Electrostatic Energy
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[{"Name":"Electrostatic Energy","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Recap E-Field And Potential Of Spherical Shell","Duration":"9m 32s","ChapterTopicVideoID":21298,"CourseChapterTopicPlaylistID":99478,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/21298.jpeg","UploadDate":"2020-04-06T14:04:58.1270000","DurationForVideoObject":"PT9M32S","Description":null,"MetaTitle":"Recap E-Field And Potential Of Spherical Shell: Video + Workbook | Proprep","MetaDescription":"Electrostatic Energy - Electrostatic Energy. 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/electrostatic-energy/electrostatic-energy/vid21378","VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.010","Text":"Hello. In this lesson,"},{"Start":"00:02.010 ","End":"00:04.410","Text":"we\u0027re going to see how to find the electric field and"},{"Start":"00:04.410 ","End":"00:07.725","Text":"potential of a charged spherical shell."},{"Start":"00:07.725 ","End":"00:09.360","Text":"Over here, we have"},{"Start":"00:09.360 ","End":"00:15.690","Text":"uniform charge density per unit area because we are speaking about a spherical shell,"},{"Start":"00:15.690 ","End":"00:18.420","Text":"and that means that in the center of the sphere,"},{"Start":"00:18.420 ","End":"00:20.220","Text":"we have 0 charge."},{"Start":"00:20.220 ","End":"00:25.080","Text":"Our charge is only on the surface area of the sphere."},{"Start":"00:25.080 ","End":"00:29.375","Text":"We can say that our Sigma is equal to"},{"Start":"00:29.375 ","End":"00:33.845","Text":"the total charge divided by the surface area of the sphere,"},{"Start":"00:33.845 ","End":"00:38.060","Text":"which is equal to 4 Pi R squared."},{"Start":"00:38.060 ","End":"00:39.845","Text":"Now in this question,"},{"Start":"00:39.845 ","End":"00:43.250","Text":"we\u0027re being asked to find the electric field."},{"Start":"00:43.250 ","End":"00:45.215","Text":"We can do this in two ways."},{"Start":"00:45.215 ","End":"00:49.390","Text":"We can either use Coulomb\u0027s law or we can use Gauss\u0027s law."},{"Start":"00:49.390 ","End":"00:53.180","Text":"Gauss we can only use if we have some sphere,"},{"Start":"00:53.180 ","End":"00:55.670","Text":"be it a full sphere or a spherical shell,"},{"Start":"00:55.670 ","End":"00:58.820","Text":"and of course, if we have a wire or a cylinder,"},{"Start":"00:58.820 ","End":"01:01.855","Text":"or a cylindrical shell, or an infinite plane."},{"Start":"01:01.855 ","End":"01:03.845","Text":"Here, because we have a spherical shell,"},{"Start":"01:03.845 ","End":"01:05.935","Text":"we can use Gauss\u0027 law."},{"Start":"01:05.935 ","End":"01:08.405","Text":"When we\u0027re dealing with Gauss\u0027 law,"},{"Start":"01:08.405 ","End":"01:13.070","Text":"we have to split up our spherical shell into two areas."},{"Start":"01:13.070 ","End":"01:17.930","Text":"That\u0027s when our radius for our Gaussian surface is"},{"Start":"01:17.930 ","End":"01:24.240","Text":"located outside or is larger than the radius of our spherical shell."},{"Start":"01:24.240 ","End":"01:28.160","Text":"That means that our Gaussian surface is located outside."},{"Start":"01:28.160 ","End":"01:34.700","Text":"The second region is when our Gaussian surface is within our spherical shell."},{"Start":"01:34.700 ","End":"01:41.610","Text":"That means our r is smaller than our R located within the spherical shell."},{"Start":"01:41.610 ","End":"01:48.050","Text":"Let\u0027s first speak about when our r is bigger than R. That"},{"Start":"01:48.050 ","End":"01:54.540","Text":"means that we\u0027re putting a Gaussian surface which is also some spherical shell."},{"Start":"01:54.540 ","End":"01:59.915","Text":"It\u0027s a 3D shape with surface area but without volume."},{"Start":"01:59.915 ","End":"02:04.565","Text":"That means because we\u0027re using Gauss\u0027 law,"},{"Start":"02:04.565 ","End":"02:10.160","Text":"that our electric field lines are going to be perpendicular to our surface"},{"Start":"02:10.160 ","End":"02:16.910","Text":"and the electric field is always going to be in the radial direction."},{"Start":"02:16.910 ","End":"02:22.460","Text":"Now, of course, we have a uniform electric field across the surface area of our shape,"},{"Start":"02:22.460 ","End":"02:25.735","Text":"and that\u0027s because we have uniform charge distribution,"},{"Start":"02:25.735 ","End":"02:27.021","Text":"and of course,"},{"Start":"02:27.021 ","End":"02:29.465","Text":"our spherical shell is symmetrical,"},{"Start":"02:29.465 ","End":"02:32.960","Text":"which means that we can use Gauss\u0027 law."},{"Start":"02:32.960 ","End":"02:40.715","Text":"We know that Gauss\u0027 law is equal to the closed loop integral of Eds."},{"Start":"02:40.715 ","End":"02:42.140","Text":"Here we said because"},{"Start":"02:42.140 ","End":"02:47.120","Text":"the electric field is in the radial direction and it\u0027s uniform throughout,"},{"Start":"02:47.120 ","End":"02:53.510","Text":"we can say that our electric field dot S is equal to this over here,"},{"Start":"02:53.510 ","End":"03:00.115","Text":"where S is the surface area of our Gaussian surface of radius"},{"Start":"03:00.115 ","End":"03:08.690","Text":"r. Then we can say that this is equal to E.4Pi r squared,"},{"Start":"03:08.690 ","End":"03:11.405","Text":"which is the surface area of our Gaussian surface."},{"Start":"03:11.405 ","End":"03:16.335","Text":"This is equal to Q_in divided by Epsilon_0."},{"Start":"03:16.335 ","End":"03:18.285","Text":"This is Gauss\u0027 law,"},{"Start":"03:18.285 ","End":"03:21.790","Text":"and then we can see that our Q_in is simply the Q."},{"Start":"03:21.790 ","End":"03:26.830","Text":"We\u0027re given the charge of our spherical shell divided by Epsilon naught."},{"Start":"03:26.830 ","End":"03:33.440","Text":"Now, what we can do is we can just isolate out our E. We\u0027ll get that Q divided"},{"Start":"03:33.440 ","End":"03:40.133","Text":"by 4Pi Epsilon_0 r squared is our electric field,"},{"Start":"03:40.133 ","End":"03:41.663","Text":"and of course, this is a vector,"},{"Start":"03:41.663 ","End":"03:43.220","Text":"so we can add in our direction."},{"Start":"03:43.220 ","End":"03:46.130","Text":"We know that our electric field is in the radial direction."},{"Start":"03:46.130 ","End":"03:48.125","Text":"Then as we know,"},{"Start":"03:48.125 ","End":"03:54.555","Text":"k is equal to 1 over 4Pi Epsilon_0."},{"Start":"03:54.555 ","End":"04:02.265","Text":"We can say that this is also equal to kQ divided by r squared in the radial direction."},{"Start":"04:02.265 ","End":"04:06.860","Text":"This is our electric field outside of our spherical shell."},{"Start":"04:06.860 ","End":"04:15.410","Text":"Now let\u0027s look at the region where our r is smaller than our R. That means that we have"},{"Start":"04:15.410 ","End":"04:25.145","Text":"a Gaussian surface within our spherical shell and a radius of r. Now again,"},{"Start":"04:25.145 ","End":"04:30.360","Text":"we can say that our Gauss\u0027 law is equal to the closed loop integral of Eds."},{"Start":"04:30.360 ","End":"04:32.330","Text":"But let\u0027s just cut to the chase."},{"Start":"04:32.330 ","End":"04:36.750","Text":"We already know it\u0027s a uniform electric field in the radial direction."},{"Start":"04:36.750 ","End":"04:42.935","Text":"We have E.S, which is equal to E multiplied by the surface area of this."},{"Start":"04:42.935 ","End":"04:47.960","Text":"That\u0027s again going to be equal to 4Pir squared,"},{"Start":"04:47.960 ","End":"04:52.290","Text":"and that\u0027s equal to Q_in divided by Epsilon_0."},{"Start":"04:52.290 ","End":"04:56.510","Text":"However, over here, our Q_in is equal to 0 because"},{"Start":"04:56.510 ","End":"05:02.520","Text":"our Q is distributed only on the surface of the spherical shell."},{"Start":"05:02.520 ","End":"05:04.290","Text":"Right now we\u0027re inside."},{"Start":"05:04.290 ","End":"05:06.615","Text":"Our Q_in is equal to 0."},{"Start":"05:06.615 ","End":"05:09.115","Text":"This is equal to 0, and therefore,"},{"Start":"05:09.115 ","End":"05:16.960","Text":"we get that our E field is equal to 0 when we\u0027re located inside the spherical shell."},{"Start":"05:16.960 ","End":"05:19.420","Text":"We can see that our electric field is equal to"},{"Start":"05:19.420 ","End":"05:22.500","Text":"0 in the region within the spherical shell,"},{"Start":"05:22.500 ","End":"05:25.180","Text":"and is equal to kQ divided by r squared in"},{"Start":"05:25.180 ","End":"05:30.475","Text":"the radial direction in the region outside of the spherical shell."},{"Start":"05:30.475 ","End":"05:37.555","Text":"Now what we\u0027re going to do is we\u0027re going to find what our potential is equal to."},{"Start":"05:37.555 ","End":"05:42.010","Text":"The equation for finding the potential is equal to the"},{"Start":"05:42.010 ","End":"05:46.585","Text":"negative integral along the electric field,"},{"Start":"05:46.585 ","End":"05:49.985","Text":"dr. Now, in this specific example,"},{"Start":"05:49.985 ","End":"05:52.550","Text":"we can see that our electric field is in"},{"Start":"05:52.550 ","End":"05:57.500","Text":"the radial direction and our dr is also in the radial direction,"},{"Start":"05:57.500 ","End":"06:00.185","Text":"which means that we can simply write this integral"},{"Start":"06:00.185 ","End":"06:05.470","Text":"as the electric field in the radial direction, dr."},{"Start":"06:05.470 ","End":"06:09.680","Text":"If we do this indefinite integral,"},{"Start":"06:09.680 ","End":"06:14.090","Text":"we\u0027ll get that our potential is given by,"},{"Start":"06:14.090 ","End":"06:18.452","Text":"in the region where r is smaller than R,"},{"Start":"06:18.452 ","End":"06:20.630","Text":"we\u0027ll get some constant C_1."},{"Start":"06:20.630 ","End":"06:24.536","Text":"In the region where r is bigger than R,"},{"Start":"06:24.536 ","End":"06:32.290","Text":"we\u0027ll get that the potential is equal to kQ divided by r plus some constant."},{"Start":"06:32.290 ","End":"06:36.150","Text":"Now let\u0027s find our constants."},{"Start":"06:36.150 ","End":"06:40.910","Text":"I can say that my potential at infinity,"},{"Start":"06:40.910 ","End":"06:44.195","Text":"so far away from my spherical shell,"},{"Start":"06:44.195 ","End":"06:46.765","Text":"my potential is meant to be equal to 0."},{"Start":"06:46.765 ","End":"06:50.930","Text":"That\u0027s of course equal to kQ divided by r,"},{"Start":"06:50.930 ","End":"06:52.460","Text":"where r is my position."},{"Start":"06:52.460 ","End":"06:56.890","Text":"My position is infinity plus C_2."},{"Start":"06:56.890 ","End":"07:01.355","Text":"Any number divided by infinity is going to tend toward 0."},{"Start":"07:01.355 ","End":"07:04.790","Text":"Then I\u0027m going to have that 0 is equal to C_2."},{"Start":"07:04.790 ","End":"07:07.708","Text":"Therefore, our C_2 is equal to 0,"},{"Start":"07:07.708 ","End":"07:10.175","Text":"and I can cross off that."},{"Start":"07:10.175 ","End":"07:15.860","Text":"Now I want to find out what my constant C_1 is equal to."},{"Start":"07:15.860 ","End":"07:20.030","Text":"Here I\u0027m going to use the idea of discontinuity."},{"Start":"07:20.030 ","End":"07:24.722","Text":"I\u0027m going to say that my electric potential,"},{"Start":"07:24.722 ","End":"07:26.330","Text":"when I\u0027m looking over here,"},{"Start":"07:26.330 ","End":"07:30.695","Text":"is simply going to be plugging in R to"},{"Start":"07:30.695 ","End":"07:35.970","Text":"this equation over here because R is my overlap point."},{"Start":"07:35.970 ","End":"07:39.635","Text":"My overlap point is the surface area of the sphere."},{"Start":"07:39.635 ","End":"07:41.480","Text":"Because on one end,"},{"Start":"07:41.480 ","End":"07:42.650","Text":"after the overlap point,"},{"Start":"07:42.650 ","End":"07:44.675","Text":"I\u0027m outside of the spherical shell,"},{"Start":"07:44.675 ","End":"07:49.515","Text":"and then in one point before R,"},{"Start":"07:49.515 ","End":"07:51.180","Text":"I\u0027m inside the spherical shell."},{"Start":"07:51.180 ","End":"07:54.840","Text":"That\u0027s my possible point for discontinuity."},{"Start":"07:54.840 ","End":"07:59.550","Text":"I can see that my kQ,"},{"Start":"07:59.550 ","End":"08:03.030","Text":"where I substitute in R instead of my r,"},{"Start":"08:03.030 ","End":"08:08.145","Text":"is going to be equal to C_1 so that I can substitute in."},{"Start":"08:08.145 ","End":"08:09.735","Text":"Let\u0027s worry about this C_1."},{"Start":"08:09.735 ","End":"08:16.350","Text":"This is simply going to be equal to kq divided by R. Now we can see that"},{"Start":"08:16.350 ","End":"08:21.650","Text":"the potential outside of our spherical shell is equal to kQ"},{"Start":"08:21.650 ","End":"08:27.530","Text":"divided by r. It depends on what our distance is away from the spherical shell."},{"Start":"08:27.530 ","End":"08:29.968","Text":"We can see that within the spherical shell,"},{"Start":"08:29.968 ","End":"08:31.655","Text":"our potential is a constant,"},{"Start":"08:31.655 ","End":"08:34.895","Text":"and it equals to kq divided by R,"},{"Start":"08:34.895 ","End":"08:39.115","Text":"where R is the radius of our spherical shell."},{"Start":"08:39.115 ","End":"08:48.575","Text":"If we want to draw a graph of our potential where these is the r axes."