that this is equal to"},{"Start":"13:44.020 ","End":"13:48.670","Text":"Sigma d divided by Epsilon naughts and as we know,"},{"Start":"13:48.670 ","End":"13:51.820","Text":"our Sigma is equal to Q divided by A,"},{"Start":"13:51.820 ","End":"13:55.900","Text":"so we\u0027ll get that our absolute value of our voltage"},{"Start":"13:55.900 ","End":"14:00.385","Text":"is equal to Qd divided by Epsilon naught A."},{"Start":"14:00.385 ","End":"14:02.437","Text":"Now something useful to note,"},{"Start":"14:02.437 ","End":"14:06.820","Text":"here we can see that our voltage is equal to,"},{"Start":"14:06.820 ","End":"14:10.540","Text":"we can see this as simply equal to the electric field."},{"Start":"14:10.540 ","End":"14:13.780","Text":"Sigma divided by epsilon naught,"},{"Start":"14:13.780 ","End":"14:15.475","Text":"which was our electric field,"},{"Start":"14:15.475 ","End":"14:19.960","Text":"multiplied by d, the distance between the 2 plates."},{"Start":"14:19.960 ","End":"14:23.245","Text":"We can say that the voltage,"},{"Start":"14:23.245 ","End":"14:29.080","Text":"but only when we\u0027re dealing with a parallel plate capacitor,"},{"Start":"14:29.080 ","End":"14:32.725","Text":"because then our electric field is uniform,"},{"Start":"14:32.725 ","End":"14:38.590","Text":"is equal to the electric field multiplied by the distance between"},{"Start":"14:38.590 ","End":"14:45.190","Text":"the plates and this is only for a parallel plate capacitor."},{"Start":"14:45.190 ","End":"14:47.605","Text":"This is useful to remember,"},{"Start":"14:47.605 ","End":"14:51.535","Text":"and only for a parallel plate capacitor."},{"Start":"14:51.535 ","End":"14:55.930","Text":"Now we can go on to step Number 4, our final step."},{"Start":"14:55.930 ","End":"15:01.405","Text":"Where we\u0027re substituting in our V and Q to our equation."},{"Start":"15:01.405 ","End":"15:08.315","Text":"That means that our capacitance for a parallel plate capacitor is going to be equal to"},{"Start":"15:08.315 ","End":"15:12.550","Text":"Q divided by V."},{"Start":"15:12.550 ","End":"15:19.405","Text":"Our V is equal to Qd divided by Epsilon naught A,"},{"Start":"15:19.405 ","End":"15:21.610","Text":"which is simply equal to,"},{"Start":"15:21.610 ","End":"15:25.910","Text":"so we can already see now that our Q\u0027s canceled out,"},{"Start":"15:25.910 ","End":"15:31.055","Text":"so we can see that this is equal to Epsilon naught A"},{"Start":"15:31.055 ","End":"15:36.515","Text":"divided by D and this is our capacitance."},{"Start":"15:36.515 ","End":"15:43.205","Text":"I\u0027m reminding you that this is just the capacitance for a parallel plate capacitor."},{"Start":"15:43.205 ","End":"15:46.805","Text":"The capacitance, sorry,"},{"Start":"15:46.805 ","End":"15:51.665","Text":"for different types of capacitors such as cylindrical or spherical capacitors,"},{"Start":"15:51.665 ","End":"15:54.760","Text":"will be a completely different equation."},{"Start":"15:54.760 ","End":"16:00.050","Text":"This is the capacitance just for a parallel plate capacitor,"},{"Start":"16:00.050 ","End":"16:04.420","Text":"and this is how we work out the capacitance by Method 1,"},{"Start":"16:04.420 ","End":"16:06.895","Text":"which is by the definition."},{"Start":"16:06.895 ","End":"16:09.339","Text":"That\u0027s the end of this lesson."},{"Start":"16:09.339 ","End":"16:16.860","Text":"In future lesson, we\u0027re going to speak about the 2nd method for calculating capacitance."}],"ID":22275},{"Watched":false,"Name":"Capacitors with Dielectric Material","Duration":"4m 18s","ChapterTopicVideoID":21495,"CourseChapterTopicPlaylistID":99472,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.660","Text":"Hello. In this lesson we\u0027re going to speak about what happens to"},{"Start":"00:03.660 ","End":"00:09.420","Text":"a capacitor when we put a dielectric material between its plates."},{"Start":"00:09.420 ","End":"00:17.280","Text":"We saw in our previous lesson that capacitance C is equal to the absolute value of"},{"Start":"00:17.280 ","End":"00:20.370","Text":"the charge on the capacitor divided"},{"Start":"00:20.370 ","End":"00:25.530","Text":"by the potential difference between the capacitor\u0027s plate."},{"Start":"00:25.530 ","End":"00:33.735","Text":"If we add in between the plates a dielectric material with dielectric constant Epsilon r,"},{"Start":"00:33.735 ","End":"00:38.912","Text":"which is also sometimes called Kappa or denoted by Kappa,"},{"Start":"00:38.912 ","End":"00:42.980","Text":"then what we\u0027ll see is that the new capacitance of"},{"Start":"00:42.980 ","End":"00:48.485","Text":"the capacitor with the dielectric material is going to be equal to Kappa,"},{"Start":"00:48.485 ","End":"00:53.480","Text":"this constant, multiplied by the original capacitance."},{"Start":"00:53.480 ","End":"00:58.590","Text":"The capacitance of the capacitor without the dielectric material."},{"Start":"00:59.240 ","End":"01:03.980","Text":"Now I\u0027m going to show how we get to this equation."},{"Start":"01:03.980 ","End":"01:07.250","Text":"If you already know it or this doesn\u0027t interest you,"},{"Start":"01:07.250 ","End":"01:10.080","Text":"you can skip to the next video."},{"Start":"01:10.430 ","End":"01:16.000","Text":"Let\u0027s take a parallel plate capacitor,"},{"Start":"01:16.000 ","End":"01:18.585","Text":"which looks something like this."},{"Start":"01:18.585 ","End":"01:21.815","Text":"The capacitance is q divided by V,"},{"Start":"01:21.815 ","End":"01:26.910","Text":"where V is the negative integral of E.dr."},{"Start":"01:28.940 ","End":"01:31.130","Text":"Here\u0027s our equation."},{"Start":"01:31.130 ","End":"01:35.645","Text":"This is how we find the voltage or the potential difference between the plates."},{"Start":"01:35.645 ","End":"01:39.080","Text":"Now let\u0027s imagine that we take the same capacitor,"},{"Start":"01:39.080 ","End":"01:43.270","Text":"but we fill it with some dielectric material."},{"Start":"01:43.270 ","End":"01:51.125","Text":"Let\u0027s imagine that this dielectric material has a dielectric constant of Kappa."},{"Start":"01:51.125 ","End":"01:55.200","Text":"That means that the new E field in-between"},{"Start":"01:55.200 ","End":"02:00.620","Text":"the capacitor plates and due to us adding in this dielectric material,"},{"Start":"02:00.620 ","End":"02:04.400","Text":"is going to be equal to the original electric field"},{"Start":"02:04.400 ","End":"02:09.750","Text":"without the dielectric material divided by Kappa."},{"Start":"02:10.400 ","End":"02:13.657","Text":"This is our new electric field,"},{"Start":"02:13.657 ","End":"02:16.925","Text":"so what is our potential difference between the plates?"},{"Start":"02:16.925 ","End":"02:19.580","Text":"V tag, the new potential difference,"},{"Start":"02:19.580 ","End":"02:21.260","Text":"is equal to, as we know,"},{"Start":"02:21.260 ","End":"02:26.055","Text":"the negative integral on the electric field."},{"Start":"02:26.055 ","End":"02:29.680","Text":"The electric field now is our E tag electric field.dr."},{"Start":"02:31.990 ","End":"02:36.520","Text":"All of this is in absolute value because"},{"Start":"02:36.520 ","End":"02:43.835","Text":"our voltage is some kind of positive number."},{"Start":"02:43.835 ","End":"02:49.145","Text":"Then we can write this simply as the integral of E tag,"},{"Start":"02:49.145 ","End":"02:52.940","Text":"which is E_0 divided by"},{"Start":"02:52.940 ","End":"02:58.160","Text":"Kappa dr. Then we"},{"Start":"02:58.160 ","End":"03:03.368","Text":"can simply say that we can take our Kappa out because it is a constant,"},{"Start":"03:03.368 ","End":"03:09.160","Text":"so 1 divided by Kappa of E_0 dr or E naught,"},{"Start":"03:09.160 ","End":"03:13.745","Text":"which is equal to just our original voltage."},{"Start":"03:13.745 ","End":"03:16.730","Text":"If we didn\u0027t have the dielectric material,"},{"Start":"03:16.730 ","End":"03:19.250","Text":"so the voltage will be just the original voltage."},{"Start":"03:19.250 ","End":"03:25.200","Text":"We can say that this is equal to 1 divided by Kappa of V_0."},{"Start":"03:27.740 ","End":"03:30.590","Text":"We can see that the voltage when we have"},{"Start":"03:30.590 ","End":"03:35.710","Text":"a dielectric material becomes smaller by a factor of Kappa."},{"Start":"03:35.710 ","End":"03:39.080","Text":"Now let\u0027s see what the capacitance is."},{"Start":"03:39.080 ","End":"03:45.110","Text":"C tag, which is equal to q divided by the voltage, which is V tag,"},{"Start":"03:45.110 ","End":"03:53.970","Text":"that is going to be equal to q divided by 1 divided by Kappa V naught."},{"Start":"03:54.080 ","End":"04:04.531","Text":"That is simply going to be equal to Kappa multiplied by q divided by V naught,"},{"Start":"04:04.531 ","End":"04:08.640","Text":"and q divided by V naught is just our original capacitance."},{"Start":"04:08.640 ","End":"04:13.160","Text":"What we have is Kappa multiplied by C naught."},{"Start":"04:13.160 ","End":"04:15.725","Text":"There we got the exact same equation."},{"Start":"04:15.725 ","End":"04:19.050","Text":"That\u0027s the end of this lesson."}],"ID":22276},{"Watched":false,"Name":"Exercise 1","Duration":"27m 46s","ChapterTopicVideoID":21504,"CourseChapterTopicPlaylistID":99472,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.995","Text":"Hello, in this lesson,"},{"Start":"00:01.995 ","End":"00:04.635","Text":"we\u0027re going to be answering the following question."},{"Start":"00:04.635 ","End":"00:07.335","Text":"A cylindrical capacitor is given."},{"Start":"00:07.335 ","End":"00:09.135","Text":"It\u0027s a length."},{"Start":"00:09.135 ","End":"00:12.270","Text":"This over here is of length L,"},{"Start":"00:12.270 ","End":"00:15.855","Text":"its inner radius is a,"},{"Start":"00:15.855 ","End":"00:20.330","Text":"and its outer radius is b."},{"Start":"00:20.330 ","End":"00:23.165","Text":"Calculate the capacitance."},{"Start":"00:23.165 ","End":"00:27.000","Text":"What we\u0027re going to do is we\u0027re going to look at this capacitor,"},{"Start":"00:27.000 ","End":"00:33.270","Text":"which is made of 2 cylindrical shells that are 1 inside of the other."},{"Start":"00:33.270 ","End":"00:38.325","Text":"We\u0027re going to use the definition for capacitance in order to calculate it."},{"Start":"00:38.325 ","End":"00:43.620","Text":"Let\u0027s just look at the capacitor from this side."},{"Start":"00:43.620 ","End":"00:46.460","Text":"A bird\u0027s eye view or we\u0027re just going to look at"},{"Start":"00:46.460 ","End":"00:51.940","Text":"its cross-section and that will make it easier for us to do this calculation."},{"Start":"00:51.940 ","End":"00:54.945","Text":"Here we can see the cross-section."},{"Start":"00:54.945 ","End":"00:57.720","Text":"It\u0027s of length L, don\u0027t forget."},{"Start":"00:57.720 ","End":"01:01.620","Text":"We have our inner radius a and outer radius b."},{"Start":"01:01.620 ","End":"01:06.080","Text":"Let\u0027s write out the steps for how we solve"},{"Start":"01:06.080 ","End":"01:11.840","Text":"this question according to the first method which we\u0027ve already learned."},{"Start":"01:12.490 ","End":"01:19.670","Text":"Here is our recipe for calculating the capacitance according to the first method."},{"Start":"01:19.670 ","End":"01:26.355","Text":"The first step that we do is we assume that there is a charge Q on the plates."},{"Start":"01:26.355 ","End":"01:30.950","Text":"Let\u0027s imagine that we have a charge plus Q on"},{"Start":"01:30.950 ","End":"01:36.775","Text":"this plate and a charge of negative Q on this outer plate."},{"Start":"01:36.775 ","End":"01:42.515","Text":"Step Number 2 is to calculate the electric field between the plates."},{"Start":"01:42.515 ","End":"01:49.160","Text":"Let\u0027s choose an arbitrary point between the plates over here and the distance to"},{"Start":"01:49.160 ","End":"01:56.957","Text":"this point is r. Now what we\u0027re going to do is we\u0027re going to use Gauss\u0027s law,"},{"Start":"01:56.957 ","End":"02:01.820","Text":"so we\u0027re going to put some Gaussian envelope over here."},{"Start":"02:01.820 ","End":"02:06.020","Text":"Now, we will calculate the electric field."},{"Start":"02:06.020 ","End":"02:12.105","Text":"E multiplied by the surface area of this shell."},{"Start":"02:12.105 ","End":"02:17.140","Text":"That\u0027s 2 Pi r multiplied by L. It\u0027s"},{"Start":"02:17.140 ","End":"02:24.595","Text":"just some arbitrary length L that fits into this length over here."},{"Start":"02:24.595 ","End":"02:26.989","Text":"This is of course,"},{"Start":"02:26.989 ","End":"02:34.020","Text":"equal to the charge enclosed divided by Epsilon naught."},{"Start":"02:34.040 ","End":"02:37.110","Text":"What is our Q_in?"},{"Start":"02:37.110 ","End":"02:43.265","Text":"Our Q_in, is our charge distribution along here."},{"Start":"02:43.265 ","End":"02:53.060","Text":"Where we\u0027ve chosen some arbitrary length l over here."},{"Start":"02:53.280 ","End":"03:04.260","Text":"We have Sigma, which is this multiplied by the surface area."},{"Start":"03:04.260 ","End":"03:10.625","Text":"That is 2 Pi multiplied by the radius that is at,"},{"Start":"03:10.625 ","End":"03:11.840","Text":"which is a,"},{"Start":"03:11.840 ","End":"03:17.220","Text":"and then multiplied by L. What is sigma?"},{"Start":"03:17.220 ","End":"03:22.040","Text":"Sigma is the total charge on that plate, which is Q,"},{"Start":"03:22.040 ","End":"03:31.450","Text":"divided by the total surface area of this."},{"Start":"03:32.660 ","End":"03:35.870","Text":"Of course, when I was speaking about the l,"},{"Start":"03:35.870 ","End":"03:39.110","Text":"I was speaking just to make it a little bit more obvious in"},{"Start":"03:39.110 ","End":"03:43.135","Text":"the inner tube, not the outer tube."},{"Start":"03:43.135 ","End":"03:50.735","Text":"It\u0027s Q divided by 2 Pi multiplied by the radius, which is a,"},{"Start":"03:50.735 ","End":"03:58.965","Text":"multiplied by the entire length of the cylinder."},{"Start":"03:58.965 ","End":"04:06.175","Text":"Then all of this is multiplied by this over here, 2 Pi al."},{"Start":"04:06.175 ","End":"04:10.430","Text":"The 2 Pi a cancels out."},{"Start":"04:10.710 ","End":"04:20.965","Text":"We\u0027re left with Q_in Ql divided by L. Therefore,"},{"Start":"04:20.965 ","End":"04:29.330","Text":"E multiplied by 2 Pi rl is equal to Q_in,"},{"Start":"04:29.330 ","End":"04:38.820","Text":"which is Ql divided by L and then Epsilon naught over here."},{"Start":"04:38.820 ","End":"04:42.180","Text":"The l cancels out from both sides"},{"Start":"04:42.180 ","End":"04:45.830","Text":"and therefore we get that the electric field is equal to"},{"Start":"04:45.830 ","End":"04:54.000","Text":"Q divided by 2 Pi Epsilon naught Lr,"},{"Start":"04:54.000 ","End":"04:58.980","Text":"and of course, it\u0027s in the radial direction."},{"Start":"04:58.980 ","End":"05:02.150","Text":"Now we\u0027re on step Number 3 to calculate"},{"Start":"05:02.150 ","End":"05:05.225","Text":"the voltage where this is the equation for voltage."},{"Start":"05:05.225 ","End":"05:09.170","Text":"Now of course, because we want the absolute value of voltage,"},{"Start":"05:09.170 ","End":"05:12.340","Text":"we can just ignore this minus sign."},{"Start":"05:12.340 ","End":"05:15.915","Text":"We can say that our voltage,"},{"Start":"05:15.915 ","End":"05:17.475","Text":"let\u0027s just make it clear,"},{"Start":"05:17.475 ","End":"05:22.775","Text":"is equal to the integral of our electric field."},{"Start":"05:22.775 ","End":"05:25.385","Text":"Which is Q divided by"},{"Start":"05:25.385 ","End":"05:30.020","Text":"2 Pi Epsilon naught"},{"Start":"05:30.020 ","End":"05:35.760","Text":"Lr in the r direction.dr vector."},{"Start":"05:35.760 ","End":"05:37.125","Text":"What is dr vector?"},{"Start":"05:37.125 ","End":"05:40.305","Text":"It\u0027s dr in the r direction."},{"Start":"05:40.305 ","End":"05:43.185","Text":"R hat.r hat is just 1."},{"Start":"05:43.185 ","End":"05:47.840","Text":"We\u0027re just left with this because of course,"},{"Start":"05:47.840 ","End":"05:56.610","Text":"our electric field is following in the radial direction, like so."},{"Start":"05:56.610 ","End":"06:02.555","Text":"Then we\u0027re integrating along the bounds of"},{"Start":"06:02.555 ","End":"06:09.210","Text":"a to b from the inner radius until the outer radius."},{"Start":"06:09.980 ","End":"06:14.045","Text":"Then, we can see that this is just the integral."},{"Start":"06:14.045 ","End":"06:16.940","Text":"We can take out the constants."},{"Start":"06:16.940 ","End":"06:22.400","Text":"We have Q divided by 2 Pi Epsilon naught L,"},{"Start":"06:22.400 ","End":"06:28.205","Text":"and what we\u0027re doing is we\u0027re integrating from a to b on 1 divided by r,"},{"Start":"06:28.205 ","End":"06:30.380","Text":"dr. As we know,"},{"Start":"06:30.380 ","End":"06:34.405","Text":"the answer to this type of integral is Q divided by"},{"Start":"06:34.405 ","End":"06:42.225","Text":"2 Pi Epsilon naught L and then the integral of 1 divided by r dr is the ln."},{"Start":"06:42.225 ","End":"06:50.490","Text":"We have ln, and then we\u0027ll have ln of b divided by a."},{"Start":"06:51.170 ","End":"06:56.570","Text":"Because we\u0027ll have the constants multiplied by ln b minus ln a,"},{"Start":"06:56.570 ","End":"06:59.660","Text":"which according to the rules when dealing with ln,"},{"Start":"06:59.660 ","End":"07:01.980","Text":"is the same as this."},{"Start":"07:02.480 ","End":"07:05.240","Text":"The fourth and final step,"},{"Start":"07:05.240 ","End":"07:09.005","Text":"we\u0027re going to sub in the values to our equation."},{"Start":"07:09.005 ","End":"07:15.950","Text":"The capacitance is going to be equal to our charge,"},{"Start":"07:15.950 ","End":"07:21.155","Text":"which we said is equal to Q divided by our voltage,"},{"Start":"07:21.155 ","End":"07:24.295","Text":"which is Q divided by"},{"Start":"07:24.295 ","End":"07:31.835","Text":"2 Pi Epsilon naught L multiplied by ln of b divided by a."},{"Start":"07:31.835 ","End":"07:34.115","Text":"The Qs canceled out,"},{"Start":"07:34.115 ","End":"07:37.010","Text":"which is what we predicted will happen."},{"Start":"07:37.010 ","End":"07:39.170","Text":"What we get here,"},{"Start":"07:39.170 ","End":"07:41.645","Text":"this is the denominator in the denominator,"},{"Start":"07:41.645 ","End":"07:43.985","Text":"so it goes up to the numerator."},{"Start":"07:43.985 ","End":"07:52.290","Text":"2 Pi Epsilon naught L divided by ln of b divided by a."},{"Start":"07:53.120 ","End":"07:55.865","Text":"That is the answer to this question."},{"Start":"07:55.865 ","End":"08:00.170","Text":"But now let\u0027s just deal with another similar question,"},{"Start":"08:00.170 ","End":"08:02.945","Text":"dealing with dielectric materials."},{"Start":"08:02.945 ","End":"08:09.900","Text":"Let\u0027s say that 1 dielectric material is added over here."},{"Start":"08:09.980 ","End":"08:16.520","Text":"All of this, and let\u0027s say that this is a width d and it"},{"Start":"08:16.520 ","End":"08:22.130","Text":"has dielectric constant, let\u0027s say K_1."},{"Start":"08:22.130 ","End":"08:28.430","Text":"Then from d up until the end of the capacitor,"},{"Start":"08:28.430 ","End":"08:30.960","Text":"so up until radius b."},{"Start":"08:31.220 ","End":"08:38.580","Text":"We have another dielectric material of constant K_2."},{"Start":"08:39.440 ","End":"08:47.090","Text":"Now the question is to calculate the capacitance in this instance."},{"Start":"08:47.090 ","End":"08:50.930","Text":"We answer these types of questions exactly the same"},{"Start":"08:50.930 ","End":"08:54.500","Text":"as how we answered to get this answer."},{"Start":"08:54.500 ","End":"08:58.715","Text":"Number 1, I\u0027m just going to extend my d to the center over here."},{"Start":"08:58.715 ","End":"09:03.865","Text":"We\u0027ll make it up just a little bit neater answer."},{"Start":"09:03.865 ","End":"09:09.125","Text":"First of all, we assume that there\u0027s the charges on the 2 plates."},{"Start":"09:09.125 ","End":"09:12.980","Text":"Then we calculate the electric field between the plates."},{"Start":"09:12.980 ","End":"09:17.600","Text":"We did that over here when there was no dielectric material,"},{"Start":"09:17.600 ","End":"09:21.410","Text":"but now there is a dielectric material. What do we do?"},{"Start":"09:21.410 ","End":"09:26.250","Text":"We just plug in the different regions."},{"Start":"09:26.750 ","End":"09:32.660","Text":"What we do is we just multiply this, I\u0027ll write it in blue."},{"Start":"09:32.660 ","End":"09:36.020","Text":"We multiply it by 1 divided"},{"Start":"09:36.020 ","End":"09:41.150","Text":"by the dielectric constant in the region where it\u0027s irrelevant."},{"Start":"09:41.150 ","End":"09:47.885","Text":"We multiply this equation by 1 divided by K_1 when we\u0027re in the region"},{"Start":"09:47.885 ","End":"09:56.795","Text":"between a and d. When we\u0027re located somewhere in here."},{"Start":"09:56.795 ","End":"09:59.920","Text":"We multiply it the same equation over here."},{"Start":"09:59.920 ","End":"10:02.785","Text":"This multiplied by 1,"},{"Start":"10:02.785 ","End":"10:07.995","Text":"divided by K_2 when we\u0027re in this region over here."},{"Start":"10:07.995 ","End":"10:19.580","Text":"This region, so when we\u0027re located between d and the outer radius b."},{"Start":"10:20.390 ","End":"10:25.180","Text":"Now let\u0027s calculate the potential difference or the voltage."},{"Start":"10:25.180 ","End":"10:27.220","Text":"Let\u0027s just rub out what we have over here."},{"Start":"10:27.220 ","End":"10:30.260","Text":"We\u0027ll do the integral again."},{"Start":"10:30.450 ","End":"10:35.880","Text":"Again, we\u0027re calculating the absolute value of the voltage."},{"Start":"10:35.880 ","End":"10:37.905","Text":"We can forget the minus."},{"Start":"10:37.905 ","End":"10:41.960","Text":"It\u0027s equal to the integral again from a to b from"},{"Start":"10:41.960 ","End":"10:47.550","Text":"the inner cylinder to the outer cylinder of E.dr."},{"Start":"10:48.970 ","End":"10:53.075","Text":"Now we have to split the integral into regions."},{"Start":"10:53.075 ","End":"10:57.775","Text":"What we have is the first region from a to d,"},{"Start":"10:57.775 ","End":"11:04.310","Text":"and then our electric field is just Q divided by"},{"Start":"11:04.310 ","End":"11:16.040","Text":"2 Pi Epsilon naught L K_1"},{"Start":"11:16.040 ","End":"11:17.380","Text":"r dr."},{"Start":"11:17.380 ","End":"11:21.475","Text":"Remember the E is in the r hat direction and"},{"Start":"11:21.475 ","End":"11:28.080","Text":"dr vector is dr in the r hat direction and r hat.r hat is 1."},{"Start":"11:28.080 ","End":"11:29.715","Text":"We can write it like this."},{"Start":"11:29.715 ","End":"11:32.359","Text":"Then we add on the second integral,"},{"Start":"11:32.359 ","End":"11:37.135","Text":"which is from where we left off from d until b."},{"Start":"11:37.135 ","End":"11:41.460","Text":"Then the electric field is Q divided by"},{"Start":"11:41.460 ","End":"11:50.200","Text":"2 Pi Epsilon naught L K_2 r dr."},{"Start":"11:52.130 ","End":"11:55.995","Text":"Again, we have all of these constants."},{"Start":"11:55.995 ","End":"12:00.870","Text":"Then we\u0027re just integrating along 1 divided by idr,"},{"Start":"12:00.870 ","End":"12:02.790","Text":"which as we know is ln."},{"Start":"12:02.790 ","End":"12:06.825","Text":"What we can do is we can just,"},{"Start":"12:06.825 ","End":"12:08.805","Text":"I\u0027m just going to write this out already."},{"Start":"12:08.805 ","End":"12:16.710","Text":"Our common factors are q divided by 2Pi Epsilon naught L. Then we"},{"Start":"12:16.710 ","End":"12:25.635","Text":"multiply this by 1 divided by Kappa 1 of ln d divided by a,"},{"Start":"12:25.635 ","End":"12:33.830","Text":"plus 1 divided by Kappa 2 of ln b divided by"},{"Start":"12:33.830 ","End":"12:38.910","Text":"d. A nice and easy"},{"Start":"12:38.910 ","End":"12:44.430","Text":"sanity check that we can do is we can imagine that Kappa 1 is equal to Kappa 2."},{"Start":"12:44.430 ","End":"12:47.370","Text":"What does that mean? That means that we just have"},{"Start":"12:47.370 ","End":"12:52.209","Text":"1 dielectric material between these 2 plates."},{"Start":"12:52.340 ","End":"13:00.585","Text":"In that case, what we can see is that our voltage is independent of d,"},{"Start":"13:00.585 ","End":"13:06.270","Text":"which makes sense because d just is the distance or the point at which"},{"Start":"13:06.270 ","End":"13:13.680","Text":"Kappa 1 joins onto Kappa 2 where they\u0027re just different materials."},{"Start":"13:13.680 ","End":"13:16.200","Text":"Obviously, if there\u0027s the same material,"},{"Start":"13:16.200 ","End":"13:21.975","Text":"d is irrelevant and then we can see that the equation works out."},{"Start":"13:21.975 ","End":"13:25.230","Text":"Now we can just plug this into the equation."},{"Start":"13:25.230 ","End":"13:30.960","Text":"C tag, k the capacitance with"},{"Start":"13:30.960 ","End":"13:37.215","Text":"the 2 dielectric materials is equal to q divided by the voltage,"},{"Start":"13:37.215 ","End":"13:46.690","Text":"which is just q divided by 2Pi Epsilon naught L multiplied by all of this."},{"Start":"13:46.970 ","End":"13:51.255","Text":"Of course, the qs cancel out."},{"Start":"13:51.255 ","End":"13:55.800","Text":"Our new capacitance is simply equal to 2Pi"},{"Start":"13:55.800 ","End":"14:01.620","Text":"Epsilon naught L divided by,1 divided by Kappa 1 ln"},{"Start":"14:01.620 ","End":"14:08.070","Text":"of d divided by a plus 1 divided by Kappa 2 ln b divided"},{"Start":"14:08.070 ","End":"14:16.335","Text":"by d. That\u0027s the onset if we have 2 dielectric materials."},{"Start":"14:16.335 ","End":"14:21.090","Text":"Now, we can do the exact same calculation."},{"Start":"14:21.090 ","End":"14:27.585","Text":"If we have just 1 dielectric material inside the capacitor,"},{"Start":"14:27.585 ","End":"14:33.570","Text":"where it\u0027s constant, Kappa is dependent on the radius,"},{"Start":"14:33.570 ","End":"14:36.760","Text":"on its distance from the center."},{"Start":"14:37.790 ","End":"14:42.600","Text":"Now let\u0027s imagine that we have some Kappa"},{"Start":"14:42.600 ","End":"14:46.800","Text":"over here which is dependent on r. Let\u0027s say that"},{"Start":"14:46.800 ","End":"14:55.905","Text":"the equation for Kappa r=k naught multiplied by r divided by b."},{"Start":"14:55.905 ","End":"15:01.140","Text":"Now, what we can do is we can take the same exact electric field."},{"Start":"15:01.140 ","End":"15:03.870","Text":"All the steps to find the electric field is the same."},{"Start":"15:03.870 ","End":"15:09.660","Text":"But this time we just multiply it by this Kappa."},{"Start":"15:09.660 ","End":"15:14.940","Text":"What we\u0027re going to have is that the new electric field is going to be equal to"},{"Start":"15:14.940 ","End":"15:22.515","Text":"q divided by 2Pi Epsilon naught lr and Kappa."},{"Start":"15:22.515 ","End":"15:26.895","Text":"Kappa naught r, and then divided by b."},{"Start":"15:26.895 ","End":"15:30.300","Text":"The b goes on to the denominator over here."},{"Start":"15:30.300 ","End":"15:34.485","Text":"Of course, this is still in the radial direction."},{"Start":"15:34.485 ","End":"15:38.310","Text":"Then, we\u0027ll just plug this into"},{"Start":"15:38.310 ","End":"15:42.525","Text":"our equation for calculating the absolute value of the voltage."},{"Start":"15:42.525 ","End":"15:47.610","Text":"We\u0027re going to be integrating again from a to b on the electric field,"},{"Start":"15:47.610 ","End":"15:53.520","Text":"which is qb divided by 2Pi Epsilon naught L,"},{"Start":"15:53.520 ","End":"16:01.140","Text":"k naught r^2 dr. Then we just do this integration."},{"Start":"16:01.140 ","End":"16:07.575","Text":"Then of course, plugging into this equation and we\u0027ll get the capacitance."},{"Start":"16:07.575 ","End":"16:10.725","Text":"The final question that we\u0027ll ask is,"},{"Start":"16:10.725 ","End":"16:13.905","Text":"let\u0027s go back to this scenario over here,"},{"Start":"16:13.905 ","End":"16:18.570","Text":"where we had 2 different dielectric materials,"},{"Start":"16:18.570 ","End":"16:24.300","Text":"1 with constant k1 and 1 with constant k2."},{"Start":"16:24.300 ","End":"16:28.270","Text":"This was the electric field that we got."},{"Start":"16:28.270 ","End":"16:32.750","Text":"Now what we want to do is we want to"},{"Start":"16:32.750 ","End":"16:39.235","Text":"calculate the surface charge distribution in this case."},{"Start":"16:39.235 ","End":"16:44.775","Text":"Where will I have surface charge distribution?"},{"Start":"16:44.775 ","End":"16:49.530","Text":"This happens in areas where my electric field changes."},{"Start":"16:49.530 ","End":"16:57.785","Text":"Where do I suspect that I have a changing electric field or a jump in the electric field?"},{"Start":"16:57.785 ","End":"17:03.155","Text":"It\u0027s over here at radius a because here I have nothing."},{"Start":"17:03.155 ","End":"17:07.430","Text":"Then suddenly I have this dielectric material over here."},{"Start":"17:07.430 ","End":"17:13.325","Text":"The next area I suspect is between the 2 dielectric materials."},{"Start":"17:13.325 ","End":"17:15.620","Text":"In this line over here d,"},{"Start":"17:15.620 ","End":"17:18.770","Text":"I can see that there\u0027s 2 different dielectric materials."},{"Start":"17:18.770 ","End":"17:20.600","Text":"I can even see in my equation for"},{"Start":"17:20.600 ","End":"17:25.455","Text":"the electric field that there\u0027s some difference over here."},{"Start":"17:25.455 ","End":"17:27.360","Text":"In my third place,"},{"Start":"17:27.360 ","End":"17:29.760","Text":"it\u0027s obviously at this edge over here"},{"Start":"17:29.760 ","End":"17:33.300","Text":"because I have a dielectric material over here and then nothing."},{"Start":"17:33.300 ","End":"17:39.060","Text":"I\u0027m assuming that my electric field will change at this interface."},{"Start":"17:39.060 ","End":"17:43.122","Text":"Let\u0027s start at my first region and that\u0027s at a."},{"Start":"17:43.122 ","End":"17:49.320","Text":"The interface between nothing and my k1 dielectric material."},{"Start":"17:49.320 ","End":"17:52.875","Text":"Let\u0027s call this Sigma at a."},{"Start":"17:52.875 ","End":"17:58.170","Text":"What I\u0027m trying to see is the jump in the electric field at this point."},{"Start":"17:58.170 ","End":"18:02.940","Text":"What I\u0027m going to do is I\u0027m going to measure a point very close to a from"},{"Start":"18:02.940 ","End":"18:08.410","Text":"the inside and another point very close to a on the outside."},{"Start":"18:08.410 ","End":"18:16.080","Text":"This will be minus and this will be plus over here for this points very close."},{"Start":"18:16.080 ","End":"18:20.865","Text":"Sigma a is going to be equal to Epsilon naught multiplied by"},{"Start":"18:20.865 ","End":"18:29.410","Text":"the electric field at a plus-minus the electric field at a minus."},{"Start":"18:29.510 ","End":"18:33.150","Text":"What is this equal to?"},{"Start":"18:33.150 ","End":"18:38.790","Text":"I have Epsilon naught multiplied by the electric field in a plus."},{"Start":"18:38.790 ","End":"18:46.050","Text":"A plus is basically a plus some very small infinitesimal number."},{"Start":"18:46.050 ","End":"18:47.990","Text":"We can just call it a,"},{"Start":"18:47.990 ","End":"18:52.155","Text":"so that\u0027s in this region of the electric field."},{"Start":"18:52.155 ","End":"18:57.195","Text":"I have Epsilon naught multiplied by q divided by"},{"Start":"18:57.195 ","End":"19:02.610","Text":"2 Pi Epsilon naught l multiplied by r,"},{"Start":"19:02.610 ","End":"19:09.700","Text":"where my r over here is a because I\u0027m in this region of a."},{"Start":"19:10.820 ","End":"19:13.515","Text":"Then, of course,"},{"Start":"19:13.515 ","End":"19:17.520","Text":"divided by k1 as well."},{"Start":"19:17.520 ","End":"19:19.845","Text":"Because I\u0027m in this region."},{"Start":"19:19.845 ","End":"19:23.415","Text":"Then I subtract the electric field in a minus."},{"Start":"19:23.415 ","End":"19:25.620","Text":"That\u0027s just before my a."},{"Start":"19:25.620 ","End":"19:28.320","Text":"Here, of course, it\u0027s empty space."},{"Start":"19:28.320 ","End":"19:29.760","Text":"I don\u0027t have an electric field,"},{"Start":"19:29.760 ","End":"19:31.755","Text":"so I can subtract 0."},{"Start":"19:31.755 ","End":"19:36.270","Text":"Now I can cancel out my Epsilon naught over here and I can see that"},{"Start":"19:36.270 ","End":"19:42.520","Text":"my Sigma at a=q divided by 2 Pi Lk1a."},{"Start":"19:45.810 ","End":"19:49.405","Text":"Now, of course, this is our total Sigma,"},{"Start":"19:49.405 ","End":"19:55.210","Text":"our total charge distribution per unit area."},{"Start":"19:55.210 ","End":"19:57.205","Text":"But what we actually want,"},{"Start":"19:57.205 ","End":"20:01.270","Text":"we know that Sigma_total is equal to"},{"Start":"20:01.270 ","End":"20:08.755","Text":"the Sigma_bound plus our Sigma_free."},{"Start":"20:08.755 ","End":"20:12.710","Text":"What we want to calculate is our Sigma_free."},{"Start":"20:13.230 ","End":"20:20.425","Text":"Here Sigma_free is the charge on the plates of the capacitor,"},{"Start":"20:20.425 ","End":"20:24.490","Text":"so here it would our plus Q and minus Q."},{"Start":"20:24.490 ","End":"20:27.250","Text":"In this case specifically, it\u0027s easy,"},{"Start":"20:27.250 ","End":"20:30.895","Text":"I can just divide it by the total surface area."},{"Start":"20:30.895 ","End":"20:34.840","Text":"However, let\u0027s imagine that I don\u0027t know that."},{"Start":"20:34.840 ","End":"20:38.710","Text":"How could I calculate my Sigma_free?"},{"Start":"20:38.710 ","End":"20:47.480","Text":"My Sigma_free, I would be able to calculate via the jump in the free electric field."},{"Start":"20:47.520 ","End":"20:56.985","Text":"Sigma_free is equal to Epsilon_naught multiplied by Delta of E_naught."},{"Start":"20:56.985 ","End":"21:00.690","Text":"What is my Delta in E_naught?"},{"Start":"21:00.690 ","End":"21:07.525","Text":"I\u0027m taking again these points around this a, over here,"},{"Start":"21:07.525 ","End":"21:11.170","Text":"where I\u0027m using the electric field and imagining that"},{"Start":"21:11.170 ","End":"21:15.740","Text":"I didn\u0027t have the dielectric material inside."},{"Start":"21:16.110 ","End":"21:21.625","Text":"What we\u0027d have is Epsilon_naught multiplied by Q"},{"Start":"21:21.625 ","End":"21:27.340","Text":"divided by 2Pi Epsilon_naught L. Then our radius,"},{"Start":"21:27.340 ","End":"21:29.380","Text":"we\u0027re located at a,"},{"Start":"21:29.380 ","End":"21:30.640","Text":"multiplied by a,"},{"Start":"21:30.640 ","End":"21:33.175","Text":"and we\u0027re imagining there\u0027s no dielectric material,"},{"Start":"21:33.175 ","End":"21:34.705","Text":"there\u0027s no Kappa,"},{"Start":"21:34.705 ","End":"21:39.325","Text":"and then we subtract the electric field at this point over here,"},{"Start":"21:39.325 ","End":"21:43.225","Text":"where of course we don\u0027t have an electric field because it\u0027s empty,"},{"Start":"21:43.225 ","End":"21:45.025","Text":"so we subtract 0."},{"Start":"21:45.025 ","End":"21:48.490","Text":"Again, we can cancel off our Epsilon_naught."},{"Start":"21:48.490 ","End":"21:53.420","Text":"What we get is just Q divided by 2PiLa."},{"Start":"21:57.030 ","End":"21:59.890","Text":"This is our Sigma_free."},{"Start":"21:59.890 ","End":"22:03.505","Text":"Then, of course, our Sigma_B,"},{"Start":"22:03.505 ","End":"22:12.565","Text":"our bound charge is just going to be equal to Sigma_total minus Sigma_free."},{"Start":"22:12.565 ","End":"22:20.620","Text":"Now let\u0027s calculate our Sigma_total at point"},{"Start":"22:20.620 ","End":"22:29.755","Text":"d. Now we\u0027re going to take 2 points very close to d,"},{"Start":"22:29.755 ","End":"22:31.360","Text":"1 point in here,"},{"Start":"22:31.360 ","End":"22:32.680","Text":"this will be d minus,"},{"Start":"22:32.680 ","End":"22:33.850","Text":"and 1 point over here,"},{"Start":"22:33.850 ","End":"22:35.830","Text":"which is d plus."},{"Start":"22:35.830 ","End":"22:41.680","Text":"We can see from the image that d minus is in the region of Kappa_1,"},{"Start":"22:41.680 ","End":"22:46.180","Text":"and d plus is in the region of Kappa_2."},{"Start":"22:46.180 ","End":"22:52.000","Text":"Our Sigma at d will be equal to Epsilon_naught multiplied"},{"Start":"22:52.000 ","End":"22:59.800","Text":"by E at d plus minus E at d minus."},{"Start":"22:59.800 ","End":"23:06.205","Text":"This will be equal to Epsilon_naught multiplied by"},{"Start":"23:06.205 ","End":"23:15.640","Text":"Q divided by 2Pi Epsilon_naught L multiplied by our radius."},{"Start":"23:15.640 ","End":"23:19.030","Text":"Because these 2 points are so close to our point d,"},{"Start":"23:19.030 ","End":"23:24.145","Text":"we can just say that our radius is d. Then we have"},{"Start":"23:24.145 ","End":"23:32.240","Text":"1 divided by Kappa_2 minus 1 divided by Kappa_1."},{"Start":"23:33.780 ","End":"23:40.780","Text":"Then of course we can cancel out our Epsilon_naught and then we\u0027re left with this."},{"Start":"23:40.780 ","End":"23:42.745","Text":"This is of course, just like before,"},{"Start":"23:42.745 ","End":"23:49.645","Text":"equal to our bound charge distribution plus our free charge distribution."},{"Start":"23:49.645 ","End":"23:53.365","Text":"Now specifically, at point d over here,"},{"Start":"23:53.365 ","End":"23:56.800","Text":"our Sigma_free is equal to 0."},{"Start":"23:56.800 ","End":"23:59.725","Text":"Why is it equal to 0?"},{"Start":"23:59.725 ","End":"24:07.495","Text":"It\u0027s equal to 0 because our free charges are given by our conductor."},{"Start":"24:07.495 ","End":"24:10.765","Text":"Which is located at a or at b."},{"Start":"24:10.765 ","End":"24:15.550","Text":"Our conductor has the free charges plus Q and minus Q."},{"Start":"24:15.550 ","End":"24:18.640","Text":"However, at this interchange d,"},{"Start":"24:18.640 ","End":"24:23.995","Text":"we have just the connection between 2 dielectric materials,"},{"Start":"24:23.995 ","End":"24:27.415","Text":"which means that we only have bound charges."},{"Start":"24:27.415 ","End":"24:28.990","Text":"Remember in a dielectric,"},{"Start":"24:28.990 ","End":"24:30.880","Text":"we only have bound charges,"},{"Start":"24:30.880 ","End":"24:33.310","Text":"there\u0027s no way for us to add in"},{"Start":"24:33.310 ","End":"24:39.115","Text":"free charges because it\u0027s not connected to anything, it\u0027s a dielectric."},{"Start":"24:39.115 ","End":"24:42.445","Text":"In this case, our Sigma_free is equal to 0,"},{"Start":"24:42.445 ","End":"24:46.165","Text":"and our Sigma_total is also in this case,"},{"Start":"24:46.165 ","End":"24:48.830","Text":"equal to our Sigma_bound."},{"Start":"24:50.790 ","End":"24:56.695","Text":"Another way to try and calculate the Sigma_free,"},{"Start":"24:56.695 ","End":"24:58.780","Text":"you could do what we did previously,"},{"Start":"24:58.780 ","End":"25:01.720","Text":"which is to try and find the jump in"},{"Start":"25:01.720 ","End":"25:06.620","Text":"the free electric field and you\u0027ll get that that is equal to 0."},{"Start":"25:07.860 ","End":"25:12.400","Text":"Now let\u0027s quickly that we understand the idea."},{"Start":"25:12.400 ","End":"25:17.620","Text":"Calculate our Sigma_total at b."},{"Start":"25:17.620 ","End":"25:21.280","Text":"Over here, so we\u0027re taking 2 points close to B,"},{"Start":"25:21.280 ","End":"25:22.945","Text":"these 2 points over here,"},{"Start":"25:22.945 ","End":"25:27.820","Text":"so this is equal to Epsilon_naught multiplied by the jump in the electric field."},{"Start":"25:27.820 ","End":"25:34.405","Text":"E at b plus minus E at b minus,"},{"Start":"25:34.405 ","End":"25:36.490","Text":"which is equal to,"},{"Start":"25:36.490 ","End":"25:40.705","Text":"so we have E at b plus is this over here."},{"Start":"25:40.705 ","End":"25:44.500","Text":"We can see that it\u0027s in empty space over here,"},{"Start":"25:44.500 ","End":"25:47.095","Text":"so the electric field is equal to 0 over there,"},{"Start":"25:47.095 ","End":"25:49.540","Text":"minus E and b minus,"},{"Start":"25:49.540 ","End":"25:58.585","Text":"so minus Epsilon_naught multiplied by Q divided by 2Pi Epsilon_naught L,"},{"Start":"25:58.585 ","End":"26:00.520","Text":"our radius is this over here."},{"Start":"26:00.520 ","End":"26:03.955","Text":"Which is so close to b that we might as well write b."},{"Start":"26:03.955 ","End":"26:06.070","Text":"Then of course multiplied by,"},{"Start":"26:06.070 ","End":"26:08.920","Text":"we\u0027re in the region of Kappa_2 over here,"},{"Start":"26:08.920 ","End":"26:11.650","Text":"so divided by Kappa_2."},{"Start":"26:11.650 ","End":"26:17.920","Text":"This is of course, equal to our Sigma_bound plus our Sigma_free."},{"Start":"26:17.920 ","End":"26:26.840","Text":"Our Sigma_free, we\u0027re just going to calculate the jump in the free electric field."},{"Start":"26:32.250 ","End":"26:37.825","Text":"The free electric field out here is equal to 0 because we\u0027re in empty space"},{"Start":"26:37.825 ","End":"26:44.185","Text":"minus Epsilon_naught multiplied by Q,"},{"Start":"26:44.185 ","End":"26:51.415","Text":"divided by the electric field if we didn\u0027t have the dielectric material inside."},{"Start":"26:51.415 ","End":"26:55.030","Text":"That just means that we don\u0027t have Kappa_2, so we just write this."},{"Start":"26:55.030 ","End":"27:01.285","Text":"So divided by 2Pi Epsilon_naught L and the radius b."},{"Start":"27:01.285 ","End":"27:05.020","Text":"Of course, we can cancel out the Epsilon_naughts."},{"Start":"27:05.020 ","End":"27:08.200","Text":"Therefore, we can scroll down a little bit more."},{"Start":"27:08.200 ","End":"27:18.980","Text":"We\u0027ll get that our Sigma_induced or our Sigma_bound is equal to Q divided by 2PibL,"},{"Start":"27:20.940 ","End":"27:23.170","Text":"this is from here,"},{"Start":"27:23.170 ","End":"27:28.190","Text":"and then we have 1 minus 1 divided by Kappa_2."},{"Start":"27:29.160 ","End":"27:37.130","Text":"Sigma_bound is obviously equal to Sigma_total minus our Sigma_f."},{"Start":"27:37.260 ","End":"27:42.250","Text":"Sigma_total is this minus our Sigma_f,"},{"Start":"27:42.250 ","End":"27:44.455","Text":"which is this, and we get this."},{"Start":"27:44.455 ","End":"27:47.540","Text":"That\u0027s the end of this lesson."}],"ID":22285},{"Watched":false,"Name":"Equation for Parallel Plate Capacitors","Duration":"2m 52s","ChapterTopicVideoID":21506,"CourseChapterTopicPlaylistID":99472,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.875","Text":"Hello. In this lesson,"},{"Start":"00:01.875 ","End":"00:04.905","Text":"we\u0027re going to be learning about the equation for"},{"Start":"00:04.905 ","End":"00:09.220","Text":"the electric field in a parallel plate capacitor."},{"Start":"00:10.070 ","End":"00:13.080","Text":"In a parallel plate capacitor,"},{"Start":"00:13.080 ","End":"00:20.025","Text":"we know that the electric field is like the electric field of infinite planes,"},{"Start":"00:20.025 ","End":"00:22.125","Text":"when dealing with infinite planes."},{"Start":"00:22.125 ","End":"00:29.100","Text":"What does that mean? That means that if I look at any point within the capacitor,"},{"Start":"00:29.100 ","End":"00:33.190","Text":"the electric field will be the same."},{"Start":"00:34.730 ","End":"00:38.775","Text":"We have this constant electric field."},{"Start":"00:38.775 ","End":"00:44.720","Text":"Now I want to calculate the voltage or the potential difference between the 2 plates."},{"Start":"00:44.720 ","End":"00:49.295","Text":"Of course, I\u0027m calculating the absolute value of the voltage,"},{"Start":"00:49.295 ","End":"00:52.670","Text":"so I can ignore the minus sign."},{"Start":"00:52.670 ","End":"00:56.510","Text":"This will be equal to the integral e.dr,"},{"Start":"00:56.510 ","End":"01:02.265","Text":"so I have to find some route."},{"Start":"01:02.265 ","End":"01:07.395","Text":"I can see that I have this straight rout over here"},{"Start":"01:07.395 ","End":"01:12.870","Text":"from 1 plate to the next plate, just straight up."},{"Start":"01:13.130 ","End":"01:19.285","Text":"Let\u0027s say that this axis over here is the y-axis."},{"Start":"01:19.285 ","End":"01:29.190","Text":"Let\u0027s say that the distance between the plates is some distance d. Then once we do e.dr,"},{"Start":"01:29.190 ","End":"01:35.280","Text":"where dr is just the distance in the y-direction."},{"Start":"01:35.280 ","End":"01:41.050","Text":"What we\u0027ll have is an integral on Edy,"},{"Start":"01:41.050 ","End":"01:43.320","Text":"once we plug in all of that."},{"Start":"01:43.320 ","End":"01:46.280","Text":"Our electric field, as we just said,"},{"Start":"01:46.280 ","End":"01:49.460","Text":"because we\u0027re dealing with infinite planes."},{"Start":"01:49.460 ","End":"01:52.640","Text":"Our electric field was constant. That\u0027s what we said."},{"Start":"01:52.640 ","End":"01:55.415","Text":"We can take it out of the integration sign."},{"Start":"01:55.415 ","End":"01:59.680","Text":"We\u0027ll get an integral on Edy."},{"Start":"01:59.680 ","End":"02:01.850","Text":"Of course we\u0027re going from,"},{"Start":"02:01.850 ","End":"02:08.030","Text":"let\u0027s say that this is at 0 up until a distance d. All of this rout,"},{"Start":"02:08.030 ","End":"02:11.735","Text":"which we\u0027ve said as a distance d. What we\u0027ll get,"},{"Start":"02:11.735 ","End":"02:17.744","Text":"is that our potential difference between the 2 plates is simply"},{"Start":"02:17.744 ","End":"02:24.180","Text":"E multiplied by the plate separation d. Then,"},{"Start":"02:24.180 ","End":"02:26.760","Text":"we get that v is equal to E.d,"},{"Start":"02:26.760 ","End":"02:32.360","Text":"and then we can just plug this into the equation for calculating the capacitance."},{"Start":"02:32.360 ","End":"02:39.320","Text":"Remember that this is only when we\u0027re dealing with parallel plate capacitor."},{"Start":"02:39.320 ","End":"02:41.614","Text":"If it\u0027s not a parallel plate capacitor,"},{"Start":"02:41.614 ","End":"02:43.910","Text":"then the E field won\u0027t be constant,"},{"Start":"02:43.910 ","End":"02:47.405","Text":"and so you can take it out of the integration sign,"},{"Start":"02:47.405 ","End":"02:50.230","Text":"and then of course, you won\u0027t get this answer."},{"Start":"02:50.230 ","End":"02:53.230","Text":"That\u0027s the end of this lesson."}],"ID":22287},{"Watched":false,"Name":"Exercise 2","Duration":"31m 16s","ChapterTopicVideoID":21505,"CourseChapterTopicPlaylistID":99472,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.100","Text":"Hello. In this lesson,"},{"Start":"00:02.100 ","End":"00:06.360","Text":"we\u0027re going to be learning about the second method to calculate capacitance."},{"Start":"00:06.360 ","End":"00:09.089","Text":"However, in order to do this, first,"},{"Start":"00:09.089 ","End":"00:13.975","Text":"we\u0027re going to explain how to connect capacitors."},{"Start":"00:13.975 ","End":"00:21.239","Text":"Let\u0027s begin with connecting the capacitors in series."},{"Start":"00:21.239 ","End":"00:30.455","Text":"Now what we want to do is we want to take these 2 capacitors in series and"},{"Start":"00:30.455 ","End":"00:40.230","Text":"form 1 capacitor that has a total capacitance that is equal to these 2 capacitors."},{"Start":"00:40.230 ","End":"00:44.990","Text":"In other words, that instead of having these 2 capacitors, in my circuit,"},{"Start":"00:44.990 ","End":"00:49.970","Text":"I could replace it with this capacitor of capacitance C_T,"},{"Start":"00:49.970 ","End":"00:56.460","Text":"and my circuit would be mathematically the exact same thing."},{"Start":"00:56.510 ","End":"01:01.939","Text":"Now I\u0027m going to give an equation and the important thing to remember with this equation"},{"Start":"01:01.939 ","End":"01:07.565","Text":"is that we can only use it if the charge is on the 2 capacitors identical."},{"Start":"01:07.565 ","End":"01:11.690","Text":"In other words, if on this plate we have a charge plus Q,"},{"Start":"01:11.690 ","End":"01:12.995","Text":"and on this plate,"},{"Start":"01:12.995 ","End":"01:14.791","Text":"we have minus Q,"},{"Start":"01:14.791 ","End":"01:19.190","Text":"that means that on this plate we also have to have a charge of plus Q,"},{"Start":"01:19.190 ","End":"01:22.375","Text":"and on this plate, a charge of minus Q."},{"Start":"01:22.375 ","End":"01:24.965","Text":"This is very important to remember,"},{"Start":"01:24.965 ","End":"01:32.370","Text":"and that\u0027s the condition that we need in order to use the equation that I will give now."},{"Start":"01:32.990 ","End":"01:36.375","Text":"The equation goes like so,"},{"Start":"01:36.375 ","End":"01:45.694","Text":"1 divided by C total is equal to 1 divided by C_1 plus 1 divided by C_2."},{"Start":"01:45.694 ","End":"01:48.619","Text":"Exactly the same equation,"},{"Start":"01:48.619 ","End":"01:54.600","Text":"just with C instead of R for resistors connected in parallel."},{"Start":"01:54.650 ","End":"01:58.989","Text":"What do we have to remember is that the charge on"},{"Start":"01:58.989 ","End":"02:03.850","Text":"this capacitor is equal to the charge on the other 2 capacitors."},{"Start":"02:03.850 ","End":"02:06.879","Text":"We can say that Q total,"},{"Start":"02:06.879 ","End":"02:13.020","Text":"the total charge on this capacitor is equal to Q_1,"},{"Start":"02:13.020 ","End":"02:15.729","Text":"it\u0027s equal to the charge on this capacitor,"},{"Start":"02:15.729 ","End":"02:21.140","Text":"and it\u0027s equal to the charge on this capacitor."},{"Start":"02:21.620 ","End":"02:25.494","Text":"In other words, the charge on this capacitor,"},{"Start":"02:25.494 ","End":"02:31.764","Text":"C_T is equal to the charge on any one of these capacitors."},{"Start":"02:31.764 ","End":"02:37.125","Text":"It is not equal to the sum of the charges on these capacitors."},{"Start":"02:37.125 ","End":"02:41.885","Text":"However, the voltage on this capacitor,"},{"Start":"02:41.885 ","End":"02:47.330","Text":"so V_T is equal to the sum of the 2 voltages."},{"Start":"02:47.330 ","End":"02:54.759","Text":"The voltage V_T is equal to V_1 plus V_2."},{"Start":"02:56.050 ","End":"02:59.224","Text":"Why is the voltage like this?"},{"Start":"02:59.224 ","End":"03:03.470","Text":"We can agree that these 2 points over"},{"Start":"03:03.470 ","End":"03:10.379","Text":"here are the same or identical to these 2 points over here."},{"Start":"03:10.580 ","End":"03:14.805","Text":"So when we calculate our voltage drops,"},{"Start":"03:14.805 ","End":"03:18.679","Text":"our voltage drop across this capacitor has to be equal"},{"Start":"03:18.679 ","End":"03:22.646","Text":"to the voltage drop across these 2 capacitors,"},{"Start":"03:22.646 ","End":"03:25.759","Text":"which means that we take the voltage drop across"},{"Start":"03:25.759 ","End":"03:32.159","Text":"the first capacitor plus the voltage drop across the second capacitor."},{"Start":"03:32.680 ","End":"03:38.669","Text":"Now let\u0027s take a look at capacitors connected in parallel."},{"Start":"03:39.260 ","End":"03:44.330","Text":"Now we have these 2 capacitors in parallel."},{"Start":"03:44.330 ","End":"03:49.500","Text":"Of course, we want to take this whole system over here,"},{"Start":"03:49.500 ","End":"03:53.254","Text":"and we want to convert it to just 1 capacitor."},{"Start":"03:53.254 ","End":"04:00.530","Text":"Where if we switched out this system of 2 capacitors in parallel with this 1 capacitor,"},{"Start":"04:00.530 ","End":"04:09.569","Text":"then nothing would change with regards to our measurements in our circuit."},{"Start":"04:09.860 ","End":"04:13.984","Text":"When the capacitors are connected in parallel,"},{"Start":"04:13.984 ","End":"04:20.390","Text":"so the voltage on each capacitor has to be equal."},{"Start":"04:20.390 ","End":"04:28.510","Text":"In this case, V total is equal to V_1 and that is also equal to V_2,"},{"Start":"04:28.510 ","End":"04:32.185","Text":"and so the equation therefore for capacitance,"},{"Start":"04:32.185 ","End":"04:39.094","Text":"so C total is just going to be equal to C_1 plus C_2."