[{"Name":"Introduction to Coulombs Law","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"1 Coulomb_s Law","Duration":"11m 21s","ChapterTopicVideoID":21425,"CourseChapterTopicPlaylistID":99473,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/21425.jpeg","UploadDate":"2020-04-21T08:42:39.7830000","DurationForVideoObject":"PT11M21S","Description":null,"MetaTitle":"1 Coulomb_s Law: Video + Workbook | Proprep","MetaDescription":"Coulombs Law - Introduction to Coulombs Law. 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/coulombs-law/introduction-to-coulombs-law/vid22259","VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.430","Text":"Hello. In today\u0027s lesson,"},{"Start":"00:02.430 ","End":"00:04.920","Text":"we\u0027re going to be learning about Coulomb\u0027s law."},{"Start":"00:04.920 ","End":"00:14.910","Text":"Now, Coulomb\u0027s law speaks about the force of interaction between charges q_1 and q_2."},{"Start":"00:14.910 ","End":"00:18.269","Text":"Coulomb\u0027s law is over here,"},{"Start":"00:18.269 ","End":"00:19.905","Text":"this is the equation in red,"},{"Start":"00:19.905 ","End":"00:26.085","Text":"and it says that the force between charges q_1 and q_2 is equal to k,"},{"Start":"00:26.085 ","End":"00:30.270","Text":"which is some constant given like so,"},{"Start":"00:30.270 ","End":"00:33.540","Text":"multiplied by the charge of q_1,"},{"Start":"00:33.540 ","End":"00:38.235","Text":"multiplied by the charge of q_2 divided by r squared,"},{"Start":"00:38.235 ","End":"00:43.260","Text":"where r is the distance between q_1 and q_2,"},{"Start":"00:43.260 ","End":"00:48.795","Text":"multiplied by the direction that our r is going in,"},{"Start":"00:48.795 ","End":"00:51.360","Text":"the unit vector for that."},{"Start":"00:51.360 ","End":"00:53.505","Text":"We can also write it,"},{"Start":"00:53.505 ","End":"00:56.745","Text":"this means the exact same thing, as kq_1q_2,"},{"Start":"00:56.745 ","End":"01:00.659","Text":"but this time divided by r cubed because here,"},{"Start":"01:00.659 ","End":"01:05.715","Text":"it\u0027s being multiplied by r vector instead of the r unit vector."},{"Start":"01:05.715 ","End":"01:13.080","Text":"Our constant k over here is equal to 1 divided by 4Pi Epsilon naught,"},{"Start":"01:13.080 ","End":"01:18.310","Text":"where Epsilon naught is also some constant."},{"Start":"01:18.520 ","End":"01:22.650","Text":"What\u0027s important to note over here is this,"},{"Start":"01:22.650 ","End":"01:24.510","Text":"our force, which is Coulomb\u0027s law,"},{"Start":"01:24.510 ","End":"01:28.845","Text":"which is equal to kq_1q_2 divided by r squared r hat,"},{"Start":"01:28.845 ","End":"01:34.155","Text":"or kq_1q_2 divided by r cubed r vector."},{"Start":"01:34.155 ","End":"01:38.415","Text":"Where our k is some constant which is given as so,"},{"Start":"01:38.415 ","End":"01:41.580","Text":"and Epsilon naught is also a constant."},{"Start":"01:41.580 ","End":"01:45.615","Text":"Now, you\u0027ll see in all of the equations either k or Epsilon naught,"},{"Start":"01:45.615 ","End":"01:48.220","Text":"and they\u0027re always going to be given."},{"Start":"01:48.830 ","End":"01:52.425","Text":"Let\u0027s take a look at this diagram over here."},{"Start":"01:52.425 ","End":"01:55.440","Text":"If we have our charges q_1 and q_2,"},{"Start":"01:55.440 ","End":"01:57.090","Text":"our r vector,"},{"Start":"01:57.090 ","End":"01:58.830","Text":"which is this one over here,"},{"Start":"01:58.830 ","End":"02:04.530","Text":"is the vector pointing from q_1 to q_2."},{"Start":"02:04.530 ","End":"02:06.795","Text":"Then this distance,"},{"Start":"02:06.795 ","End":"02:09.900","Text":"if we\u0027re going to just look at the size of our r vector,"},{"Start":"02:09.900 ","End":"02:14.860","Text":"this is r, which is simply the size of our r vector."},{"Start":"02:15.530 ","End":"02:20.250","Text":"Now, our force is also a vector quantity and it\u0027s always"},{"Start":"02:20.250 ","End":"02:24.570","Text":"going to be pointing in the same direction as our r vector."},{"Start":"02:24.570 ","End":"02:27.165","Text":"You can imagine it as some arrow,"},{"Start":"02:27.165 ","End":"02:30.760","Text":"continuation of our r vector arrow."},{"Start":"02:31.130 ","End":"02:34.320","Text":"Sometimes, in some textbooks,"},{"Start":"02:34.320 ","End":"02:40.170","Text":"this force over here is called F_2,1,"},{"Start":"02:40.170 ","End":"02:47.220","Text":"so that means the force on charge number 2 applied by charge number 1."},{"Start":"02:47.220 ","End":"02:54.255","Text":"Then of course, we\u0027ll also have a force equal and opposite in this direction,"},{"Start":"02:54.255 ","End":"03:00.310","Text":"which will be the force on charge 1 being applied by charge 2."},{"Start":"03:01.400 ","End":"03:05.625","Text":"This, of course, comes from Newton\u0027s third law."},{"Start":"03:05.625 ","End":"03:09.660","Text":"If we have a force in 1 direction,"},{"Start":"03:09.660 ","End":"03:13.680","Text":"then there will be an equal and opposite force."},{"Start":"03:13.680 ","End":"03:20.830","Text":"Newton\u0027s third law is also applicable to electricity in between charges."},{"Start":"03:20.900 ","End":"03:25.379","Text":"In the meantime, we can ignore this over here because we\u0027re specifically"},{"Start":"03:25.379 ","End":"03:29.595","Text":"speaking about the force from q_1 to q_2,"},{"Start":"03:29.595 ","End":"03:34.950","Text":"because our vector r is pointing from q_1 until q_2,"},{"Start":"03:34.950 ","End":"03:39.180","Text":"and so that means that our Coulomb force is going to"},{"Start":"03:39.180 ","End":"03:44.200","Text":"be pointing in the same direction that our r vector is pointing into."},{"Start":"03:45.230 ","End":"03:48.345","Text":"Now we\u0027ve discussed Coulomb\u0027s law,"},{"Start":"03:48.345 ","End":"03:52.110","Text":"which is just giving us the force between 2 charged particles,"},{"Start":"03:52.110 ","End":"03:56.145","Text":"but now what we\u0027re going to speak about is our electric field."},{"Start":"03:56.145 ","End":"04:00.210","Text":"Now, our electric field is very important and we\u0027re usually going to be"},{"Start":"04:00.210 ","End":"04:06.765","Text":"speaking during electricity and magnetism in the exam questions,"},{"Start":"04:06.765 ","End":"04:13.510","Text":"they\u0027re generally going to be asking you about electric field rather than the force."