},{"Start":"08:48.575 ","End":"08:51.350","Text":"At the value of R,"},{"Start":"08:51.350 ","End":"08:54.305","Text":"we\u0027re going to have this value over here,"},{"Start":"08:54.305 ","End":"09:00.200","Text":"which is going to be equal to kQ divided by R. This is going"},{"Start":"09:00.200 ","End":"09:06.320","Text":"to be a constant straight line whilst we\u0027re still inside the spherical shell."},{"Start":"09:06.320 ","End":"09:11.960","Text":"As we reach the region where our radius is larger than R,"},{"Start":"09:11.960 ","End":"09:18.500","Text":"we\u0027re going to see that our potential diminishes as a function of 1 divided by r,"},{"Start":"09:18.500 ","End":"09:22.805","Text":"and then infinity, it\u0027s going to tend to 0."},{"Start":"09:22.805 ","End":"09:30.740","Text":"This is what our graph for the potential of a spherical shell looks like."},{"Start":"09:30.740 ","End":"09:33.240","Text":"That\u0027s the end of this lesson."}],"ID":21378},{"Watched":false,"Name":"Energy Required To Build A System","Duration":"31m 31s","ChapterTopicVideoID":21450,"CourseChapterTopicPlaylistID":99478,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.055","Text":"Hello. In this lesson,"},{"Start":"00:02.055 ","End":"00:06.240","Text":"we\u0027re going to learn about the energy required to build a system."},{"Start":"00:06.240 ","End":"00:11.144","Text":"We\u0027re going to start by talking about what we learned in one of the previous chapters,"},{"Start":"00:11.144 ","End":"00:14.519","Text":"which is the potential energy,"},{"Start":"00:14.519 ","End":"00:17.039","Text":"and that was given by q,"},{"Start":"00:17.039 ","End":"00:19.920","Text":"the charge of our particle,"},{"Start":"00:19.920 ","End":"00:24.195","Text":"multiplied by our potential."},{"Start":"00:24.195 ","End":"00:29.460","Text":"What exactly does this potential energy mean or represent?"},{"Start":"00:29.460 ","End":"00:35.595","Text":"U, our potential energy is equal to the work denoted by the letter W"},{"Start":"00:35.595 ","End":"00:42.259","Text":"required in order to bring a charge q from an area of 0 potential,"},{"Start":"00:42.259 ","End":"00:44.359","Text":"where this is equal to 0."},{"Start":"00:44.359 ","End":"00:46.100","Text":"That\u0027s usually at infinity,"},{"Start":"00:46.100 ","End":"00:51.605","Text":"so far away from its current position to its current position."},{"Start":"00:51.605 ","End":"00:53.900","Text":"If its current position is,"},{"Start":"00:53.900 ","End":"00:55.680","Text":"let\u0027s say over here,"},{"Start":"00:55.680 ","End":"01:00.889","Text":"so how much work was required in order to bring the charge from some point over here,"},{"Start":"01:00.889 ","End":"01:03.700","Text":"very far away to here?"},{"Start":"01:03.700 ","End":"01:08.840","Text":"Here we are speaking about the energy of 1 charge but what happens when we want to"},{"Start":"01:08.840 ","End":"01:14.640","Text":"speak about the energy of multiple charges or other system?"},{"Start":"01:14.780 ","End":"01:19.759","Text":"The energy required to build a system is also sometimes called"},{"Start":"01:19.759 ","End":"01:24.350","Text":"the electrostatic energy or just the energy of the system,"},{"Start":"01:24.350 ","End":"01:28.710","Text":"or the energy stored within an electronic system."},{"Start":"01:28.710 ","End":"01:34.715","Text":"There\u0027s a few words for the exact same thing but the most relevant name,"},{"Start":"01:34.715 ","End":"01:38.584","Text":"which describes more accurately what we\u0027re trying to do,"},{"Start":"01:38.584 ","End":"01:42.969","Text":"is the energy required to build a system."},{"Start":"01:42.969 ","End":"01:45.589","Text":"Now, instead of asking the question,"},{"Start":"01:45.589 ","End":"01:50.915","Text":"how much work is required to bring 1 charge from infinity until its position,"},{"Start":"01:50.915 ","End":"01:57.620","Text":"here we\u0027re asking how much energy is required to bring multiple charges from infinity"},{"Start":"01:57.620 ","End":"02:04.745","Text":"to the system and taking into account the forces acting against the incoming charge q_2,"},{"Start":"02:04.745 ","End":"02:09.240","Text":"the charge is already present in the system."},{"Start":"02:09.710 ","End":"02:12.125","Text":"Let\u0027s give an example."},{"Start":"02:12.125 ","End":"02:15.500","Text":"If we had a system with 2 charges,"},{"Start":"02:15.500 ","End":"02:19.400","Text":"so we had q_1 and q_2,"},{"Start":"02:19.400 ","End":"02:23.820","Text":"and the distance between the 2 charges was I,"},{"Start":"02:23.820 ","End":"02:28.790","Text":"so we would have to speak about the energy that is required or the work"},{"Start":"02:28.790 ","End":"02:35.455","Text":"required in order to bring q_1 from infinity to here and q_2 from infinity to here."},{"Start":"02:35.455 ","End":"02:39.139","Text":"Then I would work out from the work done,"},{"Start":"02:39.139 ","End":"02:43.610","Text":"how much energy was required in order to bring these two charges and that would"},{"Start":"02:43.610 ","End":"02:49.590","Text":"be the energy of the system or the energy required to build the system."},{"Start":"02:49.850 ","End":"02:53.269","Text":"What we can see is that first I\u0027m bringing"},{"Start":"02:53.269 ","End":"02:57.290","Text":"q_1 and then because there\u0027re still no charges over here,"},{"Start":"02:57.290 ","End":"02:58.880","Text":"there\u0027s no potential,"},{"Start":"02:58.880 ","End":"03:04.630","Text":"so I need 0 work in order to bring q_1 from infinity to its position over here."},{"Start":"03:04.630 ","End":"03:08.585","Text":"That\u0027s great but then when I bring charge q_2,"},{"Start":"03:08.585 ","End":"03:11.254","Text":"I\u0027m bringing it from infinity until here,"},{"Start":"03:11.254 ","End":"03:13.565","Text":"but q_1 is already here."},{"Start":"03:13.565 ","End":"03:21.300","Text":"So I have to take into account the forces and the potential now due to q_1\u0027s presence."},{"Start":"03:21.590 ","End":"03:26.930","Text":"We\u0027re going to have 0 potential or 0 energy required in order to"},{"Start":"03:26.930 ","End":"03:32.150","Text":"bring q_1 and then q_2 is going to require energy to bring it."},{"Start":"03:32.150 ","End":"03:38.824","Text":"So its energy in order to bring charge q_2 is going to be equal to its charge,"},{"Start":"03:38.824 ","End":"03:40.220","Text":"which is q_2,"},{"Start":"03:40.220 ","End":"03:42.820","Text":"multiplied by the potential."},{"Start":"03:42.820 ","End":"03:46.980","Text":"The potential is due to q_1 being present."},{"Start":"03:46.980 ","End":"03:49.249","Text":"If q_1 is a point charge,"},{"Start":"03:49.249 ","End":"03:51.860","Text":"so the equation for the potential of"},{"Start":"03:51.860 ","End":"03:56.372","Text":"a point charge is equal to k multiplied by its charge,"},{"Start":"03:56.372 ","End":"04:02.730","Text":"so kq_1 divided by r the distance between the 2 charges."},{"Start":"04:03.620 ","End":"04:08.464","Text":"This is also going to be the energy of"},{"Start":"04:08.464 ","End":"04:12.574","Text":"the system or the energy required to build a system,"},{"Start":"04:12.574 ","End":"04:14.494","Text":"or the electrostatic energy,"},{"Start":"04:14.494 ","End":"04:16.770","Text":"however you want to call it."},{"Start":"04:17.050 ","End":"04:20.000","Text":"Now what would happen if we had our q_1 on here,"},{"Start":"04:20.000 ","End":"04:25.444","Text":"our q_2 here and we brought another charge, q_3,"},{"Start":"04:25.444 ","End":"04:34.204","Text":"and it\u0027s also a distance r away from q_1 and it\u0027s a distance r away from q_2 as well?"},{"Start":"04:34.204 ","End":"04:38.630","Text":"So what would be the energy required to bring our q_3?"},{"Start":"04:38.630 ","End":"04:40.295","Text":"Now we can rub this out."},{"Start":"04:40.295 ","End":"04:44.245","Text":"The energy required to bring our q_3,"},{"Start":"04:44.245 ","End":"04:50.600","Text":"so u_3 is going to be equal to our charge q_3 multiplied"},{"Start":"04:50.600 ","End":"04:56.845","Text":"by the potential due to our q_1 and the potential due to our q_2."},{"Start":"04:56.845 ","End":"05:01.215","Text":"That\u0027s going to be q_3 multiplied by k,"},{"Start":"05:01.215 ","End":"05:07.909","Text":"q_1 plus q_2 divided by r. This is"},{"Start":"05:07.909 ","End":"05:14.840","Text":"because it\u0027s divided by r because q_3 is an equal distance away from both q_1 and q_2."},{"Start":"05:14.840 ","End":"05:17.705","Text":"We can just add that up and then,"},{"Start":"05:17.705 ","End":"05:22.114","Text":"therefore, once we have these three charges,"},{"Start":"05:22.114 ","End":"05:26.060","Text":"the energy of the system or the energy to build the system"},{"Start":"05:26.060 ","End":"05:30.590","Text":"is going to be equal to u_2 plus u_3."},{"Start":"05:30.590 ","End":"05:32.879","Text":"Where of course u_1 was equal to"},{"Start":"05:32.879 ","End":"05:36.960","Text":"0 because there were no other charges here when q_1 came."},{"Start":"05:37.040 ","End":"05:41.529","Text":"Now we can carry on doing this technique for"},{"Start":"05:41.529 ","End":"05:47.269","Text":"every single charge that joins our little group over here."},{"Start":"05:47.269 ","End":"05:49.489","Text":"We can do that, but it\u0027s a little bit long and"},{"Start":"05:49.489 ","End":"05:52.354","Text":"tedious and it can maybe sometimes get confusing."},{"Start":"05:52.354 ","End":"05:57.770","Text":"So there\u0027s an equation that we can use when dealing with a system with multiple charges."},{"Start":"05:57.770 ","End":"06:02.005","Text":"That is the equation that we\u0027re going to learn now."},{"Start":"06:02.005 ","End":"06:06.105","Text":"The equation goes like so, the energy."},{"Start":"06:06.105 ","End":"06:11.120","Text":"So here u is the total energy of the system and that\u0027s going to be"},{"Start":"06:11.120 ","End":"06:16.435","Text":"equal to 1/2 the sum of Phi,"},{"Start":"06:16.435 ","End":"06:19.234","Text":"which has the potential of each particle,"},{"Start":"06:19.234 ","End":"06:21.290","Text":"multiplied by qi,"},{"Start":"06:21.290 ","End":"06:23.420","Text":"the charge of each particle."},{"Start":"06:23.420 ","End":"06:29.100","Text":"We\u0027re just summing along q and Phi for each particle."},{"Start":"06:29.470 ","End":"06:34.895","Text":"Now, of course, this isn\u0027t the only equation that we can use in order to find"},{"Start":"06:34.895 ","End":"06:37.819","Text":"the energy required to build"},{"Start":"06:37.819 ","End":"06:41.135","Text":"a system but this is the first equation that we\u0027re going to talk about."},{"Start":"06:41.135 ","End":"06:45.049","Text":"Now before we explain why the equation looks exactly like this,"},{"Start":"06:45.049 ","End":"06:51.240","Text":"let\u0027s just understand how to use this equation or how to substitute in values."},{"Start":"06:51.830 ","End":"06:56.809","Text":"Every using the example of this system that we have over here,"},{"Start":"06:56.809 ","End":"07:02.984","Text":"we\u0027ll multiply the charge of q_1 multiplied by its potential."},{"Start":"07:02.984 ","End":"07:09.585","Text":"Then we\u0027ll add that on to the charge of q_2 multiplied by"},{"Start":"07:09.585 ","End":"07:18.110","Text":"its potential and then we\u0027ll add onto that the charge of q_3 multiplied by its potential."},{"Start":"07:18.110 ","End":"07:24.960","Text":"We\u0027ll add all of that up and then we\u0027ll divide that answer by 2."},{"Start":"07:25.270 ","End":"07:29.434","Text":"Now, notice that what I said was,"},{"Start":"07:29.434 ","End":"07:33.500","Text":"I started from q_1 and I said we multiply its charge by its potential,"},{"Start":"07:33.500 ","End":"07:35.539","Text":"and we add that and so on and so forth."},{"Start":"07:35.539 ","End":"07:36.964","Text":"What does that mean?"},{"Start":"07:36.964 ","End":"07:41.785","Text":"That means when using this equation over here for energy,"},{"Start":"07:41.785 ","End":"07:46.100","Text":"we\u0027re not dealing with which charge came first and then this one,"},{"Start":"07:46.100 ","End":"07:47.390","Text":"and then this one, and then that."},{"Start":"07:47.390 ","End":"07:52.759","Text":"Then taking the order in which all of our charges were brought to this position."},{"Start":"07:52.759 ","End":"07:56.900","Text":"No. Straightaway, we look at each charge as if it\u0027s"},{"Start":"07:56.900 ","End":"08:00.995","Text":"the last charge that arrived in the system. What does that mean?"},{"Start":"08:00.995 ","End":"08:03.080","Text":"When we\u0027re looking at our charge q_1,"},{"Start":"08:03.080 ","End":"08:08.430","Text":"we calculate it as its charge q_1 multiplied its potential,"},{"Start":"08:08.430 ","End":"08:11.089","Text":"and when we\u0027re multiplying it by the potential,"},{"Start":"08:11.089 ","End":"08:16.645","Text":"we\u0027re taking into account our charges q_2, and q_3."},{"Start":"08:16.645 ","End":"08:22.920","Text":"As if q_1 was added into this group last,"},{"Start":"08:22.920 ","End":"08:24.339","Text":"we do the same for q_2,"},{"Start":"08:24.339 ","End":"08:26.479","Text":"as if it is the last one to be added,"},{"Start":"08:26.479 ","End":"08:28.145","Text":"and so on and so forth."},{"Start":"08:28.145 ","End":"08:31.075","Text":"Let\u0027s really do this example."},{"Start":"08:31.075 ","End":"08:33.540","Text":"I\u0027m writing it in blue."