},{"Start":"04:39.094 ","End":"04:47.360","Text":"The charge on this capacitor is going to be equal to the charge on the first capacitor,"},{"Start":"04:47.360 ","End":"04:54.690","Text":"Q_1 plus the charge on the second capacitor, Q_2."},{"Start":"04:54.710 ","End":"05:00.845","Text":"These are the 2 equations or 2 sets of equations that you need to remember."},{"Start":"05:00.845 ","End":"05:02.329","Text":"This one for series,"},{"Start":"05:02.329 ","End":"05:06.020","Text":"where the definition for capacitors connected in"},{"Start":"05:06.020 ","End":"05:10.480","Text":"series is that the charge on each capacitor is equal,"},{"Start":"05:10.480 ","End":"05:13.639","Text":"and the definition for capacitors connected in"},{"Start":"05:13.639 ","End":"05:19.120","Text":"parallel is that the voltage on each capacitor is equal."},{"Start":"05:19.120 ","End":"05:22.844","Text":"Write these out in your formula sheets."},{"Start":"05:22.844 ","End":"05:28.555","Text":"Now we\u0027ve finished describing how we connect capacitors,"},{"Start":"05:28.555 ","End":"05:34.280","Text":"so we\u0027re going to move on to the second method for calculating capacitance."},{"Start":"05:36.030 ","End":"05:42.325","Text":"The second method for calculating the capacitance is breaking up."},{"Start":"05:42.325 ","End":"05:47.275","Text":"The first step that we do is we break up the capacitor into different capacitors,"},{"Start":"05:47.275 ","End":"05:51.954","Text":"which are connected either in series or in parallel to one another."},{"Start":"05:51.954 ","End":"05:56.435","Text":"Then, we calculate the capacitance for each one."},{"Start":"05:56.435 ","End":"05:57.640","Text":"Then after all of that,"},{"Start":"05:57.640 ","End":"06:01.600","Text":"we add up all of the values for our capacitance that we calculated,"},{"Start":"06:01.600 ","End":"06:04.814","Text":"and we add it up according to the relevant formula."},{"Start":"06:04.814 ","End":"06:07.700","Text":"If the capacitors are connected in series,"},{"Start":"06:07.700 ","End":"06:11.464","Text":"we\u0027ll use the equation for"},{"Start":"06:11.464 ","End":"06:15.470","Text":"capacitors connected in series to calculate the total capacitance,"},{"Start":"06:15.470 ","End":"06:19.609","Text":"and if the capacitors were connected in parallel,"},{"Start":"06:19.609 ","End":"06:22.589","Text":"then we will use that formula."},{"Start":"06:22.850 ","End":"06:25.830","Text":"Let\u0027s look at an example."},{"Start":"06:25.830 ","End":"06:30.590","Text":"Here we have a parallel plate capacitor where"},{"Start":"06:30.590 ","End":"06:37.085","Text":"a certain section of it is filled with a dielectric material of constant Epsilon r,"},{"Start":"06:37.085 ","End":"06:41.119","Text":"and the other section is empty."},{"Start":"06:41.119 ","End":"06:48.270","Text":"Our plate dimensions are a by b and the distance between the 2 plates are d,"},{"Start":"06:48.270 ","End":"06:52.064","Text":"and we\u0027re being told that d is much smaller than a and b."},{"Start":"06:52.064 ","End":"06:54.740","Text":"What does that mean? That just means that this is"},{"Start":"06:54.740 ","End":"07:00.880","Text":"a traditional parallel plate capacitor where the 2 plates are very close to one another,"},{"Start":"07:00.880 ","End":"07:06.500","Text":"and we can see that the dielectric material fills the whole width,"},{"Start":"07:06.500 ","End":"07:08.330","Text":"so all of a."},{"Start":"07:08.330 ","End":"07:12.115","Text":"However, it doesn\u0027t fill the whole length of the capacitor,"},{"Start":"07:12.115 ","End":"07:16.924","Text":"and we can see that we have this space over here,"},{"Start":"07:16.924 ","End":"07:20.434","Text":"x, which is left empty."},{"Start":"07:20.434 ","End":"07:27.089","Text":"This distance over here is going to be b minus x."},{"Start":"07:27.310 ","End":"07:31.340","Text":"Now we want to calculate the capacitance."},{"Start":"07:31.340 ","End":"07:37.169","Text":"Let\u0027s see how we do this according to our second method."},{"Start":"07:37.880 ","End":"07:40.889","Text":"In our previous experience,"},{"Start":"07:40.889 ","End":"07:46.055","Text":"we\u0027re used to calculating the capacitance of a capacitor if it\u0027s either"},{"Start":"07:46.055 ","End":"07:51.763","Text":"has just air in-between the plates or if it\u0027s filled with a dielectric material."},{"Start":"07:51.763 ","End":"07:56.270","Text":"But we don\u0027t know how to calculate the capacitance of a capacitor when we"},{"Start":"07:56.270 ","End":"08:01.624","Text":"have part of it filled with dielectric material and part of it not."},{"Start":"08:01.624 ","End":"08:06.055","Text":"What we\u0027re going to do is we\u0027re going to cut the capacitor."},{"Start":"08:06.055 ","End":"08:08.670","Text":"That is step Number 1."},{"Start":"08:08.670 ","End":"08:12.294","Text":"We break up the capacitor into different capacitors."},{"Start":"08:12.294 ","End":"08:18.535","Text":"Soon, we\u0027ll see if these 2 different capacitors are connected in series or in parallel."},{"Start":"08:18.535 ","End":"08:20.575","Text":"We\u0027ll discuss that in a second."},{"Start":"08:20.575 ","End":"08:24.310","Text":"As I said, I know how to calculate either a capacitor filled with"},{"Start":"08:24.310 ","End":"08:28.700","Text":"dielectric material or filled with air."},{"Start":"08:28.700 ","End":"08:31.810","Text":"I\u0027m going to cut the capacitor right over here,"},{"Start":"08:31.810 ","End":"08:33.879","Text":"so then I have 1 capacitor,"},{"Start":"08:33.879 ","End":"08:35.499","Text":"this section over here,"},{"Start":"08:35.499 ","End":"08:37.485","Text":"which is just filled with air,"},{"Start":"08:37.485 ","End":"08:39.505","Text":"and I have my second capacitor,"},{"Start":"08:39.505 ","End":"08:40.824","Text":"this section over here,"},{"Start":"08:40.824 ","End":"08:44.240","Text":"which is filled with a dielectric material."},{"Start":"08:44.510 ","End":"08:48.704","Text":"Soon we\u0027ll discuss if they\u0027re connected in series or in parallel."},{"Start":"08:48.704 ","End":"08:54.475","Text":"Let\u0027s just first calculate the capacitance for each one."},{"Start":"08:54.475 ","End":"08:58.155","Text":"We\u0027re doing step Number 2 now."},{"Start":"08:58.155 ","End":"09:01.864","Text":"Let\u0027s call this capacitor C_1,"},{"Start":"09:01.864 ","End":"09:05.329","Text":"and let\u0027s call this capacitor C_2."},{"Start":"09:05.329 ","End":"09:09.215","Text":"d is still much smaller than a and b."},{"Start":"09:09.215 ","End":"09:11.992","Text":"The shape of our capacitor hasn\u0027t changed."},{"Start":"09:11.992 ","End":"09:15.100","Text":"It just means that the 2 plates are very close together."},{"Start":"09:15.100 ","End":"09:21.320","Text":"As we know, the capacitance of a parallel plate capacitor we saw in"},{"Start":"09:21.320 ","End":"09:28.156","Text":"a previous lesson is equal to Epsilon naught a divided by d,"},{"Start":"09:28.156 ","End":"09:33.190","Text":"where a is of course the surface area of each plate,"},{"Start":"09:33.190 ","End":"09:36.490","Text":"and d is the distance between the plates."},{"Start":"09:36.490 ","End":"09:42.799","Text":"Over here, the surface area of the plates is a,"},{"Start":"09:42.799 ","End":"09:45.725","Text":"the width multiplied by the length."},{"Start":"09:45.725 ","End":"09:48.455","Text":"The length of C_1 isn\u0027t b,"},{"Start":"09:48.455 ","End":"09:51.784","Text":"it\u0027s this length over here, x."},{"Start":"09:51.784 ","End":"09:57.799","Text":"Therefore we have a multiplied by x divided by the distance between the 2 plates,"},{"Start":"09:57.799 ","End":"10:01.864","Text":"which is d. Now,"},{"Start":"10:01.864 ","End":"10:05.974","Text":"let\u0027s look at our next capacitor, which is C_2."},{"Start":"10:05.974 ","End":"10:10.719","Text":"We see that C_2 is filled with a dielectric material."},{"Start":"10:10.719 ","End":"10:18.180","Text":"First, let\u0027s calculate the capacitance if C_2 wasn\u0027t filled with a dielectric material."},{"Start":"10:18.180 ","End":"10:20.340","Text":"Let\u0027s call it C_2 tag."},{"Start":"10:20.340 ","End":"10:23.460","Text":"In that case, we use the same equation over here,"},{"Start":"10:23.460 ","End":"10:26.840","Text":"so we\u0027d have Epsilon naught multiplied by the surface area,"},{"Start":"10:26.840 ","End":"10:36.320","Text":"so we have the width is a and this length over here is the total length b minus x."},{"Start":"10:36.320 ","End":"10:42.080","Text":"What we would have is a multiplied by b minus x and"},{"Start":"10:42.080 ","End":"10:47.960","Text":"then divided by the distance between the plates d. What is C_2?"},{"Start":"10:47.960 ","End":"10:51.050","Text":"C_2 is filled with a dielectric material."},{"Start":"10:51.050 ","End":"10:52.655","Text":"What do we do?"},{"Start":"10:52.655 ","End":"10:57.334","Text":"We simply take the equation for"},{"Start":"10:57.334 ","End":"11:02.614","Text":"our capacitor that we calculated with the dielectric materials, so C_2 tag,"},{"Start":"11:02.614 ","End":"11:08.960","Text":"and then we multiply it by the dielectric constant Epsilon r. Now,"},{"Start":"11:08.960 ","End":"11:11.960","Text":"this is a great equation to remember,"},{"Start":"11:11.960 ","End":"11:16.640","Text":"and as long as the dielectric material is uniform,"},{"Start":"11:16.640 ","End":"11:20.134","Text":"so it\u0027s even uniform throughout,"},{"Start":"11:20.134 ","End":"11:22.340","Text":"we can use this equation."},{"Start":"11:22.340 ","End":"11:28.580","Text":"You just calculate the capacitance of the capacitor without the dielectric material,"},{"Start":"11:28.580 ","End":"11:30.830","Text":"and then you take that capacitance and you"},{"Start":"11:30.830 ","End":"11:33.889","Text":"multiply it by the dielectric constant of your material."},{"Start":"11:33.889 ","End":"11:38.798","Text":"In this case, what we would have is Epsilon naught,"},{"Start":"11:38.798 ","End":"11:46.260","Text":"Epsilon r^ a(b minus x) divided by d."},{"Start":"11:46.830 ","End":"11:50.410","Text":"Great. Now we\u0027ve finished step number 2."},{"Start":"11:50.410 ","End":"11:54.130","Text":"We calculated the capacitance for C_1 and for C_2."},{"Start":"11:54.130 ","End":"11:56.230","Text":"Let\u0027s move on to step number 3,"},{"Start":"11:56.230 ","End":"12:01.660","Text":"which is to add up all of the values for capacitance according to the relative formula."},{"Start":"12:01.660 ","End":"12:06.009","Text":"The first thing that I need to do is I have to understand if C_1 and"},{"Start":"12:06.009 ","End":"12:11.420","Text":"C_2 are connected in series or in parallel."},{"Start":"12:12.240 ","End":"12:15.340","Text":"Just from some basic intuition,"},{"Start":"12:15.340 ","End":"12:20.455","Text":"we can see that the 2 capacitors are side-by-side."},{"Start":"12:20.455 ","End":"12:25.130","Text":"Which means that they\u0027re connected in parallel."},{"Start":"12:25.350 ","End":"12:30.085","Text":"Generally speaking, capacitors side-by-side are connected in parallel,"},{"Start":"12:30.085 ","End":"12:36.474","Text":"and 1 capacitor on top of the other capacitor is connected in series."},{"Start":"12:36.474 ","End":"12:43.225","Text":"However, this is just by intuition and we want to do this properly. Let\u0027s see."},{"Start":"12:43.225 ","End":"12:49.420","Text":"We know that the definition for a capacitor connected in"},{"Start":"12:49.420 ","End":"12:55.575","Text":"parallel is that the voltage on both the capacitance has to be the same."},{"Start":"12:55.575 ","End":"13:04.830","Text":"We know that V_1 has to be equal to V_2 when they\u0027re connected in parallel."},{"Start":"13:04.830 ","End":"13:06.989","Text":"Let\u0027s take a look over here."},{"Start":"13:06.989 ","End":"13:08.565","Text":"Let\u0027s look at C_1."},{"Start":"13:08.565 ","End":"13:14.814","Text":"Let\u0027s say that our potential over here on the upper plate is"},{"Start":"13:14.814 ","End":"13:21.070","Text":"phi 1 or phi 1 and that the potential on the lower plate is phi 2."},{"Start":"13:21.070 ","End":"13:28.660","Text":"Therefore, we can see that V_1 is equal to the potential difference."},{"Start":"13:28.660 ","End":"13:33.415","Text":"It\u0027s just going to be equal to phi 1 minus phi 2."},{"Start":"13:33.415 ","End":"13:35.499","Text":"This is the potential difference."},{"Start":"13:35.499 ","End":"13:39.024","Text":"Now, if we look at the capacitor, again,"},{"Start":"13:39.024 ","End":"13:46.225","Text":"we remember that the fact that we cut this capacitor over here was an imaginary incision."},{"Start":"13:46.225 ","End":"13:50.050","Text":"We didn\u0027t actually cut the capacitor and separate the plates."},{"Start":"13:50.050 ","End":"13:52.344","Text":"This is still 1 whole plate,"},{"Start":"13:52.344 ","End":"13:54.204","Text":"and this is also 1 whole plate."},{"Start":"13:54.204 ","End":"14:00.040","Text":"Which means that if the potential on this side of the plate is equal to Phi 1,"},{"Start":"14:00.040 ","End":"14:04.510","Text":"then the potential on this side of the plate must also be equal to Phi 1,"},{"Start":"14:04.510 ","End":"14:06.275","Text":"because it\u0027s the same plate."},{"Start":"14:06.275 ","End":"14:10.620","Text":"Similarly, if the potential on this plate is phi 2,"},{"Start":"14:10.620 ","End":"14:13.510","Text":"then further on, on the same plate,"},{"Start":"14:13.510 ","End":"14:18.145","Text":"the potential over here is also going to be equal to phi 2,"},{"Start":"14:18.145 ","End":"14:22.644","Text":"which means that the potential difference between"},{"Start":"14:22.644 ","End":"14:29.630","Text":"these 2 imaginary plates and these 2 imaginary plates is the same."},{"Start":"14:30.030 ","End":"14:35.020","Text":"Each plate of the capacitor is a conductor."},{"Start":"14:35.020 ","End":"14:40.254","Text":"Which means that the potential along each capacity,"},{"Start":"14:40.254 ","End":"14:44.215","Text":"each conductor, sorry, is going to be the same."},{"Start":"14:44.215 ","End":"14:50.125","Text":"We won\u0027t get differing potentials at different points."},{"Start":"14:50.125 ","End":"14:55.030","Text":"Of course this we learned when we were speaking about"},{"Start":"14:55.030 ","End":"15:01.940","Text":"conductors and where we saw that the electric field inside a conductor is equal to 0."},{"Start":"15:02.490 ","End":"15:06.189","Text":"If the electric field inside the conductor,"},{"Start":"15:06.189 ","End":"15:08.109","Text":"this plate is equal to 0,"},{"Start":"15:08.109 ","End":"15:13.134","Text":"that means that the potential across the plate has to be constant."},{"Start":"15:13.134 ","End":"15:15.310","Text":"That\u0027s why we get this."},{"Start":"15:15.310 ","End":"15:21.850","Text":"Therefore, we know that these 2 capacitors are connected in parallel."},{"Start":"15:21.850 ","End":"15:25.014","Text":"We\u0027ve seen that they\u0027re connected in parallel,"},{"Start":"15:25.014 ","End":"15:29.695","Text":"which means that we would have some wire"},{"Start":"15:29.695 ","End":"15:35.680","Text":"connecting the top of these 2 together and these 2 together."},{"Start":"15:35.680 ","End":"15:42.760","Text":"Then we can see that it really looks like parallel connection between capacitors."},{"Start":"15:42.760 ","End":"15:47.499","Text":"What we saw earlier on in this lesson that the equation,"},{"Start":"15:47.499 ","End":"15:52.555","Text":"the relevant formula for the total capacitance when 2 capacitors"},{"Start":"15:52.555 ","End":"15:58.240","Text":"are joined in parallel is equal to C_1 plus C_2."},{"Start":"15:58.240 ","End":"16:01.030","Text":"We\u0027re adding this plus this,"},{"Start":"16:01.030 ","End":"16:06.325","Text":"and what we get is Epsilon naught a divided by d"},{"Start":"16:06.325 ","End":"16:14.000","Text":"multiplied by x plus Epsilon r(b minus x)."},{"Start":"16:14.970 ","End":"16:20.890","Text":"Now we can do a sanity check just to check that our answer makes sense."},{"Start":"16:20.890 ","End":"16:24.580","Text":"What we can check is we can imagine that"},{"Start":"16:24.580 ","End":"16:29.605","Text":"the whole capacitor has no dielectric material inside."},{"Start":"16:29.605 ","End":"16:31.200","Text":"The whole capacitor,"},{"Start":"16:31.200 ","End":"16:33.640","Text":"the only thing separating the plates is air,"},{"Start":"16:33.640 ","End":"16:38.365","Text":"which means that we can just substitute Epsilon r for 1,"},{"Start":"16:38.365 ","End":"16:42.370","Text":"because that\u0027s the dielectric constant for air."},{"Start":"16:42.370 ","End":"16:45.909","Text":"Then we would have Epsilon a divided by d,"},{"Start":"16:45.909 ","End":"16:49.855","Text":"and then we have x plus 1(b minus x)."},{"Start":"16:49.855 ","End":"16:53.830","Text":"What we would be left with is the x\u0027s would cancel out and we\u0027ll"},{"Start":"16:53.830 ","End":"16:57.790","Text":"be left with Epsilon naught a multiplied by b,"},{"Start":"16:57.790 ","End":"17:01.525","Text":"which is the surface area of each plate,"},{"Start":"17:01.525 ","End":"17:04.990","Text":"divided by d, the distance between the 2 plates."},{"Start":"17:04.990 ","End":"17:08.290","Text":"We come back to this exact equation."},{"Start":"17:08.290 ","End":"17:12.115","Text":"Of course, that matches for a capacitor"},{"Start":"17:12.115 ","End":"17:17.140","Text":"with no dielectric material in between the plates."},{"Start":"17:17.140 ","End":"17:20.515","Text":"We calculated the capacitance."},{"Start":"17:20.515 ","End":"17:25.735","Text":"Now what I\u0027m going to do is I\u0027m going to slightly change the topic."},{"Start":"17:25.735 ","End":"17:32.920","Text":"We\u0027re going to see what the charge distribution is on the capacitor\u0027s plates."},{"Start":"17:32.920 ","End":"17:42.460","Text":"Let\u0027s imagine that we connect these 2 plates via some battery,"},{"Start":"17:42.460 ","End":"17:45.955","Text":"some voltage source, like so."},{"Start":"17:45.955 ","End":"17:50.950","Text":"Let\u0027s say that the voltage source is V_0."},{"Start":"17:50.950 ","End":"17:55.764","Text":"Now what we want to do is we want to find"},{"Start":"17:55.764 ","End":"18:01.045","Text":"the charge distribution on the capacitor plates."},{"Start":"18:01.045 ","End":"18:05.470","Text":"Every time we\u0027re working with a capacitor and we\u0027re"},{"Start":"18:05.470 ","End":"18:09.400","Text":"being asked to calculate the charge distribution."},{"Start":"18:09.400 ","End":"18:15.609","Text":"The first thing that we want to do is we want to use our equation for capacitance,"},{"Start":"18:15.609 ","End":"18:23.635","Text":"which is C is equal to q divided by V. Therefore,"},{"Start":"18:23.635 ","End":"18:25.900","Text":"we can isolate out q,"},{"Start":"18:25.900 ","End":"18:29.290","Text":"and that is equal to a CV,"},{"Start":"18:29.290 ","End":"18:32.739","Text":"which is equal to our capacitance,"},{"Start":"18:32.739 ","End":"18:35.319","Text":"multiplied by the voltage,"},{"Start":"18:35.319 ","End":"18:37.944","Text":"which over here is V_0."},{"Start":"18:37.944 ","End":"18:43.720","Text":"What\u0027s important to note is that this charge q that we just"},{"Start":"18:43.720 ","End":"18:49.255","Text":"calculated is the total charge on the whole capacitor."},{"Start":"18:49.255 ","End":"18:53.064","Text":"What does that mean? If we remember that we could split our capacitor,"},{"Start":"18:53.064 ","End":"18:55.600","Text":"where we have 1 side of the capacitor with"},{"Start":"18:55.600 ","End":"18:58.495","Text":"this dielectric material and the other side without,"},{"Start":"18:58.495 ","End":"19:02.544","Text":"so this charge q is the charge all over here."},{"Start":"19:02.544 ","End":"19:05.769","Text":"However, we can assume that because 1 side of"},{"Start":"19:05.769 ","End":"19:09.249","Text":"the capacitor has dielectric material and the other side doesn\u0027t."},{"Start":"19:09.249 ","End":"19:17.785","Text":"We can imagine that our charge isn\u0027t uniformly distributed throughout the plate."},{"Start":"19:17.785 ","End":"19:21.849","Text":"We could say that on the side of the capacitor without"},{"Start":"19:21.849 ","End":"19:25.135","Text":"the dielectric material, we have q_1."},{"Start":"19:25.135 ","End":"19:27.459","Text":"On the side with a dielectric material,"},{"Start":"19:27.459 ","End":"19:29.110","Text":"we have q_2,"},{"Start":"19:29.110 ","End":"19:32.380","Text":"and that these 2 charges are different."},{"Start":"19:32.380 ","End":"19:38.620","Text":"Which means that our charge distribution on each side will also be different."},{"Start":"19:38.620 ","End":"19:48.595","Text":"I can say that this q that I just calculated over here is equal to q_1 plus q_2."},{"Start":"19:48.595 ","End":"19:53.065","Text":"But this of course, doesn\u0027t help me to solve this question."},{"Start":"19:53.065 ","End":"19:58.810","Text":"We have 2 options to solve this kind of question."},{"Start":"19:58.810 ","End":"20:03.670","Text":"The first option is to calculate the jump in the electric field."},{"Start":"20:03.670 ","End":"20:09.280","Text":"The second option is to calculate the charge on each capacitor separately."},{"Start":"20:09.280 ","End":"20:12.115","Text":"We\u0027ll calculate the charge on this capacitor,"},{"Start":"20:12.115 ","End":"20:13.960","Text":"the charge on this capacitor,"},{"Start":"20:13.960 ","End":"20:18.984","Text":"and then we can calculate the charge distribution."},{"Start":"20:18.984 ","End":"20:23.664","Text":"I\u0027m going to solve this question in both ways."},{"Start":"20:23.664 ","End":"20:28.130","Text":"I\u0027m actually going to start from option number 2."},{"Start":"20:28.800 ","End":"20:33.230","Text":"Let\u0027s start with calculating q_1."},{"Start":"20:33.930 ","End":"20:38.530","Text":"Let\u0027s just shift over a bit to this side."},{"Start":"20:38.530 ","End":"20:44.124","Text":"What we\u0027ll have is that q_1 is equal to,"},{"Start":"20:44.124 ","End":"20:49.900","Text":"from this equation, C_1 multiplied by the voltage."},{"Start":"20:49.900 ","End":"20:55.315","Text":"The voltage and C_1 and C_2 is obviously going to be uniform."},{"Start":"20:55.315 ","End":"20:57.960","Text":"We discussed that before where we said that"},{"Start":"20:57.960 ","End":"21:03.749","Text":"the potential difference between the 2 plates is equal at every point."},{"Start":"21:03.749 ","End":"21:09.940","Text":"We saw here we have V_1 and here V_1 and here V_2, and here V_2."},{"Start":"21:09.940 ","End":"21:14.155","Text":"That\u0027s just going to be equal to C_1,"},{"Start":"21:14.155 ","End":"21:21.115","Text":"which is Epsilon naught ax divided by d multiplied by the voltage,"},{"Start":"21:21.115 ","End":"21:23.975","Text":"which is just V naught."},{"Start":"21:23.975 ","End":"21:26.875","Text":"Then we can calculate q_2,"},{"Start":"21:26.875 ","End":"21:33.055","Text":"which is of course C_2 multiplied by the voltage."},{"Start":"21:33.055 ","End":"21:35.710","Text":"C_2 was Epsilon naught,"},{"Start":"21:35.710 ","End":"21:41.664","Text":"Epsilon ra (b minus x) and then"},{"Start":"21:41.664 ","End":"21:49.675","Text":"V_0 divided by d. Now we want to calculate sigma."},{"Start":"21:49.675 ","End":"21:52.060","Text":"What we\u0027ll have, of course,"},{"Start":"21:52.060 ","End":"21:54.325","Text":"sigma is the charge distribution."},{"Start":"21:54.325 ","End":"21:59.970","Text":"Sigma 1 is just going to be the total charge on this area"},{"Start":"21:59.970 ","End":"22:06.089","Text":"over here of C_1 divided by the area."},{"Start":"22:06.089 ","End":"22:12.994","Text":"That\u0027s just going to be equal to q_1 divided by the area A_1."},{"Start":"22:12.994 ","End":"22:18.324","Text":"Which is of course this area over here,"},{"Start":"22:18.324 ","End":"22:21.205","Text":"or down here, steroid and gray."},{"Start":"22:21.205 ","End":"22:24.890","Text":"It\u0027s this, this is A_1."},{"Start":"22:26.180 ","End":"22:29.340","Text":"What is A_1?"},{"Start":"22:29.340 ","End":"22:35.770","Text":"A_1 is simply this length over here,"},{"Start":"22:35.770 ","End":"22:40.915","Text":"which we know is x multiplied by the width of the plate, which is a."},{"Start":"22:40.915 ","End":"22:43.280","Text":"We just have xa."},{"Start":"22:43.940 ","End":"22:47.910","Text":"Here we have our q_1 divided by A_1,"},{"Start":"22:47.910 ","End":"22:51.310","Text":"which we said was equal to ax."},{"Start":"22:51.650 ","End":"22:55.320","Text":"Ax cancels out."},{"Start":"22:55.320 ","End":"23:04.259","Text":"What we get is that Sigma 1 is equal to Epsilon naught V naught divided by d. Then,"},{"Start":"23:04.259 ","End":"23:10.960","Text":"Sigma 2 is equal to q_2 divided by A_2."},{"Start":"23:11.300 ","End":"23:16.345","Text":"A_2 is, of course, their surface area."},{"Start":"23:16.345 ","End":"23:19.910","Text":"Over here, A_2,"},{"Start":"23:19.910 ","End":"23:21.934","Text":"which is, of course,"},{"Start":"23:21.934 ","End":"23:23.360","Text":"this width over here,"},{"Start":"23:23.360 ","End":"23:26.690","Text":"which is a multiplied by this length over here,"},{"Start":"23:26.690 ","End":"23:28.349","Text":"which is the same as this length,"},{"Start":"23:28.349 ","End":"23:30.375","Text":"which is b minus x."},{"Start":"23:30.375 ","End":"23:37.095","Text":"What we\u0027ll have is Epsilon naught Epsilon r ab minus"},{"Start":"23:37.095 ","End":"23:44.564","Text":"x V_0 divided by d. This is q_2 divided by A_2,"},{"Start":"23:44.564 ","End":"23:50.940","Text":"so divided by a multiplied by b minus x."},{"Start":"23:50.940 ","End":"23:58.020","Text":"A and b minus x cancels out and what we\u0027re left with is that Sigma 2 is equal to"},{"Start":"23:58.020 ","End":"24:06.884","Text":"Epsilon naught Epsilon r V_0 divided by d. Here we saw the second method."},{"Start":"24:06.884 ","End":"24:11.264","Text":"We calculated the charge on each capacitor separately,"},{"Start":"24:11.264 ","End":"24:13.965","Text":"and then we found out the charge distribution."},{"Start":"24:13.965 ","End":"24:19.079","Text":"Now we\u0027re going to look at the first method where we\u0027re going to"},{"Start":"24:19.079 ","End":"24:24.675","Text":"calculate the charge distribution by calculating the jump in the E field."},{"Start":"24:24.675 ","End":"24:26.534","Text":"Now I\u0027m going to just rub"},{"Start":"24:26.534 ","End":"24:32.290","Text":"out lots of things over here to give us a little bit more space."},{"Start":"24:33.260 ","End":"24:40.035","Text":"Here was our answer for method number 2."},{"Start":"24:40.035 ","End":"24:43.680","Text":"Now we\u0027re going to solve via method number 1,"},{"Start":"24:43.680 ","End":"24:45.180","Text":"the jump in the E field."},{"Start":"24:45.180 ","End":"24:53.594","Text":"We know that the E field outside of the capacitor is equal to 0,"},{"Start":"24:53.594 ","End":"24:59.500","Text":"so all we have to do is we have to calculate the E field inside the capacitor."},{"Start":"24:59.810 ","End":"25:03.404","Text":"In order to calculate the electric field,"},{"Start":"25:03.404 ","End":"25:07.589","Text":"we can remember that our voltage or"},{"Start":"25:07.589 ","End":"25:13.875","Text":"our potential difference is equal to the integral on the electric field."},{"Start":"25:13.875 ","End":"25:17.444","Text":"Another thing that we can remember from"},{"Start":"25:17.444 ","End":"25:22.500","Text":"previous lessons is that in a parallel plate capacitor,"},{"Start":"25:22.500 ","End":"25:27.930","Text":"V = E multiplied by d,"},{"Start":"25:27.930 ","End":"25:30.525","Text":"the distance between the plates."},{"Start":"25:30.525 ","End":"25:37.109","Text":"This equation can be used only when dealing with a parallel plate capacitor,"},{"Start":"25:37.109 ","End":"25:43.930","Text":"because the electric field is uniform throughout in a parallel plate capacitor."},{"Start":"25:43.970 ","End":"25:51.510","Text":"In that case, we can isolate out our E and we get that our E field is equal to"},{"Start":"25:51.510 ","End":"25:58.859","Text":"the voltage which is V_0 divided by d. This E field is,"},{"Start":"25:58.859 ","End":"26:04.769","Text":"of course, uniform throughout whether there is a dielectric material or there isn\u0027t."},{"Start":"26:04.769 ","End":"26:08.010","Text":"Because it is just dependent on the voltage,"},{"Start":"26:08.010 ","End":"26:12.210","Text":"which we saw is the same as a uniform on"},{"Start":"26:12.210 ","End":"26:16.439","Text":"every plate divided by the distance between the 2 plates, which,"},{"Start":"26:16.439 ","End":"26:18.945","Text":"of course, is also this constant"},{"Start":"26:18.945 ","End":"26:26.130","Text":"d. This is the electric field that there will be in practice."},{"Start":"26:26.130 ","End":"26:29.820","Text":"In real life, this is the electric field that we will"},{"Start":"26:29.820 ","End":"26:34.140","Text":"feel wherever we are located within the capacitor."},{"Start":"26:34.140 ","End":"26:37.659","Text":"This is the real electric field."},{"Start":"26:37.880 ","End":"26:41.834","Text":"The electric field that I\u0027ll feel here and here."},{"Start":"26:41.834 ","End":"26:47.969","Text":"However, the electric field due to the free charges here and"},{"Start":"26:47.969 ","End":"26:55.120","Text":"here will be different because here I have this dielectric material."},{"Start":"26:55.280 ","End":"26:58.515","Text":"Let\u0027s see."},{"Start":"26:58.515 ","End":"27:04.560","Text":"Let\u0027s take a look at this electric field over here in C_1."},{"Start":"27:04.560 ","End":"27:10.090","Text":"E_0_1, here,"},{"Start":"27:10.090 ","End":"27:12.389","Text":"is going to, as we see,"},{"Start":"27:12.389 ","End":"27:14.145","Text":"we don\u0027t have a dielectric material,"},{"Start":"27:14.145 ","End":"27:18.660","Text":"so it\u0027s just going to be equal to our original E,"},{"Start":"27:18.660 ","End":"27:28.440","Text":"which is simply equal to V_0 divided by d. Then the electric field over here,"},{"Start":"27:28.440 ","End":"27:35.340","Text":"so E_0 for Capacitor 2 is going to be equal"},{"Start":"27:35.340 ","End":"27:43.050","Text":"to our dielectric constant Epsilon r multiplied by the original E field."},{"Start":"27:43.050 ","End":"27:52.815","Text":"That is going to be equal to Epsilon r V_0 divided by d. Now,"},{"Start":"27:52.815 ","End":"27:55.575","Text":"let\u0027s calculate the jump in the E field."},{"Start":"27:55.575 ","End":"27:59.080","Text":"That will be equal to our Sigmas."},{"Start":"27:59.210 ","End":"28:02.925","Text":"Let\u0027s see what Sigma 1 is."},{"Start":"28:02.925 ","End":"28:07.793","Text":"Sigma 1 is the difference,"},{"Start":"28:07.793 ","End":"28:11.520","Text":"the jump in the electric field between a point close to"},{"Start":"28:11.520 ","End":"28:14.730","Text":"the capacitor located within the capacitor and"},{"Start":"28:14.730 ","End":"28:18.525","Text":"a point next to the capacitor plate located outside."},{"Start":"28:18.525 ","End":"28:21.060","Text":"Then we multiply that by Epsilon naught."},{"Start":"28:21.060 ","End":"28:27.435","Text":"We have Epsilon naught multiplied by the E field inside the capacitor."},{"Start":"28:27.435 ","End":"28:32.834","Text":"In C_1, we saw that that is equal to V_0 divided by"},{"Start":"28:32.834 ","End":"28:37.919","Text":"d and then subtracting the E field at this point outside the capacitor,"},{"Start":"28:37.919 ","End":"28:39.480","Text":"which is equal to 0."},{"Start":"28:39.480 ","End":"28:47.070","Text":"Of course, we get that this is equal to Epsilon naught V_0 divided by d,"},{"Start":"28:47.070 ","End":"28:50.205","Text":"which is exactly what we got over here for Sigma 1,"},{"Start":"28:50.205 ","End":"28:53.070","Text":"when we used Option 2."},{"Start":"28:53.070 ","End":"28:56.355","Text":"Then Sigma 2 is the same thing."},{"Start":"28:56.355 ","End":"29:00.270","Text":"We\u0027re taking a point here and we\u0027re calculating"},{"Start":"29:00.270 ","End":"29:04.560","Text":"the electric field and a point here in calculating the electric field."},{"Start":"29:04.560 ","End":"29:08.039","Text":"These 2 points are both close to the capacitor plates,"},{"Start":"29:08.039 ","End":"29:12.795","Text":"just 1 is within the capacitor and 1 is located without the capacitor."},{"Start":"29:12.795 ","End":"29:18.359","Text":"Here we have Epsilon naught multiplied by the electric field at this point,"},{"Start":"29:18.359 ","End":"29:23.100","Text":"which is Epsilon r V_0 divided by"},{"Start":"29:23.100 ","End":"29:28.780","Text":"d minus the electric field at this point outside, which is 0."},{"Start":"29:29.030 ","End":"29:34.350","Text":"Of course, that is just equal to Epsilon naught, Epsilon r,"},{"Start":"29:34.350 ","End":"29:37.770","Text":"V_0 divided by d,"},{"Start":"29:37.770 ","End":"29:42.569","Text":"which is exactly the answer that we got when using the second method."},{"Start":"29:42.569 ","End":"29:50.054","Text":"Here, I did the calculation using the bottom plate."},{"Start":"29:50.054 ","End":"29:54.180","Text":"If I were to do the calculation using the upper plates,"},{"Start":"29:54.180 ","End":"29:58.965","Text":"so I would take these 2 points and these 2 points."},{"Start":"29:58.965 ","End":"30:03.180","Text":"All I would change over here is that my Sigma 1 would be"},{"Start":"30:03.180 ","End":"30:09.525","Text":"0 minus V_0 divided by d. This minus this."},{"Start":"30:09.525 ","End":"30:15.960","Text":"Then here I would have 0 minus the electric field over here, which would be this."},{"Start":"30:15.960 ","End":"30:19.920","Text":"What we could see is that our Sigma 1 and our Sigma 2,"},{"Start":"30:19.920 ","End":"30:23.175","Text":"we just have a minus sign in front of them."},{"Start":"30:23.175 ","End":"30:27.570","Text":"Which make sense because when dealing with the capacitor 1 plate is going to have"},{"Start":"30:27.570 ","End":"30:30.179","Text":"a positive charge distribution and"},{"Start":"30:30.179 ","End":"30:34.875","Text":"the other plate is going to have a negative charge distribution."},{"Start":"30:34.875 ","End":"30:39.700","Text":"That is exactly what we get when we do this calculation."},{"Start":"30:40.160 ","End":"30:42.930","Text":"If we want to do a sanity check,"},{"Start":"30:42.930 ","End":"30:46.710","Text":"I can take the charge distribution on"},{"Start":"30:46.710 ","End":"30:51.735","Text":"each plate and multiply it by the area that it takes up."},{"Start":"30:51.735 ","End":"30:53.670","Text":"When I add it together,"},{"Start":"30:53.670 ","End":"30:59.699","Text":"then I should get the same answer that I got for q if you"},{"Start":"30:59.699 ","End":"31:06.120","Text":"remember before where I got that q was equal to C,"},{"Start":"31:06.120 ","End":"31:10.080","Text":"this, multiplied by V_0."},{"Start":"31:10.080 ","End":"31:13.440","Text":"You can try that if you just want to check your answer."},{"Start":"31:13.440 ","End":"31:16.809","Text":"That is the end of this lesson."}],"ID":22286},{"Watched":false,"Name":"Exercise 3","Duration":"14m 55s","ChapterTopicVideoID":21511,"CourseChapterTopicPlaylistID":99472,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.870","Text":"Hello. In this lesson we\u0027re going to be answering the following question."},{"Start":"00:03.870 ","End":"00:11.130","Text":"A parallel plate capacitor is charged with a charge distribution plus and minus sigma."},{"Start":"00:11.130 ","End":"00:15.270","Text":"The area of the plate of each plate is A and"},{"Start":"00:15.270 ","End":"00:20.570","Text":"the distance between the 2 plates is d. A dielectric material is placed"},{"Start":"00:20.570 ","End":"00:24.710","Text":"between the plates and its dielectric constant is dependent on"},{"Start":"00:24.710 ","End":"00:30.785","Text":"y and is given by 1 plus y divided by d^2."},{"Start":"00:30.785 ","End":"00:34.865","Text":"The lower plate is located at y,"},{"Start":"00:34.865 ","End":"00:36.845","Text":"is equal to 0."},{"Start":"00:36.845 ","End":"00:39.650","Text":"We\u0027re being asked to calculate the capacitance."},{"Start":"00:39.650 ","End":"00:42.900","Text":"The way that we can do this is,"},{"Start":"00:42.900 ","End":"00:44.180","Text":"we have 2 methods."},{"Start":"00:44.180 ","End":"00:48.965","Text":"The first is to calculate the capacitance by calculating"},{"Start":"00:48.965 ","End":"00:57.720","Text":"the electric field and the second way is by splitting this up into different capacitors."},{"Start":"00:58.130 ","End":"01:02.320","Text":"Usually, we would say that the capacitance of"},{"Start":"01:02.320 ","End":"01:05.860","Text":"a capacitor with a dielectric material inside."},{"Start":"01:05.860 ","End":"01:11.605","Text":"C tag would be equal to the dielectric constant multiplied by"},{"Start":"01:11.605 ","End":"01:19.260","Text":"the capacitance of the capacitor if there was no dielectric material inside."},{"Start":"01:19.260 ","End":"01:26.125","Text":"Right now we cannot use this equation and why can we not use this?"},{"Start":"01:26.125 ","End":"01:32.265","Text":"It is because our dielectric constant is dependent on y."},{"Start":"01:32.265 ","End":"01:37.510","Text":"We don\u0027t have a uniform dielectric material between the capacitor plates,"},{"Start":"01:37.510 ","End":"01:41.510","Text":"and therefore, we cannot use this equation."},{"Start":"01:41.870 ","End":"01:47.665","Text":"Therefore, we\u0027re going to use the second method for calculating capacitance,"},{"Start":"01:47.665 ","End":"01:50.469","Text":"which means that we\u0027re going to split up this capacitor"},{"Start":"01:50.469 ","End":"01:53.740","Text":"into lots and lots of different capacitors,"},{"Start":"01:53.740 ","End":"01:55.720","Text":"1 on top of the other,"},{"Start":"01:55.720 ","End":"02:01.360","Text":"where we can assume that the capacitance of each sub capacitor, or sorry,"},{"Start":"02:01.360 ","End":"02:04.090","Text":"that the dielectric material between"},{"Start":"02:04.090 ","End":"02:10.414","Text":"each sub capacitor is going to be uniform because the change in y will be so small."},{"Start":"02:10.414 ","End":"02:14.650","Text":"Then we\u0027ll work out the capacitance of each sub capacitor."},{"Start":"02:14.650 ","End":"02:17.810","Text":"Then add all of them up."},{"Start":"02:18.050 ","End":"02:22.390","Text":"We know that each tiny capacitor is going to be in"},{"Start":"02:22.390 ","End":"02:28.325","Text":"the y-direction because our dielectric constant is dependent on y."},{"Start":"02:28.325 ","End":"02:33.485","Text":"That means that we\u0027ll split this up into lots of little capacitors that look"},{"Start":"02:33.485 ","End":"02:39.125","Text":"like so where the surface area of each place is still A,"},{"Start":"02:39.125 ","End":"02:41.540","Text":"but this distance over here between"},{"Start":"02:41.540 ","End":"02:47.000","Text":"the 2 imaginary plates dy is extremely, extremely small."},{"Start":"02:47.000 ","End":"02:51.695","Text":"Therefore, if this change in y is so small,"},{"Start":"02:51.695 ","End":"02:55.300","Text":"then we can look at the dielectric constant."},{"Start":"02:55.300 ","End":"03:05.010","Text":"Epsilon_r as being exactly that a constant because the change in y is so small."},{"Start":"03:06.050 ","End":"03:11.735","Text":"If you remember the general equation for the capacitance"},{"Start":"03:11.735 ","End":"03:16.760","Text":"of parallel plate capacitors is equal to this."},{"Start":"03:16.760 ","End":"03:21.160","Text":"C is equal to Epsilon_naught A."},{"Start":"03:21.160 ","End":"03:27.005","Text":"Epsilon_naught multiplied by the surface area of the plate divided by d,"},{"Start":"03:27.005 ","End":"03:30.410","Text":"the distance between the 2 plates and this is of course if"},{"Start":"03:30.410 ","End":"03:36.205","Text":"I capacitor doesn\u0027t have some dielectric material placed in the middle."},{"Start":"03:36.205 ","End":"03:43.370","Text":"In our case, we\u0027re calculating the capacitance of these tiny capacitors."},{"Start":"03:43.370 ","End":"03:45.959","Text":"We\u0027ll call this dC."},{"Start":"03:45.959 ","End":"03:51.620","Text":"The capacitance of each tiny capacitor, dC,"},{"Start":"03:51.620 ","End":"03:58.550","Text":"is equal to Epsilon_naught multiplied by the surface area of the plate,"},{"Start":"03:58.550 ","End":"03:59.885","Text":"which is still A,"},{"Start":"03:59.885 ","End":"04:05.135","Text":"and then divided by the distance between each of these imaginary plates,"},{"Start":"04:05.135 ","End":"04:09.101","Text":"which in our case over here is dy."},{"Start":"04:09.101 ","End":"04:11.855","Text":"This is, of course,"},{"Start":"04:11.855 ","End":"04:16.220","Text":"the equation when we\u0027re imagining that there\u0027s"},{"Start":"04:16.220 ","End":"04:20.820","Text":"no dielectric material between the 2 plates."},{"Start":"04:20.820 ","End":"04:24.015","Text":"Let\u0027s call this dC_naught."},{"Start":"04:24.015 ","End":"04:26.660","Text":"This would be the capacitance of"},{"Start":"04:26.660 ","End":"04:33.080","Text":"this tiny capacitor if there was no dielectric material placed inside."},{"Start":"04:33.650 ","End":"04:39.680","Text":"In that case, we can say that the capacitance of this capacitor,"},{"Start":"04:39.680 ","End":"04:43.965","Text":"including the dielectric material, is equal to."},{"Start":"04:43.965 ","End":"04:53.065","Text":"We have Epsilon_r as a function of y multiplied by our dC_naught."},{"Start":"04:53.065 ","End":"04:59.044","Text":"Multiplied by the capacitance if there was no dielectric material."},{"Start":"04:59.044 ","End":"05:01.160","Text":"Now we can substitute this in."},{"Start":"05:01.160 ","End":"05:05.930","Text":"This will be equal to 1 plus y divided by"},{"Start":"05:05.930 ","End":"05:15.575","Text":"d^2 and all of this is multiplied by this Epsilon_naught A divided by dy."},{"Start":"05:15.575 ","End":"05:19.940","Text":"Now again, remember that the only reason we can use this equation is"},{"Start":"05:19.940 ","End":"05:25.160","Text":"because we\u0027re saying that the distance between the 2 plates is so small."},{"Start":"05:25.160 ","End":"05:28.190","Text":"Because we split the capacitor up into many,"},{"Start":"05:28.190 ","End":"05:34.535","Text":"many tiny capacitors and the distance between the 2 tiny capacitor plates is so small."},{"Start":"05:34.535 ","End":"05:41.950","Text":"We can consider the dielectric constant over there as being uniform."},{"Start":"05:41.990 ","End":"05:46.890","Text":"That\u0027s the only reason we can use this equation now."},{"Start":"05:47.780 ","End":"05:55.775","Text":"Now, we have to decide how the capacitors are connected in series or in parallel."},{"Start":"05:55.775 ","End":"06:00.015","Text":"If we draw another infinitesimal capacitor over"},{"Start":"06:00.015 ","End":"06:06.365","Text":"here will look like this with a dielectric material inside."},{"Start":"06:06.365 ","End":"06:11.225","Text":"First of all, we can say that generally speaking,"},{"Start":"06:11.225 ","End":"06:17.495","Text":"when we see the sketches of a capacitor drawn one on top of the other,"},{"Start":"06:17.495 ","End":"06:23.735","Text":"then we can assume generally that the capacitors are connected in series."},{"Start":"06:23.735 ","End":"06:28.659","Text":"If you have 2 capacitors side-by-side,"},{"Start":"06:28.659 ","End":"06:33.650","Text":"if you had something like this connected like so,"},{"Start":"06:33.650 ","End":"06:39.080","Text":"then this is of course parallel connection."},{"Start":"06:39.080 ","End":"06:44.120","Text":"However, if the capacitors are one on top of the other,"},{"Start":"06:44.120 ","End":"06:47.525","Text":"drawn something like so,"},{"Start":"06:47.525 ","End":"06:50.690","Text":"then this is generally series."},{"Start":"06:50.690 ","End":"06:54.245","Text":"We can see that this is probably going to be series."},{"Start":"06:54.245 ","End":"06:58.640","Text":"But a more formal answer is,"},{"Start":"06:58.640 ","End":"07:03.340","Text":"if we look at the potential difference between these 2 points,"},{"Start":"07:03.340 ","End":"07:08.325","Text":"we can call this V total."},{"Start":"07:08.325 ","End":"07:12.005","Text":"If we put some volt meter over here,"},{"Start":"07:12.005 ","End":"07:14.800","Text":"this would measure the total voltage across"},{"Start":"07:14.800 ","End":"07:20.030","Text":"all the capacitors and we can say that v_Total is equal to"},{"Start":"07:20.030 ","End":"07:29.625","Text":"the sum of all of the voltages across each tiny capacitor, so across dv."},{"Start":"07:29.625 ","End":"07:32.645","Text":"As we go past every single capacitor,"},{"Start":"07:32.645 ","End":"07:36.275","Text":"there\u0027s some voltage drop, dv."},{"Start":"07:36.275 ","End":"07:39.560","Text":"As we go down all of the capacitors dv,"},{"Start":"07:39.560 ","End":"07:42.365","Text":"then we\u0027ll get this v_Total,"},{"Start":"07:42.365 ","End":"07:52.032","Text":"the total voltage across all of these capacitors or across the one original capacitor."},{"Start":"07:52.032 ","End":"07:56.155","Text":"Of course we know when we have capacitors in series,"},{"Start":"07:56.155 ","End":"08:05.710","Text":"then the total voltage is the sum of all of the voltages put together."},{"Start":"08:05.710 ","End":"08:12.175","Text":"This would mean that this is connected in series."},{"Start":"08:12.175 ","End":"08:18.430","Text":"Another way that we can see this is that if we add in over here,"},{"Start":"08:18.430 ","End":"08:21.385","Text":"let\u0027s say some plate."},{"Start":"08:21.385 ","End":"08:25.150","Text":"We know that this plate is going to"},{"Start":"08:25.150 ","End":"08:31.435","Text":"have a charge of plus Sigma and a charge of minus Sigma."},{"Start":"08:31.435 ","End":"08:37.750","Text":"Together, plus Sigma minus Sigma equals charge of 0."},{"Start":"08:37.750 ","End":"08:41.500","Text":"But also, we can imagine that the minus Sigma goes"},{"Start":"08:41.500 ","End":"08:45.400","Text":"up here with a plus Sigma and the plus Sigma over here,"},{"Start":"08:45.400 ","End":"08:54.145","Text":"connects to another imaginary plate that will then here have the opposite charges."},{"Start":"08:54.145 ","End":"09:01.190","Text":"Then we can see that we\u0027re getting these mini capacitor plates over here as we go on."},{"Start":"09:01.410 ","End":"09:04.390","Text":"We get these tiny capacitor plates,"},{"Start":"09:04.390 ","End":"09:07.210","Text":"which is exactly what we\u0027re doing over here."},{"Start":"09:07.210 ","End":"09:08.560","Text":"But also at the same time,"},{"Start":"09:08.560 ","End":"09:15.429","Text":"we\u0027re not changing the total charge inside the original capacitor."},{"Start":"09:15.429 ","End":"09:21.550","Text":"Whenever you have a case like this where you\u0027re not changing the charge inside of"},{"Start":"09:21.550 ","End":"09:27.385","Text":"the capacitor when you add more plate-like so that have charge distributions on them,"},{"Start":"09:27.385 ","End":"09:31.930","Text":"then that also means that you are connected in series."},{"Start":"09:31.930 ","End":"09:39.490","Text":"All right, so we\u0027ve gathered that the many capacitors are connected in series,"},{"Start":"09:39.490 ","End":"09:42.910","Text":"so how do we add capacitors connected in series?"},{"Start":"09:42.910 ","End":"09:44.590","Text":"We know that the equation,"},{"Start":"09:44.590 ","End":"09:52.750","Text":"1 divided by C_total is equal to the sum on 1 divided by C_i."},{"Start":"09:52.750 ","End":"09:56.620","Text":"We add the reciprocal of each tiny capacitor,"},{"Start":"09:56.620 ","End":"10:01.850","Text":"and then we take the reciprocal of that and we get the total capacitance."},{"Start":"10:01.950 ","End":"10:04.600","Text":"Now of course, because we\u0027ve taken"},{"Start":"10:04.600 ","End":"10:12.099","Text":"infinitesmal capacitors so what we do when we\u0027re dealing with extremely small numbers,"},{"Start":"10:12.099 ","End":"10:18.301","Text":"such as dy, so our summation turns into integration,"},{"Start":"10:18.301 ","End":"10:20.545","Text":"and then instead of 1 divided by C_i,"},{"Start":"10:20.545 ","End":"10:25.970","Text":"we integrate along one divided by dc."},{"Start":"10:27.810 ","End":"10:32.590","Text":"Now let\u0027s sub in our numbers."},{"Start":"10:32.590 ","End":"10:37.525","Text":"First of all, we\u0027re integrating from y is equal to 0 and to y is equal to"},{"Start":"10:37.525 ","End":"10:41.680","Text":"d. Then this is good that we have the"},{"Start":"10:41.680 ","End":"10:46.630","Text":"reciprocal because we don\u0027t know how to integrate when our dy is in the denominator."},{"Start":"10:46.630 ","End":"10:52.360","Text":"What we\u0027re going to have is dy divided by"},{"Start":"10:52.360 ","End":"10:57.505","Text":"Epsilon_naught A multiplied by"},{"Start":"10:57.505 ","End":"11:05.510","Text":"1 plus y divided by d^2."},{"Start":"11:07.320 ","End":"11:17.755","Text":"We know that the integral of dy divided by 1 plus y^2,"},{"Start":"11:17.755 ","End":"11:23.006","Text":"or let\u0027s just call it dx right now so that we don\u0027t get confused."},{"Start":"11:23.006 ","End":"11:33.500","Text":"Dx divided by 1 plus x^2 is equal to tan to the minus 1 or arctan of x."},{"Start":"11:33.720 ","End":"11:39.800","Text":"Let\u0027s use this in order to solve this equation."},{"Start":"11:40.650 ","End":"11:44.980","Text":"We\u0027ll plug this in here,"},{"Start":"11:44.980 ","End":"11:46.915","Text":"or we\u0027ll use it here."},{"Start":"11:46.915 ","End":"11:53.080","Text":"What we\u0027ll say is we have to use integration by substitution as well,"},{"Start":"11:53.080 ","End":"11:58.105","Text":"so let\u0027s say that x is equal to this over here,"},{"Start":"11:58.105 ","End":"12:04.615","Text":"because Epsilon_naught a is a constant so we can take it out of our integrating bounds."},{"Start":"12:04.615 ","End":"12:08.920","Text":"Then we have 1 plus this in the brackets squared,"},{"Start":"12:08.920 ","End":"12:12.220","Text":"which is the x. X is what\u0027s in the brackets,"},{"Start":"12:12.220 ","End":"12:18.445","Text":"y divided by d and then we have x^2 and then we can say that dx is"},{"Start":"12:18.445 ","End":"12:25.390","Text":"equal to 1 divided by dy, the change in y."},{"Start":"12:25.390 ","End":"12:26.890","Text":"Or in other words,"},{"Start":"12:26.890 ","End":"12:30.595","Text":"because we want to get everything in terms of dy,"},{"Start":"12:30.595 ","End":"12:34.405","Text":"and y so that we can get rid of this over here,"},{"Start":"12:34.405 ","End":"12:43.880","Text":"so what we can say is that dy is equal to d multiplied by dx."},{"Start":"12:44.610 ","End":"12:47.335","Text":"Let\u0027s take out the constants."},{"Start":"12:47.335 ","End":"12:52.195","Text":"We have 1 divided by Epsilon_naught A,"},{"Start":"12:52.195 ","End":"12:53.559","Text":"and then we integrate,"},{"Start":"12:53.559 ","End":"12:55.090","Text":"soon we\u0027ll add in the balance,"},{"Start":"12:55.090 ","End":"13:01.120","Text":"and then we have dy in the numerator so that is d multiplied by"},{"Start":"13:01.120 ","End":"13:08.740","Text":"dx divided by and then we have 1 plus y divided by d,"},{"Start":"13:08.740 ","End":"13:11.920","Text":"which we said is x^2."},{"Start":"13:11.920 ","End":"13:15.100","Text":"This is our equation and our bounds,"},{"Start":"13:15.100 ","End":"13:21.580","Text":"so we said that x is equal to y divided by d. Here at the lower bound,"},{"Start":"13:21.580 ","End":"13:23.740","Text":"we\u0027re going from y is equal to 0,"},{"Start":"13:23.740 ","End":"13:26.740","Text":"so x will be equal to 0 divided by d,"},{"Start":"13:26.740 ","End":"13:30.985","Text":"which is 0 and then the upper bound y is equal to d,"},{"Start":"13:30.985 ","End":"13:35.900","Text":"so x will be d divided by d, which is 1."},{"Start":"13:37.500 ","End":"13:41.785","Text":"Now we can continue this over here,"},{"Start":"13:41.785 ","End":"13:44.380","Text":"so d is also constant, so we can take it out."},{"Start":"13:44.380 ","End":"13:47.635","Text":"If d divided by Epsilon_naught A,"},{"Start":"13:47.635 ","End":"13:53.950","Text":"and then we have our integration of dx divided by 1 plus x^2,"},{"Start":"13:53.950 ","End":"13:56.245","Text":"which is exactly this."},{"Start":"13:56.245 ","End":"14:00.790","Text":"We saw that this is going to be equal to tan to the minus"},{"Start":"14:00.790 ","End":"14:06.940","Text":"1 of x between the bounds of 0 and 1."},{"Start":"14:06.940 ","End":"14:11.380","Text":"We\u0027re left with dx divided by 1 plus x^2,"},{"Start":"14:11.380 ","End":"14:14.080","Text":"which we saw when we integrated this."