},{"Start":"04:14.030 ","End":"04:17.490","Text":"Let\u0027s speak about the electric field now."},{"Start":"04:17.490 ","End":"04:20.445","Text":"If we\u0027re going to look at the exact same diagram,"},{"Start":"04:20.445 ","End":"04:23.310","Text":"but this time our q_2,"},{"Start":"04:23.310 ","End":"04:26.130","Text":"which was meant to be over here, isn\u0027t."},{"Start":"04:26.130 ","End":"04:29.920","Text":"It\u0027s the exact same diagram just without the q_2."},{"Start":"04:30.170 ","End":"04:36.330","Text":"Now, what I want to do in the absence of my q_2 is I want to define"},{"Start":"04:36.330 ","End":"04:45.650","Text":"some quantity which will later on help me find if I did have some test particle."},{"Start":"04:45.650 ","End":"04:49.325","Text":"Over here, this quantity will help me"},{"Start":"04:49.325 ","End":"04:55.860","Text":"define what the force between my q_1 and my test particle here will be."},{"Start":"04:56.450 ","End":"05:01.055","Text":"Now, this quantity which is going to help me define"},{"Start":"05:01.055 ","End":"05:04.355","Text":"what force there will be between"},{"Start":"05:04.355 ","End":"05:07.985","Text":"my q_1 and some particle if I place it here in the future,"},{"Start":"05:07.985 ","End":"05:13.730","Text":"is called the electric field and it is called E-field."},{"Start":"05:13.730 ","End":"05:20.720","Text":"This is equal to our force from our Coulomb\u0027s law divided by q."},{"Start":"05:20.720 ","End":"05:23.600","Text":"What we have to remember is that our force,"},{"Start":"05:23.600 ","End":"05:26.585","Text":"our Coulomb\u0027s law, is equal to qE."},{"Start":"05:26.585 ","End":"05:33.439","Text":"Here specifically, our electric field is going to be equal to our force,"},{"Start":"05:33.439 ","End":"05:41.520","Text":"which is kq_1 multiplied by some test charged particle over here,"},{"Start":"05:41.520 ","End":"05:45.180","Text":"so let\u0027s call it q, divided by r squared,"},{"Start":"05:45.180 ","End":"05:49.825","Text":"and then divided by this charged particle q."},{"Start":"05:49.825 ","End":"05:53.164","Text":"Here, in this specific example,"},{"Start":"05:53.164 ","End":"06:01.200","Text":"our electric field is going to be equal to kq_1 divided by r squared."},{"Start":"06:02.720 ","End":"06:06.000","Text":"Now, I have my E quantity,"},{"Start":"06:06.000 ","End":"06:09.075","Text":"which is this expression over here."},{"Start":"06:09.075 ","End":"06:10.745","Text":"Now, if in the future,"},{"Start":"06:10.745 ","End":"06:13.205","Text":"I were to put some charged particle"},{"Start":"06:13.205 ","End":"06:17.840","Text":"a distance r away and call this charged particle q_2,"},{"Start":"06:17.840 ","End":"06:26.640","Text":"so I would therefore get that my force is going to be equal to my new charge,"},{"Start":"06:26.640 ","End":"06:30.335","Text":"so that\u0027s q_2, multiplied by my E-field,"},{"Start":"06:30.335 ","End":"06:36.385","Text":"which was kq_1 divided by r squared in the r direction."},{"Start":"06:36.385 ","End":"06:41.930","Text":"Then we can look at this and we can see that this is our equation for the force,"},{"Start":"06:41.930 ","End":"06:44.190","Text":"which is what we saw before."},{"Start":"06:46.580 ","End":"06:53.210","Text":"The electric field is some theoretical function that describes"},{"Start":"06:53.210 ","End":"07:00.280","Text":"what will happen if we add another charge into this system."},{"Start":"07:00.280 ","End":"07:05.480","Text":"The electric field can predict that if I put a charged particle over here,"},{"Start":"07:05.480 ","End":"07:08.870","Text":"due to this charged particle over here,"},{"Start":"07:08.870 ","End":"07:13.400","Text":"I can predict what the force is going to be between the 2 particles."},{"Start":"07:13.400 ","End":"07:18.290","Text":"My Coulomb\u0027s law is just the definition of the force between"},{"Start":"07:18.290 ","End":"07:21.080","Text":"the 2 particles once another particle"},{"Start":"07:21.080 ","End":"07:25.140","Text":"is placed somewhere next to my first charged particle."},{"Start":"07:27.080 ","End":"07:31.609","Text":"Now, let\u0027s assume that our charged particle is positive."},{"Start":"07:31.609 ","End":"07:34.805","Text":"If our charged particle is going to be negatively charged,"},{"Start":"07:34.805 ","End":"07:37.970","Text":"then they\u0027ll write a negative sign."},{"Start":"07:37.970 ","End":"07:41.335","Text":"This is a positively charged particle."},{"Start":"07:41.335 ","End":"07:44.195","Text":"Whenever we have a positively charged particle,"},{"Start":"07:44.195 ","End":"07:51.020","Text":"the electric field lines point away from the particle."},{"Start":"07:51.020 ","End":"07:59.870","Text":"Because we can see that our electric field is proportional to 1 divided by r squared,"},{"Start":"07:59.870 ","End":"08:03.785","Text":"we can see that the further away we get from the charged particle,"},{"Start":"08:03.785 ","End":"08:07.810","Text":"our electric field will be weaker and weaker."},{"Start":"08:07.810 ","End":"08:11.990","Text":"That means that the closer we are to our charged particle,"},{"Start":"08:11.990 ","End":"08:14.585","Text":"the stronger the electric field is."},{"Start":"08:14.585 ","End":"08:18.635","Text":"A positively charged particles,"},{"Start":"08:18.635 ","End":"08:23.880","Text":"so the electric field lines go out and away from the particle,"},{"Start":"08:23.880 ","End":"08:28.325","Text":"and the further we get from the particle,"},{"Start":"08:28.325 ","End":"08:33.180","Text":"the lower the electric field or the weaker the electric field is."},{"Start":"08:33.580 ","End":"08:35.915","Text":"Again, just so we\u0027re clear,"},{"Start":"08:35.915 ","End":"08:40.445","Text":"if suddenly we\u0027re being told that our q_1 is a negatively charged particle,"},{"Start":"08:40.445 ","End":"08:43.730","Text":"so all of our arrows will point inwards."},{"Start":"08:43.730 ","End":"08:45.334","Text":"Ignore the orange."},{"Start":"08:45.334 ","End":"08:46.925","Text":"If our particle is negative,"},{"Start":"08:46.925 ","End":"08:48.920","Text":"all the blue arrows,"},{"Start":"08:48.920 ","End":"08:53.090","Text":"the electric field lines, will point inwards."},{"Start":"08:53.900 ","End":"08:56.850","Text":"Let\u0027s do a quick little recap."},{"Start":"08:56.850 ","End":"09:02.900","Text":"We said that Coulomb\u0027s law is the force of interaction between charges q_1 and q_2."