},{"Start":"08:33.540 ","End":"08:38.150","Text":"We\u0027re going to have, 1/2 multiplied by,"},{"Start":"08:38.150 ","End":"08:45.290","Text":"so we have our charge q_1 multiplied by the potential due to q_2 and q_3."},{"Start":"08:45.290 ","End":"08:46.790","Text":"They\u0027re both point charges,"},{"Start":"08:46.790 ","End":"08:50.874","Text":"so it\u0027s going to be kq_2 plus q_3"},{"Start":"08:50.874 ","End":"08:55.460","Text":"divided by r. Now sometimes the r\u0027s will be different but here,"},{"Start":"08:55.460 ","End":"08:57.079","Text":"specifically in this example,"},{"Start":"08:57.079 ","End":"09:00.619","Text":"they\u0027re all the same distance away from each other."},{"Start":"09:00.619 ","End":"09:03.855","Text":"So it\u0027s always going to be divided by r. Plus,"},{"Start":"09:03.855 ","End":"09:05.190","Text":"now we\u0027re on to q_2."},{"Start":"09:05.190 ","End":"09:11.485","Text":"The charge of q_2 is q_2 multiplied by its potential due to charges q_1 and q_3."},{"Start":"09:11.485 ","End":"09:12.890","Text":"Again, they\u0027re point charges,"},{"Start":"09:12.890 ","End":"09:16.485","Text":"so it\u0027s going to be kq_1 plus q_3,"},{"Start":"09:16.485 ","End":"09:20.925","Text":"and again divided by r, and plus q_3."},{"Start":"09:20.925 ","End":"09:26.939","Text":"Again, it\u0027s the last one being added and then k q_1 plus"},{"Start":"09:26.939 ","End":"09:33.849","Text":"q_2 divided by r. So now,"},{"Start":"09:33.849 ","End":"09:40.835","Text":"let\u0027s see that this answer over here is going to be equal to this answer over here."},{"Start":"09:40.835 ","End":"09:46.349","Text":"What we can see is that we\u0027re summing twice inside the square brackets."},{"Start":"09:46.349 ","End":"09:48.900","Text":"We\u0027re summing twice across our q_1,"},{"Start":"09:48.900 ","End":"09:50.600","Text":"q_2, and q_3s,"},{"Start":"09:50.600 ","End":"09:56.710","Text":"and that\u0027s why we also divide it by 2 in order to cancel out that double sum."},{"Start":"09:56.710 ","End":"10:03.340","Text":"If here we have q_1 and then multiplied by kq_2 divided by i."},{"Start":"10:03.340 ","End":"10:07.269","Text":"Here we can see that we also have q_2 multiplied by"},{"Start":"10:07.269 ","End":"10:14.155","Text":"kq_1 divided by r. Then we can look q_1,"},{"Start":"10:14.155 ","End":"10:17.660","Text":"q_3, q_3, q_1."},{"Start":"10:18.210 ","End":"10:21.100","Text":"Another highlighter color."},{"Start":"10:21.100 ","End":"10:24.430","Text":"We can see q_2 and q_3,"},{"Start":"10:24.430 ","End":"10:26.815","Text":"and here q_3 and q_2."},{"Start":"10:26.815 ","End":"10:29.065","Text":"We can see where something twice along everything."},{"Start":"10:29.065 ","End":"10:31.180","Text":"Then we divide it by 2,"},{"Start":"10:31.180 ","End":"10:34.270","Text":"and then if you want to do the algebra for this,"},{"Start":"10:34.270 ","End":"10:37.599","Text":"you\u0027re welcome to and also the algebra for this."},{"Start":"10:37.599 ","End":"10:39.190","Text":"Add up these 2."},{"Start":"10:39.190 ","End":"10:44.150","Text":"You\u0027ll see that we\u0027re going to get the exact same answer."},{"Start":"10:45.440 ","End":"10:51.625","Text":"This 1/2 is very important to not forget because as we can see from this equation,"},{"Start":"10:51.625 ","End":"10:54.130","Text":"we\u0027re summing up on the potentials twice."},{"Start":"10:54.130 ","End":"10:58.120","Text":"First-time summing on the potential of q_1, q_2, q_2,"},{"Start":"10:58.120 ","End":"11:02.980","Text":"and q_3 but then I\u0027m summing the same potential from q_2,"},{"Start":"11:02.980 ","End":"11:04.660","Text":"q_2, q_1, and q_3."},{"Start":"11:04.660 ","End":"11:11.545","Text":"We have q_1 and q_2 involved in the calculations twice and of course the same for q_3."},{"Start":"11:11.545 ","End":"11:13.929","Text":"In order to cancel that out,"},{"Start":"11:13.929 ","End":"11:17.260","Text":"because obviously one of these charges came first."},{"Start":"11:17.260 ","End":"11:20.259","Text":"In order to not have to think which charge came first,"},{"Start":"11:20.259 ","End":"11:22.930","Text":"but still take that into account."},{"Start":"11:22.930 ","End":"11:27.680","Text":"We divide it by 1/2 and then we get the same answer."},{"Start":"11:28.800 ","End":"11:32.750","Text":"Now let\u0027s take a look at an example."},{"Start":"11:32.880 ","End":"11:40.531","Text":"Here, I have a spherical shell where distributed evenly along my spherical shell,"},{"Start":"11:40.531 ","End":"11:42.294","Text":"I have a charge q,"},{"Start":"11:42.294 ","End":"11:47.799","Text":"and my spherical shell is of radius R. What I want"},{"Start":"11:47.799 ","End":"11:54.200","Text":"to do is I want to calculate the energy required in order to build this type of system."},{"Start":"11:55.380 ","End":"11:58.690","Text":"In order to find the energy to build"},{"Start":"11:58.690 ","End":"12:03.265","Text":"this spherical shell with an even charged distribution of q,"},{"Start":"12:03.265 ","End":"12:09.310","Text":"the first thing I need according to this equation is I need to know my potential."},{"Start":"12:09.310 ","End":"12:13.629","Text":"As we saw in the previous lesson and also in multiple previous lessons,"},{"Start":"12:13.629 ","End":"12:18.939","Text":"we saw that the potential of a spherical shell is going to be similar,"},{"Start":"12:18.939 ","End":"12:21.535","Text":"is going to be that of a point charge,"},{"Start":"12:21.535 ","End":"12:25.390","Text":"which means that it\u0027s going to be equal to kq divided by"},{"Start":"12:25.390 ","End":"12:30.970","Text":"r. When r is bigger than the radius of the spherical shell,"},{"Start":"12:30.970 ","End":"12:35.620","Text":"and it\u0027s going to be a constant kq divided by the radius of"},{"Start":"12:35.620 ","End":"12:40.795","Text":"the spherical shell when r is smaller than the radius of the sphere,"},{"Start":"12:40.795 ","End":"12:43.820","Text":"so this is our potential."},{"Start":"12:44.310 ","End":"12:47.049","Text":"Now we want to use this equation,"},{"Start":"12:47.049 ","End":"12:49.810","Text":"which tells me to take my potential,"},{"Start":"12:49.810 ","End":"12:52.285","Text":"multiply it by the charges,"},{"Start":"12:52.285 ","End":"12:54.324","Text":"add them all up, and then at the end,"},{"Start":"12:54.324 ","End":"12:56.530","Text":"divide everything by 2."},{"Start":"12:56.530 ","End":"13:00.835","Text":"Now of course, because we saw that this is a continuous function,"},{"Start":"13:00.835 ","End":"13:06.385","Text":"we can write that it\u0027s bigger or equal to or smaller or equal to. It doesn\u0027t matter."},{"Start":"13:06.385 ","End":"13:11.800","Text":"Now we can see that our charge is evenly distributed along the shell which"},{"Start":"13:11.800 ","End":"13:16.930","Text":"means that if we want to multiply our charge by the potential,"},{"Start":"13:16.930 ","End":"13:21.580","Text":"we can look at either region because it\u0027s bigger or equal to or smaller than equal to,"},{"Start":"13:21.580 ","End":"13:25.974","Text":"and so we can just substitute in our R and we get the same answer for either one."},{"Start":"13:25.974 ","End":"13:31.510","Text":"So our potential for Q is going to be equal to kq divided by"},{"Start":"13:31.510 ","End":"13:37.900","Text":"capital R. Let\u0027s work out our energy."},{"Start":"13:37.900 ","End":"13:43.420","Text":"It\u0027s going to be equal to 1/2 multiplied by the potential,"},{"Start":"13:43.420 ","End":"13:46.555","Text":"which is kq divided by R,"},{"Start":"13:46.555 ","End":"13:48.835","Text":"because we\u0027re located on the surface,"},{"Start":"13:48.835 ","End":"13:53.485","Text":"and then we\u0027re going to multiply that by the charge, which is q."},{"Start":"13:53.485 ","End":"13:56.439","Text":"Notice we only have one charge over here q,"},{"Start":"13:56.439 ","End":"13:59.475","Text":"so we don\u0027t have to sum up on anything else,"},{"Start":"13:59.475 ","End":"14:07.724","Text":"and then we can see that our energy is going to be equal to 1/2kq squared divided by"},{"Start":"14:07.724 ","End":"14:12.400","Text":"R. So now we can see what we did when we"},{"Start":"14:12.400 ","End":"14:17.500","Text":"take this to be just one giant point charge but if you want to look at it,"},{"Start":"14:17.500 ","End":"14:24.669","Text":"when we split up our spherical shell into tiny little pieces let\u0027s take a look at it now."},{"Start":"14:24.669 ","End":"14:30.145","Text":"I split everything up over here into tiny little charges dq."},{"Start":"14:30.145 ","End":"14:33.940","Text":"Let\u0027s plug this in to my equation,"},{"Start":"14:33.940 ","End":"14:36.490","Text":"and I\u0027m going to write in blue so that we can see."},{"Start":"14:36.490 ","End":"14:45.415","Text":"I\u0027m going to get that u is equal to 1/2 the sum of Phi qI,"},{"Start":"14:45.415 ","End":"14:52.780","Text":"which over here is going to be equal to 1/2 of the total potential,"},{"Start":"14:52.780 ","End":"15:01.374","Text":"is going to be kq divided by R. That\u0027s the potential of this entire spherical shell,"},{"Start":"15:01.374 ","End":"15:09.549","Text":"multiplied by the potential of this tiny little piece over here, which is dq."},{"Start":"15:09.549 ","End":"15:14.394","Text":"Of course, we have the sigma summation sign over here."},{"Start":"15:14.394 ","End":"15:18.880","Text":"Then kq divided by capital R is a constant."},{"Start":"15:18.880 ","End":"15:21.159","Text":"So we can take it out of sigma sign."},{"Start":"15:21.159 ","End":"15:25.749","Text":"So we\u0027re going to get 1/2kq divided by R,"},{"Start":"15:25.749 ","End":"15:29.690","Text":"and then the sum of dqs."},{"Start":"15:29.790 ","End":"15:34.810","Text":"Now what we\u0027re doing is we\u0027re summing up all of"},{"Start":"15:34.810 ","End":"15:39.400","Text":"these tiny dqs all across the surface area of this spherical shell,"},{"Start":"15:39.400 ","End":"15:41.620","Text":"and then multiplying it by this."},{"Start":"15:41.620 ","End":"15:46.435","Text":"So if we sum up all the tiny dq\u0027s is on this whole surface area of the shell,"},{"Start":"15:46.435 ","End":"15:49.975","Text":"we\u0027re simply going to get big Q."},{"Start":"15:49.975 ","End":"15:53.350","Text":"So in total, we will get the exact same answer."},{"Start":"15:53.350 ","End":"15:56.950","Text":"So we\u0027ll have kq multiplied by the q."},{"Start":"15:56.950 ","End":"15:59.500","Text":"Again. Divided by R,"},{"Start":"15:59.500 ","End":"16:03.325","Text":"we get the exact same equation that we did before."},{"Start":"16:03.325 ","End":"16:06.100","Text":"That\u0027s great now we\u0027ve spoken about"},{"Start":"16:06.100 ","End":"16:09.339","Text":"our first equation in order to find the energy of a system."},{"Start":"16:09.339 ","End":"16:12.100","Text":"Now, let\u0027s speak about the second equation,"},{"Start":"16:12.100 ","End":"16:16.390","Text":"which is used maybe more often than this equation over here,"},{"Start":"16:16.390 ","End":"16:21.565","Text":"and it is equal to Epsilon naught divided by 2,"},{"Start":"16:21.565 ","End":"16:24.160","Text":"and then the integral on"},{"Start":"16:24.160 ","End":"16:32.829","Text":"our electric field squared dv."},{"Start":"16:32.829 ","End":"16:36.039","Text":"Now let\u0027s learn how to use this equation"},{"Start":"16:36.039 ","End":"16:40.310","Text":"and how we can substitute in all of our values to get an answer."},{"Start":"16:41.310 ","End":"16:45.025","Text":"This equation, we have Epsilon naught divided by 2,"},{"Start":"16:45.025 ","End":"16:46.974","Text":"so that\u0027s a constant."},{"Start":"16:46.974 ","End":"16:48.700","Text":"Then we\u0027re integrating,"},{"Start":"16:48.700 ","End":"16:54.144","Text":"and we\u0027re integrating on everywhere all of space"},{"Start":"16:54.144 ","End":"17:00.249","Text":"from negative infinity until infinity, all of space."},{"Start":"17:00.249 ","End":"17:01.855","Text":"Here in this equation,"},{"Start":"17:01.855 ","End":"17:05.769","Text":"we were summing up the charges that we had within our system."},{"Start":"17:05.769 ","End":"17:11.187","Text":"Here, we\u0027re integrating along every single thing"},{"Start":"17:11.187 ","End":"17:17.425","Text":"and then we\u0027re integrating our electric field squared."},{"Start":"17:17.425 ","End":"17:19.480","Text":"We square our electric field,"},{"Start":"17:19.480 ","End":"17:21.640","Text":"and then we integrate dv."},{"Start":"17:21.640 ","End":"17:23.709","Text":"We\u0027re integrating along our volume,"},{"Start":"17:23.709 ","End":"17:25.599","Text":"which if that\u0027s Cartesian coordinates,"},{"Start":"17:25.599 ","End":"17:31.429","Text":"so that will be dx dy dz all over everywhere in space."},{"Start":"17:31.620 ","End":"17:33.984","Text":"When using this equation,"},{"Start":"17:33.984 ","End":"17:36.873","Text":"I don\u0027t really care what charges I have where."},{"Start":"17:36.873 ","End":"17:39.789","Text":"All I care about is my electric field."},{"Start":"17:39.789 ","End":"17:44.410","Text":"All right, so now let\u0027s use this technique in order to"},{"Start":"17:44.410 ","End":"17:50.260","Text":"find what our energy is in the same example that we have over here."},{"Start":"17:50.260 ","End":"17:53.094","Text":"The first thing we need is our electric field."