},{"Start":"14:14.080 ","End":"14:20.980","Text":"Arctan of 1 is equal to Pi divided by 4 so we have d divided by"},{"Start":"14:20.980 ","End":"14:27.925","Text":"Epsilon_naught A multiplied by Pi divided by 4,"},{"Start":"14:27.925 ","End":"14:33.460","Text":"and then when we take arctan of 0,"},{"Start":"14:33.460 ","End":"14:37.030","Text":"we get 0 minus 0."},{"Start":"14:37.030 ","End":"14:41.830","Text":"In total, we have d divided by"},{"Start":"14:41.830 ","End":"14:48.520","Text":"Epsilon_naught A multiplied by Pi divided by 4."},{"Start":"14:48.520 ","End":"14:51.010","Text":"This is the answer to the question."},{"Start":"14:51.010 ","End":"14:52.690","Text":"This is the capacitance,"},{"Start":"14:52.690 ","End":"14:53.935","Text":"and that is it."},{"Start":"14:53.935 ","End":"14:56.240","Text":"That\u0027s the end of this lesson."}],"ID":22292},{"Watched":false,"Name":"Exercise 4","Duration":"30m 25s","ChapterTopicVideoID":21507,"CourseChapterTopicPlaylistID":99472,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.039","Text":"Hello. In this question,"},{"Start":"00:02.039 ","End":"00:04.605","Text":"we\u0027re going to be answering the following question."},{"Start":"00:04.605 ","End":"00:09.119","Text":"A parallel plate capacitor was partially deformed at production."},{"Start":"00:09.119 ","End":"00:12.929","Text":"Each plate is of length and width"},{"Start":"00:12.929 ","End":"00:18.330","Text":"a and the distance between the 2 plates was originally d,"},{"Start":"00:18.330 ","End":"00:26.429","Text":"but now due to the deformation at production the upper plate has a deviation of Theta."},{"Start":"00:26.429 ","End":"00:29.880","Text":"Where we can assume that Theta is much smaller than Pi."},{"Start":"00:29.880 ","End":"00:36.120","Text":"We\u0027re being told to calculate the capacitance of this capacitor as a function of Theta."},{"Start":"00:36.120 ","End":"00:41.490","Text":"In other words, we\u0027re looking for this equation."},{"Start":"00:42.100 ","End":"00:45.864","Text":"We\u0027re used to calculating the capacitance of"},{"Start":"00:45.864 ","End":"00:51.364","Text":"a parallel plate capacitor as being equal to Epsilon Naught A"},{"Start":"00:51.364 ","End":"00:55.015","Text":"divided by the distance between"},{"Start":"00:55.015 ","End":"01:00.805","Text":"the 2 plates given that there\u0027s no dielectric material inside the capacitor."},{"Start":"01:00.805 ","End":"01:08.445","Text":"Our problem here is that here our d or distance between the plates isn\u0027t a constant."},{"Start":"01:08.445 ","End":"01:12.309","Text":"We start from right over here where the distance is d and"},{"Start":"01:12.309 ","End":"01:16.510","Text":"slowly the distance between the 2 plates increases"},{"Start":"01:16.510 ","End":"01:20.079","Text":"due to this deviation of Theta reaching a maximum over"},{"Start":"01:20.079 ","End":"01:25.520","Text":"here at this rightmost side of the capacitor."},{"Start":"01:26.270 ","End":"01:29.094","Text":"Just like in a previous example,"},{"Start":"01:29.094 ","End":"01:34.254","Text":"we see that in order to deal with these types of questions we have to split up"},{"Start":"01:34.254 ","End":"01:40.300","Text":"our capacitor into many different sub-capacitors."},{"Start":"01:40.300 ","End":"01:45.207","Text":"What we do, we can see that the distance between the plates is changing."},{"Start":"01:45.207 ","End":"01:50.965","Text":"What we\u0027re going to do is we\u0027re going to cut the capacitor over here"},{"Start":"01:50.965 ","End":"01:57.159","Text":"lengthwise where the width of each sub capacitor,"},{"Start":"01:57.159 ","End":"02:01.240","Text":"this over here, we\u0027ll call this dx."},{"Start":"02:01.240 ","End":"02:06.160","Text":"We\u0027ll say that this axis is the x-axis."},{"Start":"02:06.160 ","End":"02:10.440","Text":"This width, the dx is a very tiny width."},{"Start":"02:10.440 ","End":"02:12.014","Text":"It\u0027s infinitesimal."},{"Start":"02:12.014 ","End":"02:16.655","Text":"Of course, if we\u0027ve cut it because the capacitor has"},{"Start":"02:16.655 ","End":"02:22.340","Text":"a width so we also cut it in this direction like so."},{"Start":"02:22.340 ","End":"02:27.042","Text":"We\u0027re cutting something like this and it goes in."},{"Start":"02:27.042 ","End":"02:32.370","Text":"We just have lots and lots of slices of capacitor."},{"Start":"02:33.410 ","End":"02:38.840","Text":"Let\u0027s say that the origin is right over here on this side.."},{"Start":"02:38.840 ","End":"02:45.769","Text":"Then we can say that if we take a random sub capacitor."},{"Start":"02:45.769 ","End":"02:47.420","Text":"Let\u0027s say this one over here."},{"Start":"02:47.420 ","End":"02:55.269","Text":"Let\u0027s say that the distance from the origin to this capacitor is a distance of x."},{"Start":"02:55.269 ","End":"02:59.455","Text":"What we\u0027re going to see is that as x changes"},{"Start":"02:59.455 ","End":"03:04.020","Text":"the value for our capacitance is also going to change."},{"Start":"03:05.600 ","End":"03:12.319","Text":"Because this distance over here or the width of each sub capacitor is so small,"},{"Start":"03:12.319 ","End":"03:16.805","Text":"we can assume that the distance,"},{"Start":"03:16.805 ","End":"03:21.120","Text":"so let\u0027s say the distance,"},{"Start":"03:21.120 ","End":"03:23.055","Text":"let\u0027s draw it in black,"},{"Start":"03:23.055 ","End":"03:29.420","Text":"from this point to this point and the distance from this point to this point is the same."},{"Start":"03:29.420 ","End":"03:33.755","Text":"Because the width is changing very very slightly."},{"Start":"03:33.755 ","End":"03:37.130","Text":"Let\u0027s say that the height,"},{"Start":"03:37.130 ","End":"03:41.089","Text":"so this over here is the height."},{"Start":"03:41.089 ","End":"03:46.565","Text":"Let\u0027s call this some kind of y and it\u0027s as a function of x."},{"Start":"03:46.565 ","End":"03:51.989","Text":"We can say that our y-axis is in this direction."},{"Start":"03:51.989 ","End":"03:59.370","Text":"We can see that as our x increases our y is going to increase as well."},{"Start":"04:00.800 ","End":"04:05.464","Text":"Because this width is so small so we can consider"},{"Start":"04:05.464 ","End":"04:10.410","Text":"the height over here as being a constant,"},{"Start":"04:10.410 ","End":"04:16.700","Text":"so we can consider this as a parallel plate capacitor and the regular one and therefore,"},{"Start":"04:16.700 ","End":"04:22.900","Text":"we can use this equation over here as if the height isn\u0027t changing."},{"Start":"04:22.900 ","End":"04:26.435","Text":"The capacitance of this small capacitor,"},{"Start":"04:26.435 ","End":"04:35.149","Text":"we\u0027ll call it dC is equal to Epsilon Naught multiplied by the surface area of the plates."},{"Start":"04:35.149 ","End":"04:37.800","Text":"That\u0027s this."},{"Start":"04:37.940 ","End":"04:42.049","Text":"That means that we have to multiply by the width of this,"},{"Start":"04:42.049 ","End":"04:50.150","Text":"which is dx and by the length over here."},{"Start":"04:50.150 ","End":"04:51.890","Text":"The length is a."},{"Start":"04:51.890 ","End":"04:53.209","Text":"This hasn\u0027t changed."},{"Start":"04:53.209 ","End":"04:57.025","Text":"The only thing that\u0027s changed is on this side over here."},{"Start":"04:57.025 ","End":"05:04.475","Text":"We have dx multiplied by a and then divided by the distance between the 2 plates,"},{"Start":"05:04.475 ","End":"05:13.120","Text":"which we already said is a constant and is equal to y as a function of x."},{"Start":"05:14.150 ","End":"05:19.155","Text":"What is our y as a function of x?"},{"Start":"05:19.155 ","End":"05:21.165","Text":"Let\u0027s draw it in green."},{"Start":"05:21.165 ","End":"05:26.554","Text":"First of all, we have this distance over here,"},{"Start":"05:26.554 ","End":"05:30.669","Text":"which as we know is d. Then,"},{"Start":"05:30.669 ","End":"05:34.580","Text":"we also have to add in this over here."},{"Start":"05:34.580 ","End":"05:36.218","Text":"This is the deviation."},{"Start":"05:36.218 ","End":"05:37.580","Text":"The deviation is of course,"},{"Start":"05:37.580 ","End":"05:39.890","Text":"as a function of Theta."},{"Start":"05:39.890 ","End":"05:42.064","Text":"Let\u0027s call this Delta y,"},{"Start":"05:42.064 ","End":"05:46.899","Text":"the change in y as a function of Theta or of x."},{"Start":"05:46.899 ","End":"05:48.900","Text":"Let\u0027s write this out here."},{"Start":"05:48.900 ","End":"05:54.859","Text":"We can see that our y as a function of x is made up of the distance,"},{"Start":"05:54.859 ","End":"05:57.800","Text":"d plus our deviation,"},{"Start":"05:57.800 ","End":"06:00.630","Text":"which we called Delta y."},{"Start":"06:00.630 ","End":"06:03.675","Text":"What is Delta y equal to?"},{"Start":"06:03.675 ","End":"06:09.920","Text":"What we can see is that we have a right-angle triangle over here."},{"Start":"06:09.920 ","End":"06:15.140","Text":"This is the right angle and we have a right-angle triangle going like this."},{"Start":"06:15.140 ","End":"06:23.800","Text":"We can see that this is our angle Theta and Delta y is the opposite."},{"Start":"06:24.290 ","End":"06:26.670","Text":"Let\u0027s write it here in the meantime,"},{"Start":"06:26.670 ","End":"06:28.734","Text":"Delta y is opposite."},{"Start":"06:28.734 ","End":"06:34.289","Text":"Then we also have this length over here, which is x."},{"Start":"06:34.540 ","End":"06:37.970","Text":"Of course, x is changing so it doesn\u0027t matter"},{"Start":"06:37.970 ","End":"06:40.669","Text":"if I said that x finishes over here or over here."},{"Start":"06:40.669 ","End":"06:41.974","Text":"It\u0027s always changing."},{"Start":"06:41.974 ","End":"06:44.509","Text":"I just didn\u0027t want to draw my Delta y on"},{"Start":"06:44.509 ","End":"06:49.100","Text":"this blue line over here because the diagram would get confusing."},{"Start":"06:49.100 ","End":"06:52.730","Text":"We have Delta y divided by x,"},{"Start":"06:52.730 ","End":"06:54.870","Text":"which is opposite over adjacent,"},{"Start":"06:54.910 ","End":"07:00.910","Text":"which is equal to tan of the angle."},{"Start":"07:02.390 ","End":"07:06.085","Text":"Tan of an angle is equal to opposite over adjacent,"},{"Start":"07:06.085 ","End":"07:09.215","Text":"which is Delta y divided by x in our example."},{"Start":"07:09.215 ","End":"07:17.425","Text":"Therefore, we can say that Delta y is equal to x multiplied by the tangent of Theta."},{"Start":"07:17.425 ","End":"07:20.224","Text":"We can write this over here,"},{"Start":"07:20.224 ","End":"07:23.854","Text":"x multiplied by tan of Theta."},{"Start":"07:23.854 ","End":"07:29.030","Text":"Therefore, we can substitute this into here so we have that y as a function"},{"Start":"07:29.030 ","End":"07:37.125","Text":"of x is equal to d plus x tan of Theta."},{"Start":"07:37.125 ","End":"07:41.460","Text":"Now I\u0027ll just rub this out to give us a bit more space."},{"Start":"07:42.710 ","End":"07:48.980","Text":"Now, let\u0027s plug this into our original equation for the capacitance."},{"Start":"07:48.980 ","End":"07:50.765","Text":"We get the dC,"},{"Start":"07:50.765 ","End":"07:54.845","Text":"the capacitance of this small piece is equal to Epsilon Naught"},{"Start":"07:54.845 ","End":"08:00.709","Text":"multiplied by adx divided by our y of x,"},{"Start":"08:00.709 ","End":"08:08.070","Text":"which is equal to d plus x tan of Theta."},{"Start":"08:09.800 ","End":"08:15.350","Text":"We have the capacitance of our little capacitor over here and obviously,"},{"Start":"08:15.350 ","End":"08:21.680","Text":"we want to make lots and lots of little capacitors and then add up the total capacitance."},{"Start":"08:21.680 ","End":"08:27.215","Text":"First we have to decide how all of these capacitors are connected."},{"Start":"08:27.215 ","End":"08:33.305","Text":"First of all, we can see that the capacitors are going to be connected side-by-side."},{"Start":"08:33.305 ","End":"08:36.885","Text":"Here would be our next capacitor,"},{"Start":"08:36.885 ","End":"08:44.275","Text":"which is on the side of the original capacitor or this first one that we took."},{"Start":"08:44.275 ","End":"08:48.949","Text":"Generally speaking, when the capacitors are connected side-by-side"},{"Start":"08:48.949 ","End":"08:53.869","Text":"then we can say that they are connected in parallel."},{"Start":"08:53.869 ","End":"08:56.355","Text":"However, let\u0027s take a look."},{"Start":"08:56.355 ","End":"08:59.824","Text":"We know that each plate is a conductor,"},{"Start":"08:59.824 ","End":"09:03.094","Text":"which means that the potential difference"},{"Start":"09:03.094 ","End":"09:08.430","Text":"between the 2 plates is going to be uniform throughout."},{"Start":"09:08.870 ","End":"09:14.059","Text":"Or the potential on each plate is going to be uniform and then that"},{"Start":"09:14.059 ","End":"09:16.009","Text":"means that the potential difference"},{"Start":"09:16.009 ","End":"09:18.979","Text":"between the 2 plates is going to be uniform throughout."},{"Start":"09:18.979 ","End":"09:22.414","Text":"Which means that the potential difference between"},{"Start":"09:22.414 ","End":"09:27.109","Text":"this plate over here and this plate over here on"},{"Start":"09:27.109 ","End":"09:31.144","Text":"this sub capacitor is going to be the same as the potential difference"},{"Start":"09:31.144 ","End":"09:36.135","Text":"if we would look at a capacitor over here between these 2 points."},{"Start":"09:36.135 ","End":"09:39.724","Text":"If the potential difference is even throughout,"},{"Start":"09:39.724 ","End":"09:47.970","Text":"then whereas constant then we can say that the capacitors are joined in parallel."},{"Start":"09:49.380 ","End":"09:55.630","Text":"The total capacitance when adding capacitors that are connected in"},{"Start":"09:55.630 ","End":"10:01.719","Text":"parallel is you just add up all of the individual capacitances,"},{"Start":"10:01.719 ","End":"10:04.465","Text":"and that\u0027s it and then you get the total capacitance."},{"Start":"10:04.465 ","End":"10:10.909","Text":"Because we\u0027re dealing with these infinitesimal widths for the capacitor,"},{"Start":"10:10.909 ","End":"10:17.900","Text":"we don\u0027t add up or how we do the adding up is via an integration."},{"Start":"10:17.940 ","End":"10:22.940","Text":"We just go into integrate like so."},{"Start":"10:23.460 ","End":"10:26.095","Text":"I just rewrote this here."},{"Start":"10:26.095 ","End":"10:29.455","Text":"Now we have to add in our bounds."},{"Start":"10:29.455 ","End":"10:31.779","Text":"Here, we can say is the origin,"},{"Start":"10:31.779 ","End":"10:35.109","Text":"which means that here, x=0 and then"},{"Start":"10:35.109 ","End":"10:39.070","Text":"the maximum x is the length of this side, which is a."},{"Start":"10:39.070 ","End":"10:46.450","Text":"We\u0027re integrating from 0 over here until a over here."},{"Start":"10:46.450 ","End":"10:49.780","Text":"This is a very easy integral."},{"Start":"10:49.780 ","End":"10:53.590","Text":"We can just assume that we\u0027re just"},{"Start":"10:53.590 ","End":"10:58.206","Text":"integrating and 1 divided by x with a few other constants,"},{"Start":"10:58.206 ","End":"11:01.525","Text":"because everything else is constant, including Theta."},{"Start":"11:01.525 ","End":"11:05.950","Text":"Our Theta is also a constant angle."},{"Start":"11:05.950 ","End":"11:08.020","Text":"Let\u0027s write it out over here."},{"Start":"11:08.020 ","End":"11:16.479","Text":"All we\u0027re going to have is epsilon naught a divided by the inner derivative,"},{"Start":"11:16.479 ","End":"11:24.430","Text":"which is just tan of Theta,"},{"Start":"11:24.430 ","End":"11:29.440","Text":"and all of this is multiplied by ln(d) plus"},{"Start":"11:29.440 ","End":"11:37.430","Text":"x tan of Theta between the bounds of 0 and a."},{"Start":"11:37.530 ","End":"11:40.570","Text":"Let\u0027s substitute in the balance."},{"Start":"11:40.570 ","End":"11:44.335","Text":"We have epsilon naught a divided by tan of Theta ln."},{"Start":"11:44.335 ","End":"11:50.680","Text":"Then all we\u0027ll have is d plus instead of the x we have a,"},{"Start":"11:50.680 ","End":"11:57.189","Text":"so a tan of Theta divided by d plus instead of the x,"},{"Start":"11:57.189 ","End":"12:01.690","Text":"we have 0, so d plus 0 tan of Theta."},{"Start":"12:01.690 ","End":"12:08.620","Text":"That\u0027s just divided by d. This is the answer."},{"Start":"12:08.620 ","End":"12:10.179","Text":"In a lot of questions,"},{"Start":"12:10.179 ","End":"12:14.688","Text":"they\u0027ll ask you to use the small angle approximation."},{"Start":"12:14.688 ","End":"12:15.790","Text":"Because in the question,"},{"Start":"12:15.790 ","End":"12:19.360","Text":"we were told that Theta is much smaller than Pi."},{"Start":"12:19.360 ","End":"12:22.239","Text":"In the small angle approximation,"},{"Start":"12:22.239 ","End":"12:29.979","Text":"we know that tan of Theta is approximately equal to sin of Theta,"},{"Start":"12:29.979 ","End":"12:32.409","Text":"which again, in small angles,"},{"Start":"12:32.409 ","End":"12:36.590","Text":"sin of Theta is approximately equal to Theta."},{"Start":"12:37.200 ","End":"12:40.989","Text":"Therefore, what we can write is that"},{"Start":"12:40.989 ","End":"12:45.039","Text":"our capacitance is equal to epsilon naught a divided"},{"Start":"12:45.039 ","End":"12:52.359","Text":"by Theta multiplied by ln and I\u0027ll cancel out the d\u0027s over here,"},{"Start":"12:52.359 ","End":"13:01.880","Text":"divided by ln of 1 plus a divided by d multiplied by Theta."},{"Start":"13:02.760 ","End":"13:07.180","Text":"Now we can do a few more approximations."},{"Start":"13:07.180 ","End":"13:13.840","Text":"We can see that Theta is a very small value because we\u0027re using small angles."},{"Start":"13:13.840 ","End":"13:17.665","Text":"This coefficient multiplied by a small number,"},{"Start":"13:17.665 ","End":"13:20.860","Text":"it\u0027s still going to be a very small value."},{"Start":"13:20.860 ","End":"13:28.450","Text":"In that case, we can derive our ln equation."},{"Start":"13:28.450 ","End":"13:33.880","Text":"What we can do is we can use this,"},{"Start":"13:33.880 ","End":"13:38.995","Text":"that ln(1) plus x,"},{"Start":"13:38.995 ","End":"13:42.175","Text":"where x is approaching 0,"},{"Start":"13:42.175 ","End":"13:46.399","Text":"which is what is going on in our case."},{"Start":"13:46.650 ","End":"13:55.525","Text":"This is equal to x plus x^2 divided by 2."},{"Start":"13:55.525 ","End":"13:59.169","Text":"Of course, we have lots of other terms over here."},{"Start":"13:59.169 ","End":"14:06.370","Text":"But what we can see is that x is going to be much larger than x^2 divided by 2,"},{"Start":"14:06.370 ","End":"14:09.400","Text":"because a small number squared is even smaller."},{"Start":"14:09.400 ","End":"14:16.100","Text":"Therefore, it\u0027s enough for me to just take this first term over here."},{"Start":"14:16.410 ","End":"14:19.375","Text":"Of course, in our case,"},{"Start":"14:19.375 ","End":"14:26.930","Text":"our x we said is equal to a divided by d multiplied by Theta."},{"Start":"14:26.930 ","End":"14:30.435","Text":"Let\u0027s scroll a bit to the side."},{"Start":"14:30.435 ","End":"14:36.435","Text":"Therefore, we can substitute this into our equation."},{"Start":"14:36.435 ","End":"14:41.849","Text":"We get that the total capacitance is equal to epsilon naught multiplied"},{"Start":"14:41.849 ","End":"14:48.595","Text":"by a divided by Theta multiplied by this approximation."},{"Start":"14:48.595 ","End":"14:50.410","Text":"We\u0027re just taking the x,"},{"Start":"14:50.410 ","End":"14:58.390","Text":"where x is a divided by d multiplied by Theta."},{"Start":"14:58.390 ","End":"15:03.355","Text":"Now we can see that our Thetas cancel out."},{"Start":"15:03.355 ","End":"15:12.140","Text":"Then we can just rewrite this as epsilon naught a^2 divided by d."},{"Start":"15:12.990 ","End":"15:17.500","Text":"We can immediately see 2 things: number 1,"},{"Start":"15:17.500 ","End":"15:20.560","Text":"when we use these small angle approximations,"},{"Start":"15:20.560 ","End":"15:23.485","Text":"our capacitance is independent of Theta,"},{"Start":"15:23.485 ","End":"15:26.020","Text":"and the second thing is that we get"},{"Start":"15:26.020 ","End":"15:31.315","Text":"the exact same equation for a normal parallel plate capacitor."},{"Start":"15:31.315 ","End":"15:34.030","Text":"What we have is that the capacitance is equal to"},{"Start":"15:34.030 ","End":"15:37.225","Text":"epsilon naught multiplied by the surface area"},{"Start":"15:37.225 ","End":"15:43.375","Text":"of the plates divided by the distance between the 2 plates."},{"Start":"15:43.375 ","End":"15:49.314","Text":"What we can see is that if we take the first term of the small angle approximation,"},{"Start":"15:49.314 ","End":"15:53.544","Text":"our capacitor, even with this Theta deviation,"},{"Start":"15:53.544 ","End":"15:58.040","Text":"acts like a regular parallel plate capacitor."},{"Start":"15:58.890 ","End":"16:04.704","Text":"Now what we\u0027re going to do is we\u0027re actually going to add in the second term from"},{"Start":"16:04.704 ","End":"16:10.240","Text":"our ln approximation for these small values."},{"Start":"16:10.240 ","End":"16:14.380","Text":"That\u0027s because we know we were told by"},{"Start":"16:14.380 ","End":"16:18.955","Text":"the production line that there is this problem with this deviation of Theta."},{"Start":"16:18.955 ","End":"16:26.139","Text":"Which means that if we\u0027re looking in a slightly less precise way,"},{"Start":"16:26.139 ","End":"16:30.280","Text":"then yes, this parallel plate capacitor with a slight deviation."},{"Start":"16:30.280 ","End":"16:33.129","Text":"We know it is actually a slight deviation because we were"},{"Start":"16:33.129 ","End":"16:36.760","Text":"told here that this is a very small angle Theta."},{"Start":"16:36.760 ","End":"16:40.179","Text":"We know that it\u0027s going to act like a parallel plate capacitor."},{"Start":"16:40.179 ","End":"16:44.470","Text":"However, we want to be more precise and we want to really know"},{"Start":"16:44.470 ","End":"16:49.540","Text":"what the slight deviation does to the capacitance."},{"Start":"16:49.540 ","End":"16:52.885","Text":"Let\u0027s just take this out."},{"Start":"16:52.885 ","End":"16:57.160","Text":"Now let\u0027s calculate the capacitance when we take into account"},{"Start":"16:57.160 ","End":"17:01.960","Text":"also our next term, our x^2 term."},{"Start":"17:01.960 ","End":"17:09.235","Text":"What we\u0027ll have is epsilon naught a divided by Theta multiplied by,"},{"Start":"17:09.235 ","End":"17:14.095","Text":"so what we\u0027ll have is x."},{"Start":"17:14.095 ","End":"17:22.495","Text":"Our x is a divided by d multiplied by Theta plus our x^2 term."},{"Start":"17:22.495 ","End":"17:32.840","Text":"That will be equal to a^2 divided by d^2 multiplied by Theta^2 divided by 2."},{"Start":"17:33.420 ","End":"17:36.144","Text":"No, I made a mistake."},{"Start":"17:36.144 ","End":"17:40.029","Text":"The small angle approximation for ln(1) plus x,"},{"Start":"17:40.029 ","End":"17:45.535","Text":"when x is approaching 0 is x minus x^2 divided by 2."},{"Start":"17:45.535 ","End":"17:50.875","Text":"Here we\u0027re meant to have a minus."},{"Start":"17:50.875 ","End":"17:55.720","Text":"Now what we can do is we can join up our like terms."},{"Start":"17:55.720 ","End":"17:58.480","Text":"What we\u0027ll have is epsilon naught."},{"Start":"17:58.480 ","End":"18:05.545","Text":"Then we can see that we have a^2 divided by"},{"Start":"18:05.545 ","End":"18:10.629","Text":"d. Then this is multiplied by"},{"Start":"18:10.629 ","End":"18:16.210","Text":"1 because this Theta cancels out 1 and then minus."},{"Start":"18:16.210 ","End":"18:22.265","Text":"Then we have another a divided by d over here multiplied by Theta."},{"Start":"18:22.265 ","End":"18:25.544","Text":"Of course, don\u0027t forget to have this."},{"Start":"18:25.544 ","End":"18:35.020","Text":"Now what we have is our first-order result for the capacitance."},{"Start":"18:35.700 ","End":"18:44.259","Text":"Our zeroth-order gave us the exact capacitance of a normal parallel plate capacitor."},{"Start":"18:44.259 ","End":"18:47.784","Text":"If we take our first order,"},{"Start":"18:47.784 ","End":"18:53.755","Text":"we get the capacitance of a regular parallel plate capacitor."},{"Start":"18:53.755 ","End":"18:57.325","Text":"Then we subtract this small term over here,"},{"Start":"18:57.325 ","End":"19:03.995","Text":"which is caused by this deviation of Theta in the angle of the upper plate."},{"Start":"19:03.995 ","End":"19:06.495","Text":"We calculated the answer."},{"Start":"19:06.495 ","End":"19:10.380","Text":"Now let\u0027s look at another question that they can ask us."},{"Start":"19:10.380 ","End":"19:15.000","Text":"The next question that they can ask us is, what is Sigma?"},{"Start":"19:15.000 ","End":"19:20.780","Text":"Or what is the surface charge distribution?"},{"Start":"19:21.570 ","End":"19:27.295","Text":"Let\u0027s clear some space so that we can answer this question."},{"Start":"19:27.295 ","End":"19:32.919","Text":"The only way that we\u0027re going to get some charge distribution is"},{"Start":"19:32.919 ","End":"19:38.290","Text":"if our capacitor is connected to some kind of battery."},{"Start":"19:38.290 ","End":"19:47.080","Text":"Let\u0027s connect it to some kind of voltage source and then let\u0027s call it V_0."},{"Start":"19:47.080 ","End":"19:52.405","Text":"It\u0027s connected to V_0 over here."},{"Start":"19:52.405 ","End":"19:58.090","Text":"What we want to do is we want to find the charge density or the charge distribution,"},{"Start":"19:58.090 ","End":"19:59.739","Text":"which as we know,"},{"Start":"19:59.739 ","End":"20:03.235","Text":"because our capacitor has this deviation over here,"},{"Start":"20:03.235 ","End":"20:08.860","Text":"the charge distribution might not be uniform throughout."},{"Start":"20:08.860 ","End":"20:12.595","Text":"Of course, we have two ways to calculate this."},{"Start":"20:12.595 ","End":"20:16.345","Text":"The first way is to calculate the jump in the electric field,"},{"Start":"20:16.345 ","End":"20:22.419","Text":"which of course has to be due to the charge distribution or"},{"Start":"20:22.419 ","End":"20:29.050","Text":"the second way is to calculate the infinitesimal charges and add them all up."},{"Start":"20:29.050 ","End":"20:34.270","Text":"Let\u0027s start from the second method."},{"Start":"20:34.270 ","End":"20:37.135","Text":"Then after we show method 2,"},{"Start":"20:37.135 ","End":"20:38.890","Text":"we\u0027ll show method number 1."},{"Start":"20:38.890 ","End":"20:41.395","Text":"In method number 2,"},{"Start":"20:41.395 ","End":"20:49.180","Text":"we\u0027re going to look at the charges on infinitesimal parallel plate capacitors."},{"Start":"20:49.180 ","End":"20:54.519","Text":"Here we\u0027re looking at this capacitor and we know that"},{"Start":"20:54.519 ","End":"20:59.740","Text":"the potential difference between this plate and"},{"Start":"20:59.740 ","End":"21:06.655","Text":"this plate is the same for all the other small capacitors."},{"Start":"21:06.655 ","End":"21:14.245","Text":"That is because we already saw before that they are connected to the same voltage source."},{"Start":"21:14.245 ","End":"21:16.540","Text":"All of these sub capacitors,"},{"Start":"21:16.540 ","End":"21:17.904","Text":"in this case over here,"},{"Start":"21:17.904 ","End":"21:20.440","Text":"are connected in parallel."},{"Start":"21:20.440 ","End":"21:28.400","Text":"Which means that the voltage on each capacitor is going to be the same."},{"Start":"21:29.460 ","End":"21:33.714","Text":"In that case, we can use our equation."},{"Start":"21:33.714 ","End":"21:42.175","Text":"Remember where our charge Q is equal to the voltage divided by the capacitance."},{"Start":"21:42.175 ","End":"21:47.020","Text":"In our case, because we\u0027re dealing with these infinitesimal capacitors,"},{"Start":"21:47.020 ","End":"21:49.180","Text":"so infinitesimal charges,"},{"Start":"21:49.180 ","End":"21:54.400","Text":"what we have is that dq is equal to our voltage,"},{"Start":"21:54.400 ","End":"21:58.570","Text":"which is V naught divided by our capacitance."},{"Start":"21:58.570 ","End":"22:02.845","Text":"It\u0027s the capacitance of this small sub-capacitor."},{"Start":"22:02.845 ","End":"22:05.920","Text":"This is equal to dc."},{"Start":"22:05.920 ","End":"22:08.049","Text":"Sorry, I made a mistake,"},{"Start":"22:08.049 ","End":"22:11.080","Text":"Q is equal to Vc."},{"Start":"22:11.080 ","End":"22:18.250","Text":"That means that dq is equal to V_0dc."},{"Start":"22:18.250 ","End":"22:20.080","Text":"What is our dc?"},{"Start":"22:20.080 ","End":"22:23.365","Text":"Our dc is this."},{"Start":"22:23.365 ","End":"22:30.955","Text":"Remember in the first half of this lesson,"},{"Start":"22:30.955 ","End":"22:33.355","Text":"we said that this was dc."},{"Start":"22:33.355 ","End":"22:36.850","Text":"Let\u0027s, therefore, substitute this end."},{"Start":"22:36.850 ","End":"22:39.280","Text":"We\u0027ll carry it on over here."},{"Start":"22:39.280 ","End":"22:46.134","Text":"We\u0027ll have that dq is equal to V_0 multiplied by"},{"Start":"22:46.134 ","End":"22:56.870","Text":"Epsilon naught adx divided by d plus x tan(Theta)."},{"Start":"22:58.740 ","End":"23:02.544","Text":"Just a reminder, we can use this equation"},{"Start":"23:02.544 ","End":"23:06.069","Text":"for Epsilon naught multiplied by the surface area of"},{"Start":"23:06.069 ","End":"23:09.159","Text":"the plate divided by the distance between"},{"Start":"23:09.159 ","End":"23:13.389","Text":"the plates because we split this up into many tiny capacitors,"},{"Start":"23:13.389 ","End":"23:19.195","Text":"so we can assume that the distance between the plates over here is uniform."},{"Start":"23:19.195 ","End":"23:26.240","Text":"Because this is such a small capacitor that we can\u0027t really see this height change."},{"Start":"23:27.210 ","End":"23:32.334","Text":"This is dq but of course, we want Sigma."},{"Start":"23:32.334 ","End":"23:34.479","Text":"Sigma, as we know,"},{"Start":"23:34.479 ","End":"23:42.384","Text":"is equal to our total charge divided by the surface area."},{"Start":"23:42.384 ","End":"23:46.975","Text":"S is usually the surface area or we could write A."},{"Start":"23:46.975 ","End":"23:49.720","Text":"Therefore, in our case,"},{"Start":"23:49.720 ","End":"23:57.625","Text":"Sigma is equal to dq divided by this surface area over here."},{"Start":"23:57.625 ","End":"24:00.290","Text":"Let\u0027s call it ds."},{"Start":"24:00.840 ","End":"24:06.807","Text":"Therefore, we can say that our Sigma is equal to"},{"Start":"24:06.807 ","End":"24:13.670","Text":"V_0 Epsilon naught adx divided by,"},{"Start":"24:13.670 ","End":"24:20.830","Text":"we still have this, so we have d plus x tan(Theta),"},{"Start":"24:20.830 ","End":"24:23.545","Text":"and then we\u0027re dividing by ds."},{"Start":"24:23.545 ","End":"24:25.629","Text":"We could just put it on the denominator,"},{"Start":"24:25.629 ","End":"24:28.345","Text":"so ds is the surface area over here."},{"Start":"24:28.345 ","End":"24:35.004","Text":"We\u0027re taking the width of the plate which is a multiplied by this length over here,"},{"Start":"24:35.004 ","End":"24:38.005","Text":"which we said was dx."},{"Start":"24:38.005 ","End":"24:43.510","Text":"We can cancel out adx from this."},{"Start":"24:43.510 ","End":"24:49.149","Text":"What we get is that our Sigma is equal to V_0 multiplied by"},{"Start":"24:49.149 ","End":"24:56.180","Text":"Epsilon_0 divided by d plus x tan(Theta)."},{"Start":"24:57.810 ","End":"25:01.299","Text":"Now a little note, the only reason I could use"},{"Start":"25:01.299 ","End":"25:04.630","Text":"this equation is because I\u0027m dividing by ds,"},{"Start":"25:04.630 ","End":"25:06.985","Text":"the very small surface area."},{"Start":"25:06.985 ","End":"25:08.919","Text":"Where I can assume that in"},{"Start":"25:08.919 ","End":"25:14.200","Text":"this small surface area that my charge distribution is uniform because it\u0027s so"},{"Start":"25:14.200 ","End":"25:16.719","Text":"small that the change will be something"},{"Start":"25:16.719 ","End":"25:21.729","Text":"that we don\u0027t really have to pay much attention to."},{"Start":"25:21.729 ","End":"25:23.559","Text":"It will be such a small change."},{"Start":"25:23.559 ","End":"25:27.100","Text":"However, if I would find the total charge on"},{"Start":"25:27.100 ","End":"25:30.835","Text":"the whole plate and divide it by the total surface area of the plates,"},{"Start":"25:30.835 ","End":"25:35.169","Text":"then I wouldn\u0027t be able to use this equation because we\u0027re using"},{"Start":"25:35.169 ","End":"25:39.849","Text":"such a large surface area that I know or"},{"Start":"25:39.849 ","End":"25:44.050","Text":"at least I have a very strong assumption that"},{"Start":"25:44.050 ","End":"25:48.760","Text":"the charge distribution is not going to be uniform because,"},{"Start":"25:48.760 ","End":"25:51.805","Text":"of course, we have this deviation over here."},{"Start":"25:51.805 ","End":"25:54.145","Text":"Therefore in this large surface area,"},{"Start":"25:54.145 ","End":"25:58.554","Text":"there will be dramatic changes in the charge distribution."},{"Start":"25:58.554 ","End":"26:02.875","Text":"I can only use this because I\u0027m using the small surface area."},{"Start":"26:02.875 ","End":"26:06.144","Text":"Therefore, I can assume that in this surface area,"},{"Start":"26:06.144 ","End":"26:10.400","Text":"the charge distribution is uniform."},{"Start":"26:11.310 ","End":"26:14.544","Text":"Now let\u0027s look at the first method,"},{"Start":"26:14.544 ","End":"26:19.510","Text":"which is to calculate Sigma via the jump in the E-field."},{"Start":"26:19.510 ","End":"26:23.620","Text":"First of all, we know that when we\u0027re dealing with a capacitor,"},{"Start":"26:23.620 ","End":"26:26.035","Text":"the E-field on the outside,"},{"Start":"26:26.035 ","End":"26:29.410","Text":"therefore above and below the capacitor is equal to 0,"},{"Start":"26:29.410 ","End":"26:34.250","Text":"and we only have an E-field within the capacitor."},{"Start":"26:35.550 ","End":"26:39.699","Text":"The next thing we know and that we\u0027ve already spoken about"},{"Start":"26:39.699 ","End":"26:43.569","Text":"is that the potential difference or the voltage between"},{"Start":"26:43.569 ","End":"26:47.470","Text":"the two plates is also uniform throughout because"},{"Start":"26:47.470 ","End":"26:52.359","Text":"the capacitor plates are conductors and they\u0027re connected to this voltage source V_0."},{"Start":"26:52.359 ","End":"26:56.710","Text":"So we have a uniform potential difference."},{"Start":"26:56.710 ","End":"27:00.070","Text":"What I want to do is I want to calculate my E-field."},{"Start":"27:00.070 ","End":"27:02.935","Text":"Again, because of this deviation,"},{"Start":"27:02.935 ","End":"27:07.570","Text":"I have to use this idea of splitting up"},{"Start":"27:07.570 ","End":"27:13.600","Text":"the original capacitor into many sub-capacitors that are extremely small."},{"Start":"27:13.600 ","End":"27:15.895","Text":"Let\u0027s look at this capacitor."},{"Start":"27:15.895 ","End":"27:22.555","Text":"I\u0027m looking at this capacitor and I\u0027m considering it a regular parallel plate capacitor."},{"Start":"27:22.555 ","End":"27:29.680","Text":"Again, I can say that it\u0027s parallel because this distance dx is so small,"},{"Start":"27:29.680 ","End":"27:34.849","Text":"I can assume that there is no deviation and that the plates are,"},{"Start":"27:34.849 ","End":"27:36.920","Text":"in fact, parallel."},{"Start":"27:38.730 ","End":"27:41.845","Text":"Let\u0027s see."},{"Start":"27:41.845 ","End":"27:43.075","Text":"We\u0027re on method number 1,"},{"Start":"27:43.075 ","End":"27:45.055","Text":"scroll down a little bit."},{"Start":"27:45.055 ","End":"27:47.439","Text":"First of all, if they\u0027re parallel,"},{"Start":"27:47.439 ","End":"27:51.370","Text":"we know that in a parallel plate capacitor,"},{"Start":"27:51.370 ","End":"27:55.480","Text":"the electric field throughout is uniform."},{"Start":"27:55.480 ","End":"27:58.570","Text":"The electric field at a point over here and"},{"Start":"27:58.570 ","End":"28:02.600","Text":"the electric field at a point over here is the same."},{"Start":"28:02.670 ","End":"28:09.429","Text":"We know that the general equation"},{"Start":"28:09.429 ","End":"28:14.589","Text":"is that the voltage on the parallel plate capacitor is equal to the electric field,"},{"Start":"28:14.589 ","End":"28:19.810","Text":"which is uniform, multiplied by the distance between the plates."},{"Start":"28:19.810 ","End":"28:22.975","Text":"In our case, we have a voltage V_0,"},{"Start":"28:22.975 ","End":"28:28.570","Text":"which is equal to the electric field multiplied by the distance between the plates,"},{"Start":"28:28.570 ","End":"28:32.005","Text":"which we said was equal to yx."},{"Start":"28:32.005 ","End":"28:35.140","Text":"What is yx?"},{"Start":"28:35.140 ","End":"28:40.839","Text":"Yx is equal to this denominator over here we spoke about at the beginning of the lesson,"},{"Start":"28:40.839 ","End":"28:47.170","Text":"so d plus x tan(Theta)."},{"Start":"28:48.450 ","End":"28:59.740","Text":"We know that our Sigma is equal to Epsilon naught multiplied by the jump in the E-field."},{"Start":"28:59.740 ","End":"29:04.915","Text":"Therefore it\u0027s equal to Epsilon naught multiplied by,"},{"Start":"29:04.915 ","End":"29:10.749","Text":"we look at two very close points to the plates."},{"Start":"29:10.749 ","End":"29:15.429","Text":"One which is located in the capacitor and one which is located out of the capacitor,"},{"Start":"29:15.429 ","End":"29:19.910","Text":"and the electric field at this point inside the capacitors, of course,"},{"Start":"29:19.910 ","End":"29:27.730","Text":"this V_0 divided by d plus x tan(Theta)."},{"Start":"29:27.730 ","End":"29:31.495","Text":"Then we subtract the electric field outside the capacitor,"},{"Start":"29:31.495 ","End":"29:34.150","Text":"the electric field at this point over here,"},{"Start":"29:34.150 ","End":"29:36.460","Text":"which is of course 0."},{"Start":"29:36.460 ","End":"29:42.075","Text":"Therefore, we get that our Sigma is equal to Epsilon naught,"},{"Start":"29:42.075 ","End":"29:49.270","Text":"V naught divided by d plus x tan(Theta)."},{"Start":"29:51.900 ","End":"29:56.810","Text":"We see that we get the exact same answer."},{"Start":"29:57.630 ","End":"30:03.910","Text":"Of course, this is correct if this is the positive side of the battery and of course,"},{"Start":"30:03.910 ","End":"30:12.765","Text":"the jump and the electric field on top would lead us to a negative charge density."},{"Start":"30:12.765 ","End":"30:15.660","Text":"Then we can see that the bottom plate is positively"},{"Start":"30:15.660 ","End":"30:20.330","Text":"charged and that the top plate is negatively charged."},{"Start":"30:20.330 ","End":"30:26.540","Text":"That\u0027s it. We get the exact same answer and that is the end of the lesson."}],"ID":22288},{"Watched":false,"Name":"Stored on a Capacitor","Duration":"17m 46s","ChapterTopicVideoID":21308,"CourseChapterTopicPlaylistID":99472,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:07.360","Text":"Hello, in this lesson we\u0027re going to be speaking about the energy stored on a capacitor."},{"Start":"00:07.580 ","End":"00:12.645","Text":"The equation for the energy is stored on a capacitor is"},{"Start":"00:12.645 ","End":"00:17.310","Text":"equal to 1/2 q^2 divided by the capacitance."},{"Start":"00:17.310 ","End":"00:22.715","Text":"Where of course this is the charge on the positively charged capacitor plate."},{"Start":"00:22.715 ","End":"00:28.395","Text":"If you substitute in the equation for capacitance,"},{"Start":"00:28.395 ","End":"00:35.580","Text":"where C is equal to q divided by V. Then we can also get that this is equal to 1/2 CV^2,"},{"Start":"00:35.580 ","End":"00:40.035","Text":"and of course it\u0027s also equal to 1/2 qV."},{"Start":"00:40.035 ","End":"00:44.195","Text":"Right now we\u0027re going to use these equations in order to solve"},{"Start":"00:44.195 ","End":"00:48.380","Text":"a question involving a parallel plate capacitor,"},{"Start":"00:48.380 ","End":"00:55.630","Text":"where we have some dielectric material placed inside of part of the capacitor."},{"Start":"00:56.390 ","End":"00:58.715","Text":"In a previous lesson,"},{"Start":"00:58.715 ","End":"01:02.209","Text":"we looked at this example where we have a capacitor"},{"Start":"01:02.209 ","End":"01:05.735","Text":"where the two plates are a distance d away from one another."},{"Start":"01:05.735 ","End":"01:09.630","Text":"The width of each plate is a and the length of each plate is b,"},{"Start":"01:09.630 ","End":"01:14.690","Text":"and the distance x into the capacitor,"},{"Start":"01:14.690 ","End":"01:18.470","Text":"we have a dielectric material placed over here and we know"},{"Start":"01:18.470 ","End":"01:24.260","Text":"that d is much smaller than our size ab."},{"Start":"01:24.860 ","End":"01:30.380","Text":"In this case, we saw that the total capacitance of"},{"Start":"01:30.380 ","End":"01:35.570","Text":"this capacitor was equal to Epsilon naught multiplied by a"},{"Start":"01:35.570 ","End":"01:45.975","Text":"divided by d multiplied by x plus epsilon r multiplied by b minus x."},{"Start":"01:45.975 ","End":"01:55.380","Text":"Our question over here is to find the energy stored on the capacitor as a function of x."},{"Start":"01:55.880 ","End":"02:00.845","Text":"In this case, the energy is dependent on whether"},{"Start":"02:00.845 ","End":"02:05.585","Text":"our capacitor is connected to some battery or voltage source,"},{"Start":"02:05.585 ","End":"02:09.240","Text":"or if it is not connected."},{"Start":"02:09.290 ","End":"02:15.865","Text":"Let\u0027s take the case that our capacitor is not connected to some battery."},{"Start":"02:15.865 ","End":"02:20.635","Text":"But what we have is some charge stored on the capacitor."},{"Start":"02:20.635 ","End":"02:25.700","Text":"Let\u0027s say that the upper plate has a charge of plus Q,"},{"Start":"02:25.700 ","End":"02:30.290","Text":"and the bottom plate has a charge of negative Q."},{"Start":"02:31.360 ","End":"02:35.630","Text":"If our capacitor isn\u0027t connected to a voltage source,"},{"Start":"02:35.630 ","End":"02:40.230","Text":"that means that the charge is going to be constant."},{"Start":"02:40.230 ","End":"02:42.245","Text":"Why is the charge constant?"},{"Start":"02:42.245 ","End":"02:47.350","Text":"Because there\u0027s no wires or anything that the charge can escape from."},{"Start":"02:47.350 ","End":"02:52.390","Text":"The charge is stuck on each plate if you want to think of it like that."},{"Start":"02:52.390 ","End":"02:54.725","Text":"If we have a constant charge,"},{"Start":"02:54.725 ","End":"02:58.760","Text":"then we can use this equation because our Q isn\u0027t changing."},{"Start":"02:58.760 ","End":"03:03.290","Text":"However, if we would use an equation that links the voltage,"},{"Start":"03:03.290 ","End":"03:05.720","Text":"or the potential difference between the two plates,"},{"Start":"03:05.720 ","End":"03:09.410","Text":"we can see that the potential difference is"},{"Start":"03:09.410 ","End":"03:14.820","Text":"different depending on the dielectric material."},{"Start":"03:14.860 ","End":"03:18.410","Text":"Because the capacitor isn\u0027t connected to a source,"},{"Start":"03:18.410 ","End":"03:23.600","Text":"the potential difference at each point along the capacitor is"},{"Start":"03:23.600 ","End":"03:29.445","Text":"going to be different depending on if there\u0027s a dielectric material or not."},{"Start":"03:29.445 ","End":"03:33.905","Text":"That means we can\u0027t use these two equations because V isn\u0027t constant."},{"Start":"03:33.905 ","End":"03:38.735","Text":"But we can use this equation because our Q is constant."},{"Start":"03:38.735 ","End":"03:42.860","Text":"Our Q is constant because there\u0027s nowhere for our charge to escape,"},{"Start":"03:42.860 ","End":"03:45.935","Text":"because it isn\u0027t connected to a source."},{"Start":"03:45.935 ","End":"03:52.960","Text":"Whereas our V isn\u0027t constant because it is not connected to a source."},{"Start":"03:53.690 ","End":"04:00.170","Text":"In that case, we can say that our energy as a function of x in"},{"Start":"04:00.170 ","End":"04:07.070","Text":"the capacitor is going to be equal to1/2 multiplied by our charge squared."},{"Start":"04:07.070 ","End":"04:09.360","Text":"Let\u0027s say that we know what q is,"},{"Start":"04:09.360 ","End":"04:12.890","Text":"so Q^2 divided by our capacitance,"},{"Start":"04:12.890 ","End":"04:16.880","Text":"which is also as a function of x."},{"Start":"04:16.880 ","End":"04:18.725","Text":"Let\u0027s write it out,"},{"Start":"04:18.725 ","End":"04:23.360","Text":"substituting in what we got over here for the capacitance."},{"Start":"04:23.360 ","End":"04:30.515","Text":"This is the energy stored on a capacitor which is not connected to a voltage source."},{"Start":"04:30.515 ","End":"04:35.180","Text":"Now, let\u0027s say that the dielectric material is entering,"},{"Start":"04:35.180 ","End":"04:39.095","Text":"it\u0027s being pushed between the parallel plates."},{"Start":"04:39.095 ","End":"04:45.230","Text":"Let\u0027s find the energy stored on the capacitor"},{"Start":"04:45.230 ","End":"04:48.230","Text":"when the dielectric material is exactly beginning"},{"Start":"04:48.230 ","End":"04:51.435","Text":"to enter the plates, between the plates."},{"Start":"04:51.435 ","End":"04:55.730","Text":"That is when our x is equal to so it will begin"},{"Start":"04:55.730 ","End":"05:00.230","Text":"entering the plates when this side of the dielectric is located here."},{"Start":"05:00.230 ","End":"05:02.075","Text":"When x is over here,"},{"Start":"05:02.075 ","End":"05:05.545","Text":"which is just the length of the capacitor plates,"},{"Start":"05:05.545 ","End":"05:08.175","Text":"which is equal to b."},{"Start":"05:08.175 ","End":"05:12.560","Text":"Then we can say that the energy is going to be equal to Q^2 divided"},{"Start":"05:12.560 ","End":"05:17.110","Text":"by 2 Epsilon naught a divided by d,"},{"Start":"05:17.110 ","End":"05:20.720","Text":"and then we substitute an x,"},{"Start":"05:20.720 ","End":"05:27.755","Text":"which is b plus Epsilon r multiplied by b minus x,"},{"Start":"05:27.755 ","End":"05:31.310","Text":"which in this case x is equal to b."},{"Start":"05:31.310 ","End":"05:34.585","Text":"Then what we get is that"},{"Start":"05:34.585 ","End":"05:39.500","Text":"this term cancels out and so we just have that this is equal to Q^2"},{"Start":"05:39.500 ","End":"05:48.165","Text":"divided by 2 Epsilon naught a divided by d multiplied by b."},{"Start":"05:48.165 ","End":"05:51.665","Text":"We have twice the capacitance"},{"Start":"05:51.665 ","End":"05:56.920","Text":"Epsilon naught a multiplied by b as the surface area of the plate divided by d,"},{"Start":"05:56.920 ","End":"05:59.985","Text":"the distance between the two plates."},{"Start":"05:59.985 ","End":"06:05.060","Text":"We get that the energy is the charge squared divided by"},{"Start":"06:05.060 ","End":"06:07.835","Text":"twice the capacitance of"},{"Start":"06:07.835 ","End":"06:13.580","Text":"a parallel plate capacitor when there is no dielectric material inside the plate."},{"Start":"06:13.580 ","End":"06:15.230","Text":"Which makes sense because we said that"},{"Start":"06:15.230 ","End":"06:18.260","Text":"the dielectric material is right on the edge of the plate."},{"Start":"06:18.260 ","End":"06:21.725","Text":"It hasn\u0027t yet entered the capacitor."},{"Start":"06:21.725 ","End":"06:28.210","Text":"Similarly, I can calculate the energy stored across,"},{"Start":"06:28.210 ","End":"06:30.530","Text":"or on the capacitor when"},{"Start":"06:30.530 ","End":"06:35.495","Text":"the dielectric material completely fills our parallel plate capacitor."},{"Start":"06:35.495 ","End":"06:37.250","Text":"That means that x,"},{"Start":"06:37.250 ","End":"06:40.940","Text":"so let\u0027s look if the dielectric material fully fills the capacitor."},{"Start":"06:40.940 ","End":"06:44.210","Text":"That means that this edge of the dielectric material is"},{"Start":"06:44.210 ","End":"06:48.695","Text":"located over here where x is equal to 0."},{"Start":"06:48.695 ","End":"06:51.740","Text":"Then I can just substitute that into this equation."},{"Start":"06:51.740 ","End":"06:55.370","Text":"I\u0027ll get that it\u0027s equal to Q^2 divided by"},{"Start":"06:55.370 ","End":"07:01.775","Text":"2 Epsilon naught multiplied by Epsilon r multiplied by"},{"Start":"07:01.775 ","End":"07:07.279","Text":"ab divided by d. This is of course the equation"},{"Start":"07:07.279 ","End":"07:13.080","Text":"for the capacitance of a parallel plate capacitor that has a dielectric,"},{"Start":"07:13.080 ","End":"07:18.675","Text":"or a uniform dielectric material placed inside the entire capacitor."},{"Start":"07:18.675 ","End":"07:23.885","Text":"That\u0027s the energy stored on the capacitor."},{"Start":"07:23.885 ","End":"07:27.860","Text":"In this case that the whole capacitor is filled with this dielectric material,"},{"Start":"07:27.860 ","End":"07:32.645","Text":"and then if I subtract these two from one another,"},{"Start":"07:32.645 ","End":"07:36.590","Text":"then I will get the energy required in order"},{"Start":"07:36.590 ","End":"07:42.630","Text":"to fill the capacitor with a dielectric material."},{"Start":"07:42.940 ","End":"07:53.670","Text":"This is the total energy required to fully insert the dielectric material."},{"Start":"07:53.670 ","End":"07:58.550","Text":"Now let\u0027s look at another case where this time"},{"Start":"07:58.550 ","End":"08:04.425","Text":"our capacitor is connected to a voltage source."},{"Start":"08:04.425 ","End":"08:09.070","Text":"It\u0027s connected to a voltage source V_0."},{"Start":"08:10.340 ","End":"08:14.390","Text":"Now we\u0027re calculating the energy on the capacitor when"},{"Start":"08:14.390 ","End":"08:18.440","Text":"the capacitor is connected to a voltage source."},{"Start":"08:18.440 ","End":"08:20.465","Text":"We\u0027re taking a capacitor,"},{"Start":"08:20.465 ","End":"08:23.285","Text":"we connect it up to a voltage source and only"},{"Start":"08:23.285 ","End":"08:27.455","Text":"then after it has already been hooked up to the voltage source,"},{"Start":"08:27.455 ","End":"08:33.880","Text":"we begin pushing in or inserting this dielectric material."},{"Start":"08:34.160 ","End":"08:41.220","Text":"In the case when the capacitor is connected to a source or to a voltage source,"},{"Start":"08:41.390 ","End":"08:46.010","Text":"then we know that there will be a constant voltage."},{"Start":"08:46.010 ","End":"08:50.870","Text":"We already spoke about it that every point between the two plates will have"},{"Start":"08:50.870 ","End":"08:57.530","Text":"a uniform voltage because the battery is supplying a constant flow of charges."},{"Start":"08:57.530 ","End":"09:02.585","Text":"Before we saw that if the capacitor is not connected to a source,"},{"Start":"09:02.585 ","End":"09:04.790","Text":"then we have a constant charge."},{"Start":"09:04.790 ","End":"09:08.750","Text":"But here we\u0027re saying that when the capacitor is connected to a source,"},{"Start":"09:08.750 ","End":"09:12.980","Text":"then in this case we have a constant voltage."},{"Start":"09:12.980 ","End":"09:15.755","Text":"Now we can look at our equation."},{"Start":"09:15.755 ","End":"09:20.585","Text":"Here we see that we aren\u0027t going to use this equation because our Q is changing,"},{"Start":"09:20.585 ","End":"09:24.560","Text":"and we also won\u0027t use this equation because here we also"},{"Start":"09:24.560 ","End":"09:28.570","Text":"have Q inside and we know that this is changing."},{"Start":"09:28.570 ","End":"09:33.110","Text":"There\u0027s a constant flow of charge in order to keep the voltage constant,"},{"Start":"09:33.110 ","End":"09:36.245","Text":"but we can use this equation over here."},{"Start":"09:36.245 ","End":"09:38.810","Text":"We have the capacitance which we calculated in"},{"Start":"09:38.810 ","End":"09:43.335","Text":"a previous lesson and we know that our V is constant."