},{"Start":"09:02.900 ","End":"09:06.325","Text":"Then it\u0027s given by this equation."},{"Start":"09:06.325 ","End":"09:14.990","Text":"What is useful to remember here is that our r hat is equal to r vector divided by r,"},{"Start":"09:14.990 ","End":"09:19.920","Text":"and then you can jump between these 2 versions of the same equation."},{"Start":"09:19.920 ","End":"09:27.410","Text":"We\u0027re remembering that our k is some constant given by 1 divided by 4Pi Epsilon naught,"},{"Start":"09:27.410 ","End":"09:30.279","Text":"where Epsilon naught is also a constant."},{"Start":"09:30.279 ","End":"09:34.955","Text":"Now, our r vector is our vector pointing between"},{"Start":"09:34.955 ","End":"09:40.018","Text":"our first charged particle and until our second charged particle,"},{"Start":"09:40.018 ","End":"09:43.130","Text":"and our force that is being experienced between"},{"Start":"09:43.130 ","End":"09:48.965","Text":"these 2 charged particles is pointing in the same direction as our r vector."},{"Start":"09:48.965 ","End":"09:52.790","Text":"Then we spoke about our electric field,"},{"Start":"09:52.790 ","End":"09:55.325","Text":"which is some theoretical function,"},{"Start":"09:55.325 ","End":"10:01.775","Text":"which will define the field strength around some charged particle."},{"Start":"10:01.775 ","End":"10:06.875","Text":"Which means that it can give a prediction that if we have 1 charged particle,"},{"Start":"10:06.875 ","End":"10:09.530","Text":"and then we know the electric field around it,"},{"Start":"10:09.530 ","End":"10:16.460","Text":"then if we play some test charged particle anywhere along the electric field lines,"},{"Start":"10:16.460 ","End":"10:24.760","Text":"then we can foresee what the force will be between those 2 charged particles."},{"Start":"10:24.760 ","End":"10:29.750","Text":"We saw that it\u0027s according to this equation over here."},{"Start":"10:29.750 ","End":"10:34.910","Text":"So this is what\u0027s important to remember from the electric field equation,"},{"Start":"10:34.910 ","End":"10:38.044","Text":"which means that we also have to remember this equation over here,"},{"Start":"10:38.044 ","End":"10:41.585","Text":"and for a single charged particle,"},{"Start":"10:41.585 ","End":"10:48.014","Text":"so the electric field is simply kq divided by r squared in the r hat direction,"},{"Start":"10:48.014 ","End":"10:51.710","Text":"and of course, we can also write out the equation like this."},{"Start":"10:51.710 ","End":"10:56.830","Text":"We know that the electric field is the force per unit charge."},{"Start":"10:56.830 ","End":"10:59.720","Text":"The last 2 important things to"},{"Start":"10:59.720 ","End":"11:03.615","Text":"remember is that if we have a positively charged particle,"},{"Start":"11:03.615 ","End":"11:08.630","Text":"our electric field lines are pointing out and away from the charged particle."},{"Start":"11:08.630 ","End":"11:11.150","Text":"If we have a negatively charged particle,"},{"Start":"11:11.150 ","End":"11:16.260","Text":"the electric field lines are pointing in towards the particle."},{"Start":"11:17.630 ","End":"11:20.950","Text":"That\u0027s the end of our lesson."}],"ID":22259},{"Watched":false,"Name":"Exercise 1 - Continuous Charge Distribution - Bent Wire","Duration":"27m 41s","ChapterTopicVideoID":21426,"CourseChapterTopicPlaylistID":99473,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.045","Text":"Hello, in this lesson we\u0027re going to be dealing with a continuous charge distribution."},{"Start":"00:06.045 ","End":"00:12.765","Text":"The idea here is to break our shape into tiny little pieces."},{"Start":"00:12.765 ","End":"00:23.210","Text":"We can regard each shape as a point charge and calculate the field via Coulomb\u0027s Law."},{"Start":"00:23.210 ","End":"00:33.615","Text":"What we can do is we break this up into little pieces where each piece has a charge, dq."},{"Start":"00:33.615 ","End":"00:39.335","Text":"Then the total field that will come out of this place,"},{"Start":"00:39.335 ","End":"00:43.625","Text":"the small point charge dq is going to be k,"},{"Start":"00:43.625 ","End":"00:45.290","Text":"the charge of that piece,"},{"Start":"00:45.290 ","End":"00:54.040","Text":"which is dq, divided by its distance squared from wherever we want to look in space."},{"Start":"00:54.040 ","End":"00:56.820","Text":"Our exercise here is,"},{"Start":"00:56.820 ","End":"01:06.920","Text":"we have an infinite wire which is charged with charge density per unit length, Lambda."},{"Start":"01:06.920 ","End":"01:12.445","Text":"The wire is bent to form a semicircle of radius r at the bend."},{"Start":"01:12.445 ","End":"01:14.075","Text":"Then we\u0027re being asked,"},{"Start":"01:14.075 ","End":"01:18.545","Text":"what is the electric field at the center of the semicircle?"},{"Start":"01:18.545 ","End":"01:21.390","Text":"At this point over here."},{"Start":"01:22.220 ","End":"01:24.335","Text":"This is our plan."},{"Start":"01:24.335 ","End":"01:29.990","Text":"We\u0027re going to use this idea or what we have over here in order to solve this exercise."},{"Start":"01:29.990 ","End":"01:31.910","Text":"Now, it\u0027s good to note how we do this,"},{"Start":"01:31.910 ","End":"01:35.570","Text":"because this idea of how to solve this exercise is going to come"},{"Start":"01:35.570 ","End":"01:41.270","Text":"up a lot when dealing with electricity and magnetism."},{"Start":"01:41.270 ","End":"01:43.705","Text":"How are we going to solve this?"},{"Start":"01:43.705 ","End":"01:48.249","Text":"We\u0027re going to split up our infinite wire into three sections."},{"Start":"01:48.249 ","End":"01:50.950","Text":"We\u0027re going to have a semi-infinite wire going in"},{"Start":"01:50.950 ","End":"01:54.310","Text":"this direction another semi-infinite wire here,"},{"Start":"01:54.310 ","End":"01:57.040","Text":"and then a semicircle."},{"Start":"01:57.040 ","End":"02:00.580","Text":"Then we\u0027re going to work out the electric field at this point"},{"Start":"02:00.580 ","End":"02:03.610","Text":"here from those three areas where we can already see from"},{"Start":"02:03.610 ","End":"02:06.490","Text":"symmetry that the electric field from"},{"Start":"02:06.490 ","End":"02:10.555","Text":"here is going to be the same as the electric field from here on this point."