},{"Start":"17:53.094 ","End":"17:55.450","Text":"As we know from Gauss\u0027s law,"},{"Start":"17:55.450 ","End":"17:59.785","Text":"our electric field is equal to 0,"},{"Start":"17:59.785 ","End":"18:02.110","Text":"and because we\u0027re dealing with a spherical shell,"},{"Start":"18:02.110 ","End":"18:11.110","Text":"so it\u0027s going to be 0 and the region between r and R. When we\u0027re inside the sphere,"},{"Start":"18:11.110 ","End":"18:14.305","Text":"our electric field is going to be equal to 0,"},{"Start":"18:14.305 ","End":"18:19.602","Text":"and when we\u0027re outside the spherical shell or electric field is going to be equal to kq"},{"Start":"18:19.602 ","End":"18:25.344","Text":"divided by I squared in the r-hat direction."},{"Start":"18:25.344 ","End":"18:27.730","Text":"So that is our electric field."},{"Start":"18:27.730 ","End":"18:31.039","Text":"Now, let\u0027s find our energy."},{"Start":"18:31.140 ","End":"18:38.755","Text":"This is going to be equal to Epsilon naught divided by 2 because that\u0027s a constant,"},{"Start":"18:38.755 ","End":"18:43.900","Text":"and then our integral because we\u0027re integrating along dv."},{"Start":"18:43.900 ","End":"18:48.438","Text":"It\u0027s going to be a triple integral of e^,"},{"Start":"18:48.438 ","End":"18:50.860","Text":"so I won\u0027t substitute that in yet."},{"Start":"18:50.860 ","End":"18:53.784","Text":"E^ multiplied by."},{"Start":"18:53.784 ","End":"18:56.184","Text":"Now I have to write my dv in"},{"Start":"18:56.184 ","End":"18:59.704","Text":"spherical coordinates because we\u0027re dealing with a spherical shell."},{"Start":"18:59.704 ","End":"19:01.470","Text":"So we have our Jacobian,"},{"Start":"19:01.470 ","End":"19:07.795","Text":"which is r^ sine Phi d Phi d Theta d Phi."},{"Start":"19:07.795 ","End":"19:12.410","Text":"Our dy is going to be going all"},{"Start":"19:12.410 ","End":"19:17.870","Text":"over space and id Theta is from 0 to 2 pi and id Phi is from 0 to pi."},{"Start":"19:17.870 ","End":"19:23.780","Text":"Now when we\u0027re integrating in spherical coordinates across all of space,"},{"Start":"19:23.780 ","End":"19:29.569","Text":"we can just simply rewrite this whole equation simply as again,"},{"Start":"19:29.569 ","End":"19:31.385","Text":"Epsilon naught divided by 2,"},{"Start":"19:31.385 ","End":"19:40.370","Text":"and then the integral of E or E field^ again."},{"Start":"19:40.370 ","End":"19:41.660","Text":"Then instead of all of this,"},{"Start":"19:41.660 ","End":"19:49.180","Text":"we just multiply it by 4 pi r squared dr."},{"Start":"19:49.180 ","End":"19:53.949","Text":"When we\u0027re integrating,"},{"Start":"19:53.949 ","End":"20:00.920","Text":"our r is on everywhere and our theta 0 to 2 pi and our phi is 0 to pi."},{"Start":"20:01.860 ","End":"20:05.980","Text":"Once we do the d theta and d phi integration,"},{"Start":"20:05.980 ","End":"20:10.849","Text":"we\u0027re going to be left with 4 pi and then r^2 comes from here."},{"Start":"20:12.300 ","End":"20:17.335","Text":"Now when we\u0027re integrating from minus infinity to infinity,"},{"Start":"20:17.335 ","End":"20:18.729","Text":"what does that really mean?"},{"Start":"20:18.729 ","End":"20:27.505","Text":"It\u0027s the same as if we shift infinity back and we integrate from 0 until infinity."},{"Start":"20:27.505 ","End":"20:30.610","Text":"We\u0027ve just shifted our axes a little bit,"},{"Start":"20:30.610 ","End":"20:33.114","Text":"and this thing that we\u0027ve done,"},{"Start":"20:33.114 ","End":"20:35.800","Text":"we\u0027ll make this integral a lot easier."},{"Start":"20:35.800 ","End":"20:41.020","Text":"Here we can substitute in our bounds from 0 to infinity."},{"Start":"20:41.020 ","End":"20:43.389","Text":"Let\u0027s see how we do this."},{"Start":"20:43.389 ","End":"20:46.225","Text":"We\u0027re going to have epsilon naught divided by 2."},{"Start":"20:46.225 ","End":"20:49.340","Text":"All of this multiplied by."},{"Start":"20:49.650 ","End":"20:54.685","Text":"Now we have to substitute in our electric field squared."},{"Start":"20:54.685 ","End":"20:58.840","Text":"First of all, let\u0027s look at our electric field and we have 2 regions."},{"Start":"20:58.840 ","End":"21:00.369","Text":"What are we going to have to do?"},{"Start":"21:00.369 ","End":"21:04.690","Text":"We\u0027re going to have to split up our integral into the 2 regions."},{"Start":"21:04.690 ","End":"21:09.515","Text":"Our first integral is going to be from 0 until the capital R,"},{"Start":"21:09.515 ","End":"21:15.705","Text":"and that\u0027s going to be =0 multiply it by 4 r squared which is 0 Dr,"},{"Start":"21:15.705 ","End":"21:17.310","Text":"and that\u0027s just going to be equal to 0."},{"Start":"21:17.310 ","End":"21:18.810","Text":"We didn\u0027t really have to write it."},{"Start":"21:18.810 ","End":"21:26.090","Text":"Then plus our integral from capital R until infinity."},{"Start":"21:26.090 ","End":"21:29.559","Text":"Now we\u0027re integrating along the electric field"},{"Start":"21:29.559 ","End":"21:34.585","Text":"squared when we\u0027re outside of our spherical shell."},{"Start":"21:34.585 ","End":"21:36.939","Text":"Our electric field squared."},{"Start":"21:36.939 ","End":"21:42.160","Text":"First of all, we have 4pi r^2 from here."},{"Start":"21:42.160 ","End":"21:49.870","Text":"Then our electric field squared is going to be equal to kq divided by r squared squared."},{"Start":"21:49.870 ","End":"21:52.464","Text":"When we square our r, r-hat,"},{"Start":"21:52.464 ","End":"21:54.880","Text":"it\u0027s going to be equal to r^.r^,"},{"Start":"21:54.880 ","End":"21:56.875","Text":"which is simply =1."},{"Start":"21:56.875 ","End":"22:01.040","Text":"That will cancel out and then of course dr."},{"Start":"22:03.150 ","End":"22:06.666","Text":"This crosses out, this is =0."},{"Start":"22:06.666 ","End":"22:08.469","Text":"Then once we integrate,"},{"Start":"22:08.469 ","End":"22:13.404","Text":"so we can see that r^2 will cancel with r to the power 4 over here,"},{"Start":"22:13.404 ","End":"22:14.710","Text":"partially cancel out,"},{"Start":"22:14.710 ","End":"22:17.665","Text":"and our 4 here we\u0027ll divide by 2."},{"Start":"22:17.665 ","End":"22:21.490","Text":"Then what we can do is we can substitute in that our k is"},{"Start":"22:21.490 ","End":"22:25.855","Text":"= 1 divided by 4 pi epsilon naught."},{"Start":"22:25.855 ","End":"22:27.744","Text":"We can substitute that in,"},{"Start":"22:27.744 ","End":"22:33.414","Text":"and we\u0027ll see that once we integrate this and put in or a substitute in our bounds,"},{"Start":"22:33.414 ","End":"22:38.260","Text":"we\u0027re going to get the exact same answer that we did over here."},{"Start":"22:38.260 ","End":"22:41.350","Text":"We\u0027re going to get 1/2(kq^2) divided by"},{"Start":"22:41.350 ","End":"22:50.905","Text":"r. The trick to remember with this equation is that when we\u0027re integrating along r,"},{"Start":"22:50.905 ","End":"22:54.760","Text":"so we\u0027re integrating on all of space."},{"Start":"22:54.760 ","End":"22:59.244","Text":"Here specifically in spherical coordinates where from 0 to infinity,"},{"Start":"22:59.244 ","End":"23:02.260","Text":"but if it was in Cartesian coordinates,"},{"Start":"23:02.260 ","End":"23:05.724","Text":"so it would be something slightly differently."},{"Start":"23:05.724 ","End":"23:10.059","Text":"Now a very important note also to note about this equation,"},{"Start":"23:10.059 ","End":"23:12.445","Text":"I\u0027ll put a star over here,"},{"Start":"23:12.445 ","End":"23:18.010","Text":"and that\u0027s that this equation doesn\u0027t work if we have lots of point charges"},{"Start":"23:18.010 ","End":"23:24.950","Text":"or if we\u0027re dealing with a charge density per unit length."},{"Start":"23:25.800 ","End":"23:31.900","Text":"This note in green is that this equation dealing with the electric field won\u0027t"},{"Start":"23:31.900 ","End":"23:34.510","Text":"work if there\u0027s an E field due to a point charge"},{"Start":"23:34.510 ","End":"23:37.690","Text":"or if there\u0027s charge density per unit length."},{"Start":"23:37.690 ","End":"23:40.730","Text":"If we have a wire of some sort."},{"Start":"23:41.490 ","End":"23:44.889","Text":"If there\u0027s 1 of these 2 cases,"},{"Start":"23:44.889 ","End":"23:48.340","Text":"you have to use this equation over here."},{"Start":"23:48.340 ","End":"23:54.280","Text":"Now let\u0027s explain why these 2 cases are problem for this equation."},{"Start":"23:54.280 ","End":"23:58.090","Text":"Let\u0027s talk about the problem when we have a point charge."},{"Start":"23:58.090 ","End":"24:02.275","Text":"Here we have some kind of point charge q1."},{"Start":"24:02.275 ","End":"24:05.965","Text":"When we use this equation, as we saw before,"},{"Start":"24:05.965 ","End":"24:08.170","Text":"we\u0027re bringing some kind of charge into"},{"Start":"24:08.170 ","End":"24:12.340","Text":"an area where there\u0027s no electric field or no potential."},{"Start":"24:12.340 ","End":"24:20.050","Text":"We know that the energy required to bring this point charge from infinity to here is =0."},{"Start":"24:20.050 ","End":"24:22.030","Text":"This, we know we\u0027ve seen."},{"Start":"24:22.030 ","End":"24:23.469","Text":"There was nothing here before,"},{"Start":"24:23.469 ","End":"24:24.805","Text":"so there\u0027s no forces,"},{"Start":"24:24.805 ","End":"24:27.894","Text":"there\u0027ll be acting against bringing this point charge over here."},{"Start":"24:27.894 ","End":"24:29.755","Text":"Our energy is =0."},{"Start":"24:29.755 ","End":"24:33.280","Text":"Let\u0027s look at this equation though on the other hand."},{"Start":"24:33.280 ","End":"24:38.275","Text":"This equation says, and this is incorrect to use for this case."},{"Start":"24:38.275 ","End":"24:41.005","Text":"We have epsilon naught divided by 2,"},{"Start":"24:41.005 ","End":"24:46.554","Text":"and then we\u0027re integrating everywhere on the electric field squared."},{"Start":"24:46.554 ","End":"24:49.220","Text":"Now the electric field,"},{"Start":"24:51.390 ","End":"24:53.980","Text":"we know that the electric field for"},{"Start":"24:53.980 ","End":"24:58.180","Text":"a point charge is equal to k multiplied by its charge,"},{"Start":"24:58.180 ","End":"25:03.325","Text":"divided by r^2 in the r direction,"},{"Start":"25:03.325 ","End":"25:06.174","Text":"and then squared and then dv."},{"Start":"25:06.174 ","End":"25:14.394","Text":"What we can see when we square this is that we\u0027re going to get some non-zero value,"},{"Start":"25:14.394 ","End":"25:18.825","Text":"which when we integrate this along all of space,"},{"Start":"25:18.825 ","End":"25:23.160","Text":"we\u0027ll see that not only will we get that our energy doesn\u0027t =0,"},{"Start":"25:23.160 ","End":"25:25.890","Text":"it\u0027s going to be = infinity,"},{"Start":"25:25.890 ","End":"25:28.645","Text":"and that\u0027s not wrong."},{"Start":"25:28.645 ","End":"25:31.400","Text":"That\u0027s not right I mean, sorry."},{"Start":"25:31.920 ","End":"25:37.790","Text":"That\u0027s why we can\u0027t use this equation when dealing with a point charge."},{"Start":"25:38.280 ","End":"25:43.239","Text":"The reason this happens is that when we\u0027re using this form of the equation,"},{"Start":"25:43.239 ","End":"25:47.575","Text":"we\u0027re talking about the energy in order to build our system."},{"Start":"25:47.575 ","End":"25:50.364","Text":"When we\u0027re talking about the energy to build our system,"},{"Start":"25:50.364 ","End":"25:53.440","Text":"when we\u0027re dealing with a point charge we\u0027re speaking about the energy to build"},{"Start":"25:53.440 ","End":"25:59.140","Text":"an electron or something, which it isn\u0027t."},{"Start":"25:59.140 ","End":"26:03.114","Text":"That\u0027s why we cannot use this equation."},{"Start":"26:03.114 ","End":"26:05.590","Text":"Whereas this equation does work because it\u0027s simply talking"},{"Start":"26:05.590 ","End":"26:08.305","Text":"about bringing some point charge to this area,"},{"Start":"26:08.305 ","End":"26:10.730","Text":"which we know how to deal with."},{"Start":"26:11.640 ","End":"26:15.999","Text":"It\u0027s the exact same idea why this would work when dealing"},{"Start":"26:15.999 ","End":"26:22.340","Text":"with something with uniform charge density per unit length, so a wire."},{"Start":"26:22.920 ","End":"26:26.919","Text":"It\u0027s important to remember this and not get mixed"},{"Start":"26:26.919 ","End":"26:31.239","Text":"up by using this equation to solve this type of question."},{"Start":"26:31.239 ","End":"26:33.535","Text":"Another note on these equations,"},{"Start":"26:33.535 ","End":"26:35.365","Text":"when we\u0027re speaking about this equation,"},{"Start":"26:35.365 ","End":"26:38.110","Text":"remember that here we\u0027re speaking about our potential."},{"Start":"26:38.110 ","End":"26:41.560","Text":"Now we\u0027re assuming that when speaking about this potential,"},{"Start":"26:41.560 ","End":"26:42.970","Text":"the way to calculate it,"},{"Start":"26:42.970 ","End":"26:48.640","Text":"we\u0027re assuming and taking into account that the potential at infinity."},{"Start":"26:48.640 ","End":"26:53.154","Text":"Very far away from where our group of"},{"Start":"26:53.154 ","End":"27:00.085","Text":"charges so that the potential at infinity = 0."},{"Start":"27:00.085 ","End":"27:04.610","Text":"This equation takes that assumption into account."},{"Start":"27:05.490 ","End":"27:09.580","Text":"What happens if you\u0027ve got some really weird question?"},{"Start":"27:09.580 ","End":"27:15.895","Text":"They tell you that the potential at infinity doesn\u0027t = 0."