},{"Start":"09:43.335 ","End":"09:45.830","Text":"We can use this."},{"Start":"09:46.110 ","End":"09:48.619","Text":"We\u0027re using this equation,"},{"Start":"09:48.619 ","End":"09:54.035","Text":"the energy on the capacitor is equal to 1/2 of CV^2,"},{"Start":"09:54.035 ","End":"09:56.944","Text":"so now we can say that this is equal to 1/2."},{"Start":"09:56.944 ","End":"10:01.355","Text":"Our capacitance is Epsilon naught a divided by"},{"Start":"10:01.355 ","End":"10:07.370","Text":"d multiplied by x plus Epsilon r b minus x,"},{"Start":"10:07.370 ","End":"10:16.140","Text":"and then multiplied by the voltage squared so our voltage over here is V_0^2."},{"Start":"10:16.920 ","End":"10:19.570","Text":"Let\u0027s take a look at this case,"},{"Start":"10:19.570 ","End":"10:21.610","Text":"and let\u0027s rewrite this."},{"Start":"10:21.610 ","End":"10:25.840","Text":"Let\u0027s rewrite this as v_0^2 divided by 2."},{"Start":"10:25.840 ","End":"10:30.985","Text":"Then we have Epsilon naught a divided by d,"},{"Start":"10:30.985 ","End":"10:35.305","Text":"and we\u0027ll rewrite what we have inside the brackets as"},{"Start":"10:35.305 ","End":"10:43.795","Text":"Epsilon rb minus x multiplied by Epsilon r minus 1."},{"Start":"10:43.795 ","End":"10:46.240","Text":"We have the exact same thing written here,"},{"Start":"10:46.240 ","End":"10:49.285","Text":"except this time we\u0027re taking the x."},{"Start":"10:49.285 ","End":"10:54.490","Text":"What we can see is that as x gets smaller, or in other words,"},{"Start":"10:54.490 ","End":"11:00.340","Text":"as the capacitor is filled more and more with a dielectric material,"},{"Start":"11:00.340 ","End":"11:05.770","Text":"so our energy stored on"},{"Start":"11:05.770 ","End":"11:12.085","Text":"the capacitor is increasing because we\u0027re subtracting this over here."},{"Start":"11:12.085 ","End":"11:14.710","Text":"As x is smaller, we\u0027re subtracting less,"},{"Start":"11:14.710 ","End":"11:17.800","Text":"which means that the energy is stored on the capacitor"},{"Start":"11:17.800 ","End":"11:22.150","Text":"that is filled with a dielectric material,"},{"Start":"11:22.150 ","End":"11:24.685","Text":"so the energy stored will be greater."},{"Start":"11:24.685 ","End":"11:29.410","Text":"Of course, also the capacitance will be greater as well."},{"Start":"11:29.410 ","End":"11:34.285","Text":"On the other hand, if we look at the case over here that we had,"},{"Start":"11:34.285 ","End":"11:38.200","Text":"where our capacitor was not connected to a source,"},{"Start":"11:38.200 ","End":"11:40.525","Text":"so therefore, we had a constant charge."},{"Start":"11:40.525 ","End":"11:46.960","Text":"This time, the equation for capacitance is located in the denominator."},{"Start":"11:46.960 ","End":"11:52.525","Text":"As x decreases, as our capacitor fills with dielectric material,"},{"Start":"11:52.525 ","End":"11:55.585","Text":"the capacitance again will increase."},{"Start":"11:55.585 ","End":"11:59.845","Text":"The capacitance is only dependent on the geometric shape."},{"Start":"11:59.845 ","End":"12:02.410","Text":"However, the energy stored on"},{"Start":"12:02.410 ","End":"12:06.369","Text":"the capacitor will decrease because the capacitance is increasing,"},{"Start":"12:06.369 ","End":"12:08.755","Text":"which means the denominator is increasing,"},{"Start":"12:08.755 ","End":"12:13.820","Text":"which means that the fraction as a whole is decreasing."},{"Start":"12:14.310 ","End":"12:19.435","Text":"A quick explanation for this difference is that in this case,"},{"Start":"12:19.435 ","End":"12:22.375","Text":"where the capacitor is connected to a source,"},{"Start":"12:22.375 ","End":"12:26.110","Text":"so it\u0027s connected to the source,"},{"Start":"12:26.110 ","End":"12:30.445","Text":"which can constantly supply energy to the capacitor."},{"Start":"12:30.445 ","End":"12:34.950","Text":"It can constantly provide energy,"},{"Start":"12:34.950 ","End":"12:39.600","Text":"but like a battery provides energy to some machine,"},{"Start":"12:39.600 ","End":"12:42.440","Text":"so that makes sense, and that\u0027s why here,"},{"Start":"12:42.440 ","End":"12:44.980","Text":"the energy will increase."},{"Start":"12:44.980 ","End":"12:47.410","Text":"However, over here, in this case,"},{"Start":"12:47.410 ","End":"12:50.950","Text":"when the capacitor isn\u0027t connected to a voltage source."},{"Start":"12:50.950 ","End":"12:55.690","Text":"It\u0027s much easier for the energy to decrease because there\u0027s no influx"},{"Start":"12:55.690 ","End":"13:01.285","Text":"of energy into the system because there isn\u0027t a battery to supply anymore energy."},{"Start":"13:01.285 ","End":"13:09.025","Text":"Now, what happens if we want to calculate the work done by a capacitor?"},{"Start":"13:09.025 ","End":"13:15.175","Text":"The equation for work done is equal to delta q,"},{"Start":"13:15.175 ","End":"13:22.760","Text":"the change in the charge multiplied by the voltage of the voltage source."},{"Start":"13:23.100 ","End":"13:25.900","Text":"If we look at this case over here,"},{"Start":"13:25.900 ","End":"13:28.900","Text":"where the capacitor was connected to a source."},{"Start":"13:28.900 ","End":"13:37.340","Text":"The source is what moved the charge between 1 plate, and the other."},{"Start":"13:37.800 ","End":"13:45.069","Text":"In this case, we can say that the change in charge of the source"},{"Start":"13:45.069 ","End":"13:52.820","Text":"is equal to the negative change in charge on the capacitor plates."},{"Start":"13:53.670 ","End":"14:00.520","Text":"As the source provided more charge or took away charge from the plate,"},{"Start":"14:00.520 ","End":"14:05.450","Text":"so the inverse happens at the capacitor as well."},{"Start":"14:05.670 ","End":"14:09.340","Text":"In that case, so this case over here,"},{"Start":"14:09.340 ","End":"14:14.965","Text":"we can say that the work that the source has done is equal to"},{"Start":"14:14.965 ","End":"14:22.190","Text":"the negative change in charge on the capacitor multiplied by v_0."},{"Start":"14:24.180 ","End":"14:28.825","Text":"Now, let\u0027s look at the energy on the capacitor,"},{"Start":"14:28.825 ","End":"14:31.720","Text":"and I\u0027m not going to use the previous equations,"},{"Start":"14:31.720 ","End":"14:35.980","Text":"let\u0027s look at a more general equation."},{"Start":"14:35.980 ","End":"14:41.230","Text":"The energy on the capacitor is going to"},{"Start":"14:41.230 ","End":"14:45.775","Text":"be equal to the third equation that we saw up top at the beginning of the lesson,"},{"Start":"14:45.775 ","End":"14:51.610","Text":"which is equal to 1/2q multiplied by the voltage of the voltage source,"},{"Start":"14:51.610 ","End":"14:55.580","Text":"which in our example over here was v naught."},{"Start":"14:55.650 ","End":"15:03.775","Text":"Now, V_0 is also the voltage that the capacitor will have,"},{"Start":"15:03.775 ","End":"15:08.390","Text":"or the voltage across the capacitor at the end."},{"Start":"15:10.010 ","End":"15:15.634","Text":"We can calculate the energy change across the capacitor or on"},{"Start":"15:15.634 ","End":"15:21.445","Text":"the capacitor when it is connected to a source."},{"Start":"15:21.445 ","End":"15:27.490","Text":"We\u0027ll have the final energy multiplied by the initial energy."},{"Start":"15:27.490 ","End":"15:29.200","Text":"The final energy is this."},{"Start":"15:29.200 ","End":"15:30.700","Text":"The capacitor is fully charged,"},{"Start":"15:30.700 ","End":"15:33.430","Text":"so it has this potential difference of v_0."},{"Start":"15:33.430 ","End":"15:39.340","Text":"We\u0027ll have that this is equal to 1/2 multiplied by the change in charge"},{"Start":"15:39.340 ","End":"15:46.780","Text":"q multiplied by v_0 minus the initial energy."},{"Start":"15:46.780 ","End":"15:49.000","Text":"Before the capacitor was charged,"},{"Start":"15:49.000 ","End":"15:50.800","Text":"the voltage was equal to 0,"},{"Start":"15:50.800 ","End":"15:54.110","Text":"so that means all of this would be equal to 0."},{"Start":"15:55.140 ","End":"16:01.240","Text":"Of course, this is the charge moved across the capacitor."},{"Start":"16:01.240 ","End":"16:03.910","Text":"Now, we can look at the work,"},{"Start":"16:03.910 ","End":"16:07.975","Text":"and we can see that the work is equal to negative Delta q_cV_0,"},{"Start":"16:07.975 ","End":"16:14.335","Text":"and the change in energy is equal to 1/2 delta q_cV_0."},{"Start":"16:14.335 ","End":"16:23.425","Text":"In other words, we can say that the change in energy delta U is equal to 1/2"},{"Start":"16:23.425 ","End":"16:33.430","Text":"of the work done by the voltage source or 1/2 of the absolute value of the work done,"},{"Start":"16:33.430 ","End":"16:43.070","Text":"so the change in energy of the capacitor is equal to 1/2 of the work done by the source."},{"Start":"16:43.740 ","End":"16:47.890","Text":"What we can see over here is that we were working with"},{"Start":"16:47.890 ","End":"16:53.260","Text":"an equation that doesn\u0027t take into account the dielectric material or anything else."},{"Start":"16:53.260 ","End":"16:59.710","Text":"We\u0027re just using the charge on the capacitor and the voltage source."},{"Start":"16:59.710 ","End":"17:03.265","Text":"What we can see is that this equation is correct,"},{"Start":"17:03.265 ","End":"17:06.790","Text":"always independent of whether there"},{"Start":"17:06.790 ","End":"17:11.510","Text":"is a dielectric material between the capacitor plates or not."},{"Start":"17:12.090 ","End":"17:16.930","Text":"This is an equation to remember the energy done,"},{"Start":"17:16.930 ","End":"17:20.140","Text":"or the energy on a capacitor is"},{"Start":"17:20.140 ","End":"17:25.180","Text":"always equal to half of the work done by the voltage source."},{"Start":"17:25.180 ","End":"17:30.970","Text":"Where is the other half of the work done by the voltage source that is lost to"},{"Start":"17:30.970 ","End":"17:33.970","Text":"heat or all other things that we\u0027ll"},{"Start":"17:33.970 ","End":"17:37.345","Text":"see later on in the course and also later on in this chapter."},{"Start":"17:37.345 ","End":"17:40.180","Text":"But, it\u0027s just energy loss to something else."},{"Start":"17:40.180 ","End":"17:44.230","Text":"But, this is the equation to keep in your minds."},{"Start":"17:44.230 ","End":"17:47.000","Text":"That\u0027s the end of this lesson."}],"ID":21388},{"Watched":false,"Name":"Force on Dielectric","Duration":"20m 24s","ChapterTopicVideoID":21508,"CourseChapterTopicPlaylistID":99472,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.830","Text":"Hello. In this lesson,"},{"Start":"00:01.830 ","End":"00:04.440","Text":"we\u0027re going to be speaking about the force acting on"},{"Start":"00:04.440 ","End":"00:09.840","Text":"a dielectric material when inserted between the capacitor plates."},{"Start":"00:09.840 ","End":"00:13.050","Text":"So we\u0027re going to be looking at the example that we\u0027ve been"},{"Start":"00:13.050 ","End":"00:17.770","Text":"working with during the past couple of lessons."},{"Start":"00:17.870 ","End":"00:21.780","Text":"The first case that we\u0027re going to look at is the case where"},{"Start":"00:21.780 ","End":"00:25.845","Text":"a capacitor isn\u0027t connected to a voltage source."},{"Start":"00:25.845 ","End":"00:28.515","Text":"However, on the capacitor plates,"},{"Start":"00:28.515 ","End":"00:36.270","Text":"we have charges plus Q up top and minus Q at the bottom."},{"Start":"00:36.590 ","End":"00:39.970","Text":"In previous lessons, we calculated"},{"Start":"00:39.970 ","End":"00:44.555","Text":"that the capacitance of such a capacitor would be equal to this,"},{"Start":"00:44.555 ","End":"00:52.010","Text":"where x is the distance from the side over here of empty space of air"},{"Start":"00:52.010 ","End":"00:55.880","Text":"between the plates until we get to the section in"},{"Start":"00:55.880 ","End":"01:00.965","Text":"the capacitor which is filled with the dielectric material."},{"Start":"01:00.965 ","End":"01:05.390","Text":"We can see that the capacitance is independent of"},{"Start":"01:05.390 ","End":"01:09.560","Text":"whether the capacitor is connected to a voltage source or not."},{"Start":"01:09.560 ","End":"01:11.810","Text":"In the first example that we\u0027re looking at,"},{"Start":"01:11.810 ","End":"01:15.200","Text":"the capacitor is not connected to a voltage source,"},{"Start":"01:15.200 ","End":"01:20.600","Text":"which means that we have a constant charge,"},{"Start":"01:20.600 ","End":"01:23.755","Text":"Q, on each plate."},{"Start":"01:23.755 ","End":"01:25.910","Text":"If it\u0027s not connected to a voltage source,"},{"Start":"01:25.910 ","End":"01:28.100","Text":"then that means the charge is constant,"},{"Start":"01:28.100 ","End":"01:32.075","Text":"which means that the energy across the capacitor,"},{"Start":"01:32.075 ","End":"01:33.530","Text":"in order to calculate this,"},{"Start":"01:33.530 ","End":"01:40.590","Text":"we\u0027ll be using the equation for 1/2q^2 divided by C,"},{"Start":"01:40.590 ","End":"01:47.290","Text":"where here, our C is a function of x."},{"Start":"01:47.290 ","End":"01:48.920","Text":"We, of course,"},{"Start":"01:48.920 ","End":"01:52.920","Text":"use this equation because our q is constant."},{"Start":"01:53.300 ","End":"01:56.060","Text":"In the previous lesson,"},{"Start":"01:56.060 ","End":"02:00.965","Text":"we saw what the energy was equal to in this case."},{"Start":"02:00.965 ","End":"02:06.185","Text":"But essentially, we can see that we have our energy on the capacitor,"},{"Start":"02:06.185 ","End":"02:10.835","Text":"which is as a function of x. What does that mean?"},{"Start":"02:10.835 ","End":"02:13.445","Text":"If the energy is a function of x,"},{"Start":"02:13.445 ","End":"02:17.965","Text":"as x changes, our energy changes."},{"Start":"02:17.965 ","End":"02:21.785","Text":"If we remember from mechanics,"},{"Start":"02:21.785 ","End":"02:25.069","Text":"we remember that if our energy is changing,"},{"Start":"02:25.069 ","End":"02:28.938","Text":"then that means that some force is acting."},{"Start":"02:28.938 ","End":"02:33.860","Text":"So we know that a force is going to be acting on this dielectric material."},{"Start":"02:33.860 ","End":"02:37.775","Text":"If we remember back to our mechanics chapter,"},{"Start":"02:37.775 ","End":"02:46.290","Text":"we know that force is equal to the negative gradient of the energy."},{"Start":"02:48.320 ","End":"02:52.095","Text":"So if our energy was a constant,"},{"Start":"02:52.095 ","End":"02:54.730","Text":"it would be independent of x,"},{"Start":"02:54.730 ","End":"02:57.445","Text":"in which case when we would take the gradient of it,"},{"Start":"02:57.445 ","End":"02:59.980","Text":"we would get that it would equal 0 and that means that"},{"Start":"02:59.980 ","End":"03:03.130","Text":"a force of 0 would be acting, or no force."},{"Start":"03:03.130 ","End":"03:04.900","Text":"But here, U is changing,"},{"Start":"03:04.900 ","End":"03:07.420","Text":"so that means that we can take the gradient of U,"},{"Start":"03:07.420 ","End":"03:10.775","Text":"and we will get some value for our force."},{"Start":"03:10.775 ","End":"03:13.325","Text":"In our example, luckily,"},{"Start":"03:13.325 ","End":"03:16.195","Text":"we only have this movement in the x direction,"},{"Start":"03:16.195 ","End":"03:20.260","Text":"give a dielectric material slots in along the x direction."},{"Start":"03:20.260 ","End":"03:23.530","Text":"That means that our force in our example is going to be equal"},{"Start":"03:23.530 ","End":"03:26.935","Text":"to the negative gradient of U,"},{"Start":"03:26.935 ","End":"03:29.865","Text":"which is only acting in the x direction."},{"Start":"03:29.865 ","End":"03:38.830","Text":"You can say that this will be equal to negative dU by dx in the x direction."},{"Start":"03:39.200 ","End":"03:45.493","Text":"So we take the partial derivative with respect to x of our energy,"},{"Start":"03:45.493 ","End":"03:48.215","Text":"and of course, this will be acting in the x direction."},{"Start":"03:48.215 ","End":"03:51.660","Text":"Let\u0027s give a little bit of intuition."},{"Start":"03:51.660 ","End":"03:54.260","Text":"If this is the positively charged plate,"},{"Start":"03:54.260 ","End":"03:56.210","Text":"and this is the negatively charged plate,"},{"Start":"03:56.210 ","End":"04:02.390","Text":"we know that the electric field within a parallel plate capacitor is uniform,"},{"Start":"04:02.390 ","End":"04:04.279","Text":"and it will be, in this case,"},{"Start":"04:04.279 ","End":"04:06.424","Text":"in this downwards direction."},{"Start":"04:06.424 ","End":"04:14.285","Text":"Why then will we have a force acting in the x direction?"},{"Start":"04:14.285 ","End":"04:16.035","Text":"It sounds a bit weird."},{"Start":"04:16.035 ","End":"04:20.750","Text":"What\u0027s happening is that the force is acting in the x direction due"},{"Start":"04:20.750 ","End":"04:26.430","Text":"to the effects of the edges of the capacitor."},{"Start":"04:27.370 ","End":"04:30.080","Text":"The whole idea that we get"},{"Start":"04:30.080 ","End":"04:35.990","Text":"this uniform electric field in the y direction over here is because we said that we can"},{"Start":"04:35.990 ","End":"04:40.340","Text":"consider the electric field due to the parallel plate capacitor as"},{"Start":"04:40.340 ","End":"04:46.090","Text":"being like the electric field between 2 infinite planes."},{"Start":"04:46.090 ","End":"04:49.925","Text":"However, when we come to the edge of the capacitor,"},{"Start":"04:49.925 ","End":"04:52.130","Text":"so somewhere along here,"},{"Start":"04:52.130 ","End":"04:55.670","Text":"we no longer can consider the capacitor"},{"Start":"04:55.670 ","End":"04:59.810","Text":"as an infinite plane because here we have a plane,"},{"Start":"04:59.810 ","End":"05:02.675","Text":"but then we jump over here, and there\u0027s nothing."},{"Start":"05:02.675 ","End":"05:08.970","Text":"That means that we get some weird thing happening with the electric field over here."},{"Start":"05:09.320 ","End":"05:16.120","Text":"What actually happens is that here we have our capacitor, and in the middle,"},{"Start":"05:16.120 ","End":"05:19.450","Text":"we can consider it as infinite plane,"},{"Start":"05:19.450 ","End":"05:23.320","Text":"so we get this electric field going like so."},{"Start":"05:23.320 ","End":"05:25.315","Text":"However, on the edges,"},{"Start":"05:25.315 ","End":"05:27.100","Text":"when we\u0027re located, let\u0027s say,"},{"Start":"05:27.100 ","End":"05:28.330","Text":"at a point over here,"},{"Start":"05:28.330 ","End":"05:31.200","Text":"so we can consider it an infinite plane,"},{"Start":"05:31.200 ","End":"05:38.450","Text":"and then we get electric field lines that are going something like so."},{"Start":"05:38.450 ","End":"05:41.680","Text":"Of course, the same on the other side."},{"Start":"05:41.680 ","End":"05:46.024","Text":"We get this type of effect."},{"Start":"05:46.024 ","End":"05:49.329","Text":"What we can see is that on the edges,"},{"Start":"05:49.329 ","End":"05:54.740","Text":"our electric field lines also have not only a y component,"},{"Start":"05:54.740 ","End":"05:59.010","Text":"but they also have some x component."},{"Start":"06:00.350 ","End":"06:03.980","Text":"If this capacitor plate is the positively charged,"},{"Start":"06:03.980 ","End":"06:09.995","Text":"so that means that we have lots of positive charges along this plate like so."},{"Start":"06:09.995 ","End":"06:12.085","Text":"However, as we know,"},{"Start":"06:12.085 ","End":"06:15.874","Text":"the charges along the dielectric, or the opposite,"},{"Start":"06:15.874 ","End":"06:20.354","Text":"they polarize in the opposite way to the capacitor plates,"},{"Start":"06:20.354 ","End":"06:23.975","Text":"so here, we\u0027re going to have lots of negative charges."},{"Start":"06:23.975 ","End":"06:25.965","Text":"On the bottom plate, we have the same."},{"Start":"06:25.965 ","End":"06:30.470","Text":"So here we have lots of negative charges because this is the negatively-charged plate."},{"Start":"06:30.470 ","End":"06:33.770","Text":"In that case over here on the dielectric,"},{"Start":"06:33.770 ","End":"06:36.995","Text":"we\u0027re going to have lots of positive charges."},{"Start":"06:36.995 ","End":"06:39.920","Text":"So then what we get is this attractive force between"},{"Start":"06:39.920 ","End":"06:44.019","Text":"the positive and negative charges on both plates,"},{"Start":"06:44.019 ","End":"06:48.390","Text":"so we get this attractive pulling force."},{"Start":"06:48.520 ","End":"06:56.600","Text":"So what we can see is that the capacitor plates due to this difference in polarization,"},{"Start":"06:56.600 ","End":"06:58.340","Text":"or difference in charge between"},{"Start":"06:58.340 ","End":"07:02.480","Text":"the plates and the dielectric material next to the plates,"},{"Start":"07:02.480 ","End":"07:10.410","Text":"we will get this force which is pulling the dielectric material inside of the capacitor."},{"Start":"07:12.380 ","End":"07:18.770","Text":"This force, it\u0027s sometimes complicated to understand because there\u0027s a minus over here."},{"Start":"07:18.770 ","End":"07:25.130","Text":"If we\u0027re talking about the force on the capacitor or on the dielectric material,"},{"Start":"07:25.130 ","End":"07:31.280","Text":"so in this case, we calculate the absolute value of the force."},{"Start":"07:31.280 ","End":"07:34.555","Text":"We take the absolute value like so,"},{"Start":"07:34.555 ","End":"07:37.610","Text":"and then we add in the sign,"},{"Start":"07:37.610 ","End":"07:44.035","Text":"depending on the axes that we defined over here in our question."},{"Start":"07:44.035 ","End":"07:52.280","Text":"In our example we can see that the dielectric is going to move inside the capacitor."},{"Start":"07:52.280 ","End":"07:55.910","Text":"We can imagine that the capacitor is stationary,"},{"Start":"07:55.910 ","End":"08:01.550","Text":"and it\u0027s the dielectric material that is experiencing this force in this direction,"},{"Start":"08:01.550 ","End":"08:06.950","Text":"where before we said that this would be the positive x direction,"},{"Start":"08:06.950 ","End":"08:09.260","Text":"because here x is equal to 0,"},{"Start":"08:09.260 ","End":"08:12.925","Text":"and then here x is equal to b."},{"Start":"08:12.925 ","End":"08:16.430","Text":"Therefore, we can see that, in our example,"},{"Start":"08:16.430 ","End":"08:21.330","Text":"the force will be acting in the negative x direction."},{"Start":"08:22.520 ","End":"08:28.520","Text":"Now what we\u0027re going to do is we\u0027re going to calculate the magnitude of the force."},{"Start":"08:28.520 ","End":"08:33.755","Text":"I\u0027m going to do it in 2 different ways because I want to show you something."},{"Start":"08:33.755 ","End":"08:38.390","Text":"Most of the time, we calculate the energy as a function of"},{"Start":"08:38.390 ","End":"08:43.595","Text":"x by just plugging in our equation for the capacitance over here."},{"Start":"08:43.595 ","End":"08:52.615","Text":"What we\u0027ll get is that our energy is equal to Q^2 divided by 2 times the capacitance,"},{"Start":"08:52.615 ","End":"09:01.505","Text":"so that\u0027s Epsilon_0 a divided by d multiplied by x plus Epsilon_r b minus x."},{"Start":"09:01.505 ","End":"09:06.470","Text":"Then when I do this over here to calculate the force,"},{"Start":"09:06.470 ","End":"09:08.675","Text":"dU, by dx,"},{"Start":"09:08.675 ","End":"09:12.425","Text":"the magnitude of our force,"},{"Start":"09:12.425 ","End":"09:21.090","Text":"this is simply going to be equal to negative Q^2 1 minus Epsilon_r multiplied by d^2"},{"Start":"09:21.090 ","End":"09:25.950","Text":"divided by 2 multiplied by Epsilon_0"},{"Start":"09:25.950 ","End":"09:33.250","Text":"a x plus Epsilon_r b minus x,"},{"Start":"09:33.250 ","End":"09:36.630","Text":"and all of this, squared."},{"Start":"09:37.090 ","End":"09:40.640","Text":"So we can see that to take this derivative,"},{"Start":"09:40.640 ","End":"09:42.710","Text":"it\u0027s a little bit complicated."},{"Start":"09:42.710 ","End":"09:46.890","Text":"So now I\u0027m going to show you an easier way."},{"Start":"09:47.480 ","End":"09:50.630","Text":"In order to find the magnitude of the force,"},{"Start":"09:50.630 ","End":"09:56.315","Text":"I know that I have to take the partial derivative of U with respect to x."},{"Start":"09:56.315 ","End":"09:58.825","Text":"So if I look at my equation over here, U,"},{"Start":"09:58.825 ","End":"10:03.110","Text":"I can see that it\u0027s as a function of C,"},{"Start":"10:03.110 ","End":"10:06.545","Text":"and C is as a function of x."},{"Start":"10:06.545 ","End":"10:09.755","Text":"In order to get my U as a function of x,"},{"Start":"10:09.755 ","End":"10:13.760","Text":"I had to plug in my equation for the capacitance."},{"Start":"10:13.760 ","End":"10:17.545","Text":"My equation for a capacitance is as a function of x."},{"Start":"10:17.545 ","End":"10:21.410","Text":"So my U is as a function of the capacitance,"},{"Start":"10:21.410 ","End":"10:26.100","Text":"and my capacitance is as a function of x."},{"Start":"10:26.690 ","End":"10:32.120","Text":"In that case, what I can do is I can use the chain rule."},{"Start":"10:32.120 ","End":"10:37.220","Text":"I can say that dU by dx is equal to,"},{"Start":"10:37.220 ","End":"10:38.700","Text":"according to the chain rule,"},{"Start":"10:38.700 ","End":"10:49.514","Text":"dU by dC multiplied by dC by dx."},{"Start":"10:49.514 ","End":"10:56.917","Text":"In this case, look how easy it is to get the exact same answer,"},{"Start":"10:56.917 ","End":"11:01.780","Text":"so dU by dC over here is simply going to be equal"},{"Start":"11:01.780 ","End":"11:11.020","Text":"to negative Q^2 divided by 2C^2."},{"Start":"11:11.020 ","End":"11:15.445","Text":"I took the derivative of U over here with respect to C. Then,"},{"Start":"11:15.445 ","End":"11:19.735","Text":"I\u0027m going to multiply this by the derivative"},{"Start":"11:19.735 ","End":"11:24.700","Text":"of my capacitance equation with respect to x."},{"Start":"11:24.700 ","End":"11:34.585","Text":"So what I\u0027ll have is negative Q^2 divided by 2C^2 multiplied by dC by dx."},{"Start":"11:34.585 ","End":"11:42.475","Text":"So here, I\u0027ll have Epsilon_0 a divided by d plus Epsilon_0 a"},{"Start":"11:42.475 ","End":"11:46.660","Text":"divided by d multiplied by"},{"Start":"11:46.660 ","End":"11:55.086","Text":"negative Epsilon_r."},{"Start":"11:55.086 ","End":"11:58.720","Text":"Then in this case,"},{"Start":"11:58.720 ","End":"12:04.210","Text":"all we have to do is substitute in our C into this equation,"},{"Start":"12:04.210 ","End":"12:08.800","Text":"and we\u0027ll get this exact equation over here."},{"Start":"12:08.800 ","End":"12:15.670","Text":"Using the chain rule is sometimes much easier because to take"},{"Start":"12:15.670 ","End":"12:19.090","Text":"the derivative with respect to x over this is much more"},{"Start":"12:19.090 ","End":"12:24.140","Text":"complicated than this method of the chain rule over here."},{"Start":"12:24.870 ","End":"12:29.859","Text":"This force is acting on the dielectric,"},{"Start":"12:29.859 ","End":"12:38.810","Text":"and it\u0027s pulling or accelerating the dielectric material inside into the capacitor."},{"Start":"12:39.030 ","End":"12:44.980","Text":"If I add in over here an external F,"},{"Start":"12:44.980 ","End":"12:47.335","Text":"so let\u0027s call it F external."},{"Start":"12:47.335 ","End":"12:52.180","Text":"This force is an equal and opposite force to this so"},{"Start":"12:52.180 ","End":"12:57.070","Text":"that it will stop the dielectric material from entering the capacitor,"},{"Start":"12:57.070 ","End":"13:03.952","Text":"or it will keep the dielectric material entering but at a constant velocity,"},{"Start":"13:03.952 ","End":"13:06.560","Text":"so with a 0 acceleration."},{"Start":"13:07.020 ","End":"13:12.640","Text":"Let\u0027s imagine that we have some velocity, V_0,"},{"Start":"13:12.640 ","End":"13:16.600","Text":"that we want this dielectric to enter the capacitor with,"},{"Start":"13:16.600 ","End":"13:19.135","Text":"so a constant velocity."},{"Start":"13:19.135 ","End":"13:25.900","Text":"Therefore, we know that our external force acting on the dielectric"},{"Start":"13:25.900 ","End":"13:32.965","Text":"has to be equal to the absolute value or the magnitude of this force,"},{"Start":"13:32.965 ","End":"13:36.500","Text":"F, that we just calculated."},{"Start":"13:36.690 ","End":"13:42.010","Text":"Now, another question that might be asked is what is the power?"},{"Start":"13:42.010 ","End":"13:44.050","Text":"The power, as we know,"},{"Start":"13:44.050 ","End":"13:48.404","Text":"is equal to dw by dt,"},{"Start":"13:48.404 ","End":"13:52.480","Text":"where this is the change in work during"},{"Start":"13:52.480 ","End":"13:57.745","Text":"the change of time or the work done in a certain amount of time."},{"Start":"13:57.745 ","End":"14:04.255","Text":"But better equation to use in these types of questions is to take"},{"Start":"14:04.255 ","End":"14:12.770","Text":"the force vector and dot product it with the velocity vector."},{"Start":"14:12.780 ","End":"14:14.890","Text":"So in this case over here,"},{"Start":"14:14.890 ","End":"14:20.770","Text":"we can see that the velocity and the force are acting in the same direction,"},{"Start":"14:20.770 ","End":"14:26.620","Text":"so I would just have to take the x components and multiply them together."},{"Start":"14:26.620 ","End":"14:30.835","Text":"If we want to work out the power for external force,"},{"Start":"14:30.835 ","End":"14:34.730","Text":"so that\u0027s P external."},{"Start":"14:35.520 ","End":"14:41.260","Text":"Because I\u0027m doing the dot product between the force and the velocity,"},{"Start":"14:41.260 ","End":"14:50.500","Text":"so what we would have is the magnitude of our external force multiplied by the velocity,"},{"Start":"14:50.500 ","End":"14:56.800","Text":"which is v_0, then multiplied by negative 1 because"},{"Start":"14:56.800 ","End":"14:59.680","Text":"our velocity is acting in the negative x-direction and"},{"Start":"14:59.680 ","End":"15:04.550","Text":"our external force is acting in the positive x-direction."},{"Start":"15:04.770 ","End":"15:09.595","Text":"What we can see is that our power is negative."},{"Start":"15:09.595 ","End":"15:12.460","Text":"Because our force over here is acting in"},{"Start":"15:12.460 ","End":"15:17.545","Text":"the negative direction to the direction of travel."},{"Start":"15:17.545 ","End":"15:23.230","Text":"Another way of remembering this is to remember that the work done is equal"},{"Start":"15:23.230 ","End":"15:28.975","Text":"to the integral along the force dot product with dr,"},{"Start":"15:28.975 ","End":"15:32.125","Text":"where dr is the route taken."},{"Start":"15:32.125 ","End":"15:36.910","Text":"We can see that the route taken is in the leftwards direction,"},{"Start":"15:36.910 ","End":"15:39.025","Text":"which is the negative x-direction."},{"Start":"15:39.025 ","End":"15:45.925","Text":"However our force, its direction is in the positive x direction."},{"Start":"15:45.925 ","End":"15:49.390","Text":"When we do the dot product between these 2,"},{"Start":"15:49.390 ","End":"15:53.540","Text":"we\u0027re going to be left with negative 1."},{"Start":"15:55.350 ","End":"15:58.479","Text":"Our work is going to be negative,"},{"Start":"15:58.479 ","End":"16:02.600","Text":"which means that our power will also be negative."},{"Start":"16:03.390 ","End":"16:07.105","Text":"Now let\u0027s look at the exact same example."},{"Start":"16:07.105 ","End":"16:13.130","Text":"However, this time the capacitor is connected to a battery."},{"Start":"16:14.480 ","End":"16:18.250","Text":"Here we can see that the top plates"},{"Start":"16:18.250 ","End":"16:21.325","Text":"of the capacitor is connected to the long side of the battery,"},{"Start":"16:21.325 ","End":"16:24.460","Text":"and the bottom is connected to the short side."},{"Start":"16:24.460 ","End":"16:29.560","Text":"Now we can see that the capacitor plates,"},{"Start":"16:29.560 ","End":"16:31.567","Text":"because they are connected to a voltage source,"},{"Start":"16:31.567 ","End":"16:36.340","Text":"so the potential difference between the plates is uniform throughout or is constant,"},{"Start":"16:36.340 ","End":"16:40.180","Text":"but the charge is not going to be constant."},{"Start":"16:40.180 ","End":"16:44.785","Text":"That means that we can\u0027t use this equation because our charge isn\u0027t constant,"},{"Start":"16:44.785 ","End":"16:48.565","Text":"but our voltage our potential difference is constant so we can"},{"Start":"16:48.565 ","End":"16:52.600","Text":"use the equation that U is equal to 1/2 c,"},{"Start":"16:52.600 ","End":"16:56.875","Text":"which we know is a function of x multiplied by V^2,"},{"Start":"16:56.875 ","End":"17:00.490","Text":"where V is of course the voltage,"},{"Start":"17:00.490 ","End":"17:03.250","Text":"where of course the voltage here is V_0."},{"Start":"17:03.250 ","End":"17:09.655","Text":"Then if I want to find the magnitude of the force, so again,"},{"Start":"17:09.655 ","End":"17:18.670","Text":"I\u0027m going to take the derivative with respect to x. I\u0027ll use the chain rule again."},{"Start":"17:18.670 ","End":"17:26.320","Text":"So what I\u0027ll have is dU by dC multiplied by dC by dx,"},{"Start":"17:26.320 ","End":"17:30.640","Text":"which dU by dC in this case is simply"},{"Start":"17:30.640 ","End":"17:37.060","Text":"equal to 1/2 V_0^2."},{"Start":"17:37.060 ","End":"17:42.790","Text":"Then I\u0027m going to multiply by dC by dx,"},{"Start":"17:42.790 ","End":"17:45.280","Text":"which just like before,"},{"Start":"17:45.280 ","End":"17:52.345","Text":"is simply equal to epsilon naught a divided by d multiplied by 1 minus epsilon"},{"Start":"17:52.345 ","End":"18:01.855","Text":"r. Something small yet important to understand is that when I\u0027m trying to find the force,"},{"Start":"18:01.855 ","End":"18:07.390","Text":"in this case over here where my capacitor is connected to a voltage source."},{"Start":"18:07.390 ","End":"18:13.404","Text":"What I should remember is that I have to take the derivative of the energy."},{"Start":"18:13.404 ","End":"18:18.910","Text":"But the energy over here is the energy of the capacitor,"},{"Start":"18:18.910 ","End":"18:22.705","Text":"but also the energy of the voltage source."},{"Start":"18:22.705 ","End":"18:27.680","Text":"I have to take both of them into consideration."},{"Start":"18:27.750 ","End":"18:30.925","Text":"We know that the energy of"},{"Start":"18:30.925 ","End":"18:36.189","Text":"the voltage source is equal to twice the energy of the capacitor."},{"Start":"18:36.189 ","End":"18:39.085","Text":"We saw this in the previous lesson."},{"Start":"18:39.085 ","End":"18:43.330","Text":"Therefore, I can say that"},{"Start":"18:43.330 ","End":"18:50.365","Text":"the total energy of my system is equal to the energy that my source is providing,"},{"Start":"18:50.365 ","End":"18:55.072","Text":"so U_s, and then the energy of the capacitor,"},{"Start":"18:55.072 ","End":"18:59.290","Text":"now notice that the capacitor is taking energy away."},{"Start":"18:59.290 ","End":"19:05.545","Text":"It\u0027s using energy in order to pull this dielectric material inside."},{"Start":"19:05.545 ","End":"19:07.810","Text":"That means it has a negative sign."},{"Start":"19:07.810 ","End":"19:11.540","Text":"We\u0027re subtracting the energy of the capacitor."},{"Start":"19:11.610 ","End":"19:16.645","Text":"Again, the total energy of my system has to take into account"},{"Start":"19:16.645 ","End":"19:21.370","Text":"the energy of my source and the energy of my capacitor."},{"Start":"19:21.370 ","End":"19:24.085","Text":"My source is providing energy,"},{"Start":"19:24.085 ","End":"19:26.941","Text":"but my capacitor is using energy."},{"Start":"19:26.941 ","End":"19:30.190","Text":"It\u0027s taking energy in order to pull in this dielectric,"},{"Start":"19:30.190 ","End":"19:33.070","Text":"which means that there\u0027s a negative over here."},{"Start":"19:33.070 ","End":"19:37.375","Text":"Now we can say that the energy of my source is equal to"},{"Start":"19:37.375 ","End":"19:41.980","Text":"twice the energy of the capacitor and then minus the energy of the capacitor,"},{"Start":"19:41.980 ","End":"19:47.140","Text":"so I\u0027m left with a total energy of the system being equal to U_C,"},{"Start":"19:47.140 ","End":"19:50.265","Text":"just the energy of the capacitor."},{"Start":"19:50.265 ","End":"19:56.220","Text":"That is why I can still use this equation over here."},{"Start":"19:56.220 ","End":"20:00.120","Text":"That\u0027s why I can take the derivative just of"},{"Start":"20:00.120 ","End":"20:03.945","Text":"the energy of my capacitor without taking into account"},{"Start":"20:03.945 ","End":"20:08.220","Text":"the voltage source because it just works out that"},{"Start":"20:08.220 ","End":"20:14.860","Text":"the total energy of the entire system is equal to the energy of the capacitor."},{"Start":"20:14.860 ","End":"20:16.779","Text":"This is a small detail,"},{"Start":"20:16.779 ","End":"20:20.030","Text":"but it\u0027s important to understand this."},{"Start":"20:20.520 ","End":"20:24.620","Text":"That is the end of this lesson."}],"ID":22289},{"Watched":false,"Name":"Deriving Equation for Energy on Capacitor","Duration":"8m 46s","ChapterTopicVideoID":21496,"CourseChapterTopicPlaylistID":99472,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.875","Text":"Hello. In this lesson,"},{"Start":"00:01.875 ","End":"00:07.610","Text":"we\u0027re going to be learning how to derive the equation for the energy on a capacitor."},{"Start":"00:07.610 ","End":"00:12.870","Text":"We\u0027ve already seen that there are 3 equations for the energy on a capacitor,"},{"Start":"00:12.870 ","End":"00:16.800","Text":"where each equation is just the exact same thing,"},{"Start":"00:16.800 ","End":"00:20.160","Text":"but where we just substitute it in either for c,"},{"Start":"00:20.160 ","End":"00:23.415","Text":"q or v from this equation over here."},{"Start":"00:23.415 ","End":"00:26.700","Text":"Then we get the different forms for the energy."},{"Start":"00:26.700 ","End":"00:31.775","Text":"But let\u0027s see how we get the equation for energy in the first place."},{"Start":"00:31.775 ","End":"00:34.680","Text":"That\u0027s what we\u0027re going to be doing this lesson."},{"Start":"00:34.760 ","End":"00:38.915","Text":"There are 2 ways that we can get to this equation."},{"Start":"00:38.915 ","End":"00:43.805","Text":"Either we can say that the energy on the capacitor is equal to"},{"Start":"00:43.805 ","End":"00:51.075","Text":"1/2 of the sum on q_i, Phi_i."},{"Start":"00:51.075 ","End":"00:55.170","Text":"On the charge is multiplied by their potential."},{"Start":"00:55.170 ","End":"01:02.090","Text":"The other way to get to this equation is by integrating V_c,"},{"Start":"01:02.090 ","End":"01:06.850","Text":"the voltage on the capacitor, dq."},{"Start":"01:07.520 ","End":"01:13.925","Text":"Let\u0027s derive the equation via Method number 1 first."},{"Start":"01:13.925 ","End":"01:18.335","Text":"Here we have our capacitor."},{"Start":"01:18.335 ","End":"01:21.650","Text":"Here we\u0027re using a parallel plate capacitor."},{"Start":"01:21.650 ","End":"01:25.220","Text":"However, this equation is true for any capacitor."},{"Start":"01:25.220 ","End":"01:32.360","Text":"The top plate has a charge of plus q and the button plate has a charge of negative q."},{"Start":"01:32.360 ","End":"01:39.110","Text":"There\u0027s some therefore a potential difference over here, V_c."},{"Start":"01:39.110 ","End":"01:45.040","Text":"Now we\u0027re going to use this equation in order to find the energy on this capacitor."},{"Start":"01:45.040 ","End":"01:48.885","Text":"Let\u0027s just say that we have some potential."},{"Start":"01:48.885 ","End":"01:55.730","Text":"Let\u0027s say that our potential is equal to,"},{"Start":"01:55.730 ","End":"01:57.935","Text":"let\u0027s say 5 volts."},{"Start":"01:57.935 ","End":"02:03.200","Text":"Now, we\u0027ll see that we can make up any potential, it doesn\u0027t matter."},{"Start":"02:03.200 ","End":"02:07.090","Text":"We could have also said that this is a million volts."},{"Start":"02:07.090 ","End":"02:11.970","Text":"We\u0027ll see that our answer is independent of this."},{"Start":"02:13.150 ","End":"02:18.530","Text":"If the potential at the bottom plate is 5 volts,"},{"Start":"02:18.530 ","End":"02:26.195","Text":"then that means that the potential of the top plate has to be 5 volts plus our V_c."},{"Start":"02:26.195 ","End":"02:30.425","Text":"Because our V_c represents the potential difference between the plates."},{"Start":"02:30.425 ","End":"02:34.830","Text":"5 plus V_c minus V_c will give us 5."},{"Start":"02:34.830 ","End":"02:39.880","Text":"Here we have 5 plus V_c."},{"Start":"02:40.610 ","End":"02:44.430","Text":"Now let\u0027s plug this into our equation."},{"Start":"02:44.430 ","End":"02:49.125","Text":"We have that U_c is equal to 1/2,"},{"Start":"02:49.125 ","End":"02:54.360","Text":"multiplied by the sum of q_i multiplied by Phi_i."},{"Start":"02:54.360 ","End":"02:58.860","Text":"Our first charge is this negative q let\u0027s say."},{"Start":"02:58.860 ","End":"03:02.580","Text":"This is q1 multiplied by Phi 1."},{"Start":"03:02.580 ","End":"03:04.680","Text":"The potential at this point."},{"Start":"03:04.680 ","End":"03:08.960","Text":"The potential of the bottom plate we said is 5 volts."},{"Start":"03:08.960 ","End":"03:13.050","Text":"We\u0027ll multiply it by 5 and that\u0027s it."},{"Start":"03:13.050 ","End":"03:15.135","Text":"Then our next charge,"},{"Start":"03:15.135 ","End":"03:17.265","Text":"our q2 is this over here."},{"Start":"03:17.265 ","End":"03:22.940","Text":"Then we have plus q because it has a positive charge and multiplied by its potential."},{"Start":"03:22.940 ","End":"03:24.200","Text":"Its potential over here,"},{"Start":"03:24.200 ","End":"03:26.660","Text":"we said is 5 plus V_c."},{"Start":"03:26.660 ","End":"03:30.690","Text":"We multiply by 5 plus V_c."},{"Start":"03:30.690 ","End":"03:36.390","Text":"Now we can say that here we have q5 or 5q,"},{"Start":"03:36.390 ","End":"03:38.445","Text":"and here we have minus 5q,"},{"Start":"03:38.445 ","End":"03:39.900","Text":"so that will cancel out."},{"Start":"03:39.900 ","End":"03:47.565","Text":"It\u0027s easy to see that we are left with 1/2q multiplied by V_c,"},{"Start":"03:47.565 ","End":"03:49.605","Text":"the voltage across the capacitor."},{"Start":"03:49.605 ","End":"03:55.310","Text":"Which is exactly this equation over here, 1/2qV."},{"Start":"03:55.310 ","End":"04:02.180","Text":"1/2qV. As we can see the fact that we said that here we have 5,"},{"Start":"04:02.180 ","End":"04:06.230","Text":"we can see that it doesn\u0027t affect our equation because the only thing"},{"Start":"04:06.230 ","End":"04:10.205","Text":"that matters is the potential difference between the 2 plates,"},{"Start":"04:10.205 ","End":"04:14.280","Text":"not the potential of each individual plate."},{"Start":"04:14.960 ","End":"04:18.385","Text":"This answer is again, I\u0027m reminding you,"},{"Start":"04:18.385 ","End":"04:24.010","Text":"correct for every single capacitor whether it\u0027s a parallel plate capacitor,"},{"Start":"04:24.010 ","End":"04:25.465","Text":"a cylindrical capacitor,"},{"Start":"04:25.465 ","End":"04:29.010","Text":"spherical capacitor it doesn\u0027t matter this equation works."},{"Start":"04:29.010 ","End":"04:32.775","Text":"Now let\u0027s go on to Method number 2."},{"Start":"04:32.775 ","End":"04:39.580","Text":"Here again, we have a parallel plate capacitor."},{"Start":"04:39.830 ","End":"04:42.800","Text":"Just like in Method number 1,"},{"Start":"04:42.800 ","End":"04:46.665","Text":"it doesn\u0027t matter which capacitor we\u0027re using if it\u0027s a parallel plate,"},{"Start":"04:46.665 ","End":"04:51.250","Text":"cylindrical, spherical, but we\u0027re just going to use in this example,"},{"Start":"04:51.250 ","End":"04:53.380","Text":"a parallel plate capacitor."},{"Start":"04:53.380 ","End":"04:56.965","Text":"Let\u0027s say that over here we have a charge of"},{"Start":"04:56.965 ","End":"05:01.855","Text":"q and over here on the bottom plate we have a charge of negative q."},{"Start":"05:01.855 ","End":"05:05.194","Text":"Now let\u0027s say that from the bottom plate,"},{"Start":"05:05.194 ","End":"05:10.590","Text":"I want to move a charge of Delta q to the top plate."},{"Start":"05:11.140 ","End":"05:17.190","Text":"A charge of Delta q is leaving the bottom plate and entering the top plate."},{"Start":"05:17.190 ","End":"05:21.095","Text":"Now the charge on my bottom plate is going to be negative q"},{"Start":"05:21.095 ","End":"05:27.605","Text":"minus my charge Delta q that I\u0027ve added to the top plates."},{"Start":"05:27.605 ","End":"05:30.590","Text":"The charge of my top plate is going to be the original charge"},{"Start":"05:30.590 ","End":"05:34.745","Text":"q plus the charge that\u0027s been added from the bottom plate."},{"Start":"05:34.745 ","End":"05:37.500","Text":"Plus Delta q."},{"Start":"05:38.030 ","End":"05:44.465","Text":"The energy required for me to move this dq from the bottom plate to the top plate,"},{"Start":"05:44.465 ","End":"05:51.300","Text":"I can call that energy du and it is equal to the charge,"},{"Start":"05:51.300 ","End":"05:58.535","Text":"so dq multiplied by the voltage at that moment on the capacitor."},{"Start":"05:58.535 ","End":"06:03.950","Text":"Of course, the voltage on the capacitor is dependent on the charges of each plate."},{"Start":"06:03.950 ","End":"06:05.315","Text":"But at this moment,"},{"Start":"06:05.315 ","End":"06:08.580","Text":"this is the voltage on the capacitor."},{"Start":"06:09.050 ","End":"06:11.555","Text":"The voltage on our capacitor,"},{"Start":"06:11.555 ","End":"06:13.535","Text":"as we can see from this equation,"},{"Start":"06:13.535 ","End":"06:20.750","Text":"is dependent on q divided by c. If we rearrange this,"},{"Start":"06:20.750 ","End":"06:24.395","Text":"we see that the voltage is always dependent on the charge."},{"Start":"06:24.395 ","End":"06:33.335","Text":"We can say that our V_c is as a function of q. Let\u0027s write this out."},{"Start":"06:33.335 ","End":"06:37.355","Text":"The energy required for me to move this Delta q"},{"Start":"06:37.355 ","End":"06:41.840","Text":"from 1 plate to the other is equal to the charge that I\u0027m moving,"},{"Start":"06:41.840 ","End":"06:43.250","Text":"which here is Delta q,"},{"Start":"06:43.250 ","End":"06:46.610","Text":"multiplied by the voltage as a function of q."},{"Start":"06:46.610 ","End":"06:53.145","Text":"As we saw, that\u0027s just going to be equal to q divided by c."},{"Start":"06:53.145 ","End":"07:01.415","Text":"Now I want to add up the energy to move the total charge from 1 plate to the other."},{"Start":"07:01.415 ","End":"07:07.375","Text":"Of course, I\u0027m going to have to integrate like so."},{"Start":"07:07.375 ","End":"07:11.990","Text":"When I\u0027m integrating, obviously my Delta q becomes dq."},{"Start":"07:11.990 ","End":"07:21.180","Text":"What I\u0027m going to be doing is I\u0027m going to be integrating along q divided by c, dq."},{"Start":"07:21.180 ","End":"07:24.500","Text":"This is going to give me my total energy to move"},{"Start":"07:24.500 ","End":"07:29.150","Text":"all of the charges from the bottom plate to the top plate."},{"Start":"07:29.150 ","End":"07:37.290","Text":"What I\u0027ll get once I do this integral is q^2 divided by 2c."},{"Start":"07:41.300 ","End":"07:44.870","Text":"Because we did an indefinite integral we\u0027re meant to"},{"Start":"07:44.870 ","End":"07:48.560","Text":"have an integrating constant added onto here."},{"Start":"07:48.560 ","End":"07:53.105","Text":"But if we imagine that the capacitor wasn\u0027t charged in the first place,"},{"Start":"07:53.105 ","End":"07:57.440","Text":"so our integrating constant in that case would be equal to 0,"},{"Start":"07:57.440 ","End":"07:59.525","Text":"in which case this is what we\u0027re left with."},{"Start":"07:59.525 ","End":"08:05.570","Text":"We have q^2 divided by 2c or 1/2 q^2 divided by c,"},{"Start":"08:05.570 ","End":"08:09.810","Text":"which is exactly this equation over here."},{"Start":"08:09.810 ","End":"08:14.690","Text":"Of course we can move between these equations and the third one just by"},{"Start":"08:14.690 ","End":"08:20.460","Text":"substituting in this equation in one of its forms."},{"Start":"08:20.460 ","End":"08:22.730","Text":"Of course this equation is correct for"},{"Start":"08:22.730 ","End":"08:26.419","Text":"any type of capacitor not just the parallel plates example."},{"Start":"08:26.419 ","End":"08:30.890","Text":"Because as we can see from our equations in both methods,"},{"Start":"08:30.890 ","End":"08:32.645","Text":"both Method 1 and 2,"},{"Start":"08:32.645 ","End":"08:38.780","Text":"we haven\u0027t used anything that describes our capacitor as a parallel plate capacitor."},{"Start":"08:38.780 ","End":"08:41.590","Text":"In order to derive the equation."},{"Start":"08:41.590 ","End":"08:43.625","Text":"This is correct for any capacitor,"},{"Start":"08:43.625 ","End":"08:46.470","Text":"and that is the end of this lesson."}],"ID":22277},{"Watched":false,"Name":"Charging RC Circuit","Duration":"18m 36s","ChapterTopicVideoID":21497,"CourseChapterTopicPlaylistID":99472,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.830","Text":"Hello. In this lesson,"},{"Start":"00:01.830 ","End":"00:05.835","Text":"we\u0027re going to be learning about the charging of an IC circuit."},{"Start":"00:05.835 ","End":"00:07.800","Text":"In an IC circuit,"},{"Start":"00:07.800 ","End":"00:10.529","Text":"we have a battery or a voltage source,"},{"Start":"00:10.529 ","End":"00:12.750","Text":"we have a capacitor,"},{"Start":"00:12.750 ","End":"00:15.090","Text":"and we have a resistor."},{"Start":"00:15.090 ","End":"00:21.330","Text":"The moment that we have a capacitor which is attached to a battery or a voltage source,"},{"Start":"00:21.330 ","End":"00:26.400","Text":"then we know that the capacitor is going to be charging."},{"Start":"00:26.400 ","End":"00:32.650","Text":"What does that mean? That means that charge is going to be stored on the capacitor."},{"Start":"00:33.080 ","End":"00:37.820","Text":"The equation for charge stored on a capacitor in"},{"Start":"00:37.820 ","End":"00:42.695","Text":"a charging case is q as a function of time,"},{"Start":"00:42.695 ","End":"00:46.280","Text":"which is equal to the capacitance of the capacitor"},{"Start":"00:46.280 ","End":"00:50.120","Text":"multiplied by the voltage of the voltage source,"},{"Start":"00:50.120 ","End":"00:59.785","Text":"multiplied by 1-e^-t divided by Tau."},{"Start":"00:59.785 ","End":"01:05.750","Text":"What is tau? Tau is equal to the resistance of"},{"Start":"01:05.750 ","End":"01:11.510","Text":"the resistor multiplied by the capacitance of the capacitor."},{"Start":"01:11.510 ","End":"01:15.530","Text":"Then the current, therefore,"},{"Start":"01:15.530 ","End":"01:22.100","Text":"as a function of time in the circuit is equal to the voltage of the voltage source"},{"Start":"01:22.100 ","End":"01:29.660","Text":"divided by the resistance multiplied by e^-t divided by Tau,"},{"Start":"01:29.660 ","End":"01:33.060","Text":"where Tau is this over here."},{"Start":"01:33.710 ","End":"01:38.120","Text":"These are equations that you have to be familiar with and that you"},{"Start":"01:38.120 ","End":"01:41.810","Text":"should probably have written in your equation sheets."},{"Start":"01:41.810 ","End":"01:44.570","Text":"Now what we\u0027re going to do in this lesson is we\u0027re going to"},{"Start":"01:44.570 ","End":"01:47.345","Text":"see how to derive these equations,"},{"Start":"01:47.345 ","End":"01:50.215","Text":"so where these equations come from."},{"Start":"01:50.215 ","End":"01:55.445","Text":"The first thing I want to do is I want to know what\u0027s happening in my circuit."},{"Start":"01:55.445 ","End":"01:57.995","Text":"Whenever I want to know what\u0027s happening in my circuit,"},{"Start":"01:57.995 ","End":"02:00.900","Text":"I start with Kirchhoff\u0027s laws."