},{"Start":"02:10.555 ","End":"02:15.260","Text":"Then we\u0027ll work out the electric field due to the semicircle."},{"Start":"02:15.260 ","End":"02:21.205","Text":"Just to note, we are allowed to split this whole wire into three sections."},{"Start":"02:21.205 ","End":"02:28.535","Text":"Just like we\u0027re allowed to split up anything into small areas of point charge, dq."},{"Start":"02:28.535 ","End":"02:33.870","Text":"Then we can add everything up according to the superposition principle."},{"Start":"02:34.570 ","End":"02:40.505","Text":"Let\u0027s start off now by working out the electric field at this point"},{"Start":"02:40.505 ","End":"02:45.980","Text":"over here for this upper section of wire,"},{"Start":"02:45.980 ","End":"02:47.825","Text":"which corresponds to this section."},{"Start":"02:47.825 ","End":"02:54.135","Text":"Now what I\u0027m going to do is I\u0027m going to define my axes."},{"Start":"02:54.135 ","End":"02:58.805","Text":"What I\u0027m going to say is that my origin is at this point over here."},{"Start":"02:58.805 ","End":"03:00.605","Text":"If this is the origin,"},{"Start":"03:00.605 ","End":"03:04.475","Text":"and then this is the positive y-direction,"},{"Start":"03:04.475 ","End":"03:08.660","Text":"and this is the positive x-direction and I\u0027m defining this as the origin."},{"Start":"03:08.660 ","End":"03:09.950","Text":"We don\u0027t always have to do that,"},{"Start":"03:09.950 ","End":"03:13.630","Text":"but specifically in this question, it\u0027s a useful."},{"Start":"03:13.630 ","End":"03:15.690","Text":"Now what I\u0027m going to do is,"},{"Start":"03:15.690 ","End":"03:20.615","Text":"I\u0027m going to split my wire up into tiny pieces."},{"Start":"03:20.615 ","End":"03:23.945","Text":"Here this is an extremely small length,"},{"Start":"03:23.945 ","End":"03:27.655","Text":"and this length has a charge of dq."},{"Start":"03:27.655 ","End":"03:33.675","Text":"Now what I\u0027m going to do is I\u0027m going to see what my dq is equal to."},{"Start":"03:33.675 ","End":"03:38.450","Text":"My dq is going to be my charge density per"},{"Start":"03:38.450 ","End":"03:44.340","Text":"unit length multiplied by that unit length. That\u0027s the dl."},{"Start":"03:44.340 ","End":"03:48.545","Text":"Now I have to choose what my dl is going to be equal to."},{"Start":"03:48.545 ","End":"03:51.170","Text":"Specifically here, we can see that"},{"Start":"03:51.170 ","End":"03:55.160","Text":"my length is according to the coordinate system that we\u0027re using,"},{"Start":"03:55.160 ","End":"03:58.110","Text":"is going in the x direction."},{"Start":"03:58.110 ","End":"04:01.425","Text":"I can say that my dl=dx."},{"Start":"04:01.425 ","End":"04:08.700","Text":"Therefore my dq=Lambda dx."},{"Start":"04:08.700 ","End":"04:11.875","Text":"Now we can begin to use Coulomb\u0027s Law."},{"Start":"04:11.875 ","End":"04:17.860","Text":"Now, sometimes you can work out Coulomb\u0027s Law straight away with magnitude and direction."},{"Start":"04:17.860 ","End":"04:21.955","Text":"The force or here specifically the field."},{"Start":"04:21.955 ","End":"04:27.920","Text":"But right now I\u0027m going to use the tried and tested way of first working out"},{"Start":"04:27.920 ","End":"04:30.335","Text":"the magnitude of the electric field and then"},{"Start":"04:30.335 ","End":"04:34.665","Text":"afterwards we can speak about in which direction it\u0027s pointing."},{"Start":"04:34.665 ","End":"04:40.385","Text":"The magnitude of the tiny electric field that"},{"Start":"04:40.385 ","End":"04:47.900","Text":"is due to this point charge dq is equal to,"},{"Start":"04:47.900 ","End":"04:51.418","Text":"as we know, k multiplied by the charge,"},{"Start":"04:51.418 ","End":"04:57.240","Text":"it\u0027s dq divided by the distance^2."},{"Start":"04:58.130 ","End":"05:04.550","Text":"The distance between my point charge over here and this dot over"},{"Start":"05:04.550 ","End":"05:11.470","Text":"here is r. Now a quick note before we continue."},{"Start":"05:11.470 ","End":"05:17.770","Text":"Always remember to choose your dq or where you\u0027re going to start doing"},{"Start":"05:17.770 ","End":"05:21.910","Text":"your calculations in some general point"},{"Start":"05:21.910 ","End":"05:26.840","Text":"along whichever shape or body or object you\u0027re calculating."},{"Start":"05:26.840 ","End":"05:33.355","Text":"Because if we choose something specific such as if my dq is right at this edge over here,"},{"Start":"05:33.355 ","End":"05:40.210","Text":"then we won\u0027t be able to use a general case and find out the total electric field,"},{"Start":"05:40.210 ","End":"05:46.555","Text":"because then our r vector would simply be this r value over here,"},{"Start":"05:46.555 ","End":"05:51.980","Text":"which as we can see, is incorrect as we move along the wire."},{"Start":"05:52.940 ","End":"06:03.650","Text":"Always choose some arbitrary distance x in one of your directions."},{"Start":"06:05.060 ","End":"06:12.710","Text":"Our magnitude of the electric field caused by this tiny piece is equal to this over here."},{"Start":"06:12.710 ","End":"06:16.290","Text":"What exactly is our r^2?"},{"Start":"06:16.290 ","End":"06:21.350","Text":"This is our r vector and of course we don\u0027t really care right now about the direction."},{"Start":"06:21.350 ","End":"06:25.340","Text":"What matters to us is the magnitude of this r vector,"},{"Start":"06:25.340 ","End":"06:34.040","Text":"which means the distance between this point charge dq to the center of the semicircle."},{"Start":"06:34.040 ","End":"06:42.630","Text":"We can say that the magnitude of our r vector is going to be equal to."},{"Start":"06:42.630 ","End":"06:46.415","Text":"Here we can see that we have a right angled triangle."},{"Start":"06:46.415 ","End":"06:55.250","Text":"We\u0027ll have the square root of r^2 plus we have this arbitrary length x,"},{"Start":"06:55.250 ","End":"07:02.250","Text":"so x^2 and this is the distance between these two points."},{"Start":"07:03.410 ","End":"07:07.925","Text":"Now we can see that the electric field caused by"},{"Start":"07:07.925 ","End":"07:12.620","Text":"this charged point charge on this point over here"},{"Start":"07:12.620 ","End":"07:17.925","Text":"in space is going to go along in this direction."},{"Start":"07:17.925 ","End":"07:21.185","Text":"This is our dE vector."