},{"Start":"27:15.895 ","End":"27:18.489","Text":"Let\u0027s do this."},{"Start":"27:18.489 ","End":"27:23.290","Text":"Blue charge, this blue star, sorry, over here."},{"Start":"27:23.290 ","End":"27:29.500","Text":"If our potential at infinity does not = 0."},{"Start":"27:29.500 ","End":"27:34.345","Text":"So then what we have to do is that this phi I"},{"Start":"27:34.345 ","End":"27:42.469","Text":"will become potential at infinity,"},{"Start":"27:42.510 ","End":"27:49.660","Text":"negative r potential at infinity plus our potential wherever we are."},{"Start":"27:49.660 ","End":"27:54.475","Text":"Let\u0027s say we\u0027re at the origin or at our local place over here."},{"Start":"27:54.475 ","End":"27:58.809","Text":"That means that therefore our phi is"},{"Start":"27:58.809 ","End":"28:02.830","Text":"going to be equal to our potential at the point that we\u0027re at,"},{"Start":"28:02.830 ","End":"28:08.000","Text":"minus whatever value we have where potential at infinity."},{"Start":"28:08.400 ","End":"28:12.024","Text":"This is important to note,"},{"Start":"28:12.024 ","End":"28:16.585","Text":"although the chances of getting a question like this are pretty slim."},{"Start":"28:16.585 ","End":"28:20.739","Text":"Because the point of this equation is to find the work needed to"},{"Start":"28:20.739 ","End":"28:25.820","Text":"bring some kind of charge from infinity to wherever we are."},{"Start":"28:25.820 ","End":"28:28.210","Text":"If an infinity there is potential,"},{"Start":"28:28.210 ","End":"28:30.310","Text":"we have to take that into account as well."},{"Start":"28:30.310 ","End":"28:34.225","Text":"We just find the potential difference."},{"Start":"28:34.225 ","End":"28:40.644","Text":"These are the 2 main equations for working out the energy required to build a system."},{"Start":"28:40.644 ","End":"28:42.430","Text":"This is what you should really remember."},{"Start":"28:42.430 ","End":"28:48.984","Text":"Now, there\u0027s another 2 equations as well that are sometimes used, but very rarely."},{"Start":"28:48.984 ","End":"28:53.650","Text":"So you should also copy them as well but maybe less important,"},{"Start":"28:53.650 ","End":"28:59.720","Text":"so 1/2 the integral of phi dq."},{"Start":"28:59.720 ","End":"29:03.960","Text":"What does this equation, this equation is simply this."},{"Start":"29:03.960 ","End":"29:07.590","Text":"Here we have a sigma sign because we\u0027re"},{"Start":"29:07.590 ","End":"29:11.195","Text":"summing up point charges, something that\u0027s discrete."},{"Start":"29:11.195 ","End":"29:18.265","Text":"Here we have an integral because we\u0027re summing up something with a more continuous form."},{"Start":"29:18.265 ","End":"29:22.135","Text":"These 2 equations are pretty much the same thing."},{"Start":"29:22.135 ","End":"29:29.810","Text":"Another equation is the integral on phi as a function of q(dq)."},{"Start":"29:29.810 ","End":"29:35.950","Text":"Now this equation really deals with the literal building of the system, charge."},{"Start":"29:35.950 ","End":"29:39.730","Text":"This probably, you don\u0027t really have to know,"},{"Start":"29:39.730 ","End":"29:41.500","Text":"just know that it exists."},{"Start":"29:41.500 ","End":"29:43.899","Text":"Also this, what\u0027s in red,"},{"Start":"29:43.899 ","End":"29:45.505","Text":"you should copy to your books."},{"Start":"29:45.505 ","End":"29:48.140","Text":"This is the most important."},{"Start":"29:48.510 ","End":"29:51.700","Text":"Just before we finish the lesson,"},{"Start":"29:51.700 ","End":"29:54.010","Text":"there\u0027s 1 last thing we need to remember."},{"Start":"29:54.010 ","End":"30:03.815","Text":"Here we have epsilon naught divided by 2 multiplied by our E^2."},{"Start":"30:03.815 ","End":"30:05.860","Text":"In physics, there\u0027s a name for that,"},{"Start":"30:05.860 ","End":"30:12.370","Text":"and it\u0027s mu E or the energy density."},{"Start":"30:12.370 ","End":"30:16.269","Text":"What we can see is that we have epsilon naught divided by"},{"Start":"30:16.269 ","End":"30:20.050","Text":"2 multiplied by the e field squared dv."},{"Start":"30:20.050 ","End":"30:24.714","Text":"Any time we have something multiplied by volume,"},{"Start":"30:24.714 ","End":"30:29.485","Text":"then we know that this something is something to do with density."},{"Start":"30:29.485 ","End":"30:32.334","Text":"The density of this, and this has energy,"},{"Start":"30:32.334 ","End":"30:39.080","Text":"this is energy density or energy per unit volume."},{"Start":"30:40.380 ","End":"30:45.505","Text":"This over here is given by mu e and it\u0027s the energy density."},{"Start":"30:45.505 ","End":"30:51.175","Text":"Let\u0027s say we have some kind of cube of volume 1."},{"Start":"30:51.175 ","End":"30:55.900","Text":"It has a volume of 1 and let\u0027s say that it has an E field."},{"Start":"30:55.900 ","End":"31:01.150","Text":"We can say that the energy is stored inside this unit of volume is equal to"},{"Start":"31:01.150 ","End":"31:08.635","Text":"this mu E. u will be = mu e multiplied by the volume, which here is 1."},{"Start":"31:08.635 ","End":"31:11.575","Text":"That\u0027s what this is equal to."},{"Start":"31:11.575 ","End":"31:17.815","Text":"What this is actually is the energy stored in some tiny volume, some unit volume."},{"Start":"31:17.815 ","End":"31:22.810","Text":"Then I sum up all these unit volumes with my integral along"},{"Start":"31:22.810 ","End":"31:28.915","Text":"my E field and I\u0027ll get my total energy throughout space."},{"Start":"31:28.915 ","End":"31:32.120","Text":"That\u0027s the end of this lesson."}],"ID":22354},{"Watched":false,"Name":"Exercise 1","Duration":"21m 46s","ChapterTopicVideoID":21451,"CourseChapterTopicPlaylistID":99478,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.145","Text":"Hello. In this question,"},{"Start":"00:02.145 ","End":"00:05.700","Text":"we\u0027re given 3 concentric spherical shells."},{"Start":"00:05.700 ","End":"00:13.110","Text":"The 1st one has radius R and has charge evenly distributed along its surface area Q."},{"Start":"00:13.110 ","End":"00:17.715","Text":"The 2nd spherical shell has a radius 2R and has charge 2Q,"},{"Start":"00:17.715 ","End":"00:23.595","Text":"and the 3rd spherical shell has radius 3R and it is grounded."},{"Start":"00:23.595 ","End":"00:26.655","Text":"What does it mean when we say that something\u0027s grounded?"},{"Start":"00:26.655 ","End":"00:32.645","Text":"It means that the potential on this shell is equal to the potential at infinity,"},{"Start":"00:32.645 ","End":"00:35.090","Text":"which we said is equal to 0."},{"Start":"00:35.090 ","End":"00:38.334","Text":"That\u0027s how we\u0027ve defined it throughout all of our lessons."},{"Start":"00:38.334 ","End":"00:42.485","Text":"In this question we\u0027re being asked to find"},{"Start":"00:42.485 ","End":"00:47.845","Text":"the energy which is required in order to build this system."},{"Start":"00:47.845 ","End":"00:50.215","Text":"In order to find our U,"},{"Start":"00:50.215 ","End":"00:56.850","Text":"the first thing that we need to do is we need to work out our electric field."},{"Start":"00:57.610 ","End":"01:03.785","Text":"As we can see we have a lot of regions which we have to deal with, so let\u0027s begin."},{"Start":"01:03.785 ","End":"01:07.180","Text":"We can obviously use Gauss\u0027s law in order to solve this,"},{"Start":"01:07.180 ","End":"01:10.790","Text":"but specifically because we\u0027re dealing here with spherical shells,"},{"Start":"01:10.790 ","End":"01:15.740","Text":"we already know the equation for the E field that comes from spherical shell,"},{"Start":"01:15.740 ","End":"01:19.070","Text":"so we\u0027re just going to plug it in and"},{"Start":"01:19.070 ","End":"01:22.940","Text":"taking into account which region we\u0027re located in within the system."},{"Start":"01:22.940 ","End":"01:24.890","Text":"As a little side note,"},{"Start":"01:24.890 ","End":"01:27.050","Text":"if our inner spherical shell"},{"Start":"01:27.050 ","End":"01:31.595","Text":"was instead of a spherical shell it was a conducting sphere,"},{"Start":"01:31.595 ","End":"01:35.420","Text":"the question would be exactly identical because"},{"Start":"01:35.420 ","End":"01:41.525","Text":"a conducting sphere only has charged density along its surface area anyway,"},{"Start":"01:41.525 ","End":"01:46.355","Text":"so then we can calculate it as if it\u0027s a spherical shell as well."},{"Start":"01:46.355 ","End":"01:48.470","Text":"It would be the exact same question,"},{"Start":"01:48.470 ","End":"01:52.875","Text":"but here specifically we\u0027re only dealing with 3 spherical shells."},{"Start":"01:52.875 ","End":"01:55.780","Text":"Let\u0027s deal with,"},{"Start":"01:55.780 ","End":"01:59.060","Text":"if we\u0027re somewhere inside the inner spherical shell."},{"Start":"01:59.060 ","End":"02:00.950","Text":"Our radius is r,"},{"Start":"02:00.950 ","End":"02:03.085","Text":"that\u0027s where we\u0027re located,"},{"Start":"02:03.085 ","End":"02:09.380","Text":"so we\u0027re located at r is smaller than R. As we know,"},{"Start":"02:09.380 ","End":"02:12.950","Text":"the electric field within a spherical shell is"},{"Start":"02:12.950 ","End":"02:17.344","Text":"going to be equal to 0 because there is no Qn over here."},{"Start":"02:17.344 ","End":"02:21.045","Text":"Now let\u0027s go to the next region."},{"Start":"02:21.045 ","End":"02:27.695","Text":"We\u0027re located somewhere between our 1st spherical shell and our 2nd spherical shell."},{"Start":"02:27.695 ","End":"02:29.770","Text":"Let\u0027s write that."},{"Start":"02:29.770 ","End":"02:34.135","Text":"We\u0027re between R and 2R."},{"Start":"02:34.135 ","End":"02:37.130","Text":"That means that we\u0027re only going to experience"},{"Start":"02:37.130 ","End":"02:40.775","Text":"electric field due to the inner spherical shell,"},{"Start":"02:40.775 ","End":"02:44.330","Text":"because that\u0027s the only place where we\u0027re going to have some value for Qn."},{"Start":"02:44.330 ","End":"02:53.000","Text":"That means that our electric field over here is going to be equal to KQ divided by r^2,"},{"Start":"02:53.000 ","End":"02:56.845","Text":"and of course, in the radial direction."},{"Start":"02:56.845 ","End":"03:02.099","Text":"We\u0027re located here,"},{"Start":"03:02.099 ","End":"03:07.345","Text":"somewhere between the 2nd spherical shell and the 3rd spherical shell which is grounded."},{"Start":"03:07.345 ","End":"03:13.280","Text":"Let\u0027s write that in we\u0027re somewhere between 2R and 3R,"},{"Start":"03:13.280 ","End":"03:18.400","Text":"and the electric field over here is going to be felt due to"},{"Start":"03:18.400 ","End":"03:24.335","Text":"our 2nd spherical shell and due to our 1st or our inner a spherical shell."},{"Start":"03:24.335 ","End":"03:30.095","Text":"Because both of these have an electric field components because they both,"},{"Start":"03:30.095 ","End":"03:32.705","Text":"when relocated here, we\u0027re going to have Q in,"},{"Start":"03:32.705 ","End":"03:34.640","Text":"from here at 2Q,"},{"Start":"03:34.640 ","End":"03:36.340","Text":"and from here at Q."},{"Start":"03:36.340 ","End":"03:40.710","Text":"We\u0027re going to have K multiplied by"},{"Start":"03:40.710 ","End":"03:50.310","Text":"2Q plus Q divided by r^2 in the radial direction."},{"Start":"03:50.310 ","End":"03:52.580","Text":"Then 2Q plus Q,"},{"Start":"03:52.580 ","End":"03:55.933","Text":"we can simply just write that as 3Q,"},{"Start":"03:55.933 ","End":"03:59.160","Text":"K multiplied by that."},{"Start":"03:59.160 ","End":"04:07.955","Text":"What happens if we\u0027re located here outside of our entire system?"},{"Start":"04:07.955 ","End":"04:13.830","Text":"Right now we\u0027re outside,"},{"Start":"04:13.830 ","End":"04:15.285","Text":"we\u0027re bigger than 3R,"},{"Start":"04:15.285 ","End":"04:18.440","Text":"so because our potential over here is equal to"},{"Start":"04:18.440 ","End":"04:22.830","Text":"0 due to the outer spherical shell being grounded."},{"Start":"04:22.830 ","End":"04:26.045","Text":"That means that we have no potential difference over here,"},{"Start":"04:26.045 ","End":"04:28.565","Text":"and if there\u0027s no potential difference over here,"},{"Start":"04:28.565 ","End":"04:31.945","Text":"that means that there\u0027s no E field over here."},{"Start":"04:31.945 ","End":"04:40.820","Text":"That means that our E field is going to be equal to 0 again outside."},{"Start":"04:40.820 ","End":"04:44.135","Text":"We have our electric field,"},{"Start":"04:44.135 ","End":"04:47.840","Text":"so let\u0027s find out what our energy is."},{"Start":"04:47.840 ","End":"04:54.140","Text":"We know that our U is equal to Epsilon naught divided by 2,"},{"Start":"04:54.140 ","End":"05:01.005","Text":"and then the integration along our E field^2 dV,"},{"Start":"05:01.005 ","End":"05:04.860","Text":"so let\u0027s calculate this now."},{"Start":"05:04.860 ","End":"05:08.130","Text":"Let\u0027s plug in our values."},{"Start":"05:08.130 ","End":"05:13.425","Text":"We\u0027ll have that our U is equal to Epsilon naught divided by 2."},{"Start":"05:13.425 ","End":"05:15.780","Text":"Integrating along dV,"},{"Start":"05:15.780 ","End":"05:19.955","Text":"and remember we\u0027re integrating across all of space."