},{"Start":"02:01.070 ","End":"02:06.520","Text":"We\u0027re going to write the equation for the voltages in the circuit."},{"Start":"02:06.520 ","End":"02:07.580","Text":"Just as a reminder,"},{"Start":"02:07.580 ","End":"02:13.085","Text":"Kirchhoff\u0027s law for voltages states that the sum of all of the voltages"},{"Start":"02:13.085 ","End":"02:20.320","Text":"around a closed loop in a circuit must be equal to 0."},{"Start":"02:20.320 ","End":"02:22.240","Text":"Let\u0027s do that,"},{"Start":"02:22.240 ","End":"02:24.610","Text":"so let\u0027s start from this point over here."},{"Start":"02:24.610 ","End":"02:28.145","Text":"Here we\u0027re going from the short end is negative,"},{"Start":"02:28.145 ","End":"02:29.975","Text":"and the longer end is positive."},{"Start":"02:29.975 ","End":"02:31.520","Text":"If we go across like this,"},{"Start":"02:31.520 ","End":"02:34.580","Text":"so we have an increase in voltage."},{"Start":"02:34.580 ","End":"02:37.505","Text":"As we go across the battery,"},{"Start":"02:37.505 ","End":"02:42.530","Text":"we have an increase in voltage of the voltage of the battery, which is V_0."},{"Start":"02:43.030 ","End":"02:45.928","Text":"Now it keep on traveling,"},{"Start":"02:45.928 ","End":"02:47.705","Text":"and then I reach the resistor."},{"Start":"02:47.705 ","End":"02:52.795","Text":"First of all, let\u0027s say that this is the direction of the current."},{"Start":"02:52.795 ","End":"02:54.814","Text":"I reach the resistor,"},{"Start":"02:54.814 ","End":"02:59.630","Text":"and if I\u0027m going in the direction of the current across the resistor,"},{"Start":"02:59.630 ","End":"03:02.510","Text":"so as I pass the resistor,"},{"Start":"03:02.510 ","End":"03:05.120","Text":"I\u0027m going to have a voltage drop."},{"Start":"03:05.120 ","End":"03:09.360","Text":"I\u0027m going to have a voltage drop of IR."},{"Start":"03:10.130 ","End":"03:14.715","Text":"There we go. Now we\u0027ve gotten past the resistor."},{"Start":"03:14.715 ","End":"03:17.320","Text":"Of course, just a note over here,"},{"Start":"03:17.320 ","End":"03:22.081","Text":"if I would have defined the current to be traveling in the opposite direction,"},{"Start":"03:22.081 ","End":"03:25.445","Text":"so as I would move from this point to this point across the resistor,"},{"Start":"03:25.445 ","End":"03:30.080","Text":"I would be traveling in the negative direction to the current,"},{"Start":"03:30.080 ","End":"03:36.330","Text":"which would mean that I would have a voltage increase of plus IR."},{"Start":"03:36.330 ","End":"03:39.710","Text":"If my arrow for current was in opposite direction,"},{"Start":"03:39.710 ","End":"03:43.110","Text":"I would write plus IR over here."},{"Start":"03:44.060 ","End":"03:48.480","Text":"Now I\u0027m traveling across my capacitor."},{"Start":"03:48.480 ","End":"03:53.435","Text":"If this is the direction that we already said of the current,"},{"Start":"03:53.435 ","End":"03:57.155","Text":"so it\u0027s going in this clockwise direction,"},{"Start":"03:57.155 ","End":"04:01.910","Text":"that means that here will be the capacitor\u0027s positive plate,"},{"Start":"04:01.910 ","End":"04:05.165","Text":"and here will be the negative plate."},{"Start":"04:05.165 ","End":"04:07.220","Text":"You should remember this,"},{"Start":"04:07.220 ","End":"04:11.330","Text":"the direction that the current is traveling in the first plate."},{"Start":"04:11.330 ","End":"04:14.360","Text":"Next to where the current is, is the positive,"},{"Start":"04:14.360 ","End":"04:18.245","Text":"and the plate furthest away is the negatively charged plate."},{"Start":"04:18.245 ","End":"04:23.690","Text":"That means that we\u0027re going from a higher charge over here to a lower charge over here,"},{"Start":"04:23.690 ","End":"04:28.730","Text":"which means that we have a voltage drop when we cross the capacitor."},{"Start":"04:28.730 ","End":"04:30.845","Text":"Again, that\u0027s going to be a minus."},{"Start":"04:30.845 ","End":"04:32.570","Text":"Then the voltage of the capacitor,"},{"Start":"04:32.570 ","End":"04:36.885","Text":"let\u0027s just write it in the meantime as VC."},{"Start":"04:36.885 ","End":"04:43.280","Text":"Then we carry on traveling around the loop and we\u0027ve reach the point where we started,"},{"Start":"04:43.280 ","End":"04:45.680","Text":"that means that we finished the loop."},{"Start":"04:45.680 ","End":"04:46.900","Text":"We have a closed loop,"},{"Start":"04:46.900 ","End":"04:48.800","Text":"and according to Kirchhoff\u0027s law,"},{"Start":"04:48.800 ","End":"04:52.480","Text":"this has to be equal to 0."},{"Start":"04:53.300 ","End":"04:56.380","Text":"First of all, what is VC?"},{"Start":"04:56.380 ","End":"04:58.309","Text":"What is the voltage on the capacitor?"},{"Start":"04:58.309 ","End":"05:02.120","Text":"We already saw in the previous lesson that the voltage on"},{"Start":"05:02.120 ","End":"05:08.730","Text":"the capacitor or across the capacitor depends on the charges on the plates."},{"Start":"05:08.750 ","End":"05:15.195","Text":"We can say that VC is equal to q divided by c,"},{"Start":"05:15.195 ","End":"05:20.310","Text":"where q is of course as a function of time."},{"Start":"05:20.310 ","End":"05:25.735","Text":"Q is changing all the time as a function of time,"},{"Start":"05:25.735 ","End":"05:30.375","Text":"and therefore, our voltage across the capacitor is also changing."},{"Start":"05:30.375 ","End":"05:32.790","Text":"What is our current?"},{"Start":"05:32.790 ","End":"05:39.300","Text":"Our current is equal to the first derivative of our charge."},{"Start":"05:39.710 ","End":"05:42.930","Text":"Our charge is as a function of time,"},{"Start":"05:42.930 ","End":"05:46.115","Text":"so if we take the first derivative of that, we get current."},{"Start":"05:46.115 ","End":"05:48.140","Text":"Because that\u0027s exactly what current means,"},{"Start":"05:48.140 ","End":"05:53.140","Text":"how much charge passes a certain point in a given time."},{"Start":"05:53.140 ","End":"05:55.370","Text":"That\u0027s exactly what this means."},{"Start":"05:55.370 ","End":"05:59.510","Text":"The change in charge in a given time is current,"},{"Start":"05:59.510 ","End":"06:01.950","Text":"and that\u0027s the first derivative."},{"Start":"06:02.570 ","End":"06:07.080","Text":"Now let\u0027s plug all of this into our equation."},{"Start":"06:07.080 ","End":"06:12.225","Text":"We get that V_0 minus IR,"},{"Start":"06:12.225 ","End":"06:13.620","Text":"I is q dot,"},{"Start":"06:13.620 ","End":"06:17.865","Text":"so minus q dot R minus VC,"},{"Start":"06:17.865 ","End":"06:21.075","Text":"which is q divided by C,"},{"Start":"06:21.075 ","End":"06:24.190","Text":"is equal to 0."},{"Start":"06:24.680 ","End":"06:35.145","Text":"Now what we have before us is a differential equation where q is my variable,"},{"Start":"06:35.145 ","End":"06:36.720","Text":"V_0, R,"},{"Start":"06:36.720 ","End":"06:39.540","Text":"and C are constants that are given to me."},{"Start":"06:39.540 ","End":"06:41.810","Text":"What we\u0027re going to do is we\u0027re going to solve"},{"Start":"06:41.810 ","End":"06:48.180","Text":"this differential equation and we\u0027ll see that we\u0027ll get this equation for q."},{"Start":"06:48.200 ","End":"06:51.380","Text":"Then once we have this equation for q,"},{"Start":"06:51.380 ","End":"06:55.020","Text":"so our I is the first derivative of q."},{"Start":"06:55.020 ","End":"06:57.620","Text":"What we\u0027ll do is we\u0027ll take this equation,"},{"Start":"06:57.620 ","End":"06:59.945","Text":"we\u0027ll take the first derivative,"},{"Start":"06:59.945 ","End":"07:02.610","Text":"and we\u0027ll get this equation."},{"Start":"07:04.010 ","End":"07:12.455","Text":"Step number 2 is to solve this differential equation."},{"Start":"07:12.455 ","End":"07:20.825","Text":"First of all, let\u0027s imagine that the initial charge on the capacitor is equal to 0."},{"Start":"07:20.825 ","End":"07:22.910","Text":"Before anything was connected,"},{"Start":"07:22.910 ","End":"07:25.910","Text":"the capacitor was completely discharged."},{"Start":"07:25.910 ","End":"07:29.405","Text":"That means that q(t=0,"},{"Start":"07:29.405 ","End":"07:32.890","Text":"was equal to 0."},{"Start":"07:34.400 ","End":"07:39.635","Text":"Now what we\u0027re going to do is we\u0027re going to take this equation."},{"Start":"07:39.635 ","End":"07:43.560","Text":"We\u0027re going to put our q dot R on 1 side of"},{"Start":"07:43.560 ","End":"07:50.095","Text":"the equation and we\u0027re going to multiply everything by C to get rid of this denominator."},{"Start":"07:50.095 ","End":"07:52.245","Text":"What we\u0027re going to have,"},{"Start":"07:52.245 ","End":"07:53.870","Text":"therefore, after we\u0027ve done that,"},{"Start":"07:53.870 ","End":"07:59.765","Text":"is we\u0027ll have C multiplied by V_0 minus q."},{"Start":"07:59.765 ","End":"08:01.910","Text":"Because we multiplied by C,"},{"Start":"08:01.910 ","End":"08:08.360","Text":"which is equal to q dot multiplied by R multiplied by"},{"Start":"08:08.360 ","End":"08:13.560","Text":"C. We know that"},{"Start":"08:13.560 ","End":"08:20.145","Text":"q dot is equal to dq by dt."},{"Start":"08:20.145 ","End":"08:24.650","Text":"We\u0027re going to substitute that in over here."},{"Start":"08:24.650 ","End":"08:34.565","Text":"Then what we\u0027ll have is that all of this is equal to dq by dt multiplied by RC."},{"Start":"08:34.565 ","End":"08:37.355","Text":"Then to get rid of this denominator,"},{"Start":"08:37.355 ","End":"08:40.685","Text":"we\u0027re going to multiply everything by dt."},{"Start":"08:40.685 ","End":"08:46.970","Text":"What we\u0027re going to have is CV_0 minus q."},{"Start":"08:46.970 ","End":"08:51.335","Text":"All of this side, multiplied by dt,"},{"Start":"08:51.335 ","End":"08:58.260","Text":"which is equal to this over here, so RC dq."},{"Start":"08:59.700 ","End":"09:02.575","Text":"Now we want to isolate out our dt."},{"Start":"09:02.575 ","End":"09:08.500","Text":"We\u0027re going to subtract both sides by what is in this bracket."},{"Start":"09:08.500 ","End":"09:15.985","Text":"What we\u0027re going to get is that dt is equal to"},{"Start":"09:15.985 ","End":"09:24.710","Text":"RC divided by CV naught minus q dq."},{"Start":"09:26.160 ","End":"09:30.685","Text":"Now let\u0027s divide both sides by RC."},{"Start":"09:30.685 ","End":"09:32.680","Text":"It doesn\u0027t really make much of a difference,"},{"Start":"09:32.680 ","End":"09:34.420","Text":"but let\u0027s just do it anyway."},{"Start":"09:34.420 ","End":"09:40.310","Text":"What we\u0027ll have is 1 divided by RC dt,"},{"Start":"09:40.310 ","End":"09:49.110","Text":"which is equal to 1 divided by CV naught minus q dq."},{"Start":"09:49.110 ","End":"09:50.760","Text":"Of course, at this stage,"},{"Start":"09:50.760 ","End":"09:56.210","Text":"what we\u0027re going to do is we\u0027re going to integrate both sides."},{"Start":"09:56.580 ","End":"10:02.410","Text":"Either I can do these indefinite integrals and then I can find some"},{"Start":"10:02.410 ","End":"10:08.320","Text":"constant the end or I can do definite integrals and set in some bounds over here."},{"Start":"10:08.320 ","End":"10:10.355","Text":"Let\u0027s do it that way."},{"Start":"10:10.355 ","End":"10:12.525","Text":"On my time integral,"},{"Start":"10:12.525 ","End":"10:22.405","Text":"I\u0027m integrating from t=0 until some general time t. Then with my charges,"},{"Start":"10:22.405 ","End":"10:25.495","Text":"I\u0027m integrating from my initial charge."},{"Start":"10:25.495 ","End":"10:29.665","Text":"My charge a t=0 until"},{"Start":"10:29.665 ","End":"10:37.555","Text":"my final charge which is going to be at time t, so qt."},{"Start":"10:37.555 ","End":"10:43.870","Text":"Of course, I\u0027m reminding you that qt is what we\u0027re actually looking for."},{"Start":"10:43.870 ","End":"10:47.530","Text":"Remember we\u0027re deriving this equation which is q as a function of"},{"Start":"10:47.530 ","End":"10:53.335","Text":"t. For this integral along dt,"},{"Start":"10:53.335 ","End":"11:00.085","Text":"all we\u0027re going to have is just t divided by RC."},{"Start":"11:00.085 ","End":"11:02.560","Text":"Once we substitute this in,"},{"Start":"11:02.560 ","End":"11:04.705","Text":"this is going to be equal to."},{"Start":"11:04.705 ","End":"11:09.700","Text":"Here we have an integral of 1 divided by something minus q,"},{"Start":"11:09.700 ","End":"11:12.265","Text":"where q is our variable."},{"Start":"11:12.265 ","End":"11:15.040","Text":"What we\u0027re going to have is Ln of that,"},{"Start":"11:15.040 ","End":"11:20.830","Text":"so Ln of CV naught minus q."},{"Start":"11:20.830 ","End":"11:25.105","Text":"Then we have to multiply this by the inner derivative."},{"Start":"11:25.105 ","End":"11:28.390","Text":"The inner derivative over here is negative 1."},{"Start":"11:28.390 ","End":"11:30.700","Text":"We multiply this by a negative."},{"Start":"11:30.700 ","End":"11:34.300","Text":"Then we have to substitute in our balance,"},{"Start":"11:34.300 ","End":"11:37.445","Text":"which is q at t=0,"},{"Start":"11:37.445 ","End":"11:40.180","Text":"which we said was our initial condition over here."},{"Start":"11:40.180 ","End":"11:43.290","Text":"So q at t=0 is 0,"},{"Start":"11:43.290 ","End":"11:51.760","Text":"then q at time t. Let\u0027s substitute our bounds."},{"Start":"11:51.760 ","End":"11:57.820","Text":"We have t divided by RC= negative Ln of"},{"Start":"11:57.820 ","End":"12:06.490","Text":"CV naught minus q at time t divided by CV naught minus 0,"},{"Start":"12:06.490 ","End":"12:09.830","Text":"so minus CV naught."},{"Start":"12:11.400 ","End":"12:14.395","Text":"Now we want to get rid of this Ln."},{"Start":"12:14.395 ","End":"12:18.730","Text":"What we\u0027re going to do is we\u0027re going to multiply both sides by negative 1."},{"Start":"12:18.730 ","End":"12:25.750","Text":"Then we\u0027re going to use the exponent because that will get rid of the Ln."},{"Start":"12:25.750 ","End":"12:27.955","Text":"It\u0027s the inverse of the Ln."},{"Start":"12:27.955 ","End":"12:31.435","Text":"What we\u0027re going to have is on this side,"},{"Start":"12:31.435 ","End":"12:38.425","Text":"we\u0027ll have e^negative t divided RC."},{"Start":"12:38.425 ","End":"12:46.240","Text":"Because if I do e^Ln so we just get this fraction over here."},{"Start":"12:46.240 ","End":"12:47.710","Text":"It just gets rid of the Ln."},{"Start":"12:47.710 ","End":"12:55.765","Text":"Then that is equal to CV naught minus q as a function of t divided by CV naught,"},{"Start":"12:55.765 ","End":"12:58.945","Text":"we\u0027ll multiply both sides by CV naught."},{"Start":"12:58.945 ","End":"13:05.920","Text":"We have CV naught e^negative t divided by RC,"},{"Start":"13:05.920 ","End":"13:11.830","Text":"which is equal to CV naught minus q as a function of t."},{"Start":"13:11.830 ","End":"13:15.400","Text":"Now what I want to do is I want to isolate out my Q"},{"Start":"13:15.400 ","End":"13:18.985","Text":"as a function of t. I\u0027ll move it over to this side."},{"Start":"13:18.985 ","End":"13:22.970","Text":"Then I can see that I have CV naught here and here."},{"Start":"13:23.280 ","End":"13:31.270","Text":"I have as a common multiple is CV naught multiplied by over here."},{"Start":"13:31.270 ","End":"13:33.220","Text":"This I\u0027ve moved onto this side."},{"Start":"13:33.220 ","End":"13:43.130","Text":"I have 1 minus e^negative t divided by RC."},{"Start":"13:43.760 ","End":"13:46.560","Text":"This is my q as a function of t,"},{"Start":"13:46.560 ","End":"13:48.840","Text":"and this is exactly what we have over here."},{"Start":"13:48.840 ","End":"13:51.525","Text":"Remember, RC is tau."},{"Start":"13:51.525 ","End":"13:55.210","Text":"We got this exact same equation."},{"Start":"13:55.230 ","End":"13:59.170","Text":"This value tau, which is equal to RC,"},{"Start":"13:59.170 ","End":"14:03.925","Text":"is called the RC time constant. What does it mean?"},{"Start":"14:03.925 ","End":"14:08.050","Text":"It\u0027s the time required for the voltage to fall or"},{"Start":"14:08.050 ","End":"14:15.070","Text":"increase to V0 divided by e, the exponential function."},{"Start":"14:15.070 ","End":"14:17.710","Text":"What we\u0027re going to do now is we\u0027re going to draw"},{"Start":"14:17.710 ","End":"14:22.970","Text":"a graph of a charge on the capacitor as a function of time."},{"Start":"14:23.070 ","End":"14:27.400","Text":"Here is our graph and here is"},{"Start":"14:27.400 ","End":"14:32.870","Text":"the charge which is as a function of time and here is our time."},{"Start":"14:33.300 ","End":"14:35.650","Text":"We\u0027re using this equation,"},{"Start":"14:35.650 ","End":"14:38.660","Text":"so at time t=0."},{"Start":"14:40.260 ","End":"14:50.965","Text":"All of this e^0 is 1 and then we get that our charge at t=0 is equal to 0,"},{"Start":"14:50.965 ","End":"14:58.730","Text":"which if we remember was our initial condition that we set."},{"Start":"14:58.740 ","End":"15:07.285","Text":"The charge rarely is equal to 0 at t=0."},{"Start":"15:07.285 ","End":"15:09.535","Text":"What at infinite time?"},{"Start":"15:09.535 ","End":"15:12.220","Text":"In another 50 years,"},{"Start":"15:12.220 ","End":"15:14.425","Text":"so t is infinity."},{"Start":"15:14.425 ","End":"15:20.005","Text":"I\u0027m reminding you that if we have e to the power of negative,"},{"Start":"15:20.005 ","End":"15:26.035","Text":"let\u0027s say a, so this is equal 1 divided by e^a."},{"Start":"15:26.035 ","End":"15:31.600","Text":"Here we have e to the power of negative infinity divided by RC,"},{"Start":"15:31.600 ","End":"15:34.890","Text":"which is the same as 1 divided by e to"},{"Start":"15:34.890 ","End":"15:40.100","Text":"the infinity divided by RC or 1 divided by e to infinity."},{"Start":"15:40.230 ","End":"15:42.475","Text":"This is equal to,"},{"Start":"15:42.475 ","End":"15:45.010","Text":"so an a is approaching infinity which is"},{"Start":"15:45.010 ","End":"15:48.295","Text":"what we\u0027re doing here when t is approaching infinity."},{"Start":"15:48.295 ","End":"15:52.405","Text":"That means we have 1 divided by an infinitely large number"},{"Start":"15:52.405 ","End":"15:57.145","Text":"which means that this is approaching 0."},{"Start":"15:57.145 ","End":"16:01.105","Text":"That means that this term over here is equal to 0."},{"Start":"16:01.105 ","End":"16:04.255","Text":"What we\u0027re left with that q at t is equal to"},{"Start":"16:04.255 ","End":"16:10.345","Text":"infinity is CV naught multiplied by 1 minus 0."},{"Start":"16:10.345 ","End":"16:19.030","Text":"What does that mean? That means that there\u0027s some values over here, CV naught,"},{"Start":"16:19.030 ","End":"16:22.195","Text":"that at infinity,"},{"Start":"16:22.195 ","End":"16:29.860","Text":"the maximum charge that we can have on the capacitor plate is this value over here."},{"Start":"16:29.860 ","End":"16:33.200","Text":"Let\u0027s say it\u0027s something like this."},{"Start":"16:34.230 ","End":"16:38.530","Text":"What happens is that the charge as a function of time"},{"Start":"16:38.530 ","End":"16:44.294","Text":"from t=0 until infinity increases something like so,"},{"Start":"16:44.294 ","End":"16:47.380","Text":"where as we keep on going,"},{"Start":"16:47.380 ","End":"16:53.300","Text":"our graph is approaching this value of CV naught."},{"Start":"16:53.370 ","End":"16:57.550","Text":"Then the graph, obviously,"},{"Start":"16:57.550 ","End":"17:01.360","Text":"as we know, never actually hits CV naughts."},{"Start":"17:01.360 ","End":"17:05.920","Text":"However, it\u0027s constantly approaching and coming closer."},{"Start":"17:05.920 ","End":"17:13.690","Text":"What we say with this RC time constant is that if t is much greater than tau,"},{"Start":"17:13.690 ","End":"17:18.745","Text":"so if our time t is much greater than our RC time constant,"},{"Start":"17:18.745 ","End":"17:22.270","Text":"then we can say therefore,"},{"Start":"17:22.270 ","End":"17:31.360","Text":"that the charge is around about equal to this CV naught,"},{"Start":"17:31.360 ","End":"17:33.920","Text":"it\u0027s around about equal to this value over here."},{"Start":"17:35.130 ","End":"17:40.150","Text":"CV naught is the maximum charge I can have on my capacitor."},{"Start":"17:40.150 ","End":"17:45.505","Text":"This happens when my time t is much greater than my RC time constant."},{"Start":"17:45.505 ","End":"17:50.350","Text":"This happens when t is approximately generally it"},{"Start":"17:50.350 ","End":"17:56.380","Text":"said equal to 5 times my RC time constant, 5 times tau."},{"Start":"17:56.380 ","End":"18:00.325","Text":"Then we can say that my capacitor is fully charged and it has"},{"Start":"18:00.325 ","End":"18:05.600","Text":"a total charge of approximately CV naught."},{"Start":"18:05.890 ","End":"18:11.635","Text":"Remember this, the RC time constant and what it represents."},{"Start":"18:11.635 ","End":"18:15.485","Text":"That this is the graph of a capacitor charging."},{"Start":"18:15.485 ","End":"18:18.290","Text":"Now we saw how we derive this equation"},{"Start":"18:18.290 ","End":"18:21.680","Text":"for the charge on a capacitor as a function of time."},{"Start":"18:21.680 ","End":"18:26.405","Text":"Of course, if we take the first derivative of this,"},{"Start":"18:26.405 ","End":"18:30.710","Text":"then we\u0027ll get our function for the current as a function of time,"},{"Start":"18:30.710 ","End":"18:33.110","Text":"which you should get this."},{"Start":"18:33.110 ","End":"18:36.300","Text":"That is the end of this lesson."}],"ID":22278},{"Watched":false,"Name":"Discharging RC Circuit","Duration":"17m 44s","ChapterTopicVideoID":21498,"CourseChapterTopicPlaylistID":99472,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.860","Text":"Hello. In this lesson,"},{"Start":"00:01.860 ","End":"00:06.195","Text":"we\u0027re going to be talking about discharging an RC circuit."},{"Start":"00:06.195 ","End":"00:09.960","Text":"The last lesson we saw that we had a battery, a resistor,"},{"Start":"00:09.960 ","End":"00:15.524","Text":"and a capacitor attached in a circuit and we spoke about how we can charge the capacitor."},{"Start":"00:15.524 ","End":"00:19.250","Text":"Now, we\u0027re looking at the case where we\u0027re discharging the capacitor."},{"Start":"00:19.250 ","End":"00:24.555","Text":"The main difference to see is that we don\u0027t have a battery or a voltage source connected."},{"Start":"00:24.555 ","End":"00:29.475","Text":"We just have a resistor over here and a capacitor over here."},{"Start":"00:29.475 ","End":"00:35.430","Text":"Let\u0027s imagine that we have a charge Q naught or plus"},{"Start":"00:35.430 ","End":"00:42.730","Text":"Q naught on this plate and a charge of negative Q naught on this plate."},{"Start":"00:42.950 ","End":"00:46.625","Text":"The equation q,"},{"Start":"00:46.625 ","End":"00:50.665","Text":"the charge on the capacitor as a function of time,"},{"Start":"00:50.665 ","End":"00:58.495","Text":"when the capacitor is discharging is equal to the original charge or the initial charge,"},{"Start":"00:58.495 ","End":"01:03.970","Text":"multiplied by e to the power of negative t divided by Tau,"},{"Start":"01:03.970 ","End":"01:10.020","Text":"where Tau we saw in the previous lesson is equal to RC."},{"Start":"01:11.320 ","End":"01:14.980","Text":"This is an equation that you should be familiar with"},{"Start":"01:14.980 ","End":"01:18.525","Text":"and you should also write in your equation sheets."},{"Start":"01:18.525 ","End":"01:21.850","Text":"Let\u0027s say that in this case where this"},{"Start":"01:21.850 ","End":"01:24.735","Text":"is our positive plate and this is our negative plate,"},{"Start":"01:24.735 ","End":"01:32.150","Text":"we can see therefore that our current is going to be traveling in this direction."},{"Start":"01:32.150 ","End":"01:35.345","Text":"Why is the current traveling in this direction?"},{"Start":"01:35.345 ","End":"01:40.880","Text":"We have excess charge on the positive plate."},{"Start":"01:40.880 ","End":"01:42.590","Text":"That\u0027s what it means."},{"Start":"01:42.590 ","End":"01:45.820","Text":"We have this excess charge,"},{"Start":"01:45.820 ","End":"01:54.980","Text":"which is discharging out in this direction and then joining onto this plate,"},{"Start":"01:54.980 ","End":"01:57.920","Text":"and eventually, the charges or"},{"Start":"01:57.920 ","End":"02:02.990","Text":"the difference in charge between the two plates is going to be equal to 0."},{"Start":"02:02.990 ","End":"02:09.064","Text":"Then we will have a case where no charge is stored on the capacitor,"},{"Start":"02:09.064 ","End":"02:13.240","Text":"which means that the capacitor is fully discharged."},{"Start":"02:13.240 ","End":"02:16.970","Text":"What we\u0027re going to do now is like in the previous lesson,"},{"Start":"02:16.970 ","End":"02:21.965","Text":"we\u0027re going to derive this equation for the discharge of the capacitor,"},{"Start":"02:21.965 ","End":"02:25.895","Text":"and then afterward we\u0027re going to look at a circuit where we have"},{"Start":"02:25.895 ","End":"02:32.430","Text":"one capacitor which is charging whilst the other is discharging, and vice versa."},{"Start":"02:32.930 ","End":"02:36.410","Text":"We\u0027re going to derive this equation the exact same way that"},{"Start":"02:36.410 ","End":"02:39.485","Text":"we did in the previous lesson for a charging."},{"Start":"02:39.485 ","End":"02:48.845","Text":"Number 1, we\u0027re going to use Kirchhoff\u0027s laws where we\u0027re speaking about the voltage is."},{"Start":"02:48.845 ","End":"02:54.320","Text":"The first thing that we\u0027re going to do is we\u0027re going to choose a point over here."},{"Start":"02:54.320 ","End":"02:57.035","Text":"Then we know that as we go along the circuit,"},{"Start":"02:57.035 ","End":"03:00.805","Text":"adding or subtracting the voltage drops."},{"Start":"03:00.805 ","End":"03:04.955","Text":"Then we will return to this point."},{"Start":"03:04.955 ","End":"03:07.870","Text":"We would have completed a closed loop in a circuit."},{"Start":"03:07.870 ","End":"03:13.780","Text":"We know that that whole equation has to be equal to 0 according to Kirchhoff."},{"Start":"03:13.780 ","End":"03:15.150","Text":"We start from here,"},{"Start":"03:15.150 ","End":"03:16.530","Text":"we\u0027re going like so,"},{"Start":"03:16.530 ","End":"03:20.220","Text":"our current is going in this clockwise direction."},{"Start":"03:20.220 ","End":"03:24.154","Text":"That means that we go past"},{"Start":"03:24.154 ","End":"03:28.610","Text":"the resistor when we\u0027re going in the same direction as the current,"},{"Start":"03:28.610 ","End":"03:33.000","Text":"which means that we have to subtract IR,"},{"Start":"03:33.520 ","End":"03:36.080","Text":"because we have a voltage drop."},{"Start":"03:36.080 ","End":"03:39.950","Text":"Then we go over here and then we reach the capacitor."},{"Start":"03:39.950 ","End":"03:43.850","Text":"We can see that we\u0027re going from the negative plate to the positive plate,"},{"Start":"03:43.850 ","End":"03:47.330","Text":"and we\u0027re also going in the direction of the current,"},{"Start":"03:47.330 ","End":"03:51.635","Text":"which means that we\u0027re adding voltage over here."},{"Start":"03:51.635 ","End":"03:53.330","Text":"As we cross the capacitor,"},{"Start":"03:53.330 ","End":"03:55.985","Text":"we get a jump in the voltage."},{"Start":"03:55.985 ","End":"03:59.985","Text":"We add the voltage of the capacitor v_c,"},{"Start":"03:59.985 ","End":"04:04.185","Text":"and then we carry on and we reached the exact same point."},{"Start":"04:04.185 ","End":"04:08.070","Text":"We\u0027ve completed the loop so all of this is equal to 0."},{"Start":"04:08.070 ","End":"04:11.505","Text":"All right, so let\u0027s begin."},{"Start":"04:11.505 ","End":"04:15.680","Text":"We have the voltage across the capacitor,"},{"Start":"04:15.680 ","End":"04:17.135","Text":"which is equal to,"},{"Start":"04:17.135 ","End":"04:19.180","Text":"as we know, q,"},{"Start":"04:19.180 ","End":"04:21.140","Text":"the charge on the capacitor,"},{"Start":"04:21.140 ","End":"04:26.130","Text":"q_c divided by the capacitance of the capacitor."},{"Start":"04:26.960 ","End":"04:30.080","Text":"I\u0027ve written c over here to"},{"Start":"04:30.080 ","End":"04:33.395","Text":"highlight they were speaking about the charge on the capacitor."},{"Start":"04:33.395 ","End":"04:36.244","Text":"Now, why does this become important?"},{"Start":"04:36.244 ","End":"04:40.690","Text":"Let\u0027s take the current I."},{"Start":"04:40.690 ","End":"04:42.900","Text":"Let\u0027s write it out over here."},{"Start":"04:42.900 ","End":"04:45.725","Text":"We know that in general,"},{"Start":"04:45.725 ","End":"04:49.150","Text":"the current is equal to q dot,"},{"Start":"04:49.150 ","End":"04:53.000","Text":"the first derivative of q."},{"Start":"04:53.000 ","End":"04:55.820","Text":"However, what q are we speaking about?"},{"Start":"04:55.820 ","End":"04:59.030","Text":"We\u0027re of course speaking about the q on the capacitor,"},{"Start":"04:59.030 ","End":"05:02.390","Text":"the charge on the capacitor."},{"Start":"05:02.390 ","End":"05:05.795","Text":"But when we look at our diagram over here,"},{"Start":"05:05.795 ","End":"05:11.525","Text":"we see that the current represents the charge leaving the capacitor."},{"Start":"05:11.525 ","End":"05:17.490","Text":"We can see that the charge is exiting the capacitor and going all the way around."},{"Start":"05:18.460 ","End":"05:22.940","Text":"The charge on the capacitor obviously refers"},{"Start":"05:22.940 ","End":"05:26.675","Text":"to a positive value and increase in the charge."},{"Start":"05:26.675 ","End":"05:29.690","Text":"However, the current in our specific example over"},{"Start":"05:29.690 ","End":"05:35.040","Text":"here references the charge leaving the capacitor."},{"Start":"05:36.320 ","End":"05:41.870","Text":"Therefore, we have to add in a minus over here,"},{"Start":"05:41.870 ","End":"05:43.610","Text":"because what is the current?"},{"Start":"05:43.610 ","End":"05:47.660","Text":"The amount of charge that passes a point in a given time."},{"Start":"05:47.660 ","End":"05:50.615","Text":"But here, the current is leaving."},{"Start":"05:50.615 ","End":"05:52.505","Text":"It isn\u0027t charging on the capacitor."},{"Start":"05:52.505 ","End":"05:53.870","Text":"It\u0027s leaving the capacitor,"},{"Start":"05:53.870 ","End":"05:57.275","Text":"which means that we have to put a minus over here."},{"Start":"05:57.275 ","End":"06:06.320","Text":"Of course, the first derivative is also equal to dq by dt,"},{"Start":"06:06.320 ","End":"06:09.535","Text":"and of course with this minus over here."},{"Start":"06:09.535 ","End":"06:14.905","Text":"Now let\u0027s rewrite the equation and we\u0027ll move the IR over to the side."},{"Start":"06:14.905 ","End":"06:16.940","Text":"We get that v_c,"},{"Start":"06:16.940 ","End":"06:23.370","Text":"which is equal to q_c divided by c is equal to IR,"},{"Start":"06:23.370 ","End":"06:34.210","Text":"where I is negative dqc by dt multiplied by R, the resistance."},{"Start":"06:34.370 ","End":"06:40.700","Text":"The important thing to remember is that when we have a discharging capacitor,"},{"Start":"06:40.700 ","End":"06:43.325","Text":"that means that I is equal to"},{"Start":"06:43.325 ","End":"06:50.180","Text":"negative q dot because we\u0027re referring to the charge that is leaving the capacitor."},{"Start":"06:50.180 ","End":"06:53.855","Text":"When we have a case of a charging capacitor,"},{"Start":"06:53.855 ","End":"06:57.570","Text":"then I is equal to positive q dot."},{"Start":"06:58.250 ","End":"07:04.130","Text":"Now, in order to get the right format for our integration,"},{"Start":"07:04.130 ","End":"07:05.945","Text":"what we want to do is we want to have"},{"Start":"07:05.945 ","End":"07:11.945","Text":"one function over on this side as a function of t and then integrate"},{"Start":"07:11.945 ","End":"07:21.310","Text":"dt and on this side we want to have a function of q where we integrate according to dq."},{"Start":"07:21.310 ","End":"07:25.380","Text":"This is what we want to get."},{"Start":"07:25.380 ","End":"07:31.565","Text":"What we\u0027re going to do is we\u0027re going to divide both sides by qc/c,"},{"Start":"07:31.565 ","End":"07:35.580","Text":"and multiply both sides by dt."},{"Start":"07:35.930 ","End":"07:42.245","Text":"In that case, what we\u0027re going to have is that dt is equal to,"},{"Start":"07:42.245 ","End":"07:48.210","Text":"and then we\u0027ll have RC divided by q."},{"Start":"07:48.210 ","End":"07:52.110","Text":"Now, I just won\u0027t write out the c. We know it\u0027s the charge on"},{"Start":"07:52.110 ","End":"07:59.625","Text":"the capacitor divided by q and then we have our dq over here."},{"Start":"07:59.625 ","End":"08:02.895","Text":"Then, of course, we can integrate."},{"Start":"08:02.895 ","End":"08:07.775","Text":"We can either leave this as an indefinite integral and then we have to add in"},{"Start":"08:07.775 ","End":"08:12.680","Text":"our integrates and constant at the end or we can do a definite integral."},{"Start":"08:12.680 ","End":"08:15.920","Text":"Let\u0027s begin the definite integral."},{"Start":"08:15.920 ","End":"08:22.045","Text":"First of all, we know that our charge"},{"Start":"08:22.045 ","End":"08:29.705","Text":"q(t=0) is equal to Q naught,"},{"Start":"08:29.705 ","End":"08:31.815","Text":"as we saw over here."},{"Start":"08:31.815 ","End":"08:36.265","Text":"Now, what we can do when we plug in our integration constants."},{"Start":"08:36.265 ","End":"08:46.260","Text":"We\u0027re integrating a long time from time t=0 until some general time t. Then along q,"},{"Start":"08:46.260 ","End":"08:50.240","Text":"we\u0027re integrating from the time at t=0,"},{"Start":"08:50.240 ","End":"08:53.890","Text":"the charge at that time, which is this,"},{"Start":"08:53.890 ","End":"08:58.295","Text":"until the charge at our general time,"},{"Start":"08:58.295 ","End":"09:03.600","Text":"t. I\u0027m reminding you that this is in fact what we\u0027re trying to find."},{"Start":"09:03.600 ","End":"09:07.650","Text":"This is the exact equation that we are deriving over here."},{"Start":"09:08.600 ","End":"09:13.065","Text":"When we integrate our dt from 0 to t,"},{"Start":"09:13.065 ","End":"09:16.025","Text":"we\u0027re just going to have t over here,"},{"Start":"09:16.025 ","End":"09:18.050","Text":"which is equal to,"},{"Start":"09:18.050 ","End":"09:20.165","Text":"so here we\u0027re integrating."},{"Start":"09:20.165 ","End":"09:26.029","Text":"We have RC and then we\u0027re integrating 1 divided by q dq,"},{"Start":"09:26.029 ","End":"09:34.530","Text":"which is just going to give us ln of q from the bounds of q at t=0,"},{"Start":"09:34.530 ","End":"09:43.860","Text":"which is just Q naught until q(t)."},{"Start":"09:43.860 ","End":"09:47.430","Text":"Of course, we mustn\u0027t forget the minus sign."},{"Start":"09:47.430 ","End":"09:52.225","Text":"I forgot it now where I is equal to negative q dot."},{"Start":"09:52.225 ","End":"09:54.890","Text":"Here we added in the minus,"},{"Start":"09:54.890 ","End":"09:58.875","Text":"but here I forgot to add it in."},{"Start":"09:58.875 ","End":"10:02.000","Text":"We\u0027ll put it over here."},{"Start":"10:02.480 ","End":"10:05.600","Text":"Now when we substitute in the bounds,"},{"Start":"10:05.600 ","End":"10:15.680","Text":"so we have negative RC and then we have ln of q(t) divided by Q naught."},{"Start":"10:17.730 ","End":"10:23.740","Text":"Now what I want to do is I want to divide both sides by negative RC."},{"Start":"10:23.740 ","End":"10:28.135","Text":"I\u0027ll have negative t divided by RC,"},{"Start":"10:28.135 ","End":"10:35.080","Text":"which is equal to ln of qt divided by Q naught."},{"Start":"10:35.080 ","End":"10:40.630","Text":"Now I\u0027m going to raise everything to e with my e function,"},{"Start":"10:40.630 ","End":"10:43.225","Text":"which of course gets rid of the ln over here."},{"Start":"10:43.225 ","End":"10:48.595","Text":"I\u0027ll be left with is e to the negative t divided by RC,"},{"Start":"10:48.595 ","End":"10:54.835","Text":"which is equal to q as a function of t divided by Q naught."},{"Start":"10:54.835 ","End":"10:58.300","Text":"Now we\u0027ll multiply both sides by this Q naught."},{"Start":"10:58.300 ","End":"11:03.685","Text":"What we\u0027ll get is that q as a function of t is equal to"},{"Start":"11:03.685 ","End":"11:09.310","Text":"Q naught multiplied by e to the negative t divided by RC,"},{"Start":"11:09.310 ","End":"11:15.980","Text":"which is exactly what we have here where I\u0027m reminding that Tau is equal to RC."},{"Start":"11:16.920 ","End":"11:21.685","Text":"Now let\u0027s look on the graph at what the charge on the capacitor"},{"Start":"11:21.685 ","End":"11:26.830","Text":"looks like as a function of time."},{"Start":"11:26.830 ","End":"11:30.610","Text":"Our initial charge we said was this Q naught."},{"Start":"11:30.610 ","End":"11:32.800","Text":"This is what we started with."},{"Start":"11:32.800 ","End":"11:38.955","Text":"Then, slowly the charge is going to approach 0,"},{"Start":"11:38.955 ","End":"11:44.984","Text":"which means that slowly the charge on the capacitor is going to be equal to 0."},{"Start":"11:44.984 ","End":"11:46.800","Text":"When does this happen?"},{"Start":"11:46.800 ","End":"11:52.555","Text":"We say that when t is much greater than Tau,"},{"Start":"11:52.555 ","End":"11:54.670","Text":"where Tau is RC,"},{"Start":"11:54.670 ","End":"11:58.179","Text":"then we can say that the capacitor is completely"},{"Start":"11:58.179 ","End":"12:03.380","Text":"discharged meaning that the charge on it is 0."},{"Start":"12:04.260 ","End":"12:10.060","Text":"The value for Tau is somewhere around here."},{"Start":"12:10.060 ","End":"12:15.790","Text":"We generally say that when t is much larger than Tau,"},{"Start":"12:15.790 ","End":"12:23.875","Text":"so we\u0027re generally referring to t being equal to approximately 5 times the value of Tau."},{"Start":"12:23.875 ","End":"12:28.630","Text":"Now, let\u0027s look at the equation for the current"},{"Start":"12:28.630 ","End":"12:33.355","Text":"as a function of time when we\u0027re dealing with a discharging RC circuit."},{"Start":"12:33.355 ","End":"12:39.475","Text":"We know that current is equal to the negative in discharging"},{"Start":"12:39.475 ","End":"12:45.870","Text":"negative first derivative of the charge on the capacitor."},{"Start":"12:45.870 ","End":"12:51.510","Text":"All we have to do is we have to take our equation for the charge on the capacitor."},{"Start":"12:51.510 ","End":"12:55.215","Text":"Take the first derivative with respect to t,"},{"Start":"12:55.215 ","End":"12:57.930","Text":"and then obviously put a minus in this case of"},{"Start":"12:57.930 ","End":"13:01.755","Text":"discharging and that will give us the equation which is"},{"Start":"13:01.755 ","End":"13:08.830","Text":"equal to the initial charge Q naught divided by RC or divided by Tau,"},{"Start":"13:08.830 ","End":"13:15.890","Text":"multiplied by e to the power of negative t divided by Tau."},{"Start":"13:16.770 ","End":"13:21.925","Text":"This is another equation that you have to be familiar with and write it out in"},{"Start":"13:21.925 ","End":"13:27.250","Text":"your equation sheets and remember that this is for discharging, don\u0027t get confused."},{"Start":"13:27.250 ","End":"13:32.560","Text":"Now let\u0027s draw in blue on this graph to see the current."},{"Start":"13:32.560 ","End":"13:36.625","Text":"The current as a function of time."},{"Start":"13:36.625 ","End":"13:39.970","Text":"The initial current will be over somewhere here,"},{"Start":"13:39.970 ","End":"13:41.230","Text":"which will be equal to"},{"Start":"13:41.230 ","End":"13:50.260","Text":"the initial charge divided by RC."},{"Start":"13:50.260 ","End":"13:55.270","Text":"Then it will also gradually become smaller and"},{"Start":"13:55.270 ","End":"14:02.060","Text":"smaller and approaches 0 at t is greater than Tau."},{"Start":"14:02.760 ","End":"14:08.305","Text":"What we can see is that in this circuit as the capacitor discharges,"},{"Start":"14:08.305 ","End":"14:12.160","Text":"we have a current which starts very strong at the beginning of"},{"Start":"14:12.160 ","End":"14:19.370","Text":"the discharge and gradually weakens as the charge on the capacitor empties."},{"Start":"14:20.220 ","End":"14:27.370","Text":"Now, what we\u0027re going to do is we\u0027re going to look at another circuit."},{"Start":"14:27.370 ","End":"14:30.130","Text":"Let\u0027s take a look at this circuit."},{"Start":"14:30.130 ","End":"14:32.020","Text":"Here we have a resistor,"},{"Start":"14:32.020 ","End":"14:34.254","Text":"here we have a capacitor,"},{"Start":"14:34.254 ","End":"14:36.865","Text":"and here we have a voltage source."},{"Start":"14:36.865 ","End":"14:39.969","Text":"Then here, we have a switch."},{"Start":"14:39.969 ","End":"14:45.115","Text":"What we can do is either we can close the switch over here."},{"Start":"14:45.115 ","End":"14:52.330","Text":"Therefore, we have RC circuit with a battery source or a voltage source,"},{"Start":"14:52.330 ","End":"14:56.720","Text":"which means that our capacitor is charging."},{"Start":"14:58.020 ","End":"15:02.380","Text":"In this case, this is the positive side of the battery."},{"Start":"15:02.380 ","End":"15:06.310","Text":"The charges are going to charge the capacitor like so,"},{"Start":"15:06.310 ","End":"15:10.335","Text":"that here is our positive plate plus Q naught,"},{"Start":"15:10.335 ","End":"15:14.740","Text":"and here is our negative plate, negative Q naught."},{"Start":"15:16.430 ","End":"15:23.110","Text":"Our current in this case will be traveling in this direction."},{"Start":"15:23.820 ","End":"15:32.095","Text":"The second option, we have a case like this where the switch is like so."},{"Start":"15:32.095 ","End":"15:36.170","Text":"What we can see is that in this case,"},{"Start":"15:36.170 ","End":"15:40.110","Text":"the voltage source isn\u0027t connected over here."},{"Start":"15:40.110 ","End":"15:43.635","Text":"We can see that we have a circuit just with a capacitor and a resistor,"},{"Start":"15:43.635 ","End":"15:49.285","Text":"which means that our capacitor is going to discharge."},{"Start":"15:49.285 ","End":"15:53.665","Text":"Of course, it\u0027s going to discharge in this direction."},{"Start":"15:53.665 ","End":"15:56.095","Text":"Remember from the positive plate."},{"Start":"15:56.095 ","End":"16:01.270","Text":"Here I\u0027ve written the equation for the charging of a capacitor,"},{"Start":"16:01.270 ","End":"16:04.390","Text":"where the final charge on the capacitor,"},{"Start":"16:04.390 ","End":"16:05.920","Text":"when it\u0027s fully charged,"},{"Start":"16:05.920 ","End":"16:07.300","Text":"which we already saw last lesson,"},{"Start":"16:07.300 ","End":"16:10.330","Text":"is equal to CV_0."},{"Start":"16:10.330 ","End":"16:16.555","Text":"Let\u0027s just rewrite for a discharging capacitor,"},{"Start":"16:16.555 ","End":"16:19.540","Text":"which is what we saw just above."},{"Start":"16:19.540 ","End":"16:22.915","Text":"The charge as a function of time is equal to"},{"Start":"16:22.915 ","End":"16:29.605","Text":"the initial charge multiplied by e to the negative t divided by Tau,"},{"Start":"16:29.605 ","End":"16:32.005","Text":"where of course,"},{"Start":"16:32.005 ","End":"16:35.330","Text":"Tau is equal to RC."},{"Start":"16:35.520 ","End":"16:42.590","Text":"Of course, the final charge on a discharged capacitor is of course equal to 0."},{"Start":"16:43.230 ","End":"16:47.389","Text":"A little note which links everything."},{"Start":"16:49.590 ","End":"16:53.095","Text":"When word final charge,"},{"Start":"16:53.095 ","End":"16:54.805","Text":"at its maximal charge,"},{"Start":"16:54.805 ","End":"16:59.030","Text":"we know that is equal to C multiplied by V_0."},{"Start":"16:59.310 ","End":"17:08.185","Text":"Then if we begin discharging a fully charged capacitor,"},{"Start":"17:08.185 ","End":"17:11.450","Text":"so the initial charge is Q naught,"},{"Start":"17:12.570 ","End":"17:15.385","Text":"so this is a Q naught."},{"Start":"17:15.385 ","End":"17:19.270","Text":"What\u0027s happening?"},{"Start":"17:19.270 ","End":"17:23.500","Text":"We charge the capacitor to its maximum charge,"},{"Start":"17:23.500 ","End":"17:25.240","Text":"which is this Q final,"},{"Start":"17:25.240 ","End":"17:27.670","Text":"which we saw is equal to CV_0."},{"Start":"17:27.670 ","End":"17:30.595","Text":"Then we begin discharging."},{"Start":"17:30.595 ","End":"17:36.250","Text":"The initial charge when we begin discharge Q naught is in this case,"},{"Start":"17:36.250 ","End":"17:37.885","Text":"if it was fully charged,"},{"Start":"17:37.885 ","End":"17:40.370","Text":"it\u0027s equal to this."},{"Start":"17:41.340 ","End":"17:45.350","Text":"That is the end of this lesson."}],"ID":22279},{"Watched":false,"Name":"Exercise 5","Duration":"18m 18s","ChapterTopicVideoID":21499,"CourseChapterTopicPlaylistID":99472,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.890","Text":"Hello. In this lesson,"},{"Start":"00:01.890 ","End":"00:04.535","Text":"we\u0027re going to be answering the following question."},{"Start":"00:04.535 ","End":"00:07.140","Text":"At t is equal to 0,"},{"Start":"00:07.140 ","End":"00:10.380","Text":"the switch is switched to point A."},{"Start":"00:10.380 ","End":"00:14.655","Text":"At t is equal to 0.01,"},{"Start":"00:14.655 ","End":"00:17.790","Text":"the switch is switched to point B."},{"Start":"00:17.790 ","End":"00:20.225","Text":"Question Number 1 is,"},{"Start":"00:20.225 ","End":"00:26.400","Text":"what is the voltage across the capacitor as a function of time?"},{"Start":"00:27.290 ","End":"00:29.430","Text":"Let\u0027s start with this."},{"Start":"00:29.430 ","End":"00:34.069","Text":"We know that at t is equal to 0,"},{"Start":"00:34.069 ","End":"00:36.965","Text":"the switch over here is moved to point A."},{"Start":"00:36.965 ","End":"00:40.445","Text":"Then what we have is a capacitor, a resistor,"},{"Start":"00:40.445 ","End":"00:45.465","Text":"and a voltage source which means that we have an RC circuit,"},{"Start":"00:45.465 ","End":"00:48.940","Text":"which is going to be charging the capacitor."},{"Start":"00:50.270 ","End":"00:57.305","Text":"The voltage across the capacitor as a function of time,"},{"Start":"00:57.305 ","End":"01:03.745","Text":"is equal to the voltage on the battery."},{"Start":"01:03.745 ","End":"01:07.920","Text":"Here, I\u0027ll write battery multiplied by"},{"Start":"01:07.920 ","End":"01:15.130","Text":"1 minus e to the power of negative t divided by RC."},{"Start":"01:15.130 ","End":"01:17.390","Text":"If you haven\u0027t seen this equation yet,"},{"Start":"01:17.390 ","End":"01:21.915","Text":"please write this in your equation sheets."},{"Start":"01:21.915 ","End":"01:26.480","Text":"Now let\u0027s write out what our RC is equal to."},{"Start":"01:26.480 ","End":"01:30.395","Text":"Our resistance is over here,"},{"Start":"01:30.395 ","End":"01:35.490","Text":"100 ohms multiplied by our capacitance,"},{"Start":"01:35.490 ","End":"01:37.470","Text":"which is 50 microfarads."},{"Start":"01:37.470 ","End":"01:45.165","Text":"I\u0027m already going to write the micro as times 10 to the negative 6 farads,"},{"Start":"01:45.165 ","End":"01:51.840","Text":"and this is equal to 5 times 10 to the power of"},{"Start":"01:51.840 ","End":"01:58.430","Text":"negative 3 seconds or in other words,"},{"Start":"01:58.430 ","End":"02:02.285","Text":"it\u0027s equal to 0.005 seconds."},{"Start":"02:02.285 ","End":"02:06.775","Text":"This is, of course the RC time constant."},{"Start":"02:06.775 ","End":"02:12.530","Text":"Of course, we spoke about that approximately two lessons ago."},{"Start":"02:12.530 ","End":"02:16.280","Text":"I wrote that this is equal to Tau."},{"Start":"02:16.280 ","End":"02:20.350","Text":"Then we know that afterwards,"},{"Start":"02:20.350 ","End":"02:23.370","Text":"our switch is moved to point B,"},{"Start":"02:23.370 ","End":"02:27.160","Text":"at time t is equal to 0.01."},{"Start":"02:28.940 ","End":"02:34.010","Text":"If t is equal to 0.01,"},{"Start":"02:34.010 ","End":"02:36.020","Text":"the time when it was moved to point B,"},{"Start":"02:36.020 ","End":"02:43.645","Text":"so this is equal to twice our RC time constant."},{"Start":"02:43.645 ","End":"02:48.185","Text":"Not much time has gone by between having"},{"Start":"02:48.185 ","End":"02:53.254","Text":"this circuit over here where we have a charging circuit,"},{"Start":"02:53.254 ","End":"02:56.030","Text":"to moving the switch over to this side,"},{"Start":"02:56.030 ","End":"02:58.550","Text":"where we\u0027re dealing now with this circuit,"},{"Start":"02:58.550 ","End":"03:02.880","Text":"where we have a discharging RC circuit."},{"Start":"03:03.310 ","End":"03:08.785","Text":"Now let\u0027s write our equation for the voltage over here between"},{"Start":"03:08.785 ","End":"03:13.945","Text":"the time that the switch was switched to point A and then to point B."},{"Start":"03:13.945 ","End":"03:20.740","Text":"The total time that we were in this charging circuit so that means that t"},{"Start":"03:20.740 ","End":"03:28.185","Text":"is somewhere between 0 and 0.01 seconds,"},{"Start":"03:28.185 ","End":"03:32.325","Text":"because at 0.01, the switch was moved to B."},{"Start":"03:32.325 ","End":"03:34.584","Text":"In this time interval,"},{"Start":"03:34.584 ","End":"03:37.135","Text":"the voltage across the capacitor,"},{"Start":"03:37.135 ","End":"03:40.468","Text":"as a function of time was equal to the V_0."},{"Start":"03:40.468 ","End":"03:42.745","Text":"So the voltage on the battery,"},{"Start":"03:42.745 ","End":"03:46.675","Text":"which was equal to 10 volts multiplied by"},{"Start":"03:46.675 ","End":"03:52.355","Text":"1 minus e to the power of negative t divided by RC,"},{"Start":"03:52.355 ","End":"03:55.890","Text":"which we saw is equal to 0.005."},{"Start":"03:58.280 ","End":"04:02.150","Text":"Now let\u0027s look at what happens at the exact moment"},{"Start":"04:02.150 ","End":"04:05.705","Text":"where the switch is switched from point A to B."},{"Start":"04:05.705 ","End":"04:12.790","Text":"Now we\u0027re looking at the voltage across the capacitor at t is equal to 0.01."},{"Start":"04:12.790 ","End":"04:15.645","Text":"This is going to be still in this realm."},{"Start":"04:15.645 ","End":"04:18.425","Text":"We have the voltage of the battery,"},{"Start":"04:18.425 ","End":"04:28.500","Text":"which is 10 volts multiplied by 1 minus e to the power of negative 0.01 divided by 0.005."},{"Start":"04:29.440 ","End":"04:34.535","Text":"Then once we plug this into a calculator,"},{"Start":"04:34.535 ","End":"04:43.515","Text":"we\u0027ll get that this is approximately equal to 8.65 volts."},{"Start":"04:43.515 ","End":"04:47.630","Text":"What does that mean? This is the voltage across the capacitor at"},{"Start":"04:47.630 ","End":"04:51.860","Text":"the exact moment when the switch is switched from point A to B."},{"Start":"04:51.860 ","End":"04:55.445","Text":"Now when the switch is over here,"},{"Start":"04:55.445 ","End":"04:58.565","Text":"let\u0027s draw it again in blue."},{"Start":"04:58.565 ","End":"05:01.740","Text":"Now when the switches over here,"},{"Start":"05:02.450 ","End":"05:07.660","Text":"when we apply the equation for the voltage across"},{"Start":"05:07.660 ","End":"05:12.888","Text":"the capacitor and we have to use the initial voltage,"},{"Start":"05:12.888 ","End":"05:16.360","Text":"this is now our initial voltage."},{"Start":"05:16.430 ","End":"05:24.145","Text":"When we go into the region where t is greater than 0.01,"},{"Start":"05:24.145 ","End":"05:33.070","Text":"our initial voltage V_0 is now equal to 8.65 volts."},{"Start":"05:34.070 ","End":"05:36.765","Text":"Let\u0027s scroll down a little bit."},{"Start":"05:36.765 ","End":"05:42.690","Text":"Now we\u0027re using the equation for this circuit right now."},{"Start":"05:42.690 ","End":"05:45.090","Text":"We can see that this circuit over here,"},{"Start":"05:45.090 ","End":"05:47.660","Text":"is a discharging RC circuit."},{"Start":"05:47.660 ","End":"05:48.920","Text":"How do we know that?"},{"Start":"05:48.920 ","End":"05:53.595","Text":"We know that because there\u0027s no battery or a voltage source connected."