},{"Start":"07:21.185 ","End":"07:25.475","Text":"Now before I start integrating to find the total electric field,"},{"Start":"07:25.475 ","End":"07:28.490","Text":"at this point, the first thing that I have to do"},{"Start":"07:28.490 ","End":"07:31.910","Text":"is I have to split this up into components,"},{"Start":"07:31.910 ","End":"07:35.410","Text":"because we can see that our dq has"},{"Start":"07:35.410 ","End":"07:41.600","Text":"some component of electric field going in this direction,"},{"Start":"07:41.600 ","End":"07:43.460","Text":"in the negative y direction."},{"Start":"07:43.460 ","End":"07:49.054","Text":"It also has some component of electric field going in this direction,"},{"Start":"07:49.054 ","End":"07:52.440","Text":"which is the negative x direction."},{"Start":"07:54.320 ","End":"08:01.500","Text":"What we want to do is we can also split this into the components of our dE."},{"Start":"08:01.500 ","End":"08:04.640","Text":"We see that we have component here and a component here,"},{"Start":"08:04.640 ","End":"08:07.970","Text":"which is the exact same thing as what\u0027s happening over here."},{"Start":"08:07.970 ","End":"08:13.055","Text":"Now, we can see that this angle over here,"},{"Start":"08:13.055 ","End":"08:15.245","Text":"let\u0027s call it Alpha."},{"Start":"08:15.245 ","End":"08:23.145","Text":"Then we can say that this angle over here is also Alpha."},{"Start":"08:23.145 ","End":"08:25.850","Text":"Don\u0027t worry, Alpha isn\u0027t given in our question."},{"Start":"08:25.850 ","End":"08:30.470","Text":"But soon we\u0027re going to solve this because from our question and from the geometry,"},{"Start":"08:30.470 ","End":"08:32.935","Text":"we can find out what Alpha is."},{"Start":"08:32.935 ","End":"08:41.280","Text":"Let\u0027s begin with our DE in the x-direction."},{"Start":"08:41.280 ","End":"08:45.415","Text":"This is the x component of our electric field."},{"Start":"08:45.415 ","End":"08:52.200","Text":"That\u0027s this gray line over here or this gray arrow over here."},{"Start":"08:52.400 ","End":"08:58.150","Text":"We can see that it\u0027s pointing in the negative x direction, so write negative."},{"Start":"08:58.150 ","End":"09:02.180","Text":"Then it\u0027s going to be the magnitude of our dE."},{"Start":"09:02.180 ","End":"09:06.565","Text":"Magnitude of dE multiplied by,"},{"Start":"09:06.565 ","End":"09:08.140","Text":"so we want the x component,"},{"Start":"09:08.140 ","End":"09:10.595","Text":"which is this length over here."},{"Start":"09:10.595 ","End":"09:16.160","Text":"This is adjacent to our Alpha angle over here."},{"Start":"09:16.160 ","End":"09:21.410","Text":"We\u0027re going to be using cosine of our angle."},{"Start":"09:21.440 ","End":"09:23.660","Text":"Let\u0027s see what this is equal to."},{"Start":"09:23.660 ","End":"09:27.160","Text":"This is negative the magnitude of"},{"Start":"09:27.160 ","End":"09:33.380","Text":"our electric field multiplied by now what\u0027s cosine of the angle?"},{"Start":"09:33.380 ","End":"09:42.880","Text":"Cosine of Alpha is equal to our adjacent side divided by our hypotenuse."},{"Start":"09:42.880 ","End":"09:47.725","Text":"Our adjacent side, we can see is this arbitrary length x,"},{"Start":"09:47.725 ","End":"09:50.380","Text":"which we chose before,"},{"Start":"09:50.380 ","End":"09:52.600","Text":"divided by our hypotenuse."},{"Start":"09:52.600 ","End":"09:54.415","Text":"Now what\u0027s our hypotenuse?"},{"Start":"09:54.415 ","End":"09:57.655","Text":"It\u0027s the magnitude of our r vector,"},{"Start":"09:57.655 ","End":"09:59.560","Text":"which is what we chose over here."},{"Start":"09:59.560 ","End":"10:02.255","Text":"Let\u0027s just write instead of magnitude of our vector."},{"Start":"10:02.255 ","End":"10:03.780","Text":"Let\u0027s write this."},{"Start":"10:03.780 ","End":"10:11.820","Text":"Now let\u0027s substitute in what our dE is."},{"Start":"10:11.820 ","End":"10:15.250","Text":"We have negative, our dE=kdq divided by r^2,"},{"Start":"10:19.610 ","End":"10:24.375","Text":"and then multiplied by x divided by r,"},{"Start":"10:24.375 ","End":"10:27.660","Text":"which was our cosine Alpha."},{"Start":"10:27.660 ","End":"10:30.915","Text":"Now, we can rewrite this."},{"Start":"10:30.915 ","End":"10:32.415","Text":"What\u0027s our dq?"},{"Start":"10:32.415 ","End":"10:34.020","Text":"Well, we wrote that over here,"},{"Start":"10:34.020 ","End":"10:35.145","Text":"so let\u0027s write that as well."},{"Start":"10:35.145 ","End":"10:42.285","Text":"Negative kx divided by r^3 multiplied by our dq,"},{"Start":"10:42.285 ","End":"10:46.720","Text":"which is Lambda dx."},{"Start":"10:47.450 ","End":"10:54.090","Text":"Our last final step is substituting in our r^3."},{"Start":"10:54.090 ","End":"10:59.835","Text":"We\u0027ll have negative kx Lambda divided by,"},{"Start":"10:59.835 ","End":"11:05.145","Text":"so we\u0027re going to have r^2 plus x^2."},{"Start":"11:05.145 ","End":"11:12.705","Text":"Then half, because the square root is also equal to what\u0027s inside to the power of half."},{"Start":"11:12.705 ","End":"11:14.280","Text":"But then we also have a cubed."},{"Start":"11:14.280 ","End":"11:17.820","Text":"It\u0027s going to be to the power of 3 over 2dx."},{"Start":"11:20.150 ","End":"11:26.085","Text":"What\u0027s important to note here is that our angle over here, Alpha,"},{"Start":"11:26.085 ","End":"11:30.495","Text":"is changing as our dq moves further away,"},{"Start":"11:30.495 ","End":"11:34.095","Text":"or as our value for x is increasing."},{"Start":"11:34.095 ","End":"11:36.855","Text":"As we integrate along this wire,"},{"Start":"11:36.855 ","End":"11:38.805","Text":"our Alpha is going to be changing."},{"Start":"11:38.805 ","End":"11:43.245","Text":"That\u0027s why we have to express it with respect to x,"},{"Start":"11:43.245 ","End":"11:48.540","Text":"which works out perfectly because we\u0027re also going to be integrating with respect to x,"},{"Start":"11:48.540 ","End":"11:51.630","Text":"as can be seen by our dq."},{"Start":"11:51.630 ","End":"11:55.240","Text":"That\u0027s why this works out perfectly."},{"Start":"11:56.060 ","End":"11:59.160","Text":"Now we\u0027re ready to integrate."},{"Start":"11:59.160 ","End":"12:05.340","Text":"We can say that the total electric field or the x component of"},{"Start":"12:05.340 ","End":"12:12.720","Text":"the electric field is going to be equal to the integration of dE_x,"},{"Start":"12:12.720 ","End":"12:20.470","Text":"which is going to be equal to the integration of negative k Lambda x divided by"},{"Start":"12:20.470 ","End":"12:28.770","Text":"R^2 plus x squared^3/2 dx."