},{"Start":"05:19.955 ","End":"05:26.195","Text":"We already know that when we put in our Jacobian and spherical coordinates,"},{"Start":"05:26.195 ","End":"05:28.040","Text":"because we\u0027re dealing with a sphere,"},{"Start":"05:28.040 ","End":"05:31.950","Text":"so that should be a clue to be using spherical coordinates."},{"Start":"05:32.060 ","End":"05:37.250","Text":"We know that we\u0027re not going to have any variable for Theta,"},{"Start":"05:37.250 ","End":"05:39.445","Text":"so we\u0027re just going to have 2Pi,"},{"Start":"05:39.445 ","End":"05:41.795","Text":"multiply this 2Pi,"},{"Start":"05:41.795 ","End":"05:46.085","Text":"and then when we integrate along Phi for our sin(Phi) between 0 and Pi,"},{"Start":"05:46.085 ","End":"05:49.345","Text":"we\u0027re going to multiply everything again by 2."},{"Start":"05:49.345 ","End":"05:52.110","Text":"Let\u0027s just cut to the chase then,"},{"Start":"05:52.110 ","End":"05:55.475","Text":"what we\u0027re going to have is we\u0027re going to have Epsilon naught"},{"Start":"05:55.475 ","End":"05:59.830","Text":"divided by 2 multiplied by 4Pi,"},{"Start":"05:59.830 ","End":"06:03.390","Text":"which is what we get from our d Theta d Phi,"},{"Start":"06:03.390 ","End":"06:07.475","Text":"and then we\u0027re going to be integrating along all of space,"},{"Start":"06:07.475 ","End":"06:14.655","Text":"so that\u0027s from 0 until infinity for our E field^2"},{"Start":"06:14.655 ","End":"06:19.860","Text":"multiplied by r^2 dr."},{"Start":"06:20.810 ","End":"06:25.730","Text":"We\u0027ve also seen this equation in the previous video,"},{"Start":"06:25.730 ","End":"06:28.510","Text":"so if you don\u0027t understand what I just explained to you,"},{"Start":"06:28.510 ","End":"06:32.450","Text":"how we got to this, please go back to the previous video."},{"Start":"06:33.890 ","End":"06:37.450","Text":"I\u0027m going to substitute in my E field."},{"Start":"06:37.450 ","End":"06:42.055","Text":"As we can see, my E field is split up into different regions,"},{"Start":"06:42.055 ","End":"06:45.745","Text":"which means I\u0027m going to have to integrate along the different regions."},{"Start":"06:45.745 ","End":"06:47.920","Text":"Let\u0027s see how that works."},{"Start":"06:47.920 ","End":"06:49.450","Text":"I\u0027m going to have,"},{"Start":"06:49.450 ","End":"06:51.130","Text":"let\u0027s start canceling stuff out,"},{"Start":"06:51.130 ","End":"06:55.325","Text":"so I\u0027ll have 2 Epsilon naught multiplied by Pi,"},{"Start":"06:55.325 ","End":"06:59.140","Text":"and then this is going to be multiplied by."},{"Start":"06:59.140 ","End":"07:06.800","Text":"My first region is going to be an integration from 0 until R,"},{"Start":"07:06.800 ","End":"07:09.560","Text":"and we can see that my E field here is 0,"},{"Start":"07:09.560 ","End":"07:16.805","Text":"so 0^2 R^2 dr is simply going to be 0 and then this crosses out and it\u0027s equal to 0."},{"Start":"07:16.805 ","End":"07:21.005","Text":"Plus an integration along my next region."},{"Start":"07:21.005 ","End":"07:26.245","Text":"That is going to be from R until 2R,"},{"Start":"07:26.245 ","End":"07:30.080","Text":"and then I substitute in my E field squared, so again,"},{"Start":"07:30.080 ","End":"07:34.775","Text":"my r hat will disappear because r hat dot r hat is equal to 1."},{"Start":"07:34.775 ","End":"07:43.643","Text":"This is going to be equal to KQ divided by r^2 and it\u0027s my E field^2,"},{"Start":"07:43.643 ","End":"07:53.150","Text":"and then multiplied by this r^2 dr plus the integration over my next region,"},{"Start":"07:53.150 ","End":"07:57.680","Text":"which is from 2R until 3R,"},{"Start":"07:57.680 ","End":"08:06.515","Text":"and then the E field here is going to be equal to 3KQ divided by r^2,"},{"Start":"08:06.515 ","End":"08:12.649","Text":"because it\u0027s our E field squared multiplied by r^2 dr plus my final region,"},{"Start":"08:12.649 ","End":"08:16.290","Text":"which is from 3R until infinity,"},{"Start":"08:16.290 ","End":"08:19.590","Text":"but my electric field here is 0,"},{"Start":"08:19.590 ","End":"08:20.985","Text":"0^2 is 0 times r^2 dr,"},{"Start":"08:20.985 ","End":"08:22.290","Text":"so that\u0027s 0,"},{"Start":"08:22.290 ","End":"08:25.130","Text":"so this again crosses out."},{"Start":"08:25.130 ","End":"08:32.050","Text":"All of this is obviously multiplied by 2 Epsilon naught Pi."},{"Start":"08:33.410 ","End":"08:37.805","Text":"What we\u0027re going to do is we\u0027re not going to solve this."},{"Start":"08:37.805 ","End":"08:39.290","Text":"You\u0027re more than welcome to solve it."},{"Start":"08:39.290 ","End":"08:42.335","Text":"It\u0027s a simple integral because our r^2,"},{"Start":"08:42.335 ","End":"08:45.140","Text":"some of them cancel out and we\u0027re just integrating along"},{"Start":"08:45.140 ","End":"08:49.245","Text":"our r and then substituting in our bounds,"},{"Start":"08:49.245 ","End":"08:56.945","Text":"so our final onset that we\u0027re going to get is going to be equal to KQ^2 divided by"},{"Start":"08:56.945 ","End":"09:02.060","Text":"R. That\u0027s the answer"},{"Start":"09:02.060 ","End":"09:05.990","Text":"for the question of how much energy is required in order to build this system."},{"Start":"09:05.990 ","End":"09:09.500","Text":"I want to give another example of"},{"Start":"09:09.500 ","End":"09:15.650","Text":"a potential question or a sub-question that they could give when dealing with this."},{"Start":"09:15.650 ","End":"09:22.430","Text":"Let\u0027s imagine that they tell us that here in-between our 1st and 2nd spherical shells,"},{"Start":"09:22.430 ","End":"09:23.855","Text":"we have a point charge."},{"Start":"09:23.855 ","End":"09:26.655","Text":"Let\u0027s say it\u0027s called q_5, doesn\u0027t really matter,"},{"Start":"09:26.655 ","End":"09:31.748","Text":"and it\u0027s at a radius of 1.5"},{"Start":"09:31.748 ","End":"09:37.400","Text":"R. Then they\u0027re asking again,"},{"Start":"09:37.400 ","End":"09:42.455","Text":"given this point charge q_5 at this distance away from the center,"},{"Start":"09:42.455 ","End":"09:48.660","Text":"what is the energy now required in order to build the system?"},{"Start":"09:51.280 ","End":"09:54.700","Text":"As you\u0027ll remember from the previous lesson,"},{"Start":"09:54.700 ","End":"09:55.990","Text":"in the case like this,"},{"Start":"09:55.990 ","End":"09:57.385","Text":"where we have a point charge,"},{"Start":"09:57.385 ","End":"10:02.605","Text":"if it\u0027s just a point charge or a point charge within a system of other things,"},{"Start":"10:02.605 ","End":"10:06.065","Text":"this equation cannot be used,"},{"Start":"10:06.065 ","End":"10:10.130","Text":"and we spoke about the reason for that in the previous lesson."},{"Start":"10:10.130 ","End":"10:13.970","Text":"What do I do in a case like this?"},{"Start":"10:14.400 ","End":"10:17.095","Text":"In order to overcome this problem,"},{"Start":"10:17.095 ","End":"10:19.250","Text":"there\u0027s a nifty trick."},{"Start":"10:19.980 ","End":"10:24.355","Text":"I\u0027m going to imagine that first I built my system"},{"Start":"10:24.355 ","End":"10:27.960","Text":"of the spherical shells and I\u0027m ignoring the point charge and"},{"Start":"10:27.960 ","End":"10:35.065","Text":"the energy required in order to build that system is this over here."},{"Start":"10:35.065 ","End":"10:38.405","Text":"We just worked it out. Then I say, okay,"},{"Start":"10:38.405 ","End":"10:42.050","Text":"so now I have this and now what I\u0027m doing is I\u0027m bringing"},{"Start":"10:42.050 ","End":"10:47.510","Text":"this q_5 from infinity to this point over here,"},{"Start":"10:47.510 ","End":"10:52.340","Text":"and now I\u0027m going to see the energy required to bring this charge from infinity over"},{"Start":"10:52.340 ","End":"10:58.620","Text":"here given the system which is already here."},{"Start":"10:58.820 ","End":"11:03.905","Text":"If we remember the 2nd equation we spoke about,"},{"Start":"11:03.905 ","End":"11:13.605","Text":"to find the energy to build a system is given by 1/2 Sigma multiplied by our charge q_i,"},{"Start":"11:13.605 ","End":"11:16.185","Text":"multiplied by its potential."},{"Start":"11:16.185 ","End":"11:20.555","Text":"In this case where we have this added point charge,"},{"Start":"11:20.555 ","End":"11:25.475","Text":"so I would add to this energy of building the spherical shells system."},{"Start":"11:25.475 ","End":"11:30.650","Text":"I would simply add my charge because I only have 1 charge,"},{"Start":"11:30.650 ","End":"11:32.225","Text":"so I don\u0027t have to divide by 2,"},{"Start":"11:32.225 ","End":"11:39.985","Text":"so my charge, q_5 multiplied by the potential."},{"Start":"11:39.985 ","End":"11:45.020","Text":"Its potential when it\u0027s at this point over here,"},{"Start":"11:45.020 ","End":"11:49.881","Text":"which is at position 1.5R."},{"Start":"11:49.881 ","End":"11:56.350","Text":"That means that now I have to find its potential at that point."},{"Start":"11:56.350 ","End":"11:59.755","Text":"What do I do? How do I calculate the potential?"},{"Start":"11:59.755 ","End":"12:02.275","Text":"What I can do is I can either try and find"},{"Start":"12:02.275 ","End":"12:05.740","Text":"the general function for the potential of the system,"},{"Start":"12:05.740 ","End":"12:08.800","Text":"but that\u0027s a little bit more complicated."},{"Start":"12:08.800 ","End":"12:13.790","Text":"What I can do instead is I can find the potential difference."},{"Start":"12:15.120 ","End":"12:19.495","Text":"I want to find my potential over here."},{"Start":"12:19.495 ","End":"12:23.125","Text":"My potential at 1.5R,"},{"Start":"12:23.125 ","End":"12:28.510","Text":"and I already know that my potential over here at"},{"Start":"12:28.510 ","End":"12:35.245","Text":"radius 3R is equal to 0 because my 3rd spherical shell is grounded,"},{"Start":"12:35.245 ","End":"12:37.885","Text":"so my potential here is equal to 0."},{"Start":"12:37.885 ","End":"12:40.900","Text":"All I have to do is I have to integrate"},{"Start":"12:40.900 ","End":"12:44.200","Text":"in order to find the potential difference between this point,"},{"Start":"12:44.200 ","End":"12:50.305","Text":"and this point is I have to integrate along this blue dotted line."},{"Start":"12:50.305 ","End":"12:53.384","Text":"Let\u0027s write that here in blue."},{"Start":"12:53.384 ","End":"13:02.875","Text":"My potential at 1.5R minus my potential at 3R."},{"Start":"13:02.875 ","End":"13:07.580","Text":"Here obviously, my potential at 3R is equal to 0,"},{"Start":"13:08.340 ","End":"13:16.280","Text":"is going to be equal to the negative integral on E dot dr."},{"Start":"13:18.360 ","End":"13:23.890","Text":"Here because I know that everything is a uniform field,"},{"Start":"13:23.890 ","End":"13:25.529","Text":"so I can cross out my vectors."},{"Start":"13:25.529 ","End":"13:27.535","Text":"I\u0027m going to be integrating from"},{"Start":"13:27.535 ","End":"13:32.740","Text":"3R because I\u0027m starting here at 3R where my potential is equal to 0,"},{"Start":"13:32.740 ","End":"13:39.350","Text":"and I\u0027m going up until this point over here, which is 1.5R."},{"Start":"13:41.010 ","End":"13:49.105","Text":"We\u0027re going from our 3rd spherical shell into our 2nd spherical shell,"},{"Start":"13:49.105 ","End":"13:51.130","Text":"then for our second spherical shell,"},{"Start":"13:51.130 ","End":"13:53.185","Text":"we\u0027re going inwards as well."},{"Start":"13:53.185 ","End":"13:57.680","Text":"That\u0027s going through 2 regions."},{"Start":"13:57.840 ","End":"14:00.960","Text":"Due to the 2 different regions,"},{"Start":"14:00.960 ","End":"14:03.790","Text":"we\u0027re going to have different values for our electric field."},{"Start":"14:03.790 ","End":"14:06.475","Text":"That means that just like what we did over here,"},{"Start":"14:06.475 ","End":"14:11.140","Text":"we\u0027re going to split up our integral into 2 regions."},{"Start":"14:11.140 ","End":"14:17.650","Text":"That\u0027s going to be equal to the negative integral from 3R until 2R,"},{"Start":"14:17.650 ","End":"14:19.450","Text":"so that\u0027s this region."},{"Start":"14:19.450 ","End":"14:28.660","Text":"Then the electric fields there is 3KQ divided by r^2 dr,"},{"Start":"14:28.660 ","End":"14:35.440","Text":"and then minus plus the integral in this region,"},{"Start":"14:35.440 ","End":"14:37.390","Text":"so that\u0027s going to stay in negative."},{"Start":"14:37.390 ","End":"14:47.620","Text":"The integral from 2R until 1.5R and then the electric field over here"},{"Start":"14:47.620 ","End":"14:51.415","Text":"is given by KQ divided by"},{"Start":"14:51.415 ","End":"15:01.150","Text":"r^2 dr. Then once you find the answer to this integral and you substitute in the bounds,"},{"Start":"15:01.150 ","End":"15:05.965","Text":"you\u0027ll get your answer to what the potential is at 1.5R,"},{"Start":"15:05.965 ","End":"15:12.680","Text":"and then you just substitute that answer into here and you multiply Q5 by that."},{"Start":"15:13.440 ","End":"15:20.450","Text":"That\u0027s it. That will be the answer to this added question that they could ask you."},{"Start":"15:21.510 ","End":"15:26.110","Text":"That\u0027s the end of the question if we want to solve this,"},{"Start":"15:26.110 ","End":"15:29.020","Text":"using this equation for the energy,"},{"Start":"15:29.020 ","End":"15:32.800","Text":"but if we want to solve the entire question using this equation for the energy,"},{"Start":"15:32.800 ","End":"15:34.420","Text":"so I\u0027m going to do that now,"},{"Start":"15:34.420 ","End":"15:36.565","Text":"so you can either stop the video,"},{"Start":"15:36.565 ","End":"15:40.670","Text":"or you can carry on watching how I do it, like so."