},{"Start":"05:53.595 ","End":"05:55.775","Text":"That means that our capacitor cannot charge,"},{"Start":"05:55.775 ","End":"05:57.890","Text":"it can only discharge."},{"Start":"05:57.890 ","End":"06:04.350","Text":"Here we have a short circuit because the switch isn\u0027t closing this section over here."},{"Start":"06:04.350 ","End":"06:11.570","Text":"The equation for the voltage across the capacitor on a discharging circuit is this."},{"Start":"06:11.570 ","End":"06:14.705","Text":"Of course, it\u0027s also as a function of time."},{"Start":"06:14.705 ","End":"06:20.030","Text":"It\u0027s equal to the initial voltage V_0 multiplied"},{"Start":"06:20.030 ","End":"06:26.190","Text":"by e to the power of negative t divided by RC."},{"Start":"06:26.190 ","End":"06:31.250","Text":"Let\u0027s take a look at what our RC is equal to now."},{"Start":"06:31.250 ","End":"06:36.245","Text":"Our resistance is 50 ohms multiplied by"},{"Start":"06:36.245 ","End":"06:41.985","Text":"50 times 10 to the negative 6 farads,"},{"Start":"06:41.985 ","End":"06:50.400","Text":"which is equal to 0.0025 seconds."},{"Start":"06:50.400 ","End":"06:55.850","Text":"Now we can scroll down and we can now plug this in over here."},{"Start":"06:55.850 ","End":"07:00.260","Text":"VC as a function of t in the region where we\u0027re"},{"Start":"07:00.260 ","End":"07:05.190","Text":"discharging now is equal to V_0, which is this."},{"Start":"07:05.190 ","End":"07:13.775","Text":"It\u0027s 8.65 multiplied by e to the negative t divided by RC."},{"Start":"07:13.775 ","End":"07:15.800","Text":"Let\u0027s just write out what our t is."},{"Start":"07:15.800 ","End":"07:24.672","Text":"Actually, it\u0027s a general t divided by 0.0025."},{"Start":"07:24.672 ","End":"07:29.665","Text":"We can call this t over here."},{"Start":"07:29.665 ","End":"07:32.575","Text":"We can call it t-star."},{"Start":"07:32.575 ","End":"07:36.535","Text":"That means that also here this is t-star."},{"Start":"07:36.535 ","End":"07:38.365","Text":"What does t-star?"},{"Start":"07:38.365 ","End":"07:44.065","Text":"T-star is from the moment that the switch is at point B,"},{"Start":"07:44.065 ","End":"07:46.105","Text":"so we\u0027ll call it t-star."},{"Start":"07:46.105 ","End":"07:48.940","Text":"It\u0027s just to remind us that we\u0027re looking at"},{"Start":"07:48.940 ","End":"07:51.955","Text":"the voltage of the capacitor as a function of t but of course,"},{"Start":"07:51.955 ","End":"07:56.004","Text":"we\u0027re not going from t is equal to 0 when we\u0027re using this equation."},{"Start":"07:56.004 ","End":"07:58.330","Text":"Because at t is equal to 0,"},{"Start":"07:58.330 ","End":"08:00.295","Text":"we know that we\u0027re using this circuit,"},{"Start":"08:00.295 ","End":"08:02.755","Text":"which is a charging RC circuit."},{"Start":"08:02.755 ","End":"08:06.070","Text":"Which means that we have to use this equation or this equation,"},{"Start":"08:06.070 ","End":"08:07.465","Text":"it\u0027s a different equation."},{"Start":"08:07.465 ","End":"08:11.631","Text":"Then from t is equal to 0.01,"},{"Start":"08:11.631 ","End":"08:13.990","Text":"then we move to this circuit,"},{"Start":"08:13.990 ","End":"08:16.540","Text":"which is a discharging RC circuit,"},{"Start":"08:16.540 ","End":"08:19.285","Text":"which means that we have to use this equation."},{"Start":"08:19.285 ","End":"08:23.980","Text":"This star shows us that we\u0027re speaking"},{"Start":"08:23.980 ","End":"08:29.890","Text":"about a different time and just to remind us about this difference in the equation."},{"Start":"08:29.890 ","End":"08:32.575","Text":"A lot of the times this is called t-star,"},{"Start":"08:32.575 ","End":"08:35.500","Text":"other times it\u0027s called Delta t,"},{"Start":"08:35.500 ","End":"08:37.990","Text":"which means the time difference."},{"Start":"08:37.990 ","End":"08:41.050","Text":"We do the jump between the time"},{"Start":"08:41.050 ","End":"08:44.350","Text":"that the switches at A to the time that the switches at B."},{"Start":"08:44.350 ","End":"08:46.540","Text":"Then we only start when it\u0027s at said B."},{"Start":"08:46.540 ","End":"08:52.930","Text":"That would mean that Delta t would be equal to the time that we\u0027re at."},{"Start":"08:52.930 ","End":"08:59.635","Text":"Time t minus this over here, so minus 0.01."},{"Start":"08:59.635 ","End":"09:05.998","Text":"The time that we\u0027re at minus the timespan that the switch was at A."},{"Start":"09:05.998 ","End":"09:10.165","Text":"The switch was at A a total of 0.01 seconds,"},{"Start":"09:10.165 ","End":"09:12.325","Text":"and then it was switched to B."},{"Start":"09:12.325 ","End":"09:16.930","Text":"Here\u0027s a naught. Therefore,"},{"Start":"09:16.930 ","End":"09:19.683","Text":"I will change this equation instead of having t-star."},{"Start":"09:19.683 ","End":"09:27.970","Text":"So I will write this as 8.65 multiplied by e to the negative t star,"},{"Start":"09:27.970 ","End":"09:34.169","Text":"which we said is equal to t minus 0.01,"},{"Start":"09:34.169 ","End":"09:40.250","Text":"divided by our RC which is 0.0025."},{"Start":"09:40.860 ","End":"09:44.515","Text":"This is the answer to Question Number 1."},{"Start":"09:44.515 ","End":"09:48.970","Text":"When in t is in the region between 0 and 0.01 seconds."},{"Start":"09:48.970 ","End":"09:50.530","Text":"When the switches at A,"},{"Start":"09:50.530 ","End":"09:53.170","Text":"this is the voltage as a function of time,"},{"Start":"09:53.170 ","End":"09:55.889","Text":"and when t is greater than 0.01."},{"Start":"09:55.889 ","End":"09:58.855","Text":"So that means when the switches already at B,"},{"Start":"09:58.855 ","End":"10:03.520","Text":"this is the equation for the voltage as a function of time."},{"Start":"10:03.520 ","End":"10:05.215","Text":"That\u0027s Question 1."},{"Start":"10:05.215 ","End":"10:08.450","Text":"Let\u0027s move on to answer Question 2."},{"Start":"10:08.900 ","End":"10:11.130","Text":"Question 2 is,"},{"Start":"10:11.130 ","End":"10:16.485","Text":"what is the charge on the capacitor at t is equal to 0.02."},{"Start":"10:16.485 ","End":"10:20.190","Text":"First of all, t is equal to 0.02,"},{"Start":"10:20.190 ","End":"10:23.755","Text":"is bigger than 0.01."},{"Start":"10:23.755 ","End":"10:29.350","Text":"Which means that we\u0027re in this region over here or in other words,"},{"Start":"10:29.350 ","End":"10:34.645","Text":"we\u0027re in the region where the capacitor is discharging."},{"Start":"10:34.645 ","End":"10:43.390","Text":"What we\u0027re trying to find is the charge on the capacitor at time t is equal to 0.02."},{"Start":"10:43.390 ","End":"10:48.100","Text":"From our equation, we know that the charge on the capacitor is equal to"},{"Start":"10:48.100 ","End":"10:53.199","Text":"the capacitance multiplied by the voltage on the capacitor."},{"Start":"10:53.199 ","End":"10:55.150","Text":"Now we saw in the previous question that"},{"Start":"10:55.150 ","End":"10:58.645","Text":"the voltage on the capacitor is dependent on time."},{"Start":"10:58.645 ","End":"11:04.855","Text":"We\u0027re looking at the time t is equal to 0.02,"},{"Start":"11:04.855 ","End":"11:11.185","Text":"which we already noticed is in this region over here when the capacitor is discharging."},{"Start":"11:11.185 ","End":"11:15.070","Text":"Our capacitance is 50 microfarad,"},{"Start":"11:15.070 ","End":"11:18.640","Text":"so 50 times 10^-6 farad."},{"Start":"11:18.640 ","End":"11:22.240","Text":"Then we multiply this by the voltage at this time."},{"Start":"11:22.240 ","End":"11:29.890","Text":"That\u0027s going to be equal to 8.65 multiplied by e to the power of negative."},{"Start":"11:29.890 ","End":"11:31.975","Text":"Then we have t,"},{"Start":"11:31.975 ","End":"11:38.780","Text":"so that\u0027s 0.002 minus"},{"Start":"11:38.910 ","End":"11:46.460","Text":"0.001 divided by 0.0025."},{"Start":"11:48.630 ","End":"11:52.090","Text":"Now all we have to do is we have to plug this into"},{"Start":"11:52.090 ","End":"11:54.550","Text":"our calculator and what we\u0027ll get is that"},{"Start":"11:54.550 ","End":"11:56.635","Text":"this is approximately equal to"},{"Start":"11:56.635 ","End":"12:06.710","Text":"4.25 times 10^-4 coulombs."},{"Start":"12:08.250 ","End":"12:13.780","Text":"Remember coulombs is the unit of charge."},{"Start":"12:13.780 ","End":"12:18.850","Text":"It\u0027s the units that we give to charge and this is the answer to Question Number 2."},{"Start":"12:18.850 ","End":"12:20.830","Text":"Let\u0027s look at Question Number 3."},{"Start":"12:20.830 ","End":"12:25.070","Text":"What is the current as a function of time?"},{"Start":"12:25.590 ","End":"12:32.155","Text":"The current is also as a function of time,"},{"Start":"12:32.155 ","End":"12:37.885","Text":"and it is the same whether I see circuit is charging or discharging."},{"Start":"12:37.885 ","End":"12:48.650","Text":"It is equal to v_0 divided by R multiplied by e^-t divided by RC."},{"Start":"12:49.290 ","End":"12:52.570","Text":"Therefore, let\u0027s take a look."},{"Start":"12:52.570 ","End":"12:58.090","Text":"When we\u0027re between time 0 and 0.01,"},{"Start":"12:58.090 ","End":"13:00.695","Text":"so when we\u0027re charging."},{"Start":"13:00.695 ","End":"13:07.150","Text":"So our current is therefore going to be equal to v_0."},{"Start":"13:07.150 ","End":"13:09.490","Text":"It\u0027s the voltage of the source."},{"Start":"13:09.490 ","End":"13:12.880","Text":"That\u0027s 10 divided by the resistance,"},{"Start":"13:12.880 ","End":"13:15.325","Text":"which is this resistor over here."},{"Start":"13:15.325 ","End":"13:23.275","Text":"That\u0027s a 100 multiplied by e^-t divided by RC,"},{"Start":"13:23.275 ","End":"13:25.630","Text":"which over here RC if you remember,"},{"Start":"13:25.630 ","End":"13:29.780","Text":"and you can see over here is equal to 0.005."},{"Start":"13:30.780 ","End":"13:34.120","Text":"Then when we\u0027re located at t,"},{"Start":"13:34.120 ","End":"13:37.495","Text":"is bigger or greater than 0.01."},{"Start":"13:37.495 ","End":"13:41.695","Text":"Now we\u0027re located in this circuit over here discharging,"},{"Start":"13:41.695 ","End":"13:47.350","Text":"so v_0 is the initial voltage on the capacitor,"},{"Start":"13:47.350 ","End":"13:49.705","Text":"which we calculated in Question Number 1,"},{"Start":"13:49.705 ","End":"13:54.775","Text":"is equal to 8.65 divided by the resistance."},{"Start":"13:54.775 ","End":"14:02.185","Text":"The resistor in this circuit is 50 and then multiplied by e to the power of negative."},{"Start":"14:02.185 ","End":"14:06.550","Text":"Then we have to remember that we have to do t minus"},{"Start":"14:06.550 ","End":"14:12.646","Text":"the time interval where our capacitor was charging."},{"Start":"14:12.646 ","End":"14:18.460","Text":"T minus 0.01 divided by RC,"},{"Start":"14:18.460 ","End":"14:22.040","Text":"where RC over here was 0.0025."},{"Start":"14:23.640 ","End":"14:27.550","Text":"This is the answer to Question Number 3."},{"Start":"14:27.550 ","End":"14:33.460","Text":"Now let\u0027s go over here to Question Number 4."},{"Start":"14:33.460 ","End":"14:35.860","Text":"Question Number 4 is to draw graphs"},{"Start":"14:35.860 ","End":"14:39.505","Text":"representing the current and voltage as a function of time."},{"Start":"14:39.505 ","End":"14:47.905","Text":"Let\u0027s first draw a graph of the voltage as a function of time."},{"Start":"14:47.905 ","End":"14:58.525","Text":"First of all, between the time of 0 and let\u0027s put over here 0.01 seconds."},{"Start":"14:58.525 ","End":"15:04.010","Text":"Here is time and here is voltage on the capacitor as a function of time."},{"Start":"15:04.010 ","End":"15:13.940","Text":"Between 0-0.01 seconds, we can see that the voltage goes up to about here."},{"Start":"15:14.130 ","End":"15:16.745","Text":"Look something like this."},{"Start":"15:16.745 ","End":"15:20.060","Text":"Where this value over here,"},{"Start":"15:20.060 ","End":"15:26.150","Text":"we saw is the maximum charge that we receive in this time interval,"},{"Start":"15:26.150 ","End":"15:32.885","Text":"which we saw earlier was equal to 8.65 volts."},{"Start":"15:32.885 ","End":"15:35.300","Text":"Then at 0.01,"},{"Start":"15:35.300 ","End":"15:38.405","Text":"we begin this discharge over here,"},{"Start":"15:38.405 ","End":"15:42.035","Text":"which is going to look something like so."},{"Start":"15:42.035 ","End":"15:44.090","Text":"Where as time goes by,"},{"Start":"15:44.090 ","End":"15:48.830","Text":"the voltage across the capacitor approaches 0."},{"Start":"15:48.830 ","End":"15:53.790","Text":"It never reaches exactly 0, but it approaches."},{"Start":"15:53.940 ","End":"16:00.220","Text":"Now let\u0027s look at the graph of current versus time."},{"Start":"16:00.220 ","End":"16:04.620","Text":"Here we have time and here we have current."},{"Start":"16:04.620 ","End":"16:10.278","Text":"We saw that in the time interval between 0 and 0.01 seconds,"},{"Start":"16:10.278 ","End":"16:12.712","Text":"so when we\u0027re looking at the charging."},{"Start":"16:12.712 ","End":"16:22.295","Text":"We start at some maximum value of 0.01 in pairs."},{"Start":"16:22.295 ","End":"16:33.310","Text":"Here"},{"Start":"16:33.310 ","End":"16:33.430","Text":"is"},{"Start":"16:33.430 ","End":"16:35.878","Text":"0.1."},{"Start":"16:35.878 ","End":"16:38.675","Text":"We start at 0.1 in pairs,"},{"Start":"16:38.675 ","End":"16:46.998","Text":"and then up until we reach a time interval of 0.01 seconds,"},{"Start":"16:46.998 ","End":"16:51.539","Text":"so our current is going to be approaching 0."},{"Start":"16:51.930 ","End":"16:54.580","Text":"This is meant to be a smooth curve,"},{"Start":"16:54.580 ","End":"16:56.375","Text":"of course, approaching 0."},{"Start":"16:56.375 ","End":"17:02.880","Text":"Then we reach the timeframe where we\u0027re at a time greater than 0.01 seconds."},{"Start":"17:03.170 ","End":"17:07.550","Text":"We have a starting current of this,"},{"Start":"17:07.550 ","End":"17:17.980","Text":"so 8.65 divided by 50 is equal to 0.173 amps."},{"Start":"17:17.980 ","End":"17:21.575","Text":"We start at 0.173."},{"Start":"17:21.575 ","End":"17:28.470","Text":"We can say that that\u0027s about here, 0.173 amps."},{"Start":"17:28.470 ","End":"17:36.875","Text":"Then again, we can see that we get this slope and in time,"},{"Start":"17:36.875 ","End":"17:44.205","Text":"the current also reaches a current or approaches a current equal to 0."},{"Start":"17:44.205 ","End":"17:46.775","Text":"When we\u0027re looking at the voltage,"},{"Start":"17:46.775 ","End":"17:53.290","Text":"the voltage increases as the capacitor charges and decreases as the capacitor discharges."},{"Start":"17:53.290 ","End":"17:55.942","Text":"When we\u0027re looking at the current,"},{"Start":"17:55.942 ","End":"18:01.055","Text":"the current starts from some value and decreases."},{"Start":"18:01.055 ","End":"18:04.970","Text":"As the capacitor charges and then when the capacitor begins discharging,"},{"Start":"18:04.970 ","End":"18:10.340","Text":"the current is higher but then it also decreases and approaches 0."},{"Start":"18:10.340 ","End":"18:13.175","Text":"This is the answer to Question Number 4."},{"Start":"18:13.175 ","End":"18:16.310","Text":"We drew current and voltage as a function of time."},{"Start":"18:16.310 ","End":"18:19.230","Text":"That is the end of this lesson."}],"ID":22280},{"Watched":false,"Name":"Exercise 6","Duration":"12m 20s","ChapterTopicVideoID":21500,"CourseChapterTopicPlaylistID":99472,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.770","Text":"Hello. In this lesson,"},{"Start":"00:01.770 ","End":"00:04.485","Text":"we\u0027re going to be answering the following question."},{"Start":"00:04.485 ","End":"00:06.495","Text":"In the following circuit,"},{"Start":"00:06.495 ","End":"00:08.160","Text":"at t is equal to 0,"},{"Start":"00:08.160 ","End":"00:10.545","Text":"the switch is closed."},{"Start":"00:10.545 ","End":"00:12.690","Text":"Question number 1 is,"},{"Start":"00:12.690 ","End":"00:18.610","Text":"what is the RC time constant for this circuit?"},{"Start":"00:19.760 ","End":"00:23.835","Text":"The first thing that we can see is that we have a voltage source,"},{"Start":"00:23.835 ","End":"00:30.345","Text":"a resistor, and 2 capacitors joined in parallel."},{"Start":"00:30.345 ","End":"00:36.089","Text":"We can see that we have an RC circuit that is going to be charging."},{"Start":"00:36.089 ","End":"00:37.155","Text":"Why is it charging?"},{"Start":"00:37.155 ","End":"00:39.210","Text":"Because there\u0027s a voltage source."},{"Start":"00:39.210 ","End":"00:43.155","Text":"What we want to do is we want to change this circuit,"},{"Start":"00:43.155 ","End":"00:48.200","Text":"where we have 2 capacitors in parallel to our familiar circuit,"},{"Start":"00:48.200 ","End":"00:53.270","Text":"where we have a voltage source connected to a resistor,"},{"Start":"00:53.270 ","End":"00:58.565","Text":"connected to a capacitor and that\u0027s it."},{"Start":"00:58.565 ","End":"01:02.140","Text":"This is what we want to achieve."},{"Start":"01:02.140 ","End":"01:04.610","Text":"Where the capacitance over here is C total,"},{"Start":"01:04.610 ","End":"01:07.460","Text":"the total capacitance of this."},{"Start":"01:07.460 ","End":"01:10.955","Text":"First of all, let\u0027s work out what C total is."},{"Start":"01:10.955 ","End":"01:16.760","Text":"First of all, we\u0027re working out C total when the 2 capacitors are joined in parallel."},{"Start":"01:16.760 ","End":"01:18.695","Text":"What do we do in this case?"},{"Start":"01:18.695 ","End":"01:23.195","Text":"We just add up very simply, their capacitances."},{"Start":"01:23.195 ","End":"01:30.930","Text":"C_1 is 0.3 millifarads plus C_2,"},{"Start":"01:30.930 ","End":"01:34.785","Text":"which is 0.7 millifarads."},{"Start":"01:34.785 ","End":"01:40.080","Text":"We get a total capacitance of 1 millifarad."},{"Start":"01:42.640 ","End":"01:47.705","Text":"RC total we said is equal to 1 millifarad."},{"Start":"01:47.705 ","End":"01:52.130","Text":"Then here we have our resistance, just like before,"},{"Start":"01:52.130 ","End":"01:57.020","Text":"100 ohm and our voltage source of 10 volts."},{"Start":"01:57.020 ","End":"02:01.400","Text":"Now what we want to find is our RC time constant,"},{"Start":"02:01.400 ","End":"02:02.735","Text":"which as we know,"},{"Start":"02:02.735 ","End":"02:05.180","Text":"is called also Tau."},{"Start":"02:05.180 ","End":"02:07.070","Text":"We have our resistance,"},{"Start":"02:07.070 ","End":"02:12.330","Text":"which is 100 ohms multiplied by our capacitance,"},{"Start":"02:12.330 ","End":"02:19.025","Text":"which is 1 millifarad or 1 times 10 to the negative 3 farads,"},{"Start":"02:19.025 ","End":"02:28.980","Text":"which is going to be equal to 0.1 and the units for tau or for RC is seconds."},{"Start":"02:29.420 ","End":"02:33.855","Text":"Now let\u0027s answer question number 2."},{"Start":"02:33.855 ","End":"02:40.185","Text":"This is to calculate the voltage and charge on each capacitor at times t is equal to"},{"Start":"02:40.185 ","End":"02:49.060","Text":"0.2 seconds and at 0.8 seconds."},{"Start":"02:49.190 ","End":"02:57.395","Text":"First of all, we saw that our RC time constant is equal to 0.1 seconds."},{"Start":"02:57.395 ","End":"03:00.919","Text":"Our t of 0.8 seconds,"},{"Start":"03:00.919 ","End":"03:06.960","Text":"so t of 0.8 is bigger than"},{"Start":"03:06.960 ","End":"03:16.170","Text":"5 times Tau and 5 times Tau is 0.5 seconds."},{"Start":"03:16.170 ","End":"03:18.890","Text":"As we remember in previous lessons,"},{"Start":"03:18.890 ","End":"03:24.920","Text":"we said that anytime greater than approximately 5 times the RC time"},{"Start":"03:24.920 ","End":"03:31.470","Text":"constant means that the capacitor is at its stable state."},{"Start":"03:31.470 ","End":"03:36.605","Text":"In this case, because we have a charging circuit over here,"},{"Start":"03:36.605 ","End":"03:42.080","Text":"so stable state means that the capacitor is"},{"Start":"03:42.080 ","End":"03:47.540","Text":"fully charged and when the capacitor is fully charged,"},{"Start":"03:47.540 ","End":"03:49.475","Text":"as we saw in the previous lesson,"},{"Start":"03:49.475 ","End":"03:56.220","Text":"that means that our current is going to be equal to 0."},{"Start":"03:58.130 ","End":"04:07.400","Text":"Now what we want to do is we want to find the total voltage on our capacitance,"},{"Start":"04:07.400 ","End":"04:09.620","Text":"where we made this total capacitance,"},{"Start":"04:09.620 ","End":"04:12.830","Text":"this one capacitor made up of 2 capacitors."},{"Start":"04:12.830 ","End":"04:18.064","Text":"V_c total refers to the voltage across this capacitor."},{"Start":"04:18.064 ","End":"04:24.980","Text":"C total is equal to if where the stable state in a charging circuit."},{"Start":"04:24.980 ","End":"04:27.995","Text":"We said that that means that the capacitor is fully charged."},{"Start":"04:27.995 ","End":"04:29.825","Text":"If the capacitor is fully charged,"},{"Start":"04:29.825 ","End":"04:32.810","Text":"then the voltage across the capacitor is going to be the"},{"Start":"04:32.810 ","End":"04:36.695","Text":"same as the voltage of the voltage source,"},{"Start":"04:36.695 ","End":"04:41.230","Text":"which in this case over here is 10 volts."},{"Start":"04:41.230 ","End":"04:42.635","Text":"Then in that case,"},{"Start":"04:42.635 ","End":"04:47.930","Text":"the total charge on this C total capacitor."},{"Start":"04:47.930 ","End":"04:54.980","Text":"Total charge on C total is equal to"},{"Start":"04:54.980 ","End":"05:03.865","Text":"the capacitance of this C total multiplied by the voltage across it."},{"Start":"05:03.865 ","End":"05:07.785","Text":"Now let\u0027s just scroll a bit more to the side."},{"Start":"05:07.785 ","End":"05:10.865","Text":"What we have is the capacitance,"},{"Start":"05:10.865 ","End":"05:17.225","Text":"which is 1 times 10 to the negative 3 farads"},{"Start":"05:17.225 ","End":"05:25.560","Text":"multiplied by the voltage which is 10 volts."},{"Start":"05:25.560 ","End":"05:34.935","Text":"Then this is simply going to be equal to 10 to the power of negative 2 coulombs."},{"Start":"05:34.935 ","End":"05:38.190","Text":"Remember this C refers to coulombs,"},{"Start":"05:38.190 ","End":"05:40.930","Text":"the unit for a charge."},{"Start":"05:41.840 ","End":"05:49.385","Text":"Now, what I want to do is I want to calculate the voltage and charge on each capacitor."},{"Start":"05:49.385 ","End":"05:52.430","Text":"On this capacitor and on this one."},{"Start":"05:52.430 ","End":"05:58.235","Text":"I remember that when the capacitors are joined in parallel,"},{"Start":"05:58.235 ","End":"06:00.335","Text":"so the voltage across"},{"Start":"06:00.335 ","End":"06:07.170","Text":"the total capacitor and across each individual capacitor is the same."},{"Start":"06:08.000 ","End":"06:13.370","Text":"In parallel, that means that the voltage across"},{"Start":"06:13.370 ","End":"06:21.040","Text":"this C total capacitor is equal to the voltage on each one."},{"Start":"06:21.040 ","End":"06:24.800","Text":"It\u0027s equal to the voltage across"},{"Start":"06:24.800 ","End":"06:31.189","Text":"the first capacitor and that is also equal to the voltage across the second capacitor,"},{"Start":"06:31.189 ","End":"06:32.720","Text":"which is equal to,"},{"Start":"06:32.720 ","End":"06:37.110","Text":"as we saw over here, 10 volts."},{"Start":"06:38.810 ","End":"06:47.690","Text":"Then we can say that the charge on capacitor number 1 is equal to from our equation,"},{"Start":"06:47.690 ","End":"06:50.060","Text":"the capacitance of capacitor number 1"},{"Start":"06:50.060 ","End":"06:53.300","Text":"multiplied by the voltage on the capacitor number 1."},{"Start":"06:53.300 ","End":"06:57.950","Text":"The capacitance is 0.3 millifarads,"},{"Start":"06:57.950 ","End":"07:04.550","Text":"so times 10 to the negative 3 farads multiplied by the voltage on the capacitor 1,"},{"Start":"07:04.550 ","End":"07:07.820","Text":"which we saw is equal to 10 volts."},{"Start":"07:07.820 ","End":"07:12.720","Text":"What we get is that the charge on capacitor number 1 is equal"},{"Start":"07:12.720 ","End":"07:18.060","Text":"to 3 times 10 to the negative 3 coulombs."},{"Start":"07:18.060 ","End":"07:21.635","Text":"Then let\u0027s look at capacity number 2, the charge on it."},{"Start":"07:21.635 ","End":"07:23.405","Text":"This is going to be the capacitance on"},{"Start":"07:23.405 ","End":"07:27.265","Text":"that capacitor multiplied by the voltage on that capacitor."},{"Start":"07:27.265 ","End":"07:35.930","Text":"That is equal to 0.7 times 10 to the negative 3 farads multiplied by the voltage,"},{"Start":"07:35.930 ","End":"07:38.825","Text":"which, as we saw, is equal to 10 volts."},{"Start":"07:38.825 ","End":"07:47.230","Text":"This is going to be equal to 7 times 10 to the negative 3 coulombs."},{"Start":"07:48.080 ","End":"07:54.890","Text":"This is the answer for the charges on"},{"Start":"07:54.890 ","End":"08:01.835","Text":"each capacitor and for the voltages on each capacitor at time t is equal to 0.8 seconds,"},{"Start":"08:01.835 ","End":"08:05.945","Text":"where we saw that this is a time that is much bigger than 5 times"},{"Start":"08:05.945 ","End":"08:11.595","Text":"our RC time constant and that\u0027s what led us to this answer."},{"Start":"08:11.595 ","End":"08:17.055","Text":"Now let\u0027s look at t is equal to 0.2 seconds."},{"Start":"08:17.055 ","End":"08:22.175","Text":"We can see that this is smaller than 5 times our RC time constant."},{"Start":"08:22.175 ","End":"08:27.060","Text":"Which means that we have to use all of the equations to get to our answer."},{"Start":"08:28.640 ","End":"08:32.080","Text":"Our q total,"},{"Start":"08:32.080 ","End":"08:37.130","Text":"so the total charge on this capacitance as a function of t is"},{"Start":"08:37.130 ","End":"08:42.274","Text":"equal to the total capacitance multiplied by the initial voltage,"},{"Start":"08:42.274 ","End":"08:51.910","Text":"multiplied by 1 minus e to the power of negative t divided by RC total."},{"Start":"08:51.910 ","End":"08:54.720","Text":"Then our V total,"},{"Start":"08:54.720 ","End":"08:56.840","Text":"so the voltage on this capacitor,"},{"Start":"08:56.840 ","End":"09:05.380","Text":"as a function of time is equal to our q total divided by our C total."},{"Start":"09:05.380 ","End":"09:07.230","Text":"In other words, we\u0027re just left with"},{"Start":"09:07.230 ","End":"09:14.980","Text":"V_0 1 minus e to the power of negative t divided by RC_T."},{"Start":"09:15.110 ","End":"09:19.330","Text":"Now let\u0027s substitute in our values for this time."},{"Start":"09:19.330 ","End":"09:25.055","Text":"We have that V total at t is equal to 0.2,"},{"Start":"09:25.055 ","End":"09:26.750","Text":"is equal to,"},{"Start":"09:26.750 ","End":"09:28.160","Text":"so the initial voltage,"},{"Start":"09:28.160 ","End":"09:34.670","Text":"which is 10 volts multiplied by 1 minus e to the"},{"Start":"09:34.670 ","End":"09:41.645","Text":"negative 0.2 divided by our RC time constant,"},{"Start":"09:41.645 ","End":"09:44.300","Text":"which we saw is equal to 0.1 seconds,"},{"Start":"09:44.300 ","End":"09:50.560","Text":"so divided by 0.1 and"},{"Start":"09:50.560 ","End":"09:57.960","Text":"this is approximately equal to 8.65 volts."},{"Start":"09:59.480 ","End":"10:02.965","Text":"Just like we discussed before, in parallel,"},{"Start":"10:02.965 ","End":"10:05.995","Text":"the total voltage is equal to the voltage,"},{"Start":"10:05.995 ","End":"10:08.680","Text":"also on each individual capacitor."},{"Start":"10:08.680 ","End":"10:13.870","Text":"Therefore, V total is equal to the voltage on the first,"},{"Start":"10:13.870 ","End":"10:22.050","Text":"and it\u0027s equal to the voltage on the second capacitor and that is equal to 8.65 volts."},{"Start":"10:22.050 ","End":"10:27.220","Text":"This is the voltage on each capacitor and now we want to calculate the charge."},{"Start":"10:27.620 ","End":"10:29.800","Text":"In order to find the charge,"},{"Start":"10:29.800 ","End":"10:31.410","Text":"let us scroll down."},{"Start":"10:31.410 ","End":"10:37.674","Text":"The charge on capacitor number 1 at t is equal to 0.2 seconds,"},{"Start":"10:37.674 ","End":"10:40.000","Text":"is equal to, as we know,"},{"Start":"10:40.000 ","End":"10:49.760","Text":"the capacitance multiplied by the voltage on that capacitor at t is equal to 0.2 seconds."},{"Start":"10:49.760 ","End":"10:52.475","Text":"What is that equal to?"},{"Start":"10:52.475 ","End":"11:00.350","Text":"The capacitance is 0.3 times 10 to the negative 3 farads"},{"Start":"11:00.350 ","End":"11:08.310","Text":"multiplied by the voltage which we just calculated over here is 8.65 volts."},{"Start":"11:08.310 ","End":"11:14.955","Text":"What we get is that the charge on q_1 is equal to 2 decimal places,"},{"Start":"11:14.955 ","End":"11:22.995","Text":"2.60 times 10 to the negative 3 coulombs"},{"Start":"11:22.995 ","End":"11:32.315","Text":"and our charge on capacitor number 2 at t is equal to 0.2 seconds,"},{"Start":"11:32.315 ","End":"11:40.845","Text":"is equal to C_2 multiplied by the voltage on capacitor 2 at 0.2 seconds."},{"Start":"11:40.845 ","End":"11:43.500","Text":"This is equal to,"},{"Start":"11:43.500 ","End":"11:53.075","Text":"so the capacitance is 0.7 times 10 to the negative 3 farads multiplied by the voltage,"},{"Start":"11:53.075 ","End":"11:56.790","Text":"which is 8.65 volts."},{"Start":"11:56.790 ","End":"12:00.120","Text":"We get that it is equal to 2 decimal places,"},{"Start":"12:00.120 ","End":"12:07.690","Text":"6.01 times 10 to the negative 3 coulombs."},{"Start":"12:08.150 ","End":"12:12.170","Text":"Here\u0027s the voltage on each capacitor at t is equal to"},{"Start":"12:12.170 ","End":"12:16.130","Text":"0.2 seconds and the charges on each capacitor."},{"Start":"12:16.130 ","End":"12:18.275","Text":"We\u0027ve answered question number 2,"},{"Start":"12:18.275 ","End":"12:21.030","Text":"and that is the end of this lesson."}],"ID":22281},{"Watched":false,"Name":"Exercise 7","Duration":"22m 52s","ChapterTopicVideoID":21501,"CourseChapterTopicPlaylistID":99472,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.770","Text":"Hello. In this lesson,"},{"Start":"00:01.770 ","End":"00:04.470","Text":"we\u0027re going to be answering the following question."},{"Start":"00:04.470 ","End":"00:08.160","Text":"In the following circuit, at t=0,"},{"Start":"00:08.160 ","End":"00:13.530","Text":"the capacitor has no charge and the switch is closed."},{"Start":"00:13.530 ","End":"00:17.235","Text":"We\u0027re being asked to calculate the charge on the capacitor"},{"Start":"00:17.235 ","End":"00:21.315","Text":"and the current in each resistor as a function of time."},{"Start":"00:21.315 ","End":"00:25.680","Text":"We can assume that we have a value for V, for R_1,"},{"Start":"00:25.680 ","End":"00:32.835","Text":"for R_2, and for C. How are we going to begin?"},{"Start":"00:32.835 ","End":"00:34.320","Text":"In a question like this,"},{"Start":"00:34.320 ","End":"00:37.051","Text":"we begin with Kirchhoff\u0027s laws,"},{"Start":"00:37.051 ","End":"00:39.720","Text":"so that means we write equations for all of"},{"Start":"00:39.720 ","End":"00:44.380","Text":"the currents and all of the voltages in the circuit."},{"Start":"00:45.320 ","End":"00:48.300","Text":"Let\u0027s begin. First of all,"},{"Start":"00:48.300 ","End":"00:54.725","Text":"our switch over here is closed and let\u0027s imagine that here we have a current,"},{"Start":"00:54.725 ","End":"00:57.425","Text":"and let\u0027s call this current I_2,"},{"Start":"00:57.425 ","End":"01:00.820","Text":"because it goes past resistor R_2,"},{"Start":"01:00.820 ","End":"01:04.400","Text":"this is the same current, I_2."},{"Start":"01:04.400 ","End":"01:07.550","Text":"Then the current that goes past Resistor 1,"},{"Start":"01:07.550 ","End":"01:10.220","Text":"at this node the current I_2 splits."},{"Start":"01:10.220 ","End":"01:11.935","Text":"Pass Resistor 1,"},{"Start":"01:11.935 ","End":"01:16.970","Text":"we have I_1 and in this direction after this node,"},{"Start":"01:16.970 ","End":"01:20.770","Text":"let\u0027s call the current through here I_3."},{"Start":"01:20.770 ","End":"01:24.650","Text":"Now let\u0027s begin to write our equations."},{"Start":"01:25.040 ","End":"01:32.015","Text":"Now let\u0027s begin by writing our first equation for the voltages."},{"Start":"01:32.015 ","End":"01:35.540","Text":"Let\u0027s begin from this point over here,"},{"Start":"01:35.540 ","End":"01:37.340","Text":"just before the battery,"},{"Start":"01:37.340 ","End":"01:41.630","Text":"and then let\u0027s go and the route via the 2 resistors,"},{"Start":"01:41.630 ","End":"01:43.070","Text":"could have chosen any routes,"},{"Start":"01:43.070 ","End":"01:46.735","Text":"but I\u0027m just going to choose this route over here."},{"Start":"01:46.735 ","End":"01:52.765","Text":"First of all, we go across the voltage source,"},{"Start":"01:52.765 ","End":"01:57.920","Text":"the battery and we\u0027re going from the shorter side to the longer side,"},{"Start":"01:57.920 ","End":"02:00.755","Text":"which means we\u0027re going from the negative to the positive side,"},{"Start":"02:00.755 ","End":"02:08.520","Text":"which means that we have an addition of voltage v. Then we carry on going,"},{"Start":"02:08.520 ","End":"02:13.490","Text":"we reach this resistor R_2 and we travel across it,"},{"Start":"02:13.490 ","End":"02:17.905","Text":"which means that we have to take down I_2."},{"Start":"02:17.905 ","End":"02:20.970","Text":"Then we carry on across this resistor,"},{"Start":"02:20.970 ","End":"02:25.820","Text":"so we subtract I_1, I_1."},{"Start":"02:25.820 ","End":"02:28.970","Text":"Then we reach our starting point,"},{"Start":"02:28.970 ","End":"02:30.650","Text":"which according to Kirchhoff,"},{"Start":"02:30.650 ","End":"02:35.750","Text":"means that all of this has to be equal to 0."},{"Start":"02:36.020 ","End":"02:41.340","Text":"Let\u0027s call this equation number 1."},{"Start":"02:41.340 ","End":"02:44.780","Text":"Now, let\u0027s take another route,"},{"Start":"02:44.780 ","End":"02:46.265","Text":"so let\u0027s take this,"},{"Start":"02:46.265 ","End":"02:50.660","Text":"so that we\u0027re including also our capacitor over here."},{"Start":"02:50.660 ","End":"02:55.160","Text":"If this is I_3 and this is its direction of travel,"},{"Start":"02:55.160 ","End":"02:58.770","Text":"and we also know that our capacitor is charging in"},{"Start":"02:58.770 ","End":"03:03.900","Text":"the circuit because we have a capacitor resistors and a voltage source,"},{"Start":"03:03.900 ","End":"03:06.435","Text":"so this is a charging RC circuit."},{"Start":"03:06.435 ","End":"03:09.065","Text":"If this is the direction of the current,"},{"Start":"03:09.065 ","End":"03:15.800","Text":"so this must be the positive plates of the capacitor and this is the negative plates."},{"Start":"03:15.800 ","End":"03:20.180","Text":"Our second equation, let\u0027s begin over here."},{"Start":"03:20.180 ","End":"03:22.250","Text":"As I cross the capacitor,"},{"Start":"03:22.250 ","End":"03:26.675","Text":"I\u0027m going from the positive side to the negative side,"},{"Start":"03:26.675 ","End":"03:29.975","Text":"which means that I have a voltage drop."},{"Start":"03:29.975 ","End":"03:34.835","Text":"From the equation that c is equal to q divided by V,"},{"Start":"03:34.835 ","End":"03:40.070","Text":"I know that the voltage on this capacitor is equal to negative because it\u0027s a"},{"Start":"03:40.070 ","End":"03:45.785","Text":"drop q divided by c. This is the voltage on the capacitor."},{"Start":"03:45.785 ","End":"03:49.820","Text":"Then I keep traveling and then I reach this resistor and I see"},{"Start":"03:49.820 ","End":"03:53.840","Text":"that I\u0027m crossing the resistor in the negative direction to the current,"},{"Start":"03:53.840 ","End":"03:59.750","Text":"so that means that I add I_1, R_1."},{"Start":"03:59.750 ","End":"04:03.890","Text":"Then I carry on traveling and I reached my starting point,"},{"Start":"04:03.890 ","End":"04:08.460","Text":"so all of this according to Kirchhoff is equal to 0."},{"Start":"04:08.530 ","End":"04:15.575","Text":"Then my third equation is going to be my current equation,"},{"Start":"04:15.575 ","End":"04:18.740","Text":"so I know that I need to"},{"Start":"04:18.740 ","End":"04:22.430","Text":"work on this node over here because that\u0027s where my current splits."},{"Start":"04:22.430 ","End":"04:31.605","Text":"I can see that I_2 comes into this node and then I_1 and I_3 come out of this node,"},{"Start":"04:31.605 ","End":"04:35.965","Text":"so I can see that I_2 is equal to I_1 plus I_3."},{"Start":"04:35.965 ","End":"04:38.250","Text":"Now I have 3 equations,"},{"Start":"04:38.250 ","End":"04:40.430","Text":"but I have 4 unknowns."},{"Start":"04:40.430 ","End":"04:42.215","Text":"I have I_1, I_2,"},{"Start":"04:42.215 ","End":"04:46.235","Text":"I_3, and also I have q."},{"Start":"04:46.235 ","End":"04:49.948","Text":"But I know that q is the charge on my capacitor,"},{"Start":"04:49.948 ","End":"04:53.390","Text":"and the charge on a capacitor is dependent on the current"},{"Start":"04:53.390 ","End":"04:57.940","Text":"flowing to it and the current flowing to it is I_3."},{"Start":"04:57.940 ","End":"05:03.080","Text":"My 4th equation is that I_3 is equal"},{"Start":"05:03.080 ","End":"05:07.896","Text":"to the time derivative of the charge on my capacitor,"},{"Start":"05:07.896 ","End":"05:11.040","Text":"so that is equal to q dot."},{"Start":"05:11.620 ","End":"05:20.660","Text":"As we remember, current is the amount of charge that goes past in some time frame"},{"Start":"05:20.660 ","End":"05:24.830","Text":"and we can see that that is what\u0027s happening over here and then"},{"Start":"05:24.830 ","End":"05:30.010","Text":"the charge is just building up on our capacitor plates."},{"Start":"05:30.010 ","End":"05:36.500","Text":"It\u0027s this current which is affecting the charge on the capacitor plates and it\u0027s equal to"},{"Start":"05:36.500 ","End":"05:41.930","Text":"q dot and because capacitor\u0027s charging it\u0027s"},{"Start":"05:41.930 ","End":"05:47.750","Text":"equal to positive q dot and if we were dealing with a discharging RC circuit,"},{"Start":"05:47.750 ","End":"05:51.540","Text":"then it would be equal to negative q dot."},{"Start":"05:52.160 ","End":"05:56.775","Text":"Now I have 4 equations and 4 unknowns."},{"Start":"05:56.775 ","End":"05:59.195","Text":"I want to start playing with the equations,"},{"Start":"05:59.195 ","End":"06:06.945","Text":"so that I eventually get an equation just as a function of q with q dot."},{"Start":"06:06.945 ","End":"06:11.120","Text":"What I\u0027m going to do is I\u0027m going to plug in"},{"Start":"06:11.120 ","End":"06:15.950","Text":"my I_2 in order to get equations just with I_1 and I_3,"},{"Start":"06:15.950 ","End":"06:18.395","Text":"then I\u0027m going to plug in instead of I_3,"},{"Start":"06:18.395 ","End":"06:21.685","Text":"the q dot and that\u0027s what I\u0027m going to do."},{"Start":"06:21.685 ","End":"06:28.085","Text":"Now I\u0027m taking equation number 3 and I\u0027m plugging it in to equation number 1."},{"Start":"06:28.085 ","End":"06:32.495","Text":"We have V minus I_2, R_2."},{"Start":"06:32.495 ","End":"06:35.945","Text":"I_2 is equal to I_1 plus I_3,"},{"Start":"06:35.945 ","End":"06:39.545","Text":"so we\u0027ll have minus I_1,"},{"Start":"06:39.545 ","End":"06:46.460","Text":"R_2, minus I_3, I_2."},{"Start":"06:46.460 ","End":"06:49.265","Text":"I just already opened up the brackets."},{"Start":"06:49.265 ","End":"06:52.370","Text":"Minus I_1, R_1,"},{"Start":"06:52.370 ","End":"06:55.125","Text":"so minus I_1,"},{"Start":"06:55.125 ","End":"06:58.215","Text":"R_1 and all of this is equal to 0."},{"Start":"06:58.215 ","End":"07:00.570","Text":"Let\u0027s just look over here."},{"Start":"07:00.570 ","End":"07:04.955","Text":"From this equation, if I just move everything to the other side of the equals sign,"},{"Start":"07:04.955 ","End":"07:15.300","Text":"I can see that I_1 is equal to q divided by c divided by R_1."},{"Start":"07:15.300 ","End":"07:17.480","Text":"This is this equation,"},{"Start":"07:17.480 ","End":"07:22.500","Text":"and let\u0027s call this equation equation number 5."},{"Start":"07:22.820 ","End":"07:28.430","Text":"Now what I would do is I want to plug in equation number 2,"},{"Start":"07:28.430 ","End":"07:33.530","Text":"so what we just did over here into this equation over here,"},{"Start":"07:33.530 ","End":"07:35.510","Text":"which we call it number 5."},{"Start":"07:35.510 ","End":"07:37.985","Text":"What I have is V minus,"},{"Start":"07:37.985 ","End":"07:41.645","Text":"so here I have I_1 and here I have I_1."},{"Start":"07:41.645 ","End":"07:48.115","Text":"I saw that my I_1 is equal to q divided by CR_1,"},{"Start":"07:48.115 ","End":"07:53.600","Text":"and this is multiplied by I_2 plus R_1."},{"Start":"07:53.600 ","End":"07:56.725","Text":"The plus cancels out with a minus over here."},{"Start":"07:56.725 ","End":"08:00.240","Text":"Then minus I_3,"},{"Start":"08:00.240 ","End":"08:04.290","Text":"I_2 and all of this is equal to 0."},{"Start":"08:04.290 ","End":"08:06.845","Text":"Now instead of I_3,"},{"Start":"08:06.845 ","End":"08:09.350","Text":"I\u0027m going to substitute in q dot."},{"Start":"08:09.350 ","End":"08:12.980","Text":"I\u0027m also going to put in 4 over here,"},{"Start":"08:12.980 ","End":"08:15.080","Text":"so instead of I_3,"},{"Start":"08:15.080 ","End":"08:17.980","Text":"I have q dot."},{"Start":"08:17.980 ","End":"08:20.910","Text":"Now I can see that V, R_1,"},{"Start":"08:20.910 ","End":"08:22.440","Text":"R_2 and C,"},{"Start":"08:22.440 ","End":"08:24.570","Text":"so V, R_1,"},{"Start":"08:24.570 ","End":"08:31.930","Text":"R_2 and C are given to me and I just have an equation dealing with variables q and q dot."},{"Start":"08:31.930 ","End":"08:34.805","Text":"What I have over here is a differential equation,"},{"Start":"08:34.805 ","End":"08:36.890","Text":"which I\u0027m going to solve in just a minute,"},{"Start":"08:36.890 ","End":"08:40.380","Text":"but I\u0027m just going to write the onset in the meantime over here."},{"Start":"08:40.380 ","End":"08:42.470","Text":"The onset to the differential equation,"},{"Start":"08:42.470 ","End":"08:52.115","Text":"q is a function of t is equal to R_1CV divided by R_1 plus R_2"},{"Start":"08:52.115 ","End":"08:58.340","Text":"multiplied by 1 minus e to the power of negative R_1 plus"},{"Start":"08:58.340 ","End":"09:06.020","Text":"R_2 divided by R_1 multiplied by I_2 multiplied by C,"},{"Start":"09:06.020 ","End":"09:12.440","Text":"and all of this multiplied by t. Now let\u0027s take a look at what is going on."},{"Start":"09:12.440 ","End":"09:16.730","Text":"First of all, I can see that when t=0,"},{"Start":"09:16.730 ","End":"09:19.100","Text":"so e to the power of all of this is equal to 0,"},{"Start":"09:19.100 ","End":"09:20.810","Text":"which means this is equal to 1."},{"Start":"09:20.810 ","End":"09:26.075","Text":"Therefore, I can see that the charge at this time is equal to 0,"},{"Start":"09:26.075 ","End":"09:30.400","Text":"which was our initial condition that we were told in the question."},{"Start":"09:30.400 ","End":"09:34.440","Text":"What about when t is approaching infinity,"},{"Start":"09:34.440 ","End":"09:38.370","Text":"so after a very long time."},{"Start":"09:38.370 ","End":"09:43.385","Text":"What we\u0027ll see is that our charge approaches,"},{"Start":"09:43.385 ","End":"09:47.255","Text":"this will approach 0, all of this,"},{"Start":"09:47.255 ","End":"09:55.460","Text":"and so our charge will approach I_1 CV divided by R_1 plus R_2,"},{"Start":"09:55.460 ","End":"09:58.820","Text":"which means that our capacitor will reach this charge,"},{"Start":"09:58.820 ","End":"10:03.005","Text":"which means it is fully charged and here our current,"},{"Start":"10:03.005 ","End":"10:05.000","Text":"which we know is equal to q dot,"},{"Start":"10:05.000 ","End":"10:06.155","Text":"so as we can see,"},{"Start":"10:06.155 ","End":"10:08.850","Text":"this is just constants."},{"Start":"10:08.850 ","End":"10:12.085","Text":"When we work out q dot dq by dt,"},{"Start":"10:12.085 ","End":"10:14.450","Text":"we see that this will be equal to 0,"},{"Start":"10:14.450 ","End":"10:16.730","Text":"which means that at this case,"},{"Start":"10:16.730 ","End":"10:22.475","Text":"the current as t is approaching infinity is equal to 0,"},{"Start":"10:22.475 ","End":"10:26.000","Text":"which is exactly what we would expect because when a capacitor is"},{"Start":"10:26.000 ","End":"10:29.755","Text":"fully charged and it acts like an open circuit."},{"Start":"10:29.755 ","End":"10:34.730","Text":"Of course here speaking about this current over here, I_3."},{"Start":"10:34.730 ","End":"10:38.210","Text":"If I_3 is equal to 0,"},{"Start":"10:38.210 ","End":"10:41.935","Text":"then that means that I_1 is equal to I_2"},{"Start":"10:41.935 ","End":"10:46.745","Text":"because we just have this 1 route around the circuit."},{"Start":"10:46.745 ","End":"10:51.710","Text":"We can say that I_1 is equal to I_2,"},{"Start":"10:51.710 ","End":"10:54.215","Text":"which is, let\u0027s just call this I,"},{"Start":"10:54.215 ","End":"10:55.865","Text":"the current in the circuit."},{"Start":"10:55.865 ","End":"10:59.540","Text":"Then we know that in series,"},{"Start":"10:59.540 ","End":"11:03.469","Text":"these 2 resistors are now connected in series."},{"Start":"11:03.469 ","End":"11:09.460","Text":"The total resistance is equal to R_1 plus R_2."},{"Start":"11:09.460 ","End":"11:13.440","Text":"We just add on the resistances."},{"Start":"11:13.440 ","End":"11:20.570","Text":"Therefore, we can say that the current I is simply equal"},{"Start":"11:20.570 ","End":"11:22.430","Text":"to the voltage which we\u0027re given in"},{"Start":"11:22.430 ","End":"11:27.905","Text":"the question V divided by the total resistance of the second,"},{"Start":"11:27.905 ","End":"11:32.130","Text":"which is I_1 plus R_2."},{"Start":"11:32.370 ","End":"11:37.585","Text":"In that case, I can say that the voltage,"},{"Start":"11:37.585 ","End":"11:43.165","Text":"therefore across resistor R1 is simply equal to,"},{"Start":"11:43.165 ","End":"11:48.400","Text":"this is equal to I R1."},{"Start":"11:48.400 ","End":"11:53.065","Text":"Sorry, the voltage across resistor 1 is equal to I R1,"},{"Start":"11:53.065 ","End":"12:01.810","Text":"which is simply equal to V divided by R1 plus R2 multiplied by R1."},{"Start":"12:01.810 ","End":"12:08.020","Text":"Of course, the voltage across resistor 2 will be the same just with R2 over here."},{"Start":"12:08.020 ","End":"12:10.960","Text":"Why did I calculate this though?"},{"Start":"12:10.960 ","End":"12:17.020","Text":"As we can see, my resistor R1 is connected in parallel to my capacitor."},{"Start":"12:17.020 ","End":"12:21.355","Text":"As we know that things when they are connected in parallel,"},{"Start":"12:21.355 ","End":"12:27.710","Text":"it means that we have the same voltage across both of these components."},{"Start":"12:27.870 ","End":"12:36.700","Text":"In that case, the voltage across R1 is equal to the voltage across the capacitor."},{"Start":"12:36.700 ","End":"12:39.115","Text":"Of course, it\u0027s the voltage across the capacitor,"},{"Start":"12:39.115 ","End":"12:41.155","Text":"not equal to the capacitance."},{"Start":"12:41.155 ","End":"12:43.390","Text":"This is the voltage across the capacitor."},{"Start":"12:43.390 ","End":"12:46.360","Text":"It\u0027s the same voltage across the resistor."},{"Start":"12:46.360 ","End":"12:48.760","Text":"Then we can say that the voltage across"},{"Start":"12:48.760 ","End":"12:52.270","Text":"the capacitor from our usual capacitor equations is equal"},{"Start":"12:52.270 ","End":"12:58.465","Text":"to q divided by c divided by the capacitance."},{"Start":"12:58.465 ","End":"13:01.930","Text":"We\u0027re being told in the question that we\u0027re given the capacitance."},{"Start":"13:01.930 ","End":"13:05.050","Text":"Therefore we can isolate out q and say that it\u0027s"},{"Start":"13:05.050 ","End":"13:08.800","Text":"equal to V c multiplied by the capacitance,"},{"Start":"13:08.800 ","End":"13:12.040","Text":"which will simply be equal to"},{"Start":"13:12.040 ","End":"13:19.930","Text":"VR1C divided by R1 plus R2,"},{"Start":"13:19.930 ","End":"13:24.190","Text":"which is exactly what we got over here."},{"Start":"13:24.190 ","End":"13:27.070","Text":"Here we can see that this is going to"},{"Start":"13:27.070 ","End":"13:30.055","Text":"be the charge on the capacitor when it\u0027s fully charged."},{"Start":"13:30.055 ","End":"13:38.844","Text":"We also got that when using this equation over here and taking t as approaching infinity."},{"Start":"13:38.844 ","End":"13:44.785","Text":"It\u0027s nice to see that we can get to this in both ways."},{"Start":"13:44.785 ","End":"13:49.675","Text":"Now what I want to do is I want to see what my currents are."},{"Start":"13:49.675 ","End":"13:54.070","Text":"First of all, let\u0027s start with my current I3."},{"Start":"13:54.070 ","End":"13:58.905","Text":"I know that this is equal to q dot. Here\u0027s my q."},{"Start":"13:58.905 ","End":"14:02.880","Text":"Let\u0027s take the derivative of it with respect to t. First of all,"},{"Start":"14:02.880 ","End":"14:04.365","Text":"if I open up the brackets,"},{"Start":"14:04.365 ","End":"14:06.540","Text":"all of this multiplied by 1 is a constant,"},{"Start":"14:06.540 ","End":"14:08.710","Text":"so it drops away."},{"Start":"14:08.910 ","End":"14:11.275","Text":"Then I have over here,"},{"Start":"14:11.275 ","End":"14:14.755","Text":"I have to take the derivative of my exponent function."},{"Start":"14:14.755 ","End":"14:17.844","Text":"The minus and the minus here will become a positive."},{"Start":"14:17.844 ","End":"14:21.730","Text":"Then I have R1 plus R2,"},{"Start":"14:21.730 ","End":"14:24.370","Text":"which will cancel out over here."},{"Start":"14:24.370 ","End":"14:27.610","Text":"Then this R1 will cancel out with this R1 and"},{"Start":"14:27.610 ","End":"14:31.585","Text":"this c will cancel with this c. What we\u0027re left with is"},{"Start":"14:31.585 ","End":"14:40.220","Text":"V over here divided by the R2 that was left over here."},{"Start":"14:40.530 ","End":"14:44.395","Text":"This is of course, multiplied by our exponent,"},{"Start":"14:44.395 ","End":"14:51.685","Text":"which is e to the negative R1 plus R2 divided by R1,"},{"Start":"14:51.685 ","End":"14:59.860","Text":"R2 c multiplied by t. Now let\u0027s calculate I1."},{"Start":"14:59.860 ","End":"15:10.640","Text":"I1 we saw over here is equal to this q divided by CR1."},{"Start":"15:11.070 ","End":"15:14.350","Text":"We have our q over here,"},{"Start":"15:14.350 ","End":"15:16.495","Text":"and then we divide by C, R1."},{"Start":"15:16.495 ","End":"15:18.400","Text":"We divide by C and R1."},{"Start":"15:18.400 ","End":"15:24.955","Text":"All we\u0027re left with is V divided by R1 plus R2 multiplied by"},{"Start":"15:24.955 ","End":"15:32.860","Text":"1 minus e to the power of negative R1 plus R2 divided by R1,"},{"Start":"15:32.860 ","End":"15:37.330","Text":"R2C, and t over here."},{"Start":"15:37.330 ","End":"15:46.105","Text":"Then of course, we have that I2 is simply equal to I1 plus I3."},{"Start":"15:46.105 ","End":"15:50.240","Text":"So we can just add these 2 up and that\u0027s it."},{"Start":"15:51.140 ","End":"15:53.670","Text":"Now we\u0027ve solved the question."},{"Start":"15:53.670 ","End":"16:00.080","Text":"The question was to find the charge on the capacitor as a function of time."},{"Start":"16:00.080 ","End":"16:03.670","Text":"Also to calculate the currents in the circuit,"},{"Start":"16:03.670 ","End":"16:05.500","Text":"which we of course did."},{"Start":"16:05.500 ","End":"16:10.900","Text":"What I\u0027m going to do is I\u0027m now going to erase everything and I\u0027m going to"},{"Start":"16:10.900 ","End":"16:16.660","Text":"solve our differential equation that we had over here in order to get this answer."},{"Start":"16:16.660 ","End":"16:19.735","Text":"If you don\u0027t want to see me solve the question,"},{"Start":"16:19.735 ","End":"16:22.520","Text":"then you can end the video now."