},{"Start":"12:28.770 ","End":"12:33.310","Text":"The bounds are going to be from zero until infinity."},{"Start":"12:33.310 ","End":"12:36.515","Text":"Because we\u0027re beginning at zero over here,"},{"Start":"12:36.515 ","End":"12:40.640","Text":"x=0 because remember, our origin is at this point over here."},{"Start":"12:40.640 ","End":"12:45.270","Text":"Then the end of the wire is, obviously, infinity."},{"Start":"12:45.980 ","End":"12:50.310","Text":"Now let\u0027s solve this integration."},{"Start":"12:50.310 ","End":"12:55.170","Text":"Now, this is a very common integral."},{"Start":"12:55.170 ","End":"13:00.120","Text":"When we have some constants multiplied by"},{"Start":"13:00.120 ","End":"13:06.520","Text":"x divided by a constant plus x squared^3/2."},{"Start":"13:06.950 ","End":"13:10.365","Text":"You should memorize how to solve this."},{"Start":"13:10.365 ","End":"13:13.710","Text":"Now, what we can do is, we can say,"},{"Start":"13:13.710 ","End":"13:16.125","Text":"we can define a new variable, u,"},{"Start":"13:16.125 ","End":"13:20.770","Text":"and say that that is equal to what is inside the brackets, R^2 plus x^2."},{"Start":"13:21.650 ","End":"13:25.390","Text":"Therefore our du=2xdx."},{"Start":"13:29.300 ","End":"13:35.160","Text":"Then that means that we can rewrite our integral from zero up until infinity,"},{"Start":"13:35.160 ","End":"13:42.190","Text":"of negative k Lambda, multiplied by"},{"Start":"13:50.360 ","End":"13:56.640","Text":"dt divided by 2 divided by t^3/2. Then we"},{"Start":"13:56.640 ","End":"14:01.305","Text":"can just integrate and substitute in our bounds,"},{"Start":"14:01.305 ","End":"14:08.295","Text":"and we will get that our answer is negative k Lambda divided by"},{"Start":"14:08.295 ","End":"14:17.220","Text":"R. This is the electric field in the x direction."},{"Start":"14:17.220 ","End":"14:22.665","Text":"Now what we want to do is we want to find the electric field in the y direction."},{"Start":"14:22.665 ","End":"14:25.680","Text":"Let\u0027s write this really quickly."},{"Start":"14:25.680 ","End":"14:28.725","Text":"Imagine that instead of x we have y."},{"Start":"14:28.725 ","End":"14:32.295","Text":"That means that instead of having cosine of Alpha,"},{"Start":"14:32.295 ","End":"14:35.300","Text":"we\u0027ll have sine of Alpha,"},{"Start":"14:35.300 ","End":"14:40.400","Text":"which means that instead of x divided by r will have"},{"Start":"14:40.400 ","End":"14:50.230","Text":"R divided by r. Then we\u0027ll get some value we can just solve here."},{"Start":"14:51.250 ","End":"14:57.290","Text":"Then, all we have to do is simply integrate to find our electric field in"},{"Start":"14:57.290 ","End":"15:03.630","Text":"the y direction when we substitute in these values over here."},{"Start":"15:03.770 ","End":"15:10.035","Text":"But before we start doing that and writing everything out and doing the integration,"},{"Start":"15:10.035 ","End":"15:15.080","Text":"we can notice that in this question specifically, we have symmetry."},{"Start":"15:15.080 ","End":"15:19.160","Text":"We can see that our y component of the electric field over here,"},{"Start":"15:19.160 ","End":"15:24.705","Text":"due to the semi-infinite wires cancels out."},{"Start":"15:24.705 ","End":"15:28.335","Text":"Let\u0027s take a look at why that is happening."},{"Start":"15:28.335 ","End":"15:34.544","Text":"We can see that this dq is applying an electric field in this direction."},{"Start":"15:34.544 ","End":"15:38.160","Text":"Then if we take exactly opposite it, point over here,"},{"Start":"15:38.160 ","End":"15:44.625","Text":"this point charge dq is applying an electric field in this direction."},{"Start":"15:44.625 ","End":"15:48.930","Text":"Then we can see that if we split this up into components,"},{"Start":"15:48.930 ","End":"15:53.295","Text":"so we have the y over here and the x over here,"},{"Start":"15:53.295 ","End":"15:56.430","Text":"and then here we have the y like so and the x like so."},{"Start":"15:56.430 ","End":"16:01.710","Text":"We can see that the y components of the electric field will cancel with each other."},{"Start":"16:01.710 ","End":"16:04.440","Text":"However, the x components will add up."},{"Start":"16:04.440 ","End":"16:08.295","Text":"We\u0027ll have double the electric field along the x-axis,"},{"Start":"16:08.295 ","End":"16:15.375","Text":"and zero electric field along the y-axis for these two semi-infinite wires."},{"Start":"16:15.375 ","End":"16:20.080","Text":"That\u0027s great. We\u0027ve saved ourselves a lot of time."},{"Start":"16:20.090 ","End":"16:29.850","Text":"In that case, let\u0027s call this dE tag and E_x tag because this is just due to one wire."},{"Start":"16:29.850 ","End":"16:35.130","Text":"Then we can write that our total electric field in the x direction is"},{"Start":"16:35.130 ","End":"16:41.010","Text":"simply going to be 2 times our E_x tag."},{"Start":"16:41.010 ","End":"16:44.970","Text":"Once in the x direction from the upper wire"},{"Start":"16:44.970 ","End":"16:48.825","Text":"and once again in the x direction from the lower wire."},{"Start":"16:48.825 ","End":"16:55.755","Text":"That is going to be negative 2k Lambda divided by R,"},{"Start":"16:55.755 ","End":"16:57.960","Text":"and our E_y,"},{"Start":"16:57.960 ","End":"17:01.875","Text":"our electric field in the y direction as we can see,"},{"Start":"17:01.875 ","End":"17:03.570","Text":"is going to be canceled out."},{"Start":"17:03.570 ","End":"17:09.090","Text":"That\u0027s going to be equal to zero."},{"Start":"17:09.090 ","End":"17:12.810","Text":"Now, all we have to do is work out what the electric field on"},{"Start":"17:12.810 ","End":"17:17.400","Text":"this point over here is going to be due to this semicircle."},{"Start":"17:17.400 ","End":"17:23.560","Text":"Here we can see our semicircular bend."},{"Start":"17:23.560 ","End":"17:29.155","Text":"Now I\u0027ve drawn it a little bit bigger so that our diagram will be slightly clearer."},{"Start":"17:29.155 ","End":"17:31.705","Text":"Just like before with infinite wires,"},{"Start":"17:31.705 ","End":"17:38.110","Text":"what I\u0027m going to do is I\u0027m going to split up my bend into small areas."