},{"Start":"15:41.250 ","End":"15:47.545","Text":"We\u0027re using this technique over here in order to find what our U is equal to."},{"Start":"15:47.545 ","End":"15:51.010","Text":"I\u0027ve left in our calculation for the electric field"},{"Start":"15:51.010 ","End":"15:54.970","Text":"that we did a few minutes ago because we\u0027re going to use that as well."},{"Start":"15:54.970 ","End":"15:59.860","Text":"We have our values for Q and all we have to do is we have to"},{"Start":"15:59.860 ","End":"16:05.035","Text":"find our potential in all the different regions that we have here."},{"Start":"16:05.035 ","End":"16:10.520","Text":"We can write that our potential is going to be equal to."},{"Start":"16:10.530 ","End":"16:17.185","Text":"What we can do is we can either integrate here for each region,"},{"Start":"16:17.185 ","End":"16:21.865","Text":"which is how we get from our electric field to our potential,"},{"Start":"16:21.865 ","End":"16:25.270","Text":"but also we know what the potential is within spherical shells,"},{"Start":"16:25.270 ","End":"16:29.575","Text":"we\u0027ve gone through this question a few times so we can just fill this in straight away."},{"Start":"16:29.575 ","End":"16:32.410","Text":"We know that we\u0027re going to have some C_1,"},{"Start":"16:32.410 ","End":"16:37.435","Text":"some constant in the region within the inner sphere."},{"Start":"16:37.435 ","End":"16:43.915","Text":"Then we\u0027re going to have KQ divided by r plus another constant."},{"Start":"16:43.915 ","End":"16:48.520","Text":"When we\u0027re located between the 2 spheres,"},{"Start":"16:48.520 ","End":"16:55.660","Text":"then we\u0027re going to have 3KQ divided by R plus C_3"},{"Start":"16:55.660 ","End":"16:58.840","Text":"when we\u0027re located between the 2nd"},{"Start":"16:58.840 ","End":"17:02.980","Text":"and the 3rd sphere and we\u0027re going to have another constant,"},{"Start":"17:02.980 ","End":"17:08.870","Text":"C_4 when we\u0027re located outside of our system."},{"Start":"17:10.370 ","End":"17:17.033","Text":"What we want to do, we want to find out what these constants are equal to."},{"Start":"17:17.033 ","End":"17:20.925","Text":"We\u0027ll start with our C_4."},{"Start":"17:20.925 ","End":"17:23.745","Text":"We say that our potential at infinity,"},{"Start":"17:23.745 ","End":"17:31.225","Text":"or anywhere outside of 3R is going to be equal to 0."},{"Start":"17:31.225 ","End":"17:36.865","Text":"Potential at infinity is 0 and because our 3rd spherical shell is grounded,"},{"Start":"17:36.865 ","End":"17:39.385","Text":"we know that the potential added is equal to 0."},{"Start":"17:39.385 ","End":"17:42.205","Text":"We can also just write that and therefore,"},{"Start":"17:42.205 ","End":"17:45.115","Text":"we get that our C_4 is equal to 0."},{"Start":"17:45.115 ","End":"17:47.950","Text":"Let\u0027s go on to find what our C_3 is,"},{"Start":"17:47.950 ","End":"17:51.970","Text":"we can do that this is greater or equal than and we can put,"},{"Start":"17:51.970 ","End":"17:54.745","Text":"but this is also greater or equal than."},{"Start":"17:54.745 ","End":"17:58.285","Text":"In order to find our potential here,"},{"Start":"17:58.285 ","End":"18:03.547","Text":"so we substitute in 3R into our equation over here,"},{"Start":"18:03.547 ","End":"18:09.530","Text":"because we want to find our potential right over here."},{"Start":"18:11.130 ","End":"18:16.810","Text":"We\u0027ll get 3KQ divided by 3R"},{"Start":"18:16.810 ","End":"18:23.780","Text":"plus C_3 and this is again meant to be equal to 0."},{"Start":"18:24.600 ","End":"18:34.045","Text":"That means therefore that S3 is equal to negative KQ divided by R,"},{"Start":"18:34.045 ","End":"18:37.190","Text":"because our 3s here cancel out."},{"Start":"18:37.680 ","End":"18:42.505","Text":"Then I\u0027m going to find C_2 and C_1 in the exact same way,"},{"Start":"18:42.505 ","End":"18:45.280","Text":"so I\u0027ve done that here, so I\u0027m just going to fill"},{"Start":"18:45.280 ","End":"18:48.820","Text":"in what all of these constants are equal to."},{"Start":"18:48.820 ","End":"18:54.715","Text":"Great so now we have the potential for each region,"},{"Start":"18:54.715 ","End":"18:58.720","Text":"so now what I\u0027m going to do is I\u0027m going to use this equation over here and"},{"Start":"18:58.720 ","End":"19:03.295","Text":"multiply every single charge by its potential,"},{"Start":"19:03.295 ","End":"19:05.840","Text":"so let\u0027s do that."},{"Start":"19:06.750 ","End":"19:13.240","Text":"My U is going to equal 1/2 multiplied by,"},{"Start":"19:13.240 ","End":"19:21.835","Text":"so my charge in the atmosphere is equal to Q multiplied by where it is in the region."},{"Start":"19:21.835 ","End":"19:24.535","Text":"It\u0027s located between 0 and R,"},{"Start":"19:24.535 ","End":"19:28.937","Text":"so we can multiply it by KQ divided by R."},{"Start":"19:28.937 ","End":"19:33.340","Text":"Because we have that everything is larger or equal to so"},{"Start":"19:33.340 ","End":"19:36.220","Text":"we could have also multiplied it by here and just substitute"},{"Start":"19:36.220 ","End":"19:42.100","Text":"an N R here in the denominator and we can see that we get the exact same answer,"},{"Start":"19:42.100 ","End":"19:47.854","Text":"but it doesn\u0027t really matter, so then plus our next charge, which is 2Q."},{"Start":"19:47.854 ","End":"19:51.620","Text":"2Q is located at 2,"},{"Start":"19:51.620 ","End":"19:54.940","Text":"so we can use either this equation here, this equation here,"},{"Start":"19:54.940 ","End":"20:03.730","Text":"where in each place we substitute in 2R because that\u0027s its location into the denominator."},{"Start":"20:03.730 ","End":"20:08.005","Text":"Let\u0027s use this equation over here because it\u0027s easier to work with instead of minusing,"},{"Start":"20:08.005 ","End":"20:14.035","Text":"so I\u0027ll have 2Q multiplied by KQ divided by 2R."},{"Start":"20:14.035 ","End":"20:21.910","Text":"Then we\u0027ll have our charge located on the outer sphere."},{"Start":"20:21.910 ","End":"20:26.635","Text":"Let\u0027s call our charge on the outer sphere question-mark."},{"Start":"20:26.635 ","End":"20:29.110","Text":"We can try and work it out,"},{"Start":"20:29.110 ","End":"20:31.315","Text":"but it doesn\u0027t really matter and this is why,"},{"Start":"20:31.315 ","End":"20:34.720","Text":"so let\u0027s say that it\u0027s called charge question mark and imagine,"},{"Start":"20:34.720 ","End":"20:38.170","Text":"oh no, we\u0027re going to have to work it out soon, let\u0027s leave it here."},{"Start":"20:38.170 ","End":"20:40.450","Text":"But now we have to multiply it by"},{"Start":"20:40.450 ","End":"20:44.215","Text":"its potential and its potential because it\u0027s located at 3R,"},{"Start":"20:44.215 ","End":"20:46.795","Text":"its potential here is equal to 0,"},{"Start":"20:46.795 ","End":"20:49.240","Text":"so it\u0027s anyway going to cancel out."},{"Start":"20:49.240 ","End":"20:51.500","Text":"It doesn\u0027t really matter."},{"Start":"20:52.620 ","End":"20:55.210","Text":"Once we simplify this out,"},{"Start":"20:55.210 ","End":"20:56.500","Text":"you can do the algebra,"},{"Start":"20:56.500 ","End":"21:01.330","Text":"will get that answer is KQ^2 divided by R,"},{"Start":"21:01.330 ","End":"21:02.725","Text":"which if you remember,"},{"Start":"21:02.725 ","End":"21:05.545","Text":"is also the answer that we got when using"},{"Start":"21:05.545 ","End":"21:10.550","Text":"the 1st equation dealing with the electric field."},{"Start":"21:10.950 ","End":"21:16.615","Text":"We can use both techniques and we\u0027ll get to the exact same answer."},{"Start":"21:16.615 ","End":"21:19.780","Text":"All we did for this equation over here is"},{"Start":"21:19.780 ","End":"21:23.110","Text":"we found our electric field and all of the regions,"},{"Start":"21:23.110 ","End":"21:28.240","Text":"then we integrated our electric field,"},{"Start":"21:28.240 ","End":"21:31.960","Text":"in order to get our potential."},{"Start":"21:31.960 ","End":"21:39.460","Text":"Then we got our potential and then all we did is we multiplied by 1/2 the charge,"},{"Start":"21:39.460 ","End":"21:44.155","Text":"wherever it is, multiplied by its potential at that position,"},{"Start":"21:44.155 ","End":"21:47.030","Text":"that\u0027s the end of this lesson."}],"ID":22355},{"Watched":false,"Name":"Exercise 2","Duration":"15m 55s","ChapterTopicVideoID":21452,"CourseChapterTopicPlaylistID":99478,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.055","Text":"Hello. In this lesson,"},{"Start":"00:02.055 ","End":"00:05.280","Text":"we\u0027re going to be answering the following question."},{"Start":"00:05.280 ","End":"00:07.980","Text":"2 spherical drops of water,"},{"Start":"00:07.980 ","End":"00:12.510","Text":"each have a radius R and a uniform surface charge Q."},{"Start":"00:12.510 ","End":"00:17.865","Text":"The 2 water droplets are joined together to form a new, larger water droplet."},{"Start":"00:17.865 ","End":"00:22.065","Text":"The charge is still uniformly distributed on the surface."},{"Start":"00:22.065 ","End":"00:24.288","Text":"Question number 1,"},{"Start":"00:24.288 ","End":"00:29.730","Text":"what are the 2 smaller droplets\u0027 electrical potential energy"},{"Start":"00:29.730 ","End":"00:33.555","Text":"before being joined together?"},{"Start":"00:33.555 ","End":"00:38.225","Text":"We have 2 charged spheres,"},{"Start":"00:38.225 ","End":"00:46.940","Text":"each one with a radius R and a positive charge Q on the surface, like so."},{"Start":"00:46.940 ","End":"00:49.500","Text":"We want to find the energy."},{"Start":"00:49.500 ","End":"00:59.680","Text":"U_T is equal to 2 times the energy U of each droplet."},{"Start":"01:00.290 ","End":"01:05.560","Text":"Let\u0027s write here droplet."},{"Start":"01:05.980 ","End":"01:13.774","Text":"U total is equal to 2 times the energy of each droplet,"},{"Start":"01:13.774 ","End":"01:23.360","Text":"where the energy comes from this charge that is distributed on a spherical surface."},{"Start":"01:23.360 ","End":"01:33.420","Text":"That is equal to 1/2 times K multiplied by the charge Q times the charge here, Q,"},{"Start":"01:33.420 ","End":"01:42.575","Text":"so Q^2 divided by the radius R. Then we can cross these 2 out,"},{"Start":"01:42.575 ","End":"01:48.350","Text":"and this will be equal to simply KQ^2 divided by"},{"Start":"01:48.350 ","End":"01:55.670","Text":"R. This is the electric potential energy of the 2 drops together."},{"Start":"01:55.670 ","End":"01:59.460","Text":"Let\u0027s answer Question number 2."},{"Start":"01:59.810 ","End":"02:05.825","Text":"We\u0027re being asked, what is the electric potential energy of the large droplet?"},{"Start":"02:05.825 ","End":"02:11.835","Text":"The large droplet is when the 2 smaller droplets are joined together."},{"Start":"02:11.835 ","End":"02:20.060","Text":"We have this larger droplets where the total charge is Q plus Q,"},{"Start":"02:20.060 ","End":"02:23.780","Text":"so it\u0027s positive 2Q."},{"Start":"02:23.780 ","End":"02:26.870","Text":"The first thing that we need to do"},{"Start":"02:26.870 ","End":"02:31.550","Text":"is we have to find out the radius of this water droplet."},{"Start":"02:31.550 ","End":"02:36.248","Text":"This radius is not going to be just 2R,"},{"Start":"02:36.248 ","End":"02:38.910","Text":"that\u0027s not the answer."},{"Start":"02:38.910 ","End":"02:40.895","Text":"How are we going to work this out?"},{"Start":"02:40.895 ","End":"02:45.895","Text":"We\u0027re going to use the idea of conservation of volume."},{"Start":"02:45.895 ","End":"02:50.675","Text":"If this has a certain volume and this has a certain volume,"},{"Start":"02:50.675 ","End":"02:52.175","Text":"when we join the 2,"},{"Start":"02:52.175 ","End":"02:54.080","Text":"we\u0027re just going to be adding up,"},{"Start":"02:54.080 ","End":"02:57.995","Text":"scalar addition of the volumes."},{"Start":"02:57.995 ","End":"03:01.565","Text":"The volume of this system is conserved."},{"Start":"03:01.565 ","End":"03:05.940","Text":"Then through that, we\u0027re going to find out the radius."},{"Start":"03:07.700 ","End":"03:14.830","Text":"V tag is the volume of this new larger water droplet."},{"Start":"03:14.830 ","End":"03:23.160","Text":"It is equal to 2 times the volume of the smaller water droplet."},{"Start":"03:23.160 ","End":"03:26.960","Text":"We have the volume of this water droplet V and we\u0027re"},{"Start":"03:26.960 ","End":"03:32.280","Text":"multiplying it by 2 because we have 2 droplets joining together."},{"Start":"03:32.360 ","End":"03:34.640","Text":"The droplets are identical,"},{"Start":"03:34.640 ","End":"03:36.870","Text":"so they have the same volume."},{"Start":"03:37.900 ","End":"03:43.955","Text":"We know that the volume of a sphere is equal to"},{"Start":"03:43.955 ","End":"03:52.680","Text":"4Pi R here tag cubed divided by 3."},{"Start":"03:52.680 ","End":"03:58.220","Text":"This is equal to 2 times the volume of the smaller droplets,"},{"Start":"03:58.220 ","End":"04:06.870","Text":"which is 4Pi the radius of the smaller droplet is R^3 divided by 3."