},{"Start":"16:23.250 ","End":"16:26.890","Text":"We\u0027re going to solve this differential equation by"},{"Start":"16:26.890 ","End":"16:30.655","Text":"using the idea of separating out variables."},{"Start":"16:30.655 ","End":"16:36.370","Text":"First of all our initial condition is that q at time t is equal to 0,"},{"Start":"16:36.370 ","End":"16:38.095","Text":"is equal to 0."},{"Start":"16:38.095 ","End":"16:40.780","Text":"This is where we have to remember and let\u0027s begin."},{"Start":"16:40.780 ","End":"16:43.855","Text":"Here\u0027s our equation that we want to solve."},{"Start":"16:43.855 ","End":"16:49.360","Text":"First of all, I\u0027m going to convert my q dot to be equal to dq"},{"Start":"16:49.360 ","End":"16:55.060","Text":"by dt and I\u0027m going to move it to the other side of the equation."},{"Start":"16:55.060 ","End":"17:02.515","Text":"What I\u0027m going to have is V minus R1 plus R2"},{"Start":"17:02.515 ","End":"17:11.830","Text":"divided by C R1 multiplied by q. I just rearrange this a little bit."},{"Start":"17:11.830 ","End":"17:20.140","Text":"Is equal to R2 multiplied by dq by dt."},{"Start":"17:20.140 ","End":"17:24.490","Text":"Now, just to shorten everything,"},{"Start":"17:24.490 ","End":"17:27.790","Text":"I\u0027m going to call all of this over here,"},{"Start":"17:27.790 ","End":"17:29.665","Text":"the whole coefficient of q,"},{"Start":"17:29.665 ","End":"17:33.640","Text":"just going to call it k, so that I don\u0027t have to keep writing it out."},{"Start":"17:33.640 ","End":"17:37.570","Text":"Now I\u0027m going to multiply both sides of the equation by dt."},{"Start":"17:37.570 ","End":"17:42.820","Text":"What I\u0027ll have is dt multiplied by V minus k,"},{"Start":"17:42.820 ","End":"17:48.670","Text":"q is equal to R2 dq."},{"Start":"17:48.670 ","End":"17:50.470","Text":"Here I have my dq,"},{"Start":"17:50.470 ","End":"17:53.410","Text":"which means that anything with my variable q"},{"Start":"17:53.410 ","End":"17:56.665","Text":"attached to it has to be on this side of the equation."},{"Start":"17:56.665 ","End":"18:00.925","Text":"That means dividing both sides of the equation by V minus kq."},{"Start":"18:00.925 ","End":"18:06.145","Text":"What I\u0027ll be left with is that dt is equal to"},{"Start":"18:06.145 ","End":"18:13.585","Text":"R2dq divided by V minus kq."},{"Start":"18:13.585 ","End":"18:19.850","Text":"Now at this stage, I can add in my integration signs."},{"Start":"18:20.070 ","End":"18:22.735","Text":"Let\u0027s plug in our bounce."},{"Start":"18:22.735 ","End":"18:26.500","Text":"We can do either an indefinite"},{"Start":"18:26.500 ","End":"18:30.655","Text":"integral without bounds or a definite integral with bounds. Let\u0027s do bounds."},{"Start":"18:30.655 ","End":"18:36.460","Text":"We\u0027re going from t is equal to 0 until t is equal to some t. Generally,"},{"Start":"18:36.460 ","End":"18:40.390","Text":"if we have a t and the bounds and we\u0027re integrating by dt,"},{"Start":"18:40.390 ","End":"18:45.310","Text":"so we can say dt tag just to make it clear."},{"Start":"18:45.310 ","End":"18:50.770","Text":"Then we have to integrate the qs along the same bounds."},{"Start":"18:50.770 ","End":"18:59.170","Text":"At time 0, we take our charge at t is equal to 0 at which we were told is equal to 0,"},{"Start":"18:59.170 ","End":"19:06.400","Text":"and then we take our charge at t is equal to t. Again,"},{"Start":"19:06.400 ","End":"19:12.700","Text":"we have q over here and the bounds and we\u0027re integrating along dq."},{"Start":"19:12.700 ","End":"19:21.050","Text":"We can just add in this q tag over here just to make it clear that these are different."},{"Start":"19:21.660 ","End":"19:26.935","Text":"When we integrate dt tag from 0 to t,"},{"Start":"19:26.935 ","End":"19:32.350","Text":"we\u0027re left with just t. First of all,"},{"Start":"19:32.350 ","End":"19:35.740","Text":"we have this constant R2."},{"Start":"19:35.740 ","End":"19:41.485","Text":"Then we\u0027re integrating dq tag divided by V minus kq tag."},{"Start":"19:41.485 ","End":"19:44.560","Text":"We know that we\u0027re going to have a line over here, but first,"},{"Start":"19:44.560 ","End":"19:48.040","Text":"we have to multiply by the inner derivative,"},{"Start":"19:48.040 ","End":"19:51.288","Text":"which was divide by the inner derivative,"},{"Start":"19:51.288 ","End":"19:54.190","Text":"so we\u0027re dividing by negative k,"},{"Start":"19:54.190 ","End":"19:59.153","Text":"so we\u0027ll add in a negative over here and k. Then we have a Lan,"},{"Start":"19:59.153 ","End":"20:06.804","Text":"and our Lan is of V minus k q tag"},{"Start":"20:06.804 ","End":"20:15.775","Text":"between the bounds of 0 and q at time t. Now we can plug in our bounds."},{"Start":"20:15.775 ","End":"20:23.710","Text":"If t is equal to negative R2 divided by k multiplied by V Lan of."},{"Start":"20:23.710 ","End":"20:32.266","Text":"Here I have V minus kq as a function of t divided by V minus k times 0,"},{"Start":"20:32.266 ","End":"20:42.480","Text":"so just divided by V. The next stage is to divide both sides by negative R2"},{"Start":"20:42.480 ","End":"20:49.155","Text":"divided by k. What we\u0027re left with is negative k divided by R2"},{"Start":"20:49.155 ","End":"20:57.090","Text":"multiplied by t. This is equal to Lan of V minus kq."},{"Start":"20:57.090 ","End":"21:00.030","Text":"I just won\u0027t write as a function of t. We know that it is"},{"Start":"21:00.030 ","End":"21:03.925","Text":"divided by V. Then in order to get rid of the Lan,"},{"Start":"21:03.925 ","End":"21:07.435","Text":"I\u0027m going to apply my exponent function."},{"Start":"21:07.435 ","End":"21:16.180","Text":"Then I\u0027ll have e to the negative kt divided by R2 is equal to V minus"},{"Start":"21:16.180 ","End":"21:21.370","Text":"kq divided by V. Now I multiply both sides by V. I"},{"Start":"21:21.370 ","End":"21:27.100","Text":"have ve to the negative k divided by R2t,"},{"Start":"21:27.100 ","End":"21:30.400","Text":"which is equal to V minus kq."},{"Start":"21:30.400 ","End":"21:32.650","Text":"Of course, I want to isolate out my q,"},{"Start":"21:32.650 ","End":"21:38.470","Text":"so I have that q as a function of t is equal to."},{"Start":"21:38.470 ","End":"21:42.804","Text":"Now I\u0027ll have V as a common factor."},{"Start":"21:42.804 ","End":"21:53.110","Text":"Then I have it as 1 minus e to the negative k divided by R2t."},{"Start":"21:53.110 ","End":"21:57.115","Text":"Then of course I have to divide everything by"},{"Start":"21:57.115 ","End":"22:02.775","Text":"k. Now let\u0027s plug in what my k was equal to."},{"Start":"22:02.775 ","End":"22:07.680","Text":"I have q as a function of t is equal to V divided by,"},{"Start":"22:07.680 ","End":"22:09.840","Text":"so my k was this."},{"Start":"22:09.840 ","End":"22:18.040","Text":"I\u0027ll have VCR1 divided by R1 plus R2."},{"Start":"22:18.630 ","End":"22:25.540","Text":"Then I have 1 minus e to the power of negative k. Negative"},{"Start":"22:25.540 ","End":"22:33.400","Text":"R1 plus R2 divided by CR1."},{"Start":"22:33.400 ","End":"22:40.810","Text":"Then from here we have R2 multiplied by t. That\u0027s it."},{"Start":"22:40.810 ","End":"22:42.205","Text":"If you go back earlier,"},{"Start":"22:42.205 ","End":"22:47.500","Text":"you\u0027ll see that this is the exact equation for q that we got."},{"Start":"22:47.500 ","End":"22:53.210","Text":"That\u0027s it. Now we solve the question completely and that is the end of this lesson."}],"ID":22282},{"Watched":false,"Name":"Exercise 8","Duration":"21m 38s","ChapterTopicVideoID":21502,"CourseChapterTopicPlaylistID":99472,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.965","Text":"Hello. In this lesson,"},{"Start":"00:01.965 ","End":"00:04.560","Text":"we\u0027re going to be answering the following question."},{"Start":"00:04.560 ","End":"00:06.345","Text":"In the following circuit,"},{"Start":"00:06.345 ","End":"00:15.765","Text":"a capacitor C_1 is charged with a charge Q_0 before the switch S is moved to a."},{"Start":"00:15.765 ","End":"00:20.570","Text":"Question number 1 is to write an equation which can"},{"Start":"00:20.570 ","End":"00:25.675","Text":"be solved to give us the charge on C_1 as a function of time."},{"Start":"00:25.675 ","End":"00:30.755","Text":"What we\u0027re doing is we\u0027re actually going to solve an equation for this circuit."},{"Start":"00:30.755 ","End":"00:34.585","Text":"We\u0027re going to write an equation for this circuit and later on we\u0027ll solve it."},{"Start":"00:34.585 ","End":"00:41.412","Text":"First of all, if there\u0027s a charge Q naught on capacitors C_1,"},{"Start":"00:41.412 ","End":"00:50.295","Text":"we can say that here is the positive plates and here is the negative plate."},{"Start":"00:50.295 ","End":"00:54.050","Text":"Why is that? Because here we have a voltage source and here is"},{"Start":"00:54.050 ","End":"00:58.150","Text":"the long side of the voltage source and this is the closest plate."},{"Start":"00:58.150 ","End":"01:00.660","Text":"This is the positive plate."},{"Start":"01:02.660 ","End":"01:07.060","Text":"Our switch is closed so it\u0027s moved to point a."},{"Start":"01:07.060 ","End":"01:10.595","Text":"What are we going to do is we\u0027re going to write an equation for"},{"Start":"01:10.595 ","End":"01:17.165","Text":"the voltages in this circuit and of course we want to do this as a function of time."},{"Start":"01:17.165 ","End":"01:21.655","Text":"Any general time not at t=0."},{"Start":"01:21.655 ","End":"01:25.445","Text":"What does that mean for the charge on capacitor 1?"},{"Start":"01:25.445 ","End":"01:27.812","Text":"We know that right at the start,"},{"Start":"01:27.812 ","End":"01:29.600","Text":"capacitor 1 has charged Q_0."},{"Start":"01:29.600 ","End":"01:33.800","Text":"But we want to write an equation for later on"},{"Start":"01:33.800 ","End":"01:38.950","Text":"when the charge is slightly discharged or the capacitor is discharged."},{"Start":"01:38.950 ","End":"01:42.585","Text":"Here we\u0027re going to say that C_1 has a charge,"},{"Start":"01:42.585 ","End":"01:45.225","Text":"Q_1 as a function of t,"},{"Start":"01:45.225 ","End":"01:47.910","Text":"and that C_2 has a charge,"},{"Start":"01:47.910 ","End":"01:54.710","Text":"Q_2 as a function of t. This, of course,"},{"Start":"01:54.710 ","End":"01:59.750","Text":"is still the positive plates and this is the negative plate and let\u0027s look at C_2,"},{"Start":"01:59.750 ","End":"02:02.395","Text":"which will be the positive and negative plates."},{"Start":"02:02.395 ","End":"02:06.494","Text":"When our capacitor, C_1 begins discharging,"},{"Start":"02:06.494 ","End":"02:11.840","Text":"all the positives will travel along here and of course,"},{"Start":"02:11.840 ","End":"02:16.355","Text":"the switch is closed over here so down here to a C_2 capacitor."},{"Start":"02:16.355 ","End":"02:23.310","Text":"Which means that this will be the positive plate and this will be the negative plate."},{"Start":"02:24.100 ","End":"02:30.515","Text":"Let\u0027s define a direction of travel for the current."},{"Start":"02:30.515 ","End":"02:33.350","Text":"If this is the positive plate of the capacitor,"},{"Start":"02:33.350 ","End":"02:39.050","Text":"so the current when the capacitor begins discharging will flow in this direction,"},{"Start":"02:39.050 ","End":"02:41.930","Text":"then it goes down the switch over here."},{"Start":"02:41.930 ","End":"02:47.420","Text":"Of course, this branch over here with the voltage source is not connected to the rest of"},{"Start":"02:47.420 ","End":"02:53.690","Text":"the circuit anymore so we don\u0027t have to go down there and it goes like so,"},{"Start":"02:53.690 ","End":"02:56.400","Text":"and then reaches the capacity."},{"Start":"02:57.560 ","End":"03:00.335","Text":"Our equation for the voltages,"},{"Start":"03:00.335 ","End":"03:01.700","Text":"let\u0027s write it now."},{"Start":"03:01.700 ","End":"03:07.565","Text":"Let\u0027s start from this corner over here so we\u0027re traveling and traveling and then we"},{"Start":"03:07.565 ","End":"03:13.805","Text":"cross this capacitor C_1 and we go from the negative plate to the positive plate."},{"Start":"03:13.805 ","End":"03:18.965","Text":"Which means that we have a jump in the voltage so we\u0027re adding voltage V_1."},{"Start":"03:18.965 ","End":"03:21.005","Text":"Then we carry on traveling,"},{"Start":"03:21.005 ","End":"03:28.920","Text":"then we cross the resistor so we have a voltage drop so we have negative IR."},{"Start":"03:28.940 ","End":"03:34.145","Text":"Then we carry on traveling and we reach our second capacitor."},{"Start":"03:34.145 ","End":"03:37.880","Text":"This time we\u0027re traveling from the positive plate to"},{"Start":"03:37.880 ","End":"03:42.920","Text":"the negative plate so we have a voltage drop again, negative V_2."},{"Start":"03:42.920 ","End":"03:44.240","Text":"Now we carry on traveling,"},{"Start":"03:44.240 ","End":"03:45.875","Text":"we cross another resistor,"},{"Start":"03:45.875 ","End":"03:49.240","Text":"so negative IR,"},{"Start":"03:49.240 ","End":"03:52.365","Text":"and then we get to our starting points."},{"Start":"03:52.365 ","End":"03:54.365","Text":"According to Kirchhoff\u0027s laws,"},{"Start":"03:54.365 ","End":"03:57.360","Text":"this has to be equal to 0."},{"Start":"03:58.910 ","End":"04:06.740","Text":"Now we know that the voltage across a capacitor from this whole chapter,"},{"Start":"04:06.740 ","End":"04:09.980","Text":"we\u0027ve been dealing with this equation so we know that that\u0027s equal to the charge"},{"Start":"04:09.980 ","End":"04:13.850","Text":"on the capacitor divided by the capacitance."},{"Start":"04:13.850 ","End":"04:16.670","Text":"Now let\u0027s plug this in, so V_1,"},{"Start":"04:16.670 ","End":"04:23.120","Text":"the voltage across capacitor C_1 will equal to the charge on it at a general time,"},{"Start":"04:23.120 ","End":"04:27.545","Text":"q_1 t divided by its capacitance C_1,"},{"Start":"04:27.545 ","End":"04:29.810","Text":"and then we have negative IR,"},{"Start":"04:29.810 ","End":"04:31.715","Text":"and here we have another negative IR,"},{"Start":"04:31.715 ","End":"04:37.140","Text":"so negative 2IR and then negative V_2."},{"Start":"04:37.140 ","End":"04:40.880","Text":"We\u0027re subtracting the voltage across capacitor C_2,"},{"Start":"04:40.880 ","End":"04:45.915","Text":"which is negative q_2 as a function of"},{"Start":"04:45.915 ","End":"04:52.300","Text":"t divided by C_2 and all of this is equal to 0."},{"Start":"04:52.750 ","End":"04:57.950","Text":"What we can see is that we have three unknowns: q_1,"},{"Start":"04:57.950 ","End":"05:00.815","Text":"q_2, and I."},{"Start":"05:00.815 ","End":"05:07.310","Text":"But we know that q_1 and q_2 are connected to one another."},{"Start":"05:07.310 ","End":"05:13.085","Text":"First of all, we know that we have a charge conservation."},{"Start":"05:13.085 ","End":"05:16.475","Text":"When we have a charge conservation,"},{"Start":"05:16.475 ","End":"05:21.415","Text":"that means that if we start off with a circuit that has a charge,"},{"Start":"05:21.415 ","End":"05:24.335","Text":"Q_0 or Q naught."},{"Start":"05:24.335 ","End":"05:29.990","Text":"Then over time, as this capacitor discharges and this capacitor charges,"},{"Start":"05:29.990 ","End":"05:32.960","Text":"the total charge Q naught is conserved,"},{"Start":"05:32.960 ","End":"05:39.335","Text":"but it\u0027s just spread out in some way or another across these two capacitors."},{"Start":"05:39.335 ","End":"05:46.010","Text":"Therefore, we can write that Q naught is equal to q_1 at time"},{"Start":"05:46.010 ","End":"05:52.605","Text":"t plus q_2 at time t. Therefore,"},{"Start":"05:52.605 ","End":"05:57.515","Text":"I can say that q_2 as a function of time,"},{"Start":"05:57.515 ","End":"06:04.680","Text":"is equal to Q naught minus q_1 as a function of time."},{"Start":"06:04.680 ","End":"06:09.325","Text":"This is our equation linking q_2 to q_1,"},{"Start":"06:09.325 ","End":"06:15.425","Text":"which means that now I just have two unknowns: I and q_1,"},{"Start":"06:15.425 ","End":"06:19.780","Text":"and my I is also linked to q_1."},{"Start":"06:19.780 ","End":"06:26.270","Text":"What\u0027s going on capacitor C_1 is discharging and as it discharges,"},{"Start":"06:26.270 ","End":"06:30.775","Text":"it\u0027s causing a current to flow through the circuit,"},{"Start":"06:30.775 ","End":"06:35.000","Text":"so the current that is flowing through the circuit is connected to"},{"Start":"06:35.000 ","End":"06:41.450","Text":"the changing charge on my capacitor C_1 and that is linked to q_1,"},{"Start":"06:41.450 ","End":"06:45.550","Text":"of course, because q_1 represents the charge in the capacitor."},{"Start":"06:45.550 ","End":"06:50.460","Text":"Therefore, I can say that my current is equal to"},{"Start":"06:50.460 ","End":"06:59.270","Text":"negative q_1 dot y negative because the capacitor is discharging."},{"Start":"06:59.270 ","End":"07:02.465","Text":"When the capacitor is discharging,"},{"Start":"07:02.465 ","End":"07:05.990","Text":"then I is equal to negative q_1 dot and if"},{"Start":"07:05.990 ","End":"07:09.920","Text":"the capacitor was charging then it would be equal to positive q dot."},{"Start":"07:09.920 ","End":"07:14.690","Text":"Let\u0027s say here the capacitor C_2 is charging so I could"},{"Start":"07:14.690 ","End":"07:20.205","Text":"say that I is equal to positive q_2 dots."},{"Start":"07:20.205 ","End":"07:23.660","Text":"But I don\u0027t need to write this out because I have this equation linking the"},{"Start":"07:23.660 ","End":"07:27.080","Text":"two so it doesn\u0027t matter. I can rub this out."},{"Start":"07:27.080 ","End":"07:30.770","Text":"But it\u0027s important to remember if a capacitor is discharging and"},{"Start":"07:30.770 ","End":"07:34.594","Text":"we\u0027re basing the current on the charge on this capacitor,"},{"Start":"07:34.594 ","End":"07:39.650","Text":"then we have to add a minus over here Of course,"},{"Start":"07:39.650 ","End":"07:42.890","Text":"the current is linked to the change in charge."},{"Start":"07:42.890 ","End":"07:46.115","Text":"That\u0027s why this equation is correct."},{"Start":"07:46.115 ","End":"07:54.200","Text":"If we would draw a graph representing the charge on the capacitor as a function of time,"},{"Start":"07:54.200 ","End":"07:57.695","Text":"because it\u0027s discharging, it would look like so."},{"Start":"07:57.695 ","End":"08:03.980","Text":"Which means that if we took the derivative of this or the gradient of this graph,"},{"Start":"08:03.980 ","End":"08:07.960","Text":"we would have a negative gradient because it\u0027s going down."},{"Start":"08:07.960 ","End":"08:12.440","Text":"That\u0027s another clue why we have to have a negative over here."},{"Start":"08:12.440 ","End":"08:19.955","Text":"Conversely, if we were to draw q_2 as a function of time, here it\u0027s charging."},{"Start":"08:19.955 ","End":"08:21.875","Text":"The graph would look like this."},{"Start":"08:21.875 ","End":"08:25.010","Text":"Here we can see we have a positive gradient of the graph and that\u0027s"},{"Start":"08:25.010 ","End":"08:29.100","Text":"why I would equal positive q_2 dot."},{"Start":"08:30.020 ","End":"08:35.390","Text":"Now let\u0027s plug everything into this equation over here."},{"Start":"08:35.390 ","End":"08:38.090","Text":"I\u0027m going to stop writing as a function of t,"},{"Start":"08:38.090 ","End":"08:40.560","Text":"we know that it is."},{"Start":"08:40.560 ","End":"08:50.310","Text":"We have q_1 divided by C_1 minus 2 times i so minus"},{"Start":"08:50.310 ","End":"08:55.410","Text":"2 times negative q_1 dot multiplied"},{"Start":"08:55.410 ","End":"09:01.005","Text":"by R minus q_2 divided by C_2,"},{"Start":"09:01.005 ","End":"09:02.940","Text":"and this is equal to 0."},{"Start":"09:02.940 ","End":"09:05.240","Text":"First of all, we have a minus and a minus here,"},{"Start":"09:05.240 ","End":"09:10.745","Text":"so it becomes plus and of course my q_2."},{"Start":"09:10.745 ","End":"09:20.280","Text":"Let\u0027s plug this in so my q_2 is equal to Q naught minus q_1."},{"Start":"09:20.440 ","End":"09:25.775","Text":"Now let\u0027s get all my q_1s together."},{"Start":"09:25.775 ","End":"09:30.940","Text":"What I have is a q_1 divided by C_1,"},{"Start":"09:30.940 ","End":"09:35.820","Text":"and then I have minus and a minus,"},{"Start":"09:35.820 ","End":"09:40.020","Text":"so plus q_1 divided by C_2."},{"Start":"09:40.020 ","End":"09:43.510","Text":"Then I have plus"},{"Start":"09:44.120 ","End":"09:51.345","Text":"2Rq_1 dot and then minus Q naught divided by C_2,"},{"Start":"09:51.345 ","End":"09:54.525","Text":"and this is all equal to 0."},{"Start":"09:54.525 ","End":"09:57.755","Text":"Now what I\u0027m going to do is here,"},{"Start":"09:57.755 ","End":"10:00.535","Text":"I\u0027m going to make a common denominator."},{"Start":"10:00.535 ","End":"10:07.330","Text":"What I have is C_1 plus C_2 divided by C_1."},{"Start":"10:07.330 ","End":"10:10.505","Text":"C2 came at a common denominator,"},{"Start":"10:10.505 ","End":"10:13.745","Text":"and all of this is multiplied by q_1."},{"Start":"10:13.745 ","End":"10:19.200","Text":"Then we have plus 2Rq_1 dot"},{"Start":"10:19.200 ","End":"10:24.960","Text":"minus Q naught divided by C_2 and this is all equal to 0."},{"Start":"10:24.960 ","End":"10:29.750","Text":"Now we\u0027re going to do is I\u0027m going to divide everything by this 2R over"},{"Start":"10:29.750 ","End":"10:34.765","Text":"here so here I have in the denominator 2R,"},{"Start":"10:34.765 ","End":"10:37.310","Text":"and also here I have 2R."},{"Start":"10:38.820 ","End":"10:42.925","Text":"This is the answer to question Number 1."},{"Start":"10:42.925 ","End":"10:47.920","Text":"To write an equation which can be solved to give us the charge on C1."},{"Start":"10:47.920 ","End":"10:50.620","Text":"Now let\u0027s move on to question number 2."},{"Start":"10:50.620 ","End":"10:52.690","Text":"I\u0027m just going to leave us with this equation,"},{"Start":"10:52.690 ","End":"10:55.280","Text":"I\u0027m going to rub out everything else."},{"Start":"10:55.470 ","End":"10:58.420","Text":"Question number 2 is to solve"},{"Start":"10:58.420 ","End":"11:03.910","Text":"this equation and calculate the charge on each capacitor as a function of time."},{"Start":"11:03.910 ","End":"11:07.990","Text":"First of all, what I can do is I can see over here"},{"Start":"11:07.990 ","End":"11:12.970","Text":"that I have a bunch of constants as the coefficient for Q1,"},{"Start":"11:12.970 ","End":"11:15.100","Text":"and here I also have constants."},{"Start":"11:15.100 ","End":"11:20.200","Text":"In order to make my life a bit easier and not have to rewrite all of this all the time,"},{"Start":"11:20.200 ","End":"11:24.940","Text":"let\u0027s call all of these constants over here 1 divided by Tau."},{"Start":"11:24.940 ","End":"11:31.180","Text":"Where Tau is, of course, our time constant and has units for time, so it\u0027s seconds."},{"Start":"11:31.180 ","End":"11:34.720","Text":"I will speak later about how I know that this is 1 divided by Tau."},{"Start":"11:34.720 ","End":"11:40.225","Text":"Let\u0027s call this over here some kind of constant a."},{"Start":"11:40.225 ","End":"11:42.250","Text":"Now let\u0027s rewrite our equation."},{"Start":"11:42.250 ","End":"11:47.185","Text":"I have here 1 divided by Tau multiplied by q1"},{"Start":"11:47.185 ","End":"11:54.890","Text":"plus q1 dot minus A is equal to 0."},{"Start":"11:55.290 ","End":"11:58.855","Text":"Now, when solving differential equations,"},{"Start":"11:58.855 ","End":"12:04.480","Text":"what we want to do is we want to solve by separating out our variables."},{"Start":"12:04.480 ","End":"12:06.520","Text":"Our variables are q,"},{"Start":"12:06.520 ","End":"12:08.560","Text":"and of course q is as a function of t,"},{"Start":"12:08.560 ","End":"12:17.035","Text":"so we also have variable t. What we do is we take our Q1 dot to 1 side of the equation."},{"Start":"12:17.035 ","End":"12:19.450","Text":"We have q1 dot,"},{"Start":"12:19.450 ","End":"12:26.275","Text":"which is equal to A minus q1 divided by Tau,"},{"Start":"12:26.275 ","End":"12:32.755","Text":"and of course, q1 dot is also equal to dq1 by dt."},{"Start":"12:32.755 ","End":"12:36.220","Text":"That\u0027s what q1 dot means, dq1 by dt."},{"Start":"12:36.220 ","End":"12:40.645","Text":"Now, what I can do is I can multiply both sides by this dt."},{"Start":"12:40.645 ","End":"12:44.725","Text":"What I\u0027ll have is"},{"Start":"12:44.725 ","End":"12:53.660","Text":"A minus q1 divided by Tau dt is equal to dq1."},{"Start":"12:54.240 ","End":"12:57.535","Text":"Of course, my q1 is my variable,"},{"Start":"12:57.535 ","End":"13:03.865","Text":"so my q1 has to be on the side of dq1 because I\u0027m soon going to integrate."},{"Start":"13:03.865 ","End":"13:08.185","Text":"What I do is I divide everything by what\u0027s inside the brackets."},{"Start":"13:08.185 ","End":"13:12.040","Text":"What I\u0027m left with is that dt is equal to"},{"Start":"13:12.040 ","End":"13:20.845","Text":"dq1 divided by A minus q1 divided by Tau."},{"Start":"13:20.845 ","End":"13:24.654","Text":"At this stage, I can add in my integral signs."},{"Start":"13:24.654 ","End":"13:31.000","Text":"Now, I can either solve this by using definite integrals or indefinite integrals."},{"Start":"13:31.000 ","End":"13:32.830","Text":"I\u0027m going to use the definite integral,"},{"Start":"13:32.830 ","End":"13:34.435","Text":"which means that it has bounds."},{"Start":"13:34.435 ","End":"13:38.590","Text":"I\u0027m integrating from a time t is equal to 0 until"},{"Start":"13:38.590 ","End":"13:44.320","Text":"some general time t. Because t is in my bounds,"},{"Start":"13:44.320 ","End":"13:48.265","Text":"generally, we\u0027ll say instead of integrating along dt,"},{"Start":"13:48.265 ","End":"13:50.665","Text":"we\u0027ll see we\u0027re integrating along dt tag,"},{"Start":"13:50.665 ","End":"13:54.730","Text":"just so that it\u0027s obvious that this isn\u0027t right now a variable."},{"Start":"13:54.730 ","End":"13:56.530","Text":"The same over here, we\u0027ll see,"},{"Start":"13:56.530 ","End":"14:02.110","Text":"we\u0027re integrating from our initial charge."},{"Start":"14:02.110 ","End":"14:05.260","Text":"Our charge at t is equal to 0,"},{"Start":"14:05.260 ","End":"14:07.300","Text":"which we were told in the question,"},{"Start":"14:07.300 ","End":"14:09.370","Text":"our charge Q1 at t is equal to 0,"},{"Start":"14:09.370 ","End":"14:17.330","Text":"or is equal to Q naught up until the charge that it will be at some time t,"},{"Start":"14:18.480 ","End":"14:22.705","Text":"and of course this is q1(t)."},{"Start":"14:22.705 ","End":"14:24.220","Text":"Again, over here,"},{"Start":"14:24.220 ","End":"14:27.130","Text":"because q1(t) is in the balance,"},{"Start":"14:27.130 ","End":"14:30.010","Text":"we generally put that we\u0027re integrating"},{"Start":"14:30.010 ","End":"14:34.750","Text":"a long dq1 tag so here we also have a tag over here."},{"Start":"14:34.750 ","End":"14:37.163","Text":"Now we begin the integration."},{"Start":"14:37.163 ","End":"14:42.925","Text":"If we integrate along with the t=2,"},{"Start":"14:42.925 ","End":"14:45.325","Text":"and now we have the integral of this."},{"Start":"14:45.325 ","End":"14:51.340","Text":"What we have is a bunch of constants and q1 tag in the denominator."},{"Start":"14:51.340 ","End":"14:53.005","Text":"We know it\u0027s going to be a Ln."},{"Start":"14:53.005 ","End":"14:57.620","Text":"First of all, we have to multiply by the inner derivative."},{"Start":"14:58.770 ","End":"15:00.940","Text":"The A is a constant,"},{"Start":"15:00.940 ","End":"15:07.675","Text":"so it cancels out and then we have negative 1 divided by Tau."},{"Start":"15:07.675 ","End":"15:13.330","Text":"As it\u0027s in the denominator so we get here negative Tau."},{"Start":"15:13.330 ","End":"15:19.690","Text":"Because what we\u0027re doing is we\u0027re multiplying by negative 1 divided by 1 divided by Tau,"},{"Start":"15:19.690 ","End":"15:21.310","Text":"which is just negative Tau."},{"Start":"15:21.310 ","End":"15:24.955","Text":"Then we have Ln of"},{"Start":"15:24.955 ","End":"15:30.745","Text":"A minus q1 tag divided by"},{"Start":"15:30.745 ","End":"15:37.975","Text":"Tau between the bounds of Q naught until q1(t)."},{"Start":"15:37.975 ","End":"15:40.150","Text":"Now we\u0027ll plug in the balance."},{"Start":"15:40.150 ","End":"15:44.515","Text":"What we\u0027ll have is that t is equal to negative Tau Ln"},{"Start":"15:44.515 ","End":"15:53.715","Text":"of A minus q1(t) divided by Tau,"},{"Start":"15:53.715 ","End":"16:00.540","Text":"divided by A minus Q naught divided by Tau."},{"Start":"16:00.540 ","End":"16:09.700","Text":"Now, what we\u0027re going to do, over"},{"Start":"16:09.700 ","End":"16:12.010","Text":"here in these brackets,"},{"Start":"16:12.010 ","End":"16:14.170","Text":"we\u0027re going to multiply everything by Tau to"},{"Start":"16:14.170 ","End":"16:16.675","Text":"get rid of this inner denominator because it\u0027s ugly."},{"Start":"16:16.675 ","End":"16:23.425","Text":"What we\u0027ll have is negative Tau multiplied by Ln (Tau A"},{"Start":"16:23.425 ","End":"16:33.955","Text":"minus q1(t) divided by Tau A minus Q naught)."},{"Start":"16:33.955 ","End":"16:35.935","Text":"This looks a bit nicer."},{"Start":"16:35.935 ","End":"16:40.630","Text":"Now we\u0027re going to divide both sides by negative Tau."},{"Start":"16:40.630 ","End":"16:49.300","Text":"What we\u0027ll have is negative t divided by Tau and then this is going to be equal to"},{"Start":"16:49.300 ","End":"16:59.545","Text":"Ln (Tau A minus q1(t) divided by Tau a minus Q naught."},{"Start":"16:59.545 ","End":"17:02.200","Text":"Then we\u0027re going to use our exponent function,"},{"Start":"17:02.200 ","End":"17:10.495","Text":"which will get rid of the Ln on this side so e^ negative t divided by Tau is equal to"},{"Start":"17:10.495 ","End":"17:20.035","Text":"Tau A minus q1(t) divided by Tau A minus Q naught."},{"Start":"17:20.035 ","End":"17:24.850","Text":"Now of course, we want to isolate our q1(t) because I\u0027m"},{"Start":"17:24.850 ","End":"17:26.980","Text":"reminding you our question was to"},{"Start":"17:26.980 ","End":"17:29.665","Text":"calculate the charge on each capacitor as a function of time."},{"Start":"17:29.665 ","End":"17:32.665","Text":"The charge on capacitor number 1 is a function of time,"},{"Start":"17:32.665 ","End":"17:37.765","Text":"is q1(t), so I\u0027ll multiply both sides by this denominator over here."},{"Start":"17:37.765 ","End":"17:45.700","Text":"What we\u0027ll have is Tau A minus Q naught and all of this multiplied"},{"Start":"17:45.700 ","End":"17:54.260","Text":"by e to the negative t divided by tau is equal to Tau A minus q1(t)."},{"Start":"17:54.450 ","End":"17:59.365","Text":"Then, all we have to do is just isolate out our q1(t)."},{"Start":"17:59.365 ","End":"18:07.285","Text":"What we\u0027ll get is that q1(t) is equal to Tau A minus"},{"Start":"18:07.285 ","End":"18:16.690","Text":"Tau A minus Q naught multiplied by e to the negative t divided by Tau."},{"Start":"18:16.690 ","End":"18:23.222","Text":"Then, of course, our q2 as a function of t is from the last question,"},{"Start":"18:23.222 ","End":"18:32.890","Text":"we saw that it was equal to Q naught minus q1(t) so it\u0027s just going to be equal to that,"},{"Start":"18:32.890 ","End":"18:35.390","Text":"we just subtract to this."},{"Start":"18:36.540 ","End":"18:43.340","Text":"This is the answer to question number 2 and now let\u0027s move on to question number 3."},{"Start":"18:43.380 ","End":"18:47.485","Text":"Just remember that where we have Tau and A,"},{"Start":"18:47.485 ","End":"18:51.250","Text":"what they mean so a was this over here,"},{"Start":"18:51.250 ","End":"18:56.800","Text":"and 1 over Tau is this over here and I just wrote again."},{"Start":"18:56.800 ","End":"18:58.480","Text":"Question number 3 is,"},{"Start":"18:58.480 ","End":"19:01.825","Text":"what are the currents through each resistor as a function of time."},{"Start":"19:01.825 ","End":"19:05.575","Text":"We saw before that the equation for current is"},{"Start":"19:05.575 ","End":"19:09.640","Text":"equal to the negative time derivative of q1,"},{"Start":"19:09.640 ","End":"19:14.980","Text":"or it\u0027s also equal to the positive time derivative of q2."},{"Start":"19:14.980 ","End":"19:17.575","Text":"We already spoke about that."},{"Start":"19:17.575 ","End":"19:23.545","Text":"That means that we just have to take the negative derivative of this."},{"Start":"19:23.545 ","End":"19:34.165","Text":"Let\u0027s say that I is equal to negative d by dt(q1) so"},{"Start":"19:34.165 ","End":"19:37.765","Text":"that is of Tau A minus"},{"Start":"19:37.765 ","End":"19:46.630","Text":"Tau A minus Q naught e to the negative t divided by Tau."},{"Start":"19:46.630 ","End":"19:52.280","Text":"First of all, we can replace all the minuses."},{"Start":"19:53.280 ","End":"20:00.955","Text":"Here, we can add a minus and here we can add a plus so here we have a plus."},{"Start":"20:00.955 ","End":"20:08.472","Text":"Therefore, we can say that I is equal to d by dt of"},{"Start":"20:08.472 ","End":"20:13.695","Text":"negative Tau A plus"},{"Start":"20:13.695 ","End":"20:22.835","Text":"Tau A minus Q naught multiplied by e to the negative t divided by Tau."},{"Start":"20:22.835 ","End":"20:25.090","Text":"First of all, Tau and A,"},{"Start":"20:25.090 ","End":"20:29.305","Text":"as we saw previously, are just constants."},{"Start":"20:29.305 ","End":"20:31.315","Text":"When we take the time derivative of them,"},{"Start":"20:31.315 ","End":"20:35.095","Text":"that is going to be equal to 0."},{"Start":"20:35.095 ","End":"20:42.325","Text":"Then what we have over here is Tau A minus Q naught, which are constants."},{"Start":"20:42.325 ","End":"20:49.480","Text":"Then we have our function e over here that has a t in it over here so we"},{"Start":"20:49.480 ","End":"20:57.055","Text":"can take the derivative of this with respect to t. Of course,"},{"Start":"20:57.055 ","End":"21:00.910","Text":"first we take the inner derivative over here."},{"Start":"21:00.910 ","End":"21:08.200","Text":"We have to multiply by negative 1 divided by Tau so we have negative 1"},{"Start":"21:08.200 ","End":"21:15.100","Text":"over Tau and then we multiply this by the brackets over here,"},{"Start":"21:15.100 ","End":"21:19.495","Text":"multiplied by Tau A minus Q naught,"},{"Start":"21:19.495 ","End":"21:25.400","Text":"and then multiplied by e to the negative t divided by Tau."},{"Start":"21:25.710 ","End":"21:29.000","Text":"This is the final answer for the current."},{"Start":"21:29.000 ","End":"21:34.745","Text":"I just took the minus over here and switched it around and that is it."},{"Start":"21:34.745 ","End":"21:39.510","Text":"We\u0027ve answered all of the questions and that\u0027s the end of this lesson."}],"ID":22283},{"Watched":false,"Name":"Exercise 9","Duration":"11m 14s","ChapterTopicVideoID":21503,"CourseChapterTopicPlaylistID":99472,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.380","Text":"Hello. In this lesson we\u0027re going to be answering the following question,"},{"Start":"00:04.380 ","End":"00:09.180","Text":"2 spheres of radius a and b are placed far"},{"Start":"00:09.180 ","End":"00:14.880","Text":"apart and have charges plus Q and minus Q respectively."},{"Start":"00:14.880 ","End":"00:21.750","Text":"Question Number 1 is to calculate the total electrostatic energy of the system."},{"Start":"00:21.750 ","End":"00:27.168","Text":"That means we\u0027re calculating the total energy of these 2 spheres,"},{"Start":"00:27.168 ","End":"00:30.240","Text":"so you total and because we\u0027re told in"},{"Start":"00:30.240 ","End":"00:33.690","Text":"the question that they\u0027re placed far apart or very far apart,"},{"Start":"00:33.690 ","End":"00:37.640","Text":"we can assume that the energy of"},{"Start":"00:37.640 ","End":"00:42.740","Text":"each sphere isn\u0027t affecting the energy of the other sphere."},{"Start":"00:42.740 ","End":"00:48.260","Text":"Therefore our u total is going to be the energy of a sphere of"},{"Start":"00:48.260 ","End":"00:54.070","Text":"radius a plus the energy of sphere of radius b."},{"Start":"00:54.070 ","End":"00:57.590","Text":"A little reminder, the electrostatic energy is"},{"Start":"00:57.590 ","End":"01:00.725","Text":"the energy needed in order to build a system."},{"Start":"01:00.725 ","End":"01:03.590","Text":"Because we said these systems are so far apart,"},{"Start":"01:03.590 ","End":"01:07.050","Text":"we can consider each one on its own."},{"Start":"01:07.050 ","End":"01:10.160","Text":"We\u0027re calculating the electrostatic energy of"},{"Start":"01:10.160 ","End":"01:15.090","Text":"this sphere plus the electrostatic energy of this sphere."},{"Start":"01:16.070 ","End":"01:20.240","Text":"We have these spheres and we\u0027re assuming in"},{"Start":"01:20.240 ","End":"01:24.565","Text":"this chapter that we\u0027re only dealing with conductors."},{"Start":"01:24.565 ","End":"01:27.830","Text":"In that case, if these 2 spheres are conductors,"},{"Start":"01:27.830 ","End":"01:29.045","Text":"then as we know,"},{"Start":"01:29.045 ","End":"01:33.620","Text":"all the charges on the conductor are on the outside"},{"Start":"01:33.620 ","End":"01:38.915","Text":"or on the surface and the net charge in the middle is equal to 0."},{"Start":"01:38.915 ","End":"01:41.435","Text":"If all the charges are just on the surface,"},{"Start":"01:41.435 ","End":"01:45.660","Text":"then it\u0027s as if we\u0027re looking at a spherical shell."},{"Start":"01:46.250 ","End":"01:48.720","Text":"These spheres are conductors,"},{"Start":"01:48.720 ","End":"01:53.885","Text":"therefore all of the charges or the net charge is only on the surface and therefore,"},{"Start":"01:53.885 ","End":"01:57.110","Text":"we can consider it as just a spherical shell"},{"Start":"01:57.110 ","End":"02:00.940","Text":"and so we want to find the energy of the spherical shell."},{"Start":"02:00.940 ","End":"02:05.480","Text":"We know a few equations to calculate the energy in general."},{"Start":"02:05.480 ","End":"02:12.440","Text":"We\u0027ve seen that we can calculate that by using half of the sum of Qi,"},{"Start":"02:12.440 ","End":"02:16.760","Text":"phi i where each Qi is of course"},{"Start":"02:16.760 ","End":"02:22.085","Text":"the charge and phi i is the potential at that point where the charge"},{"Start":"02:22.085 ","End":"02:27.170","Text":"is or we can alternatively use the equation Epsilon_0 divided by"},{"Start":"02:27.170 ","End":"02:33.640","Text":"2 multiplied by the integral of e^2 dv."},{"Start":"02:34.160 ","End":"02:40.615","Text":"We\u0027re going to solve this using both of the equations so that you can get used to both."},{"Start":"02:40.615 ","End":"02:45.940","Text":"Let\u0027s calculate the electrostatic energy of the sphere of radius a."},{"Start":"02:45.940 ","End":"02:47.815","Text":"Let\u0027s start off with this."},{"Start":"02:47.815 ","End":"02:53.110","Text":"We have a 1/2, and then of course we just have 1 charge which is Q."},{"Start":"02:53.110 ","End":"02:55.870","Text":"Our Qi is just Q,"},{"Start":"02:55.870 ","End":"03:00.125","Text":"and then multiplied by the potential of this total charge Q."},{"Start":"03:00.125 ","End":"03:01.835","Text":"Remember the whole sphere,"},{"Start":"03:01.835 ","End":"03:06.315","Text":"or the whole spherical shell has this charge Q."},{"Start":"03:06.315 ","End":"03:10.295","Text":"That means that we have to multiply this by"},{"Start":"03:10.295 ","End":"03:16.400","Text":"the potential at the points where all of the charges are,"},{"Start":"03:16.400 ","End":"03:18.245","Text":"which is on the spherical shell."},{"Start":"03:18.245 ","End":"03:20.495","Text":"We\u0027re on the surface of the sphere."},{"Start":"03:20.495 ","End":"03:24.390","Text":"The potential at a radius of a,"},{"Start":"03:24.530 ","End":"03:28.010","Text":"so here is naught and then at a radius a,"},{"Start":"03:28.010 ","End":"03:31.490","Text":"if we go all around the sphere in 3-dimensions,"},{"Start":"03:31.490 ","End":"03:33.620","Text":"then we get that."},{"Start":"03:33.620 ","End":"03:38.630","Text":"This is equal to 1/2 multiplied by Q."},{"Start":"03:38.630 ","End":"03:42.440","Text":"What is the potential of a spherical shell?"},{"Start":"03:42.440 ","End":"03:46.895","Text":"We know that this is the same as the potential of a point charge."},{"Start":"03:46.895 ","End":"03:49.025","Text":"This, you should know already."},{"Start":"03:49.025 ","End":"03:50.810","Text":"It\u0027s the potential of a point charge,"},{"Start":"03:50.810 ","End":"03:54.860","Text":"so we multiply this by KQ divided by the radius,"},{"Start":"03:54.860 ","End":"03:56.345","Text":"where here the radius is a."},{"Start":"03:56.345 ","End":"04:01.600","Text":"This is the potential of a point charge and therefore,"},{"Start":"04:01.600 ","End":"04:06.875","Text":"we get that the energy of this sphere over here,"},{"Start":"04:06.875 ","End":"04:12.870","Text":"a is KQ^2 divided by 2a."},{"Start":"04:13.970 ","End":"04:20.220","Text":"This was the first method and now let\u0027s look at the second method."},{"Start":"04:20.220 ","End":"04:28.610","Text":"The energy is equal to Epsilon_0 divided by 2 multiplied by the integral of e^2 dv,"},{"Start":"04:28.610 ","End":"04:30.745","Text":"where of course V is volume."},{"Start":"04:30.745 ","End":"04:34.460","Text":"First of all, we\u0027re dealing with a spherical shell we already saw."},{"Start":"04:34.460 ","End":"04:39.650","Text":"The electric field inside the spherical shell is equal to 0 and"},{"Start":"04:39.650 ","End":"04:42.740","Text":"the electric field outside of this spherical shell is in"},{"Start":"04:42.740 ","End":"04:46.040","Text":"the radial direction and it is equal to,"},{"Start":"04:46.040 ","End":"04:47.420","Text":"as we\u0027ve already seen,"},{"Start":"04:47.420 ","End":"04:51.865","Text":"KQ divided by r^2."},{"Start":"04:51.865 ","End":"04:56.780","Text":"Now what we\u0027re doing is we\u0027re integrating on the electric field squared,"},{"Start":"04:56.780 ","End":"05:04.015","Text":"so that\u0027s KQ divided by I^2 and then dv."},{"Start":"05:04.015 ","End":"05:07.320","Text":"We\u0027re using spherical coordinates."},{"Start":"05:07.320 ","End":"05:12.360","Text":"What it\u0027s going to be is I^2 sine Theta,"},{"Start":"05:12.360 ","End":"05:14.640","Text":"d Theta, d Phi,"},{"Start":"05:14.640 ","End":"05:17.280","Text":"d r. However,"},{"Start":"05:17.280 ","End":"05:23.250","Text":"we can see that Theta and Phi are constants over here."},{"Start":"05:23.250 ","End":"05:26.720","Text":"Our electric field is only dependent on r. Therefore"},{"Start":"05:26.720 ","End":"05:31.910","Text":"we don\u0027t have to integrate along d Theta and d Phi."},{"Start":"05:31.910 ","End":"05:34.685","Text":"What we can just do is take them as constants."},{"Start":"05:34.685 ","End":"05:44.550","Text":"Then we will be left with 4 Pi r^2 dr. We get"},{"Start":"05:44.550 ","End":"05:46.700","Text":"4 Pi from integrating"},{"Start":"05:46.700 ","End":"05:55.350","Text":"sine Theta d Theta d Phi by the constants and then we\u0027re integrating,"},{"Start":"05:55.350 ","End":"05:57.960","Text":"so from 0 to a radius of a."},{"Start":"05:57.960 ","End":"05:59.740","Text":"We know that our E field is 0,"},{"Start":"05:59.740 ","End":"06:01.530","Text":"so there\u0027s nothing to sum up from there."},{"Start":"06:01.530 ","End":"06:04.940","Text":"Then we\u0027re integrating from a radius of a and up"},{"Start":"06:04.940 ","End":"06:09.745","Text":"until infinitely far away from the sphere."},{"Start":"06:09.745 ","End":"06:16.145","Text":"This is a very easy integral everything here is constants aside from the r\u0027s."},{"Start":"06:16.145 ","End":"06:18.890","Text":"Here we\u0027ll have 1 divided by r to the power of 4,"},{"Start":"06:18.890 ","End":"06:21.250","Text":"but here we have r^2."},{"Start":"06:21.250 ","End":"06:25.790","Text":"You just have to take all of the constants out and then remember"},{"Start":"06:25.790 ","End":"06:30.260","Text":"over here that Epsilon_0 is equal to 1"},{"Start":"06:30.260 ","End":"06:36.560","Text":"divided by 4 Pi k. If you substitute that in and you do"},{"Start":"06:36.560 ","End":"06:39.030","Text":"this integral it\u0027s meant to"},{"Start":"06:39.030 ","End":"06:42.980","Text":"be relatively easy and you\u0027ll see that you\u0027ll get the same answer,"},{"Start":"06:42.980 ","End":"06:44.945","Text":"KQ^2 divided by 2a."},{"Start":"06:44.945 ","End":"06:50.460","Text":"I suggest you pause the video and just give it a go to see that you get the same answer."},{"Start":"06:51.430 ","End":"07:01.415","Text":"Then you can repeat Method 1 or Method 2 in order to find the energy for sphere b,"},{"Start":"07:01.415 ","End":"07:05.614","Text":"but we can see that the total energy from this equation,"},{"Start":"07:05.614 ","End":"07:13.820","Text":"we can see that we\u0027re going to have K divided by 2 multiplied by 4a,"},{"Start":"07:13.820 ","End":"07:19.260","Text":"Q^2 divided by a and"},{"Start":"07:19.260 ","End":"07:25.095","Text":"then what we\u0027ll have is plus Q^2 divided by b for the sphere."},{"Start":"07:25.095 ","End":"07:34.190","Text":"We can see, we just plug that in and of course this is just equal to KQ^2 divided by 2,"},{"Start":"07:34.190 ","End":"07:42.115","Text":"and then multiplied by b plus a divided by ab."},{"Start":"07:42.115 ","End":"07:45.970","Text":"When we make a common denominator."},{"Start":"07:47.270 ","End":"07:50.615","Text":"This is the total energy,"},{"Start":"07:50.615 ","End":"07:54.230","Text":"electrostatic energy of the system and we\u0027ve answered question Number 1."},{"Start":"07:54.230 ","End":"07:56.600","Text":"Now let\u0027s move on to question Number 2."},{"Start":"07:56.600 ","End":"07:59.450","Text":"Using our answer to question Number 1,"},{"Start":"07:59.450 ","End":"08:03.870","Text":"let\u0027s calculate the capacitance of the system."},{"Start":"08:05.600 ","End":"08:11.690","Text":"We already know that the energy on a capacitor UC is equal"},{"Start":"08:11.690 ","End":"08:17.480","Text":"to 1/2 of Q^2 divided by the capacitance and of course,"},{"Start":"08:17.480 ","End":"08:20.090","Text":"there\u0027s other ways of writing out this equation."},{"Start":"08:20.090 ","End":"08:23.614","Text":"We already saw these examples before."},{"Start":"08:23.614 ","End":"08:27.725","Text":"Here specifically we don\u0027t have the voltage or we don\u0027t know the voltage,"},{"Start":"08:27.725 ","End":"08:33.360","Text":"but we do have the energy and we do have the charge."},{"Start":"08:33.370 ","End":"08:36.650","Text":"Which means that this version of the equation for"},{"Start":"08:36.650 ","End":"08:39.860","Text":"the energy on the capacitor is easier to use."},{"Start":"08:39.860 ","End":"08:43.400","Text":"If we say that the energy on the capacitor is equal to this,"},{"Start":"08:43.400 ","End":"08:45.230","Text":"and it\u0027s also equal to this,"},{"Start":"08:45.230 ","End":"08:47.210","Text":"what we just calculated."},{"Start":"08:47.210 ","End":"08:55.280","Text":"KQ^2 divided by 2 multiplied by b plus a divided by ab."},{"Start":"08:55.280 ","End":"09:00.260","Text":"What we can do is we can multiply both sides by 2."},{"Start":"09:00.260 ","End":"09:03.615","Text":"We can divide both sides by Q^2,"},{"Start":"09:03.615 ","End":"09:04.965","Text":"and then we\u0027re left with this."},{"Start":"09:04.965 ","End":"09:08.025","Text":"Here we have 1 divided by c is equal to all of this."},{"Start":"09:08.025 ","End":"09:10.470","Text":"What we want to do is we want to find the capacitance,"},{"Start":"09:10.470 ","End":"09:14.300","Text":"so we take the reciprocal and then we\u0027ll get that this is equal"},{"Start":"09:14.300 ","End":"09:19.920","Text":"to a plus b divided by K multiplied"},{"Start":"09:19.920 ","End":"09:25.980","Text":"by a multiplied by"},{"Start":"09:25.980 ","End":"09:31.660","Text":"b divided by K multiplied by b plus a."},{"Start":"09:32.440 ","End":"09:38.700","Text":"This is the capacitance of the system and it\u0027s the answer to question Number 2."},{"Start":"09:38.700 ","End":"09:41.110","Text":"Now let\u0027s answer Question number 3."},{"Start":"09:41.110 ","End":"09:45.145","Text":"If the spheres were to be joined via a very long wire,"},{"Start":"09:45.145 ","End":"09:49.915","Text":"so like so of resistance R,"},{"Start":"09:49.915 ","End":"09:56.230","Text":"what would be the RC time constant for the discharge of the system."},{"Start":"09:56.230 ","End":"10:00.355","Text":"I\u0027m reminding you of the RC time constant we saw is"},{"Start":"10:00.355 ","End":"10:06.350","Text":"equal to Tau and it\u0027s just equal to the resistance multiplied by the capacitance."},{"Start":"10:06.680 ","End":"10:10.105","Text":"Of course, this question is super easy to solve."},{"Start":"10:10.105 ","End":"10:12.060","Text":"Let\u0027s just do it right over here."},{"Start":"10:12.060 ","End":"10:13.965","Text":"We\u0027re just calculating this."},{"Start":"10:13.965 ","End":"10:18.400","Text":"What we can see is that we have a capacitor and we have a resistor."},{"Start":"10:18.400 ","End":"10:23.080","Text":"It\u0027s just like a regular discharge RC circuit where we have"},{"Start":"10:23.080 ","End":"10:28.880","Text":"a capacitor and a resistor and then our capacitor just discharges."},{"Start":"10:28.880 ","End":"10:32.335","Text":"Usually we have to calculate"},{"Start":"10:32.335 ","End":"10:38.620","Text":"some whole thing for all of the equations and the circuit or not specifically here,"},{"Start":"10:38.620 ","End":"10:42.340","Text":"if we have the capacitance and the resistance of the resistor."},{"Start":"10:42.340 ","End":"10:46.060","Text":"Here we also, we have the capacitance we just calculated over here."},{"Start":"10:46.060 ","End":"10:49.575","Text":"We\u0027re being told the resistance in the question."},{"Start":"10:49.575 ","End":"10:53.315","Text":"We just multiply the resistance by our capacitance over here,"},{"Start":"10:53.315 ","End":"10:58.520","Text":"so R multiplied by ab divided by"},{"Start":"10:58.520 ","End":"11:05.915","Text":"K multiplied by a plus b."},{"Start":"11:05.915 ","End":"11:09.980","Text":"That\u0027s the answer to question Number 3 for the RC time"},{"Start":"11:09.980 ","End":"11:15.450","Text":"constant of this system and that is the end of this lesson."}],"ID":22284},{"Watched":false,"Name":"Exercise 10","Duration":"10m 13s","ChapterTopicVideoID":21509,"CourseChapterTopicPlaylistID":99472,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.920","Text":"Hello. In this lesson,"},{"Start":"00:01.920 ","End":"00:05.550","Text":"we\u0027re going to learn how to solve and also actually"},{"Start":"00:05.550 ","End":"00:09.600","Text":"solve a question involving an infinite ladder circuit,"},{"Start":"00:09.600 ","End":"00:12.870","Text":"where in this circuit, it has capacitors."},{"Start":"00:12.870 ","End":"00:18.780","Text":"We\u0027ve already seen infinite ladder circuit with resistors and so on."},{"Start":"00:18.780 ","End":"00:22.230","Text":"This time, we\u0027re dealing with capacitors."},{"Start":"00:22.230 ","End":"00:23.730","Text":"We\u0027re being asked to calculate"},{"Start":"00:23.730 ","End":"00:26.955","Text":"the total capacitance or the following infinite ladder circuit,"},{"Start":"00:26.955 ","End":"00:33.120","Text":"and assume that the capacitance C_1 and C_2 of each capacitor is given."},{"Start":"00:33.120 ","End":"00:37.095","Text":"As we\u0027ve seen before when we were dealing with resistors,"},{"Start":"00:37.095 ","End":"00:42.555","Text":"is that what do we want to do is we want to take all of this over here,"},{"Start":"00:42.555 ","End":"00:48.930","Text":"so this whole circuit and replace it by 1 capacitor with a capacitance of C total,"},{"Start":"00:48.930 ","End":"00:51.150","Text":"the total capacitance of everything."},{"Start":"00:51.150 ","End":"00:53.790","Text":"Now, the trick over here when dealing with"},{"Start":"00:53.790 ","End":"00:58.785","Text":"infinite ladder circuits is to remember that we\u0027re dealing with infinite."},{"Start":"00:58.785 ","End":"01:00.690","Text":"If we\u0027re dealing with infinity,"},{"Start":"01:00.690 ","End":"01:03.810","Text":"infinity minus 1 is still infinity."