},{"Start":"17:38.110 ","End":"17:41.500","Text":"This length is going to be dl."},{"Start":"17:41.500 ","End":"17:45.670","Text":"Now careful, this dl is not going to be in terms of"},{"Start":"17:45.670 ","End":"17:50.001","Text":"dx because we can see that we have circular coordinates,"},{"Start":"17:50.001 ","End":"17:55.030","Text":"which means that as we move along this circular bend,"},{"Start":"17:55.030 ","End":"17:59.020","Text":"we\u0027re going to have x and y components."},{"Start":"17:59.020 ","End":"18:05.180","Text":"In the meantime, let\u0027s just leave it at this length is dl."},{"Start":"18:05.340 ","End":"18:07.960","Text":"Now we can draw a line,"},{"Start":"18:07.960 ","End":"18:11.860","Text":"this section dl with some charge dq,"},{"Start":"18:11.860 ","End":"18:14.770","Text":"is going to be applying"},{"Start":"18:14.770 ","End":"18:23.800","Text":"an electric field dE on this point over here."},{"Start":"18:23.800 ","End":"18:28.640","Text":"Then, if we say that this is our origin,"},{"Start":"18:28.640 ","End":"18:36.715","Text":"let\u0027s draw that this is the y direction and this is the x direction for over here."},{"Start":"18:36.715 ","End":"18:39.295","Text":"We can therefore say,"},{"Start":"18:39.295 ","End":"18:41.350","Text":"if we carry this one down,"},{"Start":"18:41.350 ","End":"18:45.370","Text":"that this angle over here is Theta."},{"Start":"18:45.370 ","End":"18:47.725","Text":"Doesn\u0027t matter what Theta is yet."},{"Start":"18:47.725 ","End":"18:53.870","Text":"We can also say that this angle over here is also Theta."},{"Start":"18:55.350 ","End":"18:58.780","Text":"Now what do we want to work out is,"},{"Start":"18:58.780 ","End":"19:03.140","Text":"what is our dq for over here?"},{"Start":"19:04.650 ","End":"19:11.455","Text":"Our dq for the section of the semicircle is again going to be equal to Lambda dl."},{"Start":"19:11.455 ","End":"19:16.555","Text":"Now this time we\u0027re working in polar coordinates,"},{"Start":"19:16.555 ","End":"19:20.050","Text":"because we\u0027re always going to have some kind of"},{"Start":"19:20.050 ","End":"19:24.310","Text":"x or y component so it\u0027s easier to work in polar coordinates."},{"Start":"19:24.310 ","End":"19:29.905","Text":"Our dl is going to be equal to Rd Theta."},{"Start":"19:29.905 ","End":"19:35.785","Text":"Now here I\u0027ve written a capital R because the radius of the semicircle is constant."},{"Start":"19:35.785 ","End":"19:37.270","Text":"However, in some questions,"},{"Start":"19:37.270 ","End":"19:40.630","Text":"maybe you\u0027ll have some spiral or something else,"},{"Start":"19:40.630 ","End":"19:41.785","Text":"in which case you can write"},{"Start":"19:41.785 ","End":"19:48.475","Text":"a small lowercase rd Theta to represent that the R is also changing."},{"Start":"19:48.475 ","End":"19:53.830","Text":"But here specifically the radius is constant and all that\u0027s changing is our angle Theta."},{"Start":"19:53.830 ","End":"19:57.700","Text":"Now again, we\u0027re going to choose our section here,"},{"Start":"19:57.700 ","End":"19:59.440","Text":"our dl or our dq,"},{"Start":"19:59.440 ","End":"20:02.545","Text":"not to be at some endpoint,"},{"Start":"20:02.545 ","End":"20:08.320","Text":"but at some general arbitrary position along our curve so that we can find"},{"Start":"20:08.320 ","End":"20:15.340","Text":"some general answer or a general solution to this problem."},{"Start":"20:15.340 ","End":"20:22.075","Text":"Now again, we\u0027re going to be finding the magnitude of our dE."},{"Start":"20:22.075 ","End":"20:24.970","Text":"Our dE is a vector quantity,"},{"Start":"20:24.970 ","End":"20:28.585","Text":"which means there\u0027s meant to be an arrow on top over here."},{"Start":"20:28.585 ","End":"20:31.090","Text":"However, it\u0027s sometimes easier to first work out"},{"Start":"20:31.090 ","End":"20:35.200","Text":"the magnitude and then later find out the direction."},{"Start":"20:35.200 ","End":"20:41.440","Text":"The magnitude is again going to be equal to k dq"},{"Start":"20:41.440 ","End":"20:47.650","Text":"divided by the distance between our point charge and this point over here,"},{"Start":"20:47.650 ","End":"20:50.830","Text":"which is the center of the semicircle."},{"Start":"20:50.830 ","End":"20:54.880","Text":"We know that the distance is always going to be R,"},{"Start":"20:54.880 ","End":"20:57.670","Text":"the radius of the semicircle."},{"Start":"20:57.670 ","End":"21:01.790","Text":"We\u0027re going to divide this by R^2."},{"Start":"21:01.860 ","End":"21:05.125","Text":"Now let\u0027s substitute in what our dq is."},{"Start":"21:05.125 ","End":"21:14.140","Text":"We have k Lambda Rd Theta divided by R^2."},{"Start":"21:14.140 ","End":"21:20.995","Text":"We can cross off one of the Rs so we\u0027ll have k Lambda d Theta divided by"},{"Start":"21:20.995 ","End":"21:31.285","Text":"R. Now what we want to do is we want to find our dE in the x direction."},{"Start":"21:31.285 ","End":"21:39.890","Text":"That is going to be equal to the magnitude of our dE multiplied by,"},{"Start":"21:40.080 ","End":"21:46.195","Text":"here because our Theta is relative to the y-axis,"},{"Start":"21:46.195 ","End":"21:50.103","Text":"usually in the x direction we\u0027ll write cosine of the angle,"},{"Start":"21:50.103 ","End":"21:54.760","Text":"however, here, because our Theta is relative to the y-axis,"},{"Start":"21:54.760 ","End":"22:00.535","Text":"we\u0027re going to write multiplied by sine of the angle, which is sine Theta."},{"Start":"22:00.535 ","End":"22:03.700","Text":"As we can see, if we split up these components,"},{"Start":"22:03.700 ","End":"22:12.770","Text":"we have our x component going this way and our y component going this way."},{"Start":"22:12.770 ","End":"22:17.010","Text":"Which means that our x component is in the positive x direction,"},{"Start":"22:17.010 ","End":"22:20.295","Text":"which means that this is fine."},{"Start":"22:20.295 ","End":"22:30.895","Text":"Then we can simply write this out as k Lambda sine Theta d Theta divided by"},{"Start":"22:30.895 ","End":"22:36.715","Text":"R. Now we can"},{"Start":"22:36.715 ","End":"22:41.665","Text":"sum up all of these little dq\u0027s along our semicircle."