},{"Start":"04:07.480 ","End":"04:13.100","Text":"We can divide both sides by 4Pi divided by 3,"},{"Start":"04:13.100 ","End":"04:16.640","Text":"so these cross out."},{"Start":"04:16.640 ","End":"04:22.160","Text":"Then what we\u0027re left with is that R^3 the new"},{"Start":"04:22.160 ","End":"04:27.515","Text":"radius^3 is equal to 2R^3,"},{"Start":"04:27.515 ","End":"04:29.885","Text":"the radius of the smaller droplets."},{"Start":"04:29.885 ","End":"04:34.835","Text":"Therefore we\u0027ll get the R tag is equal to"},{"Start":"04:34.835 ","End":"04:40.865","Text":"the cubed root of 2 multiplied by R,"},{"Start":"04:40.865 ","End":"04:43.980","Text":"the radius of the smaller droplet."},{"Start":"04:44.540 ","End":"04:50.840","Text":"Let\u0027s work out the electric potential energy."},{"Start":"04:50.840 ","End":"04:56.945","Text":"Again, it\u0027s distributed uniformly on the surface of the sphere."},{"Start":"04:56.945 ","End":"05:03.314","Text":"That means that the energy is going to be equal"},{"Start":"05:03.314 ","End":"05:09.255","Text":"to 1/2 of KQ^2 Q tag."},{"Start":"05:09.255 ","End":"05:11.830","Text":"This is the new charge over here."},{"Start":"05:11.830 ","End":"05:13.420","Text":"Let\u0027s call it Q tag."},{"Start":"05:13.420 ","End":"05:20.255","Text":"We\u0027ll plug in the 2Q in a second divided by R tag over here."},{"Start":"05:20.255 ","End":"05:25.310","Text":"So that is equal to K. Then Q tag is equal"},{"Start":"05:25.310 ","End":"05:32.640","Text":"to 2Q^2 divided by 2R tag,"},{"Start":"05:32.640 ","End":"05:42.750","Text":"where R tag is the cubed root of 2 multiplied by R. This"},{"Start":"05:42.750 ","End":"05:49.650","Text":"is going to be equal to 4KQ^2 divided by 2"},{"Start":"05:49.650 ","End":"05:58.035","Text":"multiplied by the cubed root of 2 multiplied by R. We can cross this out with this."},{"Start":"05:58.035 ","End":"06:06.845","Text":"The final answer is equal to 2KQ^2 divided by"},{"Start":"06:06.845 ","End":"06:11.375","Text":"the cubed root of 2 multiplied by"},{"Start":"06:11.375 ","End":"06:17.825","Text":"R. Let\u0027s answer Question number 3."},{"Start":"06:17.825 ","End":"06:19.445","Text":"Question number 3 is,"},{"Start":"06:19.445 ","End":"06:23.300","Text":"what is the electric potential energy of the system"},{"Start":"06:23.300 ","End":"06:28.000","Text":"of 2 droplets right before they were joined together?"},{"Start":"06:28.000 ","End":"06:33.440","Text":"That means that the droplets are almost touching each other, but they aren\u0027t."},{"Start":"06:33.440 ","End":"06:38.330","Text":"Assume that the charge distribution is still uniform."},{"Start":"06:38.330 ","End":"06:45.120","Text":"I\u0027m going to scroll down and let\u0027s explain this question."},{"Start":"06:45.550 ","End":"06:52.390","Text":"What we did in Question 1 was to find the total energy of these 2 drops,"},{"Start":"06:52.390 ","End":"06:55.470","Text":"needed to build these 2 drops up."},{"Start":"06:55.470 ","End":"07:01.415","Text":"Both of these water drops or water droplets were very far away from each other."},{"Start":"07:01.415 ","End":"07:04.790","Text":"Each was at 1 end of infinity."},{"Start":"07:04.790 ","End":"07:10.230","Text":"What I calculated was the energy to build up this water droplet."},{"Start":"07:11.770 ","End":"07:18.680","Text":"Then I joined those 2 energies together and these 2 droplets are so far away from each"},{"Start":"07:18.680 ","End":"07:27.220","Text":"other that they aren\u0027t acting or influencing one another."},{"Start":"07:27.380 ","End":"07:29.870","Text":"In Question 3,"},{"Start":"07:29.870 ","End":"07:31.640","Text":"scroll up a little bit."},{"Start":"07:31.640 ","End":"07:34.640","Text":"I\u0027m being asked to find the energy of"},{"Start":"07:34.640 ","End":"07:38.390","Text":"the system of the 2 droplets right before they are joined."},{"Start":"07:38.390 ","End":"07:41.495","Text":"That means that they\u0027re almost touching each other."},{"Start":"07:41.495 ","End":"07:48.480","Text":"What I\u0027ve done is I\u0027m now moving these 2 droplets towards one another."},{"Start":"07:48.480 ","End":"07:56.614","Text":"Let\u0027s draw that. I have a system can imagine these are 2 spheres,"},{"Start":"07:56.614 ","End":"08:02.195","Text":"each one with a radius R and where the positive charge Q,"},{"Start":"08:02.195 ","End":"08:05.900","Text":"uniformly distributed on the surface."},{"Start":"08:05.900 ","End":"08:08.590","Text":"They\u0027re almost touching."},{"Start":"08:08.590 ","End":"08:13.220","Text":"That means that my total energy for the system,"},{"Start":"08:13.220 ","End":"08:22.171","Text":"so U total is going to be equal to the energy of the droplets."},{"Start":"08:22.171 ","End":"08:27.340","Text":"Let\u0027s just call this U total of the whole system."},{"Start":"08:27.340 ","End":"08:30.760","Text":"It\u0027s the total energy of the 2 droplets together,"},{"Start":"08:30.760 ","End":"08:34.490","Text":"which is what we found in Question number 1."},{"Start":"08:34.560 ","End":"08:40.705","Text":"The energy needed or the work done in"},{"Start":"08:40.705 ","End":"08:47.545","Text":"order to bring these 2 droplets together or close to one another."},{"Start":"08:47.545 ","End":"08:54.590","Text":"U_c, U to bring the drops c closer to one another."},{"Start":"08:55.920 ","End":"09:01.480","Text":"We know what the energy to build the droplets is."},{"Start":"09:01.480 ","End":"09:04.555","Text":"We work that out in Question number 1."},{"Start":"09:04.555 ","End":"09:07.570","Text":"The energy to bring them closer,"},{"Start":"09:07.570 ","End":"09:09.235","Text":"to bring the droplets together,"},{"Start":"09:09.235 ","End":"09:10.810","Text":"we have to find out."},{"Start":"09:10.810 ","End":"09:17.450","Text":"This can be a little bit complicated so let\u0027s explain the logic behind it."},{"Start":"09:19.080 ","End":"09:26.000","Text":"Let\u0027s call this Droplet 1 and this is Droplet 2."},{"Start":"09:26.040 ","End":"09:35.895","Text":"We know from Newton that the force that Droplet number 1 applies on Droplet number 2."},{"Start":"09:35.895 ","End":"09:37.773","Text":"Or the size of the force,"},{"Start":"09:37.773 ","End":"09:38.970","Text":"if we take out the direction,"},{"Start":"09:38.970 ","End":"09:43.260","Text":"just the magnitude of the force is equal to the magnitude of"},{"Start":"09:43.260 ","End":"09:49.879","Text":"the force that Droplet number 2 applies onto Droplet number 1."},{"Start":"09:51.210 ","End":"09:55.795","Text":"The second thing that we have to realize is because we\u0027re being told"},{"Start":"09:55.795 ","End":"10:00.025","Text":"that the charge is uniformly distributed on the surface."},{"Start":"10:00.025 ","End":"10:05.515","Text":"That means that if we look outside our spherical droplets,"},{"Start":"10:05.515 ","End":"10:07.195","Text":"outside of one of them,"},{"Start":"10:07.195 ","End":"10:09.460","Text":"then the electric field,"},{"Start":"10:09.460 ","End":"10:19.240","Text":"and therefore the force experienced by some object or a charge outside"},{"Start":"10:19.240 ","End":"10:25.990","Text":"of the droplet is going to be the same electric field or force that that object or charge"},{"Start":"10:25.990 ","End":"10:32.830","Text":"outside would experience where the droplets just a single point charge."},{"Start":"10:32.830 ","End":"10:36.850","Text":"I\u0027ve written down that the E field,"},{"Start":"10:36.850 ","End":"10:39.265","Text":"the electric field of each droplet,"},{"Start":"10:39.265 ","End":"10:40.510","Text":"is uniform."},{"Start":"10:40.510 ","End":"10:42.280","Text":"This is important."},{"Start":"10:42.280 ","End":"10:51.090","Text":"Therefore, the charge outside or a charge outside of the droplet located,"},{"Start":"10:51.090 ","End":"10:53.190","Text":"let\u0027s say over here,"},{"Start":"10:53.190 ","End":"11:02.620","Text":"will experience an E field of a point charge located at the center of the droplet."},{"Start":"11:02.620 ","End":"11:05.575","Text":"In other words, we can consider,"},{"Start":"11:05.575 ","End":"11:11.830","Text":"because the charge is uniformly distributed along the surface of the sphere."},{"Start":"11:11.830 ","End":"11:21.410","Text":"We can consider each sphere as just point charges of charge Q, each one."},{"Start":"11:21.450 ","End":"11:25.480","Text":"Because the E field from"},{"Start":"11:25.480 ","End":"11:30.430","Text":"these point charges is the same as"},{"Start":"11:30.430 ","End":"11:35.860","Text":"the E field outside of the sphere."},{"Start":"11:35.860 ","End":"11:42.350","Text":"Where the location, with the radius greater than I."},{"Start":"11:42.750 ","End":"11:47.810","Text":"These 2 cases are the same."},{"Start":"11:49.110 ","End":"11:52.075","Text":"These are equivalent."},{"Start":"11:52.075 ","End":"11:54.520","Text":"Therefore, let\u0027s scroll down a little bit."},{"Start":"11:54.520 ","End":"11:57.940","Text":"Our final scenario is when we"},{"Start":"11:57.940 ","End":"12:04.525","Text":"have both of these very close together that they\u0027re almost touching but aren\u0027t."},{"Start":"12:04.525 ","End":"12:13.600","Text":"There is approximately a distance of R from the radius of each sphere between them."},{"Start":"12:13.600 ","End":"12:17.510","Text":"This total distance over here."},{"Start":"12:19.020 ","End":"12:23.690","Text":"This distance is equal to 2R."},{"Start":"12:24.090 ","End":"12:31.030","Text":"Here\u0027s the outline of the spherical or the sphere around each"},{"Start":"12:31.030 ","End":"12:37.570","Text":"of these point charges with the radius R so the distance between these 2 charges is 2R."},{"Start":"12:37.570 ","End":"12:43.135","Text":"Therefore, the energy is equal to what we already know."},{"Start":"12:43.135 ","End":"12:45.130","Text":"We can also calculate it,"},{"Start":"12:45.130 ","End":"12:50.260","Text":"but we already know that this is equal to KQ1 multiplied by"},{"Start":"12:50.260 ","End":"12:56.115","Text":"Q2 divided by the R distance between the 2 charges."},{"Start":"12:56.115 ","End":"13:01.515","Text":"In our case, this is equal to K and each one has a charge of Q,"},{"Start":"13:01.515 ","End":"13:07.900","Text":"so Q^2 and the distance between the 2 is 2R."},{"Start":"13:10.110 ","End":"13:15.865","Text":"This is this. Let\u0027s write the final answer."},{"Start":"13:15.865 ","End":"13:24.550","Text":"The total energy of the system is equal to U T,"},{"Start":"13:24.550 ","End":"13:27.415","Text":"which is what we got in Question number 1."},{"Start":"13:27.415 ","End":"13:32.180","Text":"That was KQ^2 divided by R."},{"Start":"13:32.180 ","End":"13:38.850","Text":"KQ^2 divided by R plus the energy to bring the droplets together."},{"Start":"13:38.850 ","End":"13:44.930","Text":"U C, which is K Q squared divided by 2R."},{"Start":"13:44.940 ","End":"13:53.870","Text":"In total, this is equal to 3KQ^2 divided by 2R."},{"Start":"13:55.050 ","End":"13:57.970","Text":"This is the answer to Question 3."},{"Start":"13:57.970 ","End":"14:01.135","Text":"Let\u0027s look at Question 4."},{"Start":"14:01.135 ","End":"14:06.250","Text":"What is the relationship between your answers to Question 2 and 3?"},{"Start":"14:06.250 ","End":"14:07.840","Text":"Before I typed it wrong,"},{"Start":"14:07.840 ","End":"14:10.825","Text":"it was between Question 1 and 3, I corrected it."},{"Start":"14:10.825 ","End":"14:12.655","Text":"Let\u0027s take a look."},{"Start":"14:12.655 ","End":"14:15.400","Text":"Questions 2 and 3."},{"Start":"14:15.400 ","End":"14:18.820","Text":"Here\u0027s what we got 2 and here\u0027s what we got in 3."},{"Start":"14:18.820 ","End":"14:20.815","Text":"When we ask for the relationship,"},{"Start":"14:20.815 ","End":"14:23.245","Text":"we\u0027re just asking for the ratio."},{"Start":"14:23.245 ","End":"14:28.330","Text":"We\u0027re asking for the ratio here between Question 2."},{"Start":"14:28.330 ","End":"14:33.010","Text":"The energy in Question 2 and the energy that we received in Question 3."},{"Start":"14:33.010 ","End":"14:50.080","Text":"In Question 2, this is meant to be 2 divided"},{"Start":"14:50.080 ","End":"14:59.905","Text":"by 3 root 2 multiplied by KQ^2 divided by R."},{"Start":"14:59.905 ","End":"15:06.835","Text":"The energy that we go in 3 is 3 divided by 2 multiplied by"},{"Start":"15:06.835 ","End":"15:13.810","Text":"KQ^2 divided by R. Then we can cancel this out."},{"Start":"15:13.810 ","End":"15:22.870","Text":"We get a ratio of 2 divided by the cubed root of 2 divided by 3 over 2."},{"Start":"15:22.870 ","End":"15:26.155","Text":"Which once we plug this into the calculator,"},{"Start":"15:26.155 ","End":"15:31.370","Text":"is approximately equal to 1.058."},{"Start":"15:33.810 ","End":"15:36.820","Text":"This is a ratio or relationships,"},{"Start":"15:36.820 ","End":"15:38.635","Text":"so it doesn\u0027t have any units."},{"Start":"15:38.635 ","End":"15:41.545","Text":"This is the answer to Question number 4."},{"Start":"15:41.545 ","End":"15:45.580","Text":"We can see that both of the values that we got in"},{"Start":"15:45.580 ","End":"15:50.620","Text":"Question 2 and Question 3 are pretty similar to one another."},{"Start":"15:50.620 ","End":"15:52.915","Text":"This is the ratio between them."},{"Start":"15:52.915 ","End":"15:55.760","Text":"That\u0027s the end of this lesson."}],"ID":22356}],"Thumbnail":null,"ID":99478}]

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