},{"Start":"01:03.810 ","End":"01:05.370","Text":"That\u0027s the whole premise."},{"Start":"01:05.370 ","End":"01:09.000","Text":"That means that if we take out"},{"Start":"01:09.000 ","End":"01:13.275","Text":"1 of the units over here or 1 of the rungs in this ladder,"},{"Start":"01:13.275 ","End":"01:19.785","Text":"the capacitance of all the other rungs is still going to be equal to C total."},{"Start":"01:19.785 ","End":"01:23.295","Text":"If we ignore these first 2 capacitors,"},{"Start":"01:23.295 ","End":"01:26.400","Text":"and we just take all of this,"},{"Start":"01:26.400 ","End":"01:28.995","Text":"so it\u0027s still taking infinity."},{"Start":"01:28.995 ","End":"01:33.150","Text":"The capacitance of all of these capacitors together also has to be"},{"Start":"01:33.150 ","End":"01:38.440","Text":"equal to C total because we\u0027re dealing with infinity over here."},{"Start":"01:39.770 ","End":"01:44.565","Text":"Just like in the previous infinite ladder circuits,"},{"Start":"01:44.565 ","End":"01:48.030","Text":"what we\u0027re going to do is we\u0027re going to replace everything here in"},{"Start":"01:48.030 ","End":"01:53.880","Text":"blue with a capacitance of C total."},{"Start":"01:53.880 ","End":"02:00.240","Text":"Then we\u0027re going to add the C total unto these 2 capacitors,"},{"Start":"02:00.240 ","End":"02:04.815","Text":"which were located in the first rung of this infinite ladder circuit."},{"Start":"02:04.815 ","End":"02:06.525","Text":"In other words,"},{"Start":"02:06.525 ","End":"02:08.850","Text":"let\u0027s just draw out our new circuit."},{"Start":"02:08.850 ","End":"02:10.770","Text":"What we\u0027re going to have is over here,"},{"Start":"02:10.770 ","End":"02:13.395","Text":"we have our voltage source,"},{"Start":"02:13.395 ","End":"02:16.985","Text":"so it goes something like so."},{"Start":"02:16.985 ","End":"02:22.365","Text":"Then what we have is here,"},{"Start":"02:22.365 ","End":"02:26.440","Text":"we have our capacitor C_1,"},{"Start":"02:26.440 ","End":"02:33.400","Text":"here, we have our capacitor C_2."},{"Start":"02:35.630 ","End":"02:38.790","Text":"Then joined onto this,"},{"Start":"02:38.790 ","End":"02:41.770","Text":"we have here in blue,"},{"Start":"02:41.960 ","End":"02:46.545","Text":"which we said has a capacitance of C total."},{"Start":"02:46.545 ","End":"02:51.370","Text":"This is our new circuit, 3 capacitors."},{"Start":"02:51.890 ","End":"02:55.080","Text":"Then of course, from my first diagram,"},{"Start":"02:55.080 ","End":"03:03.760","Text":"we remember that the total capacitance of all of this is also equal to C total."},{"Start":"03:04.760 ","End":"03:08.745","Text":"Now, we want to find the total capacitance of this circuit,"},{"Start":"03:08.745 ","End":"03:12.180","Text":"which we know is going to be equal to C total."},{"Start":"03:12.180 ","End":"03:18.975","Text":"The first thing we can see is that as the voltage travels around the circuit,"},{"Start":"03:18.975 ","End":"03:20.670","Text":"we can see over here,"},{"Start":"03:20.670 ","End":"03:24.735","Text":"we have a node where the voltage splits."},{"Start":"03:24.735 ","End":"03:30.210","Text":"But what we can see is that capacitor C_2 and this capacitor over here,"},{"Start":"03:30.210 ","End":"03:33.000","Text":"C total, are connected in series,"},{"Start":"03:33.000 ","End":"03:35.235","Text":"they\u0027re connected by the same wire."},{"Start":"03:35.235 ","End":"03:39.880","Text":"C_2 and C total are in series."},{"Start":"03:41.270 ","End":"03:44.040","Text":"If they\u0027re in series,"},{"Start":"03:44.040 ","End":"03:51.315","Text":"then the equation to add up the capacitors is 1 divided by C tilde."},{"Start":"03:51.315 ","End":"03:59.790","Text":"Let\u0027s call the capacitance of C_2 and C_T as C tilde,"},{"Start":"03:59.790 ","End":"04:03.240","Text":"and this is equal to 1 divided the capacitance,"},{"Start":"04:03.240 ","End":"04:09.840","Text":"C_2 plus 1 divided by the capacitance of this one, so C total."},{"Start":"04:09.840 ","End":"04:12.930","Text":"Then what we get is that C tilde,"},{"Start":"04:12.930 ","End":"04:15.105","Text":"the capacitance of C_2 and C_T,"},{"Start":"04:15.105 ","End":"04:16.545","Text":"which are in series,"},{"Start":"04:16.545 ","End":"04:25.590","Text":"is simply equal to C multiplied by C total divided by C plus C total."},{"Start":"04:25.590 ","End":"04:28.170","Text":"Just rearrange this equation."},{"Start":"04:28.170 ","End":"04:31.440","Text":"Now, what we can see is that C tilde,"},{"Start":"04:31.440 ","End":"04:37.000","Text":"all of this over here is connected in parallel to C_1."},{"Start":"04:37.790 ","End":"04:42.975","Text":"We can see that they\u0027re in parallel because we have a node over here which splits,"},{"Start":"04:42.975 ","End":"04:47.160","Text":"and we can see that we\u0027re going to have the same voltage across"},{"Start":"04:47.160 ","End":"04:52.320","Text":"this branch over here as we will across this branch over here,"},{"Start":"04:52.320 ","End":"04:54.855","Text":"which means that they are in parallel."},{"Start":"04:54.855 ","End":"05:02.920","Text":"Therefore, we can say that the total capacitance of C_1 and C tilde,"},{"Start":"05:03.280 ","End":"05:09.390","Text":"let\u0027s call that C-star is simply equal"},{"Start":"05:09.390 ","End":"05:16.020","Text":"to the capacitance C_1 plus the capacitance C tilde."},{"Start":"05:16.020 ","End":"05:20.140","Text":"That\u0027s how you add capacitors in parallel."},{"Start":"05:20.960 ","End":"05:26.145","Text":"This is going to equal to C_1 plus C tilde,"},{"Start":"05:26.145 ","End":"05:27.810","Text":"which is C,"},{"Start":"05:27.810 ","End":"05:32.895","Text":"C total divided by C plus C total,"},{"Start":"05:32.895 ","End":"05:35.830","Text":"sorry, and of course, this is C_2."},{"Start":"05:38.300 ","End":"05:44.205","Text":"Of course, this C-star incorporates C_1, C_2,"},{"Start":"05:44.205 ","End":"05:53.205","Text":"and C total, which means that C-star is equal to C total."},{"Start":"05:53.205 ","End":"05:59.265","Text":"Then we can say therefore that this is equal to C_1 plus C_2,"},{"Start":"05:59.265 ","End":"06:04.060","Text":"C_T divided by C_2 plus C_T."},{"Start":"06:05.750 ","End":"06:09.810","Text":"Now, what we\u0027re going to do is we\u0027re going to multiply"},{"Start":"06:09.810 ","End":"06:14.055","Text":"both sides by this denominator over here, C_2 plus C_T."},{"Start":"06:14.055 ","End":"06:18.765","Text":"What we\u0027re going to have is"},{"Start":"06:18.765 ","End":"06:27.975","Text":"C total multiplied by C_2 plus C total squared,"},{"Start":"06:27.975 ","End":"06:33.495","Text":"which is equal to C_1 C_2 plus C_1"},{"Start":"06:33.495 ","End":"06:40.840","Text":"C total plus C_2, C total."},{"Start":"06:40.880 ","End":"06:45.000","Text":"Then what we\u0027re going to do is we can see that we"},{"Start":"06:45.000 ","End":"06:50.220","Text":"have over here C_2 C total and here C_2 C total,"},{"Start":"06:50.220 ","End":"06:54.525","Text":"so we\u0027ll subtract from both sides C_2 C total."},{"Start":"06:54.525 ","End":"06:59.865","Text":"Then we\u0027ll move these 2 over here to the other side of the equation."},{"Start":"06:59.865 ","End":"07:09.270","Text":"What we have is C total squared minus C_1 C_2 minus C_1,"},{"Start":"07:09.270 ","End":"07:14.685","Text":"C_T, and all of this is equal to 0."},{"Start":"07:14.685 ","End":"07:17.445","Text":"I\u0027m just going to switch these 2 around."},{"Start":"07:17.445 ","End":"07:23.070","Text":"Now, what we see is we have a quadratic equation where our variable is C total."},{"Start":"07:23.070 ","End":"07:29.850","Text":"It\u0027s as if we have x^2 minus C_1x minus some constant."},{"Start":"07:29.850 ","End":"07:35.580","Text":"Write out our equation for our quadratic equation solver, so C total."},{"Start":"07:35.580 ","End":"07:39.450","Text":"Then, of course, we always have 2 answers to these types of questions,"},{"Start":"07:39.450 ","End":"07:41.550","Text":"and this is equal to,"},{"Start":"07:41.550 ","End":"07:46.560","Text":"so we have C_1 plus or minus the square root of"},{"Start":"07:46.560 ","End":"07:52.050","Text":"C_1^2 plus 4C_2 C_1,"},{"Start":"07:52.050 ","End":"07:56.565","Text":"or 4C_1 C_2 rather,"},{"Start":"07:56.565 ","End":"08:01.380","Text":"4 times this multiplied by the coefficient of this,"},{"Start":"08:01.380 ","End":"08:03.180","Text":"which is 1, of course."},{"Start":"08:03.180 ","End":"08:07.180","Text":"Then this is divided by 2."},{"Start":"08:07.460 ","End":"08:12.345","Text":"I\u0027m reminding you that this is our quadratic formula."},{"Start":"08:12.345 ","End":"08:13.890","Text":"We have negative b,"},{"Start":"08:13.890 ","End":"08:15.885","Text":"which was negative C_1,"},{"Start":"08:15.885 ","End":"08:20.309","Text":"so it just becomes C_1 plus minus the square root of b squared"},{"Start":"08:20.309 ","End":"08:25.050","Text":"minus 4 times the coefficient of a,"},{"Start":"08:25.050 ","End":"08:28.200","Text":"which is 1 multiplied by our constant C,"},{"Start":"08:28.200 ","End":"08:31.545","Text":"which is just negative C_1 C_2,"},{"Start":"08:31.545 ","End":"08:33.105","Text":"so we get a plus,"},{"Start":"08:33.105 ","End":"08:35.025","Text":"and all of this is divided by 2a,"},{"Start":"08:35.025 ","End":"08:39.040","Text":"where a over here is equal to 1."},{"Start":"08:40.190 ","End":"08:49.170","Text":"The first thing that we need to remember is that C total has to be bigger than 0."},{"Start":"08:49.170 ","End":"08:52.755","Text":"We can\u0027t have a negative capacitance."},{"Start":"08:52.755 ","End":"08:57.705","Text":"If we see here we have C_1 and we have an option for subtracting the square root"},{"Start":"08:57.705 ","End":"09:02.925","Text":"of C_1 squared plus 4 times C_1 C_2."},{"Start":"09:02.925 ","End":"09:07.830","Text":"What we can see is what is inside of this square root sign is going to be bigger"},{"Start":"09:07.830 ","End":"09:13.350","Text":"than C_1 because we\u0027re taking the square root of C_1^2."},{"Start":"09:13.350 ","End":"09:16.020","Text":"If we have C_1 minus the square root of C_1^2,"},{"Start":"09:16.020 ","End":"09:18.210","Text":"we just get 0."},{"Start":"09:18.210 ","End":"09:24.260","Text":"But what we have is actually C_1^2 plus all of this,"},{"Start":"09:24.260 ","End":"09:28.730","Text":"which of course are positive because C_1 and C_2 are positive values."},{"Start":"09:28.730 ","End":"09:33.170","Text":"As we said, our capacitance is always going to be greater or equal than 0."},{"Start":"09:33.170 ","End":"09:37.375","Text":"What we can see is that C_1 minus all of this,"},{"Start":"09:37.375 ","End":"09:39.870","Text":"all of this is greater than C_1,"},{"Start":"09:39.870 ","End":"09:44.500","Text":"so C_1 minus all of this will be less than 0,"},{"Start":"09:44.500 ","End":"09:46.880","Text":"which we cannot have."},{"Start":"09:46.880 ","End":"09:54.110","Text":"Therefore, our C total is simply going to be equal to C_1 plus the square root,"},{"Start":"09:54.110 ","End":"09:57.209","Text":"C_1 squared plus 4C_1,"},{"Start":"09:57.209 ","End":"10:03.450","Text":"C_2, and all of this is divided by 2."},{"Start":"10:03.610 ","End":"10:10.085","Text":"This is the total capacitance of this infinite ladder circuit."},{"Start":"10:10.085 ","End":"10:13.260","Text":"That is the end of this lesson."}],"ID":22290},{"Watched":false,"Name":"Exercise 10 continue","Duration":"32m 27s","ChapterTopicVideoID":21510,"CourseChapterTopicPlaylistID":99472,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.860","Text":"Hello, in this lesson,"},{"Start":"00:01.860 ","End":"00:04.655","Text":"we\u0027re going to be answering the following question."},{"Start":"00:04.655 ","End":"00:06.554","Text":"In the following system,"},{"Start":"00:06.554 ","End":"00:10.848","Text":"all of the capacitors are of equal capacitance."},{"Start":"00:10.848 ","End":"00:17.325","Text":"Question number 1 is to calculate the total capacitance of the system."},{"Start":"00:17.325 ","End":"00:20.339","Text":"What we can see here with this dot,"},{"Start":"00:20.339 ","End":"00:21.399","Text":"dot, dots over here,"},{"Start":"00:21.399 ","End":"00:27.000","Text":"it shows us that we have an infinite ladder system of capacitors,"},{"Start":"00:27.000 ","End":"00:30.945","Text":"and we\u0027re trying to find the total capacitance of this infinite ladder,"},{"Start":"00:30.945 ","End":"00:35.790","Text":"we\u0027ve seen that the trick goes something like so."},{"Start":"00:35.790 ","End":"00:40.020","Text":"We say that the capacitance of everything,"},{"Start":"00:40.020 ","End":"00:43.725","Text":"of all of the capacitors is equal to C total,"},{"Start":"00:43.725 ","End":"00:50.285","Text":"the total resistance of all of the capacitors in the system connected like set 1."},{"Start":"00:50.285 ","End":"00:57.105","Text":"Then the trick is to say that we have these rungs,"},{"Start":"00:57.105 ","End":"00:59.240","Text":"so specifically, in each question,"},{"Start":"00:59.240 ","End":"01:01.580","Text":"you have to see what the pattern is,"},{"Start":"01:01.580 ","End":"01:02.825","Text":"and where the rungs are."},{"Start":"01:02.825 ","End":"01:06.409","Text":"In this system, we can see that we have 2 capacitors like so,"},{"Start":"01:06.409 ","End":"01:07.940","Text":"one on top of the other,"},{"Start":"01:07.940 ","End":"01:12.150","Text":"and then we have a 3rd capacitor over here."},{"Start":"01:13.010 ","End":"01:16.710","Text":"This pattern repeats 2 capacitors,"},{"Start":"01:16.710 ","End":"01:17.894","Text":"one on top of the other,"},{"Start":"01:17.894 ","End":"01:19.365","Text":"and 3rd over here,"},{"Start":"01:19.365 ","End":"01:22.155","Text":"two on top of each other, and the 3rd over here."},{"Start":"01:22.155 ","End":"01:27.423","Text":"What we do is we take away the first rung in the series,"},{"Start":"01:27.423 ","End":"01:30.529","Text":"so that means that we take all of this,"},{"Start":"01:30.529 ","End":"01:34.244","Text":"and all the other capacitors."},{"Start":"01:34.244 ","End":"01:37.370","Text":"Then we say that the total capacitance of"},{"Start":"01:37.370 ","End":"01:43.234","Text":"this entire section is also equal to C total, so why is that?"},{"Start":"01:43.234 ","End":"01:46.550","Text":"If we have infinity or infinity minus 1,"},{"Start":"01:46.550 ","End":"01:48.334","Text":"it\u0027s also equal to infinity,"},{"Start":"01:48.334 ","End":"01:49.925","Text":"there is no difference."},{"Start":"01:49.925 ","End":"01:56.460","Text":"That means that the capacitance of all of these infinite rungs is equal to C total."},{"Start":"01:56.460 ","End":"01:58.115","Text":"If we take out one of the rungs,"},{"Start":"01:58.115 ","End":"02:00.994","Text":"it doesn\u0027t make a difference because there are so many rungs"},{"Start":"02:00.994 ","End":"02:05.490","Text":"that the capacitance of all of this is equal to C total."},{"Start":"02:05.810 ","End":"02:10.390","Text":"We\u0027ve seen this trick in the past few lessons and also in"},{"Start":"02:10.390 ","End":"02:14.529","Text":"dealing with the infinite ladder circuit with the resistors."},{"Start":"02:14.529 ","End":"02:19.285","Text":"What we do now, is we redraw our circuits with"},{"Start":"02:19.285 ","End":"02:25.960","Text":"our first rung connected but instead of drawing all of these infinite capacitors,"},{"Start":"02:25.960 ","End":"02:30.715","Text":"we just add in a capacitor called C total."},{"Start":"02:30.715 ","End":"02:39.370","Text":"What we can see is that C total is connected in parallel to this capacitor over here."},{"Start":"02:39.370 ","End":"02:43.230","Text":"Let\u0027s draw it. First,"},{"Start":"02:43.230 ","End":"02:48.610","Text":"we\u0027ll draw these 2 capacitors that are one on top of the other."},{"Start":"02:48.610 ","End":"02:52.130","Text":"Here we have this capacitor,"},{"Start":"02:52.130 ","End":"02:58.829","Text":"and here we have this bottom capacitor and then what we have is that these"},{"Start":"02:58.829 ","End":"03:06.518","Text":"2 join on to the 3rd capacitor over here in the rung, so that\u0027s this."},{"Start":"03:06.518 ","End":"03:11.209","Text":"Then we can see that this carries on like so."},{"Start":"03:11.209 ","End":"03:15.985","Text":"Then it connects to all of this in the green loop."},{"Start":"03:15.985 ","End":"03:19.910","Text":"The inner loop, which we\u0027ve called C total."},{"Start":"03:19.910 ","End":"03:28.102","Text":"We connect this like so onto this capacitor C total."},{"Start":"03:28.102 ","End":"03:32.539","Text":"This is now what our circuit looks like."},{"Start":"03:32.539 ","End":"03:35.449","Text":"Now of course, as we know, all of this,"},{"Start":"03:35.449 ","End":"03:36.995","Text":"this has our system,"},{"Start":"03:36.995 ","End":"03:40.235","Text":"all of this is also equal to C total."},{"Start":"03:40.235 ","End":"03:44.120","Text":"Okay, that\u0027s what we drew in this blue loop, the outer loop."},{"Start":"03:44.120 ","End":"03:46.580","Text":"I\u0027ll just redraw it over here,"},{"Start":"03:46.580 ","End":"03:51.685","Text":"so all of this is also equal to C total."},{"Start":"03:51.685 ","End":"03:53.580","Text":"Now what we do is,"},{"Start":"03:53.580 ","End":"03:56.210","Text":"add up all of the capacitors according to"},{"Start":"03:56.210 ","End":"04:00.009","Text":"the rules for adding capacitors in series and in parallel,"},{"Start":"04:00.009 ","End":"04:04.400","Text":"and then we say that all of that is equal to C total."},{"Start":"04:04.400 ","End":"04:14.322","Text":"The first thing that we\u0027re going to do is we\u0027re going to add up these 2 capacitors."},{"Start":"04:14.322 ","End":"04:18.165","Text":"This, we are going to cause C Tilde."},{"Start":"04:18.165 ","End":"04:24.484","Text":"We can see that these 2 capacitors are connected in parallel to one another."},{"Start":"04:24.484 ","End":"04:27.589","Text":"How do we know that they\u0027re connected in parallel?"},{"Start":"04:27.589 ","End":"04:30.140","Text":"Both of their top plates are connected by"},{"Start":"04:30.140 ","End":"04:34.055","Text":"the same wire and both of their bottom plates are connected by the same wire."},{"Start":"04:34.055 ","End":"04:41.045","Text":"Which means that the potential difference across each capacitor is going to be the same,"},{"Start":"04:41.045 ","End":"04:46.650","Text":"which is by definition connected in parallel."},{"Start":"04:46.650 ","End":"04:50.764","Text":"When we add capacitors in parallel,"},{"Start":"04:50.764 ","End":"04:54.190","Text":"so C Tilde is just equal to the capacitance of this one,"},{"Start":"04:54.190 ","End":"04:59.880","Text":"so C plus the capacitance of this one, C total."},{"Start":"04:59.880 ","End":"05:02.700","Text":"Now we have this circuit,"},{"Start":"05:02.700 ","End":"05:08.029","Text":"so these two capacitors and then our C Tilde over here,"},{"Start":"05:08.029 ","End":"05:11.854","Text":"which is made up of these two capacitors in parallel."},{"Start":"05:11.854 ","End":"05:15.560","Text":"Now we\u0027re adding up all of these capacitors and of course,"},{"Start":"05:15.560 ","End":"05:18.259","Text":"they\u0027re connected in series."},{"Start":"05:18.259 ","End":"05:20.270","Text":"We know that the total,"},{"Start":"05:20.270 ","End":"05:22.069","Text":"once we add up all of these capacitors,"},{"Start":"05:22.069 ","End":"05:24.985","Text":"then we\u0027ve included all of the capacitors and the system"},{"Start":"05:24.985 ","End":"05:29.240","Text":"and so that means all the capacitors inside this blue loop,"},{"Start":"05:29.240 ","End":"05:32.469","Text":"and we said that that is equal to C total."},{"Start":"05:32.469 ","End":"05:37.549","Text":"Therefore, we know that when we add capacitors in series,"},{"Start":"05:37.549 ","End":"05:41.240","Text":"so we have 1 divided by the total capacitance of everything,"},{"Start":"05:41.240 ","End":"05:43.610","Text":"which here we said is C total,"},{"Start":"05:43.610 ","End":"05:49.545","Text":"so that is equal to 1 divided by the capacitance of this capacitor,"},{"Start":"05:49.545 ","End":"05:55.005","Text":"plus 1 divided by the capacitance of this capacitor,"},{"Start":"05:55.005 ","End":"06:01.910","Text":"plus 1 divided by the capacitance of this bottom capacitor."},{"Start":"06:01.910 ","End":"06:06.795","Text":"What we\u0027re left with is 2 divided by C, okay?"},{"Start":"06:06.795 ","End":"06:10.224","Text":"Plus 1 divided by C Tilde."},{"Start":"06:10.224 ","End":"06:15.050","Text":"We already said that when we have something like so,"},{"Start":"06:15.050 ","End":"06:22.324","Text":"then what we can do very easily is have at the top."},{"Start":"06:22.324 ","End":"06:28.040","Text":"We have the multiplication of both of the capacitances,"},{"Start":"06:28.040 ","End":"06:32.764","Text":"divided by the addition of the capacitances."},{"Start":"06:32.764 ","End":"06:36.419","Text":"What I\u0027m going to do is I\u0027m going to take away the 2 from"},{"Start":"06:36.419 ","End":"06:41.795","Text":"the numerator and I\u0027m going to put the 2 down here in the denominator."},{"Start":"06:41.795 ","End":"06:44.414","Text":"It\u0027s the same thing."},{"Start":"06:44.414 ","End":"06:49.160","Text":"When the 2 is in the denominator of the denominator it moves up to the numerator,"},{"Start":"06:49.160 ","End":"06:51.475","Text":"so this means the same thing."},{"Start":"06:51.475 ","End":"06:59.660","Text":"Now, what I can say is that the total capacitance is equal to,"},{"Start":"06:59.660 ","End":"07:05.570","Text":"so now we have C divided by 2 multiplied by C Tilde."},{"Start":"07:05.570 ","End":"07:07.385","Text":"What is C Tilde?"},{"Start":"07:07.385 ","End":"07:12.000","Text":"C Tilde is C plus C total"},{"Start":"07:12.000 ","End":"07:18.090","Text":"divided by C divided by 2 plus C Tilde."},{"Start":"07:18.090 ","End":"07:19.635","Text":"What\u0027s the Tilde?"},{"Start":"07:19.635 ","End":"07:23.289","Text":"C plus C total."},{"Start":"07:23.460 ","End":"07:27.969","Text":"Now, I\u0027m going to open up the brackets and I\u0027m"},{"Start":"07:27.969 ","End":"07:32.049","Text":"going to move these 2 over here down to the denominator,"},{"Start":"07:32.049 ","End":"07:39.655","Text":"so that I\u0027ll be left with is C^2 plus CC total divided by 2,"},{"Start":"07:39.655 ","End":"07:41.920","Text":"so this is the 2 from here,"},{"Start":"07:41.920 ","End":"07:51.640","Text":"multiplied by C divided by 2 plus C is 3C divided by 2 plus C total."},{"Start":"07:51.640 ","End":"07:56.395","Text":"Now, we\u0027ll also open up brackets over here in the denominator,"},{"Start":"07:56.395 ","End":"08:02.710","Text":"so I\u0027ll be left with C^2 plus C"},{"Start":"08:02.710 ","End":"08:09.987","Text":"divided by 3C plus 2C_T."},{"Start":"08:09.987 ","End":"08:16.825","Text":"Now, we\u0027ll multiply both sides by the denominator over here."},{"Start":"08:16.825 ","End":"08:21.580","Text":"What we\u0027ll have is 3C multiplied by C_T"},{"Start":"08:21.580 ","End":"08:29.620","Text":"plus 2C_T multiplied by C_T,"},{"Start":"08:29.620 ","End":"08:36.830","Text":"so 2C_T^2, which is equal to C^2 plus CC_T."},{"Start":"08:36.830 ","End":"08:42.475","Text":"Now we\u0027ll subtract from both sides CC_T."},{"Start":"08:42.475 ","End":"08:45.740","Text":"Here we\u0027ll have a 2."},{"Start":"08:45.810 ","End":"08:49.479","Text":"Now let\u0027s move everything to one side of the equation,"},{"Start":"08:49.479 ","End":"08:54.760","Text":"and we\u0027ll see that we\u0027re left with a quadratic equation where our variable is C_T."},{"Start":"08:54.760 ","End":"09:00.399","Text":"We have 2C_T squared plus"},{"Start":"09:00.399 ","End":"09:09.100","Text":"2CC_T minus C^2 is equal to 0."},{"Start":"09:09.100 ","End":"09:15.415","Text":"Now we use a quadratic formula to solve what C_T is equal to."},{"Start":"09:15.415 ","End":"09:21.655","Text":"Our quadratic formula is this over here."},{"Start":"09:21.655 ","End":"09:23.590","Text":"We have negative b,"},{"Start":"09:23.590 ","End":"09:25.765","Text":"so b is 2C."},{"Start":"09:25.765 ","End":"09:32.049","Text":"Negative 2C plus minus the square root of b^2,"},{"Start":"09:32.049 ","End":"09:37.734","Text":"so 2C^2 is 4C^2 minus"},{"Start":"09:37.734 ","End":"09:44.350","Text":"4 multiplied by a. a is 2 multiplied by C,"},{"Start":"09:44.350 ","End":"09:47.575","Text":"which is negative C^2."},{"Start":"09:47.575 ","End":"09:50.530","Text":"The negative and the negative here become a positive."},{"Start":"09:50.530 ","End":"09:52.525","Text":"Then we have C^2."},{"Start":"09:52.525 ","End":"09:56.755","Text":"Then we have divided by 2 times a,"},{"Start":"09:56.755 ","End":"10:02.030","Text":"where a is 2 so we have here 4."},{"Start":"10:02.130 ","End":"10:05.890","Text":"Of course here we have 4 times 2."},{"Start":"10:05.890 ","End":"10:10.640","Text":"I didn\u0027t put the multiplication sign, so we have 8."},{"Start":"10:10.710 ","End":"10:15.640","Text":"We have negative 2 divided by 4,"},{"Start":"10:15.640 ","End":"10:20.560","Text":"so negative C over 2 plus minus"},{"Start":"10:20.560 ","End":"10:26.710","Text":"the square root of 4C^2 plus 8C^2 is 12C^2."},{"Start":"10:26.710 ","End":"10:29.440","Text":"We have the square root of 12,"},{"Start":"10:29.440 ","End":"10:31.329","Text":"and then we have C^2,"},{"Start":"10:31.329 ","End":"10:33.580","Text":"so the square root of C^2 is C,"},{"Start":"10:33.580 ","End":"10:37.640","Text":"and all of this is divided by 4."},{"Start":"10:38.460 ","End":"10:41.289","Text":"We can see that we have two answers,"},{"Start":"10:41.289 ","End":"10:44.275","Text":"one for the plus and one for the minus."},{"Start":"10:44.275 ","End":"10:47.859","Text":"We know that our capacitance has to be a positive value."},{"Start":"10:47.859 ","End":"10:50.230","Text":"It has to be greater than 0."},{"Start":"10:50.230 ","End":"10:52.510","Text":"I changed it back to this answer."},{"Start":"10:52.510 ","End":"10:55.614","Text":"We can see that here we have negative 2C,"},{"Start":"10:55.614 ","End":"10:59.529","Text":"and then if we subtract something else,"},{"Start":"10:59.529 ","End":"11:01.810","Text":"it\u0027s going to be even more in the negative,"},{"Start":"11:01.810 ","End":"11:05.050","Text":"which means that we can have the answer with the minus."},{"Start":"11:05.050 ","End":"11:08.409","Text":"Also, if here we had positive 2C and"},{"Start":"11:08.409 ","End":"11:12.010","Text":"we subtracted the square root of 12 multiplied by C,"},{"Start":"11:12.010 ","End":"11:18.680","Text":"we\u0027d still have a value for capacitance which is less than 0."},{"Start":"11:19.590 ","End":"11:22.210","Text":"That\u0027s just a note but anyway,"},{"Start":"11:22.210 ","End":"11:25.855","Text":"we can take the negative answer."},{"Start":"11:25.855 ","End":"11:31.180","Text":"What we have is that our capacitance C total is equal to"},{"Start":"11:31.180 ","End":"11:39.610","Text":"negative 2C plus the square root of 12 multiplied by C and divided by 4."},{"Start":"11:39.610 ","End":"11:43.630","Text":"Then we can cancel out the 4 over here."},{"Start":"11:43.630 ","End":"11:51.310","Text":"Here we can divide by 2 and then what we\u0027ll have is over here,"},{"Start":"11:51.310 ","End":"11:54.505","Text":"plus the square root of 12C."},{"Start":"11:54.505 ","End":"11:57.790","Text":"If here we also want to have it divided by 2,"},{"Start":"11:57.790 ","End":"12:04.135","Text":"what we\u0027ll have to do is have the square root of 12 divided by"},{"Start":"12:04.135 ","End":"12:14.574","Text":"2 multiplied by C. Then we can take C out as a common factor or a common multiplier."},{"Start":"12:14.574 ","End":"12:22.405","Text":"Then what we have is C divided by 2 multiplied by,"},{"Start":"12:22.405 ","End":"12:26.050","Text":"first of all, we have a negative 1 over here."},{"Start":"12:26.050 ","End":"12:30.100","Text":"Then the square root of 12 divided by 2 is the"},{"Start":"12:30.100 ","End":"12:34.225","Text":"same as the square root of 12 divided by 4."},{"Start":"12:34.225 ","End":"12:37.765","Text":"If we put in the divided by 2 into the square root sign,"},{"Start":"12:37.765 ","End":"12:39.879","Text":"12 divided by 4 is 3,"},{"Start":"12:39.879 ","End":"12:44.150","Text":"so then we\u0027re left with the square root of 3."},{"Start":"12:46.080 ","End":"12:49.435","Text":"This is the answer to Question number 1,"},{"Start":"12:49.435 ","End":"12:54.140","Text":"to calculate the total capacitance of the system."},{"Start":"12:55.190 ","End":"13:01.110","Text":"Question number 2 is to calculate the charge on each capacitor."},{"Start":"13:01.110 ","End":"13:06.460","Text":"If we are told that the system is connected to a voltage source V_1."},{"Start":"13:07.760 ","End":"13:16.420","Text":"Let\u0027s connect our system to voltage source V_1 over here."},{"Start":"13:17.850 ","End":"13:22.959","Text":"The trick in these types of questions is we want to show that"},{"Start":"13:22.959 ","End":"13:29.660","Text":"our system acts or exhibits the properties of a geometric progression."},{"Start":"13:29.730 ","End":"13:35.980","Text":"Let\u0027s call the charge on each upper capacitor Q_n,"},{"Start":"13:35.980 ","End":"13:40.045","Text":"where n is going from 1 until infinity."},{"Start":"13:40.045 ","End":"13:45.745","Text":"Here Q_1, here we have a charge of Q_2,"},{"Start":"13:45.745 ","End":"13:50.125","Text":"here we have a charge of Q_3, and so on."},{"Start":"13:50.125 ","End":"13:52.510","Text":"So just to remind you,"},{"Start":"13:52.510 ","End":"13:55.070","Text":"a geometric progression,"},{"Start":"13:55.290 ","End":"13:57.655","Text":"it looks like this;"},{"Start":"13:57.655 ","End":"14:04.810","Text":"a_n is equal to a_n minus 1 multiplied by q."},{"Start":"14:04.810 ","End":"14:08.065","Text":"Then if I know q and I know a,"},{"Start":"14:08.065 ","End":"14:13.640","Text":"then I can find the charge on every single one of the capacitors."},{"Start":"14:13.830 ","End":"14:22.780","Text":"All I need to know is q and the first component of this progression a_1."},{"Start":"14:22.780 ","End":"14:24.970","Text":"In the previous question,"},{"Start":"14:24.970 ","End":"14:30.219","Text":"we saw how we can take away the first rung in this infinite ladder,"},{"Start":"14:30.219 ","End":"14:40.615","Text":"and take all of this over here and say that this is equal to the capacitance of C_T."},{"Start":"14:40.615 ","End":"14:44.890","Text":"That\u0027s how we got this diagram over here."},{"Start":"14:44.890 ","End":"14:52.330","Text":"Then we also said because of the beauty of infinite circuits,"},{"Start":"14:52.330 ","End":"14:55.390","Text":"so this, the total over here,"},{"Start":"14:55.390 ","End":"14:59.035","Text":"the total capacitance is also equal to C_T,"},{"Start":"14:59.035 ","End":"15:06.709","Text":"where of course right now we\u0027re looking at this as if it\u0027s connected to a voltage source."},{"Start":"15:07.410 ","End":"15:12.175","Text":"It\u0027s the capacitance, not including the voltage source."},{"Start":"15:12.175 ","End":"15:14.319","Text":"Then in that case,"},{"Start":"15:14.319 ","End":"15:17.740","Text":"we can say that the total charge,"},{"Start":"15:17.740 ","End":"15:21.340","Text":"q_T is equal to the total capacitance,"},{"Start":"15:21.340 ","End":"15:29.630","Text":"which is C_T multiplied by our voltage source, which is V_1."},{"Start":"15:30.510 ","End":"15:39.684","Text":"The next thing that we saw is we saw that these 2 capacitors are connected in parallel,"},{"Start":"15:39.684 ","End":"15:45.890","Text":"and we call that capacitance C tilde, if you remember."},{"Start":"15:46.290 ","End":"15:48.400","Text":"Then what we had is"},{"Start":"15:48.400 ","End":"15:57.925","Text":"this capacitor C tilde and this capacitor C over here connected in series."},{"Start":"15:57.925 ","End":"16:01.599","Text":"What happens when capacitors are connected in series?"},{"Start":"16:01.599 ","End":"16:06.625","Text":"That means that the charge on each capacitor is equal."},{"Start":"16:06.625 ","End":"16:10.869","Text":"That means that q_T,"},{"Start":"16:10.869 ","End":"16:14.350","Text":"the total charge on this capacitor C_T"},{"Start":"16:14.350 ","End":"16:19.690","Text":"is equal to the charge on this capacitor over here,"},{"Start":"16:19.690 ","End":"16:22.525","Text":"let\u0027s say that it has a charge q_1,"},{"Start":"16:22.525 ","End":"16:27.020","Text":"it\u0027s equal to the charge on this capacitor C tilde over here."},{"Start":"16:27.020 ","End":"16:35.237","Text":"Let\u0027s call it q_2 or q tilde even."},{"Start":"16:35.237 ","End":"16:39.145","Text":"It\u0027s equal to the charge on this capacitor,"},{"Start":"16:39.145 ","End":"16:41.690","Text":"let\u0027s call it q_3."},{"Start":"16:42.210 ","End":"16:45.339","Text":"When capacitors are connected in series,"},{"Start":"16:45.339 ","End":"16:48.679","Text":"the charge on each capacitor is the same."},{"Start":"16:50.400 ","End":"16:57.670","Text":"Q_1 we already saw is equal to q total."},{"Start":"16:57.670 ","End":"17:03.550","Text":"Let\u0027s write out that V_1 is equal to Q_1,"},{"Start":"17:03.550 ","End":"17:06.234","Text":"this capacity over here,"},{"Start":"17:06.234 ","End":"17:09.100","Text":"divided by c total,"},{"Start":"17:09.100 ","End":"17:11.815","Text":"where of course c total, we know what this is."},{"Start":"17:11.815 ","End":"17:17.349","Text":"Now let\u0027s look at what q_2 is equal to, because of course,"},{"Start":"17:17.349 ","End":"17:21.790","Text":"Q_2 is integrated in this capacitor,"},{"Start":"17:21.790 ","End":"17:25.750","Text":"C tilde which is made up of 2 capacitors."},{"Start":"17:25.750 ","End":"17:30.549","Text":"What I\u0027m going to do is I\u0027m going to attach at this point over"},{"Start":"17:30.549 ","End":"17:35.274","Text":"here and this point over here, a volt meter."},{"Start":"17:35.274 ","End":"17:40.780","Text":"I\u0027m going to say that the voltage that I\u0027m measuring over here is V_2."},{"Start":"17:40.780 ","End":"17:43.209","Text":"You can\u0027t really see I wrote it too small,"},{"Start":"17:43.209 ","End":"17:44.665","Text":"but this is V_2."},{"Start":"17:44.665 ","End":"17:48.220","Text":"This is of course equal to the voltage on the second rung,"},{"Start":"17:48.220 ","End":"17:49.765","Text":"so this rung over here."},{"Start":"17:49.765 ","End":"17:54.590","Text":"It\u0027s also equal to the voltage across this capacitor."},{"Start":"17:55.350 ","End":"17:59.860","Text":"Now what I do is I do the exact same trick that I did in"},{"Start":"17:59.860 ","End":"18:03.865","Text":"order to get my Q_1. What does that mean?"},{"Start":"18:03.865 ","End":"18:08.840","Text":"Now I\u0027m going to take all of this over here,"},{"Start":"18:09.300 ","End":"18:14.245","Text":"and this over here is of course all equal to C total because it\u0027s"},{"Start":"18:14.245 ","End":"18:18.445","Text":"an infinite ladder series minus the 2 first rungs,"},{"Start":"18:18.445 ","End":"18:21.220","Text":"it\u0027s still an infinite ladder series."},{"Start":"18:21.220 ","End":"18:24.205","Text":"Then I have the same system that I have here,"},{"Start":"18:24.205 ","End":"18:26.320","Text":"except instead of V_1,"},{"Start":"18:26.320 ","End":"18:29.350","Text":"I have over here V_2."},{"Start":"18:29.350 ","End":"18:37.450","Text":"Then also, instead of here having q_1, I have q_2."},{"Start":"18:37.450 ","End":"18:39.805","Text":"Then here I have q_4,"},{"Start":"18:39.805 ","End":"18:43.550","Text":"and here I have the same C tilde."},{"Start":"18:44.520 ","End":"18:47.485","Text":"Just like with my Q_1,"},{"Start":"18:47.485 ","End":"18:52.149","Text":"I can now say that V_2 is equal to"},{"Start":"18:52.149 ","End":"19:00.940","Text":"Q_2 over here divided by C total for the exact same reason that I did before,"},{"Start":"19:00.940 ","End":"19:05.539","Text":"because the capacitors are connected in series."},{"Start":"19:05.790 ","End":"19:11.440","Text":"This of course, will go on to infinity so therefore,"},{"Start":"19:11.440 ","End":"19:19.550","Text":"what I can write is that V_n is equal to Q_n divided by C total."},{"Start":"19:20.760 ","End":"19:27.040","Text":"What we\u0027ve seen, because the capacitors are in series q_1,"},{"Start":"19:27.040 ","End":"19:29.394","Text":"or the top capacitor over here,"},{"Start":"19:29.394 ","End":"19:33.730","Text":"will have the same charge as the capacitor directly under it."},{"Start":"19:33.730 ","End":"19:39.699","Text":"This capacitor Q_2 will have the same charge as this capacitor under,"},{"Start":"19:39.699 ","End":"19:44.185","Text":"and the charge in capacitor c_3 over here,"},{"Start":"19:44.185 ","End":"19:47.394","Text":"where Q_3 is going to be the same as down here."},{"Start":"19:47.394 ","End":"19:53.140","Text":"But the problem is calculating the charge on these capacitors in the middle,"},{"Start":"19:53.140 ","End":"19:56.140","Text":"but that we\u0027ll do later."},{"Start":"19:56.140 ","End":"20:02.155","Text":"Our first port of call is to see that our charge on each capacitor is changing."},{"Start":"20:02.155 ","End":"20:05.154","Text":"Because our voltage across"},{"Start":"20:05.154 ","End":"20:09.114","Text":"each rung is going to be changing because as the voltage travels,"},{"Start":"20:09.114 ","End":"20:12.715","Text":"there\u0027s voltage drops across the capacitors."},{"Start":"20:12.715 ","End":"20:16.075","Text":"V_1 is not equal to V_2."},{"Start":"20:16.075 ","End":"20:20.980","Text":"We know what our V_1 and c total is equal to."},{"Start":"20:20.980 ","End":"20:24.355","Text":"We know what our Q_1 is equal to but here,"},{"Start":"20:24.355 ","End":"20:27.085","Text":"we don\u0027t know what our Q_2 is equal to."},{"Start":"20:27.085 ","End":"20:33.084","Text":"We don\u0027t know what our V_2 is equal to so if we can calculate what our V_2 is equal to,"},{"Start":"20:33.084 ","End":"20:36.865","Text":"then we can know what our Q_2 is equal to."},{"Start":"20:36.865 ","End":"20:40.160","Text":"Let\u0027s try and do that now."},{"Start":"20:40.830 ","End":"20:45.760","Text":"As we said, if this charge over here is Q_1,"},{"Start":"20:45.760 ","End":"20:49.345","Text":"then under here we also have a charge Q_1."},{"Start":"20:49.345 ","End":"20:52.150","Text":"What am I going to do is we\u0027re going to look at"},{"Start":"20:52.150 ","End":"20:57.094","Text":"this 1 circuit where we go across this capacitor over here,"},{"Start":"20:57.094 ","End":"20:58.950","Text":"down this voltage source,"},{"Start":"20:58.950 ","End":"21:01.619","Text":"this wire over here, not voltage source,"},{"Start":"21:01.619 ","End":"21:08.914","Text":"volt meter over here or a voltage source and across this capacitor and back."},{"Start":"21:08.914 ","End":"21:12.880","Text":"All of that we know has to be equal to V_1."},{"Start":"21:12.880 ","End":"21:15.025","Text":"We have V_1,"},{"Start":"21:15.025 ","End":"21:20.050","Text":"so we go up V_1 and then we cross this capacitor so we have"},{"Start":"21:20.050 ","End":"21:27.535","Text":"a voltage drop of Q_1 divided by C. That\u0027s the voltage over here."},{"Start":"21:27.535 ","End":"21:33.580","Text":"Then we go down over here so we add on V_2,"},{"Start":"21:33.580 ","End":"21:36.310","Text":"and then we cross this capacitor where we just saw it has"},{"Start":"21:36.310 ","End":"21:40.075","Text":"the same voltage as this capacitor over here."},{"Start":"21:40.075 ","End":"21:44.080","Text":"Then we have another plus Q_1 divided by"},{"Start":"21:44.080 ","End":"21:49.585","Text":"C. Then we come back to over here where we cross,"},{"Start":"21:49.585 ","End":"21:52.970","Text":"so all of this is equal to V_1."},{"Start":"21:55.250 ","End":"21:58.265","Text":"This is of course, just catch off?"},{"Start":"21:58.265 ","End":"22:01.660","Text":"If you rearrange it, then you can say it\u0027s equal to 0."},{"Start":"22:01.660 ","End":"22:04.810","Text":"What we can do is now we can isolate V_2."},{"Start":"22:04.810 ","End":"22:08.785","Text":"V-2 is simply equal to V_1 minus"},{"Start":"22:08.785 ","End":"22:18.039","Text":"2Q_1 divided by C. Now I know what V_2 is,"},{"Start":"22:18.039 ","End":"22:20.455","Text":"so I can also know what Q_2 is."},{"Start":"22:20.455 ","End":"22:26.170","Text":"Now I want to keep doing this over and over for each of the rungs."},{"Start":"22:26.170 ","End":"22:32.830","Text":"Here, if this is Q_2 then the charge over here is also Q_2 and if this is Q_3,"},{"Start":"22:32.830 ","End":"22:35.425","Text":"then the charge over here is also Q_3."},{"Start":"22:35.425 ","End":"22:39.745","Text":"What I can say, it\u0027s the same relationship that I have between V_1 and V_2,"},{"Start":"22:39.745 ","End":"22:44.665","Text":"I\u0027m going to have between V_2 and V_3."},{"Start":"22:44.665 ","End":"22:50.665","Text":"If I put some volt meter over here V_3,"},{"Start":"22:50.665 ","End":"22:52.465","Text":"joined it over here."},{"Start":"22:52.465 ","End":"22:55.390","Text":"I\u0027m going to have the same relationship."},{"Start":"22:55.390 ","End":"23:05.125","Text":"What I can say is that V_2 is equal to Q_2 this time divided by C plus V_3."},{"Start":"23:05.125 ","End":"23:07.315","Text":"Because now I\u0027m going through here."},{"Start":"23:07.315 ","End":"23:12.910","Text":"Then plus Q_2 divided by C and therefore,"},{"Start":"23:12.910 ","End":"23:17.844","Text":"we\u0027ll get that V_3 is equal to V_2 minus"},{"Start":"23:17.844 ","End":"23:24.144","Text":"2Q_2 divided by C. In other words,"},{"Start":"23:24.144 ","End":"23:32.020","Text":"I can say that the relationship between V_n and the next term so let\u0027s say V_2 and V_3."},{"Start":"23:32.020 ","End":"23:43.360","Text":"V_n is equal to V_n plus 1 plus 2 times Q_n"},{"Start":"23:43.360 ","End":"23:49.900","Text":"divided by C. We already"},{"Start":"23:49.900 ","End":"23:57.790","Text":"saw that V_n over here is equal to Q_n divided by C total."},{"Start":"23:57.790 ","End":"24:00.475","Text":"I can plug that in over here."},{"Start":"24:00.475 ","End":"24:09.775","Text":"I have Q_n divided by C total which is equal to V_n plus 1,"},{"Start":"24:09.775 ","End":"24:16.600","Text":"which is simply going to be Q_n plus 1 divided by C total."},{"Start":"24:16.600 ","End":"24:18.430","Text":"Just like we saw,"},{"Start":"24:18.430 ","End":"24:24.025","Text":"and then we add 2Q_n divided by"},{"Start":"24:24.025 ","End":"24:31.344","Text":"C. The next thing that we\u0027re going to do,"},{"Start":"24:31.344 ","End":"24:34.735","Text":"let\u0027s move over a little bit to the side."},{"Start":"24:34.735 ","End":"24:38.230","Text":"We\u0027re going to try and get our Q_ns to 1 side so we"},{"Start":"24:38.230 ","End":"24:42.250","Text":"can get a common multiple so what we\u0027ll have"},{"Start":"24:42.250 ","End":"24:50.080","Text":"is Q_n divided by C total minus 2Q_n divided by C,"},{"Start":"24:50.080 ","End":"24:56.740","Text":"is equal to Q_n plus 1 divided by C total."},{"Start":"24:56.740 ","End":"25:01.150","Text":"Then we can say that we have Q_n multiplied by 1"},{"Start":"25:01.150 ","End":"25:06.280","Text":"divided by C total minus 2 divided by C,"},{"Start":"25:06.280 ","End":"25:13.970","Text":"which is equal to Q_n plus 1 divided by C total."},{"Start":"25:13.970 ","End":"25:18.115","Text":"Now we\u0027re going to divide both sides by n"},{"Start":"25:18.115 ","End":"25:22.974","Text":"because I want my n plus 1 together with my n terms."},{"Start":"25:22.974 ","End":"25:24.625","Text":"I want my Q\u0027s together,"},{"Start":"25:24.625 ","End":"25:27.625","Text":"and multiply both sides by C_T."},{"Start":"25:27.625 ","End":"25:33.865","Text":"What I\u0027ll have is a C_T multiplied by 1 divided by C_T,"},{"Start":"25:33.865 ","End":"25:36.175","Text":"minus 2 divided by C,"},{"Start":"25:36.175 ","End":"25:42.800","Text":"which is equal to Q_n plus 1, divided by Q_n."},{"Start":"25:43.080 ","End":"25:45.865","Text":"Now I\u0027ll open up the brackets,"},{"Start":"25:45.865 ","End":"25:49.630","Text":"so I have C_T divided by C_T which is 1,"},{"Start":"25:49.630 ","End":"25:53.274","Text":"minus 2 times C_T,"},{"Start":"25:53.274 ","End":"25:54.760","Text":"divided by C,"},{"Start":"25:54.760 ","End":"26:00.220","Text":"which is equal to Q_n plus 1, divided by Q_n."},{"Start":"26:00.220 ","End":"26:04.719","Text":"Now we can plug in our C_T,"},{"Start":"26:04.719 ","End":"26:10.480","Text":"so we have 1 minus 2 times our C_T,"},{"Start":"26:10.480 ","End":"26:18.549","Text":"so we have 2 multiplied by C divided by 2 multiplied by root 3 minus 1."},{"Start":"26:18.549 ","End":"26:20.125","Text":"Then of course,"},{"Start":"26:20.125 ","End":"26:23.874","Text":"all of this is divided by C."},{"Start":"26:23.874 ","End":"26:29.350","Text":"So then we can see this 2 cancels out with this 2 and this C cancels out with this C,"},{"Start":"26:29.350 ","End":"26:34.197","Text":"and this is equal to Q_n plus 1 divided by Q_n."},{"Start":"26:34.197 ","End":"26:38.665","Text":"What we have is 1 minus minus 1,"},{"Start":"26:38.665 ","End":"26:45.220","Text":"so this will cancel out because then it becomes,1 minus minus 1,"},{"Start":"26:45.220 ","End":"26:46.569","Text":"so 1 plus 1."},{"Start":"26:46.569 ","End":"26:51.205","Text":"So what we have is 2 minus root 3,"},{"Start":"26:51.205 ","End":"26:57.290","Text":"is equal to Q_n plus 1 divided by Q_n."},{"Start":"26:58.020 ","End":"27:01.640","Text":"These ones don\u0027t cancel out."},{"Start":"27:02.490 ","End":"27:09.115","Text":"What we saw before, we can have a_n is equal to a_n minus 1,"},{"Start":"27:09.115 ","End":"27:16.015","Text":"or we could also just write this as a_n plus 1 is just equal to a_n,"},{"Start":"27:16.015 ","End":"27:18.160","Text":"multiplied by q,"},{"Start":"27:18.160 ","End":"27:23.605","Text":"so now what we can have is a_n plus 1 and then divide both sides by a_n,"},{"Start":"27:23.605 ","End":"27:26.275","Text":"and then we\u0027re left with q."},{"Start":"27:26.275 ","End":"27:30.550","Text":"Here we have a_n plus 1 divided by a_n."},{"Start":"27:30.550 ","End":"27:33.295","Text":"This is equal to q."},{"Start":"27:33.295 ","End":"27:37.760","Text":"The multiplier in the geometric progression."},{"Start":"27:37.830 ","End":"27:41.125","Text":"I just wrote what this is equal to."},{"Start":"27:41.125 ","End":"27:43.870","Text":"We got that from this equation over here,"},{"Start":"27:43.870 ","End":"27:45.699","Text":"so now what I want to do is,"},{"Start":"27:45.699 ","End":"27:48.084","Text":"I want to write this equation in this form,"},{"Start":"27:48.084 ","End":"27:51.423","Text":"where I have that Q_n is equal to."},{"Start":"27:51.423 ","End":"27:54.279","Text":"So the charge on the nth capacitor,"},{"Start":"27:54.279 ","End":"27:56.154","Text":"on the top or on the bottom,"},{"Start":"27:56.154 ","End":"27:57.895","Text":"is equal to,"},{"Start":"27:57.895 ","End":"28:01.390","Text":"the charge on the first capacitor which I know,"},{"Start":"28:01.390 ","End":"28:03.730","Text":"from this equation My V_1 is given,"},{"Start":"28:03.730 ","End":"28:08.500","Text":"my C_T I\u0027ve already calculated so I can know what my Q_1 is equal to,"},{"Start":"28:08.500 ","End":"28:12.979","Text":"multiplied by my geometric progression multiplier"},{"Start":"28:13.050 ","End":"28:19.570","Text":"to the power of n. Let\u0027s just plug this in,"},{"Start":"28:19.570 ","End":"28:23.755","Text":"so I have that Q_n is equal to Q_1."},{"Start":"28:23.755 ","End":"28:25.240","Text":"So what is Q_1?"},{"Start":"28:25.240 ","End":"28:30.294","Text":"It\u0027s equal to V_1 multiplied by C_T,"},{"Start":"28:30.294 ","End":"28:37.930","Text":"where C_T is C divided by 2 root 3 minus 1,"},{"Start":"28:37.930 ","End":"28:42.790","Text":"and then multiplied by Q^n,"},{"Start":"28:42.790 ","End":"28:47.515","Text":"where Q is 2 minus root 3,"},{"Start":"28:47.515 ","End":"28:52.270","Text":"and all of this is to the power of n. This is going to be"},{"Start":"28:52.270 ","End":"28:58.405","Text":"the charge on each one of these capacitors and also on the bottom capacitors,"},{"Start":"28:58.405 ","End":"29:00.519","Text":"and now all that\u0027s left to do is calculate"},{"Start":"29:00.519 ","End":"29:04.610","Text":"the charge on these capacitors over here in the middle."},{"Start":"29:04.740 ","End":"29:08.229","Text":"Earlier when we were looking at this,"},{"Start":"29:08.229 ","End":"29:13.870","Text":"we said that the volt meter that we will put over here,"},{"Start":"29:13.870 ","End":"29:18.310","Text":"will be measuring the voltage across this rung,"},{"Start":"29:18.310 ","End":"29:26.420","Text":"so these capacitors, but it\u0027s also measuring the voltage across this capacitor over here."},{"Start":"29:26.640 ","End":"29:30.280","Text":"Let\u0027s call this capacitor."},{"Start":"29:30.280 ","End":"29:34.720","Text":"Let\u0027s say that it has charge Q_1 tag."},{"Start":"29:34.720 ","End":"29:39.430","Text":"Let\u0027s say that this capacitor has charge Q_2 tag."},{"Start":"29:39.430 ","End":"29:41.950","Text":"Then what we can say,"},{"Start":"29:41.950 ","End":"29:48.055","Text":"is that the charge Q_n tag is equal to,"},{"Start":"29:48.055 ","End":"29:52.885","Text":"so we have the capacitance of the capacitor itself which were given,"},{"Start":"29:52.885 ","End":"29:55.320","Text":"multiplied by the voltage."},{"Start":"29:55.320 ","End":"29:57.854","Text":"What is the voltage across this capacitor?"},{"Start":"29:57.854 ","End":"30:03.794","Text":"It\u0027s the same as the voltage across the next rung in the series,"},{"Start":"30:03.794 ","End":"30:08.685","Text":"multiplied by the voltage of n plus 1."},{"Start":"30:08.685 ","End":"30:14.925","Text":"Let\u0027s say Q_1 tag is C multiplied by V_2,"},{"Start":"30:14.925 ","End":"30:17.920","Text":"which is V_n plus 1."},{"Start":"30:18.380 ","End":"30:22.185","Text":"What is V(n+1)?"},{"Start":"30:22.185 ","End":"30:25.745","Text":"We saw it over here."},{"Start":"30:25.745 ","End":"30:27.909","Text":"It\u0027s simply equal to,"},{"Start":"30:27.909 ","End":"30:30.535","Text":"so we have C, multiplied by,"},{"Start":"30:30.535 ","End":"30:35.379","Text":"V(n+1) is simply Q(n+1),"},{"Start":"30:35.379 ","End":"30:36.939","Text":"if we just substitute instead of n,"},{"Start":"30:36.939 ","End":"30:41.470","Text":"n plus 1, divided by C_T."},{"Start":"30:41.470 ","End":"30:44.859","Text":"Then we have C,"},{"Start":"30:44.859 ","End":"30:54.130","Text":"and then we have multiplied by 2 divided by C,"},{"Start":"30:54.130 ","End":"30:58.540","Text":"multiplied by root 3 minus 1,"},{"Start":"30:58.540 ","End":"31:03.040","Text":"multiplied by Q_n, plus 1."},{"Start":"31:03.040 ","End":"31:05.830","Text":"This is C divided by C_T,"},{"Start":"31:05.830 ","End":"31:08.184","Text":"multiplied by Q_n plus 1."},{"Start":"31:08.184 ","End":"31:11.215","Text":"So our C\u0027s cancel out,"},{"Start":"31:11.215 ","End":"31:18.685","Text":"and then what we\u0027re left with is 2Q_n plus 1,"},{"Start":"31:18.685 ","End":"31:24.470","Text":"divided by root 3 minus 1."},{"Start":"31:26.550 ","End":"31:31.675","Text":"Then instead of Q(n+1),"},{"Start":"31:31.675 ","End":"31:35.575","Text":"I can substitute n over here,"},{"Start":"31:35.575 ","End":"31:38.710","Text":"what I got over here and so therefore,"},{"Start":"31:38.710 ","End":"31:43.465","Text":"what I\u0027ll get is that Q(n) tag,"},{"Start":"31:43.465 ","End":"31:45.805","Text":"so that\u0027s all the capacitors in the middle,"},{"Start":"31:45.805 ","End":"31:52.960","Text":"is simply going to be equal to C multiplied by our initial voltage source V_1,"},{"Start":"31:52.960 ","End":"32:01.300","Text":"multiplied by 2 minus root 3 to the power of n plus 1."},{"Start":"32:01.300 ","End":"32:07.190","Text":"I just played around with my Q_n plus 1 over here."},{"Start":"32:09.390 ","End":"32:12.580","Text":"This is the answer to question number 2,"},{"Start":"32:12.580 ","End":"32:17.485","Text":"this is the charge on the top and bottoms capacitors,"},{"Start":"32:17.485 ","End":"32:21.175","Text":"and this is the charge on the metal capacitors,"},{"Start":"32:21.175 ","End":"32:28.010","Text":"and this is the total capacitance of our system and that is the end of the lesson."}],"ID":22291}],"Thumbnail":null,"ID":99472}]

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