},{"Start":"22:41.665 ","End":"22:44.320","Text":"We\u0027ll get that our E_x,"},{"Start":"22:44.320 ","End":"22:48.400","Text":"our electric field in the x direction is going to"},{"Start":"22:48.400 ","End":"22:53.410","Text":"be the integral of k Lambda divided by R"},{"Start":"22:53.410 ","End":"22:58.255","Text":"multiplied by sine Theta d Theta and"},{"Start":"22:58.255 ","End":"23:03.595","Text":"our bounds are going to be going from Theta is equal to zero."},{"Start":"23:03.595 ","End":"23:05.485","Text":"That\u0027s over here."},{"Start":"23:05.485 ","End":"23:09.745","Text":"Up until Theta\u0027s equal to half a circle,"},{"Start":"23:09.745 ","End":"23:13.700","Text":"which is equal to Pi radians."},{"Start":"23:13.830 ","End":"23:18.025","Text":"Just to clarify the bounds,"},{"Start":"23:18.025 ","End":"23:20.410","Text":"when our Theta is equal to zero,"},{"Start":"23:20.410 ","End":"23:24.310","Text":"then our dl or our dq is located right over here,"},{"Start":"23:24.310 ","End":"23:29.125","Text":"and then we go 180 degrees this way,"},{"Start":"23:29.125 ","End":"23:33.430","Text":"anticlockwise, which is Pi radians,"},{"Start":"23:33.430 ","End":"23:38.180","Text":"this way, and then our Theta will be equal to Pi."},{"Start":"23:39.660 ","End":"23:44.500","Text":"Once we solve this integral and substitute in our bounds,"},{"Start":"23:44.500 ","End":"23:51.460","Text":"we\u0027ll get that our electric field in the x direction for our semicircle is equal to"},{"Start":"23:51.460 ","End":"23:54.940","Text":"2k Lambda divided by"},{"Start":"23:54.940 ","End":"24:01.914","Text":"R. Now what we want to do is find the electric field in our y direction."},{"Start":"24:01.914 ","End":"24:05.980","Text":"Instead of x, we write y and instead of sine Theta,"},{"Start":"24:05.980 ","End":"24:08.440","Text":"we\u0027ll write cosine Theta."},{"Start":"24:08.440 ","End":"24:11.785","Text":"Then we\u0027ll have a cosine Theta over here as well."},{"Start":"24:11.785 ","End":"24:14.740","Text":"Then we\u0027ll change this around."},{"Start":"24:14.740 ","End":"24:17.094","Text":"However, we can see that from symmetry,"},{"Start":"24:17.094 ","End":"24:22.075","Text":"our electric field in the y direction will cancel out."},{"Start":"24:22.075 ","End":"24:23.980","Text":"From this dl,"},{"Start":"24:23.980 ","End":"24:26.274","Text":"we have our dE in this direction."},{"Start":"24:26.274 ","End":"24:34.690","Text":"Then if we were to take dl from exactly the opposite side of our semicircle,"},{"Start":"24:34.690 ","End":"24:41.695","Text":"we\u0027ll see that we have our dE pointing in this direction."},{"Start":"24:41.695 ","End":"24:44.695","Text":"Then we can split this up again,"},{"Start":"24:44.695 ","End":"24:48.219","Text":"so we\u0027re going to have our y component and our x component."},{"Start":"24:48.219 ","End":"24:52.720","Text":"Then we can see that our y components are equal and opposite, so they cancel out,"},{"Start":"24:52.720 ","End":"24:58.767","Text":"and our x components are simply going to be twice as strong."},{"Start":"24:58.767 ","End":"25:03.003","Text":"Because we have them from both sections,"},{"Start":"25:03.003 ","End":"25:06.910","Text":"they\u0027re going to be from opposite slices of the semicircle,"},{"Start":"25:06.910 ","End":"25:12.530","Text":"the x component is going to add into the electric field."},{"Start":"25:13.020 ","End":"25:15.970","Text":"This factor of 2, of course,"},{"Start":"25:15.970 ","End":"25:19.300","Text":"comes from these two arrows,"},{"Start":"25:19.300 ","End":"25:21.940","Text":"the two x components."},{"Start":"25:21.940 ","End":"25:28.345","Text":"That means that our E_y is again going to be equal to zero."},{"Start":"25:28.345 ","End":"25:32.410","Text":"Now what we want to know is from all of the three sections,"},{"Start":"25:32.410 ","End":"25:35.050","Text":"so from our two wires and from our semicircle,"},{"Start":"25:35.050 ","End":"25:39.800","Text":"what is the total electric field at this point over here?"},{"Start":"25:40.560 ","End":"25:45.490","Text":"Now we\u0027re going to use the superposition principle and we\u0027re"},{"Start":"25:45.490 ","End":"25:50.900","Text":"going to see what our E total is equal to."},{"Start":"25:51.360 ","End":"25:54.940","Text":"We can see that from the two wires,"},{"Start":"25:54.940 ","End":"25:57.970","Text":"these two parallel lines from the two wires,"},{"Start":"25:57.970 ","End":"26:03.050","Text":"we have zero force in the y direction."},{"Start":"26:03.330 ","End":"26:07.090","Text":"Let\u0027s write E_tot in the x and E_tot in the y."},{"Start":"26:07.090 ","End":"26:10.570","Text":"We have zero in the y direction and"},{"Start":"26:10.570 ","End":"26:16.420","Text":"the electric field in the y direction due to the semicircle,"},{"Start":"26:16.420 ","End":"26:19.465","Text":"c for semicircle,"},{"Start":"26:19.465 ","End":"26:21.515","Text":"we also have zero."},{"Start":"26:21.515 ","End":"26:24.840","Text":"Our E total in the y direction,"},{"Start":"26:24.840 ","End":"26:27.615","Text":"that\u0027s either up or down, is equal to zero."},{"Start":"26:27.615 ","End":"26:32.235","Text":"Now let\u0027s look at our total electric field in the x direction."},{"Start":"26:32.235 ","End":"26:35.970","Text":"We can see that from the two parallel wires,"},{"Start":"26:35.970 ","End":"26:45.640","Text":"we\u0027re going to have negative 2k Lambda divided by R. The electric field in"},{"Start":"26:45.640 ","End":"26:51.625","Text":"the x direction due to the semicircle is equal to 2k Lambda"},{"Start":"26:51.625 ","End":"27:00.310","Text":"divided by R. These both cancel out as well giving zero."},{"Start":"27:00.310 ","End":"27:02.695","Text":"Therefore, we get that"},{"Start":"27:02.695 ","End":"27:10.130","Text":"the total electric field at this point in the middle of the semicircle is equal to zero."},{"Start":"27:10.590 ","End":"27:12.835","Text":"In the y direction,"},{"Start":"27:12.835 ","End":"27:15.565","Text":"everything cancels out because of symmetry,"},{"Start":"27:15.565 ","End":"27:21.560","Text":"and then the x direction everything cancels out because of this semicircle over here."},{"Start":"27:21.590 ","End":"27:28.365","Text":"Therefore, we can gather from this that the electric field due to"},{"Start":"27:28.365 ","End":"27:37.630","Text":"a semicircle is equal to the electric field generated by two semi-infinite wires."},{"Start":"27:38.510 ","End":"27:41.890","Text":"That\u0027s the end of this lesson."}],"ID":22260},{"Watched":false,"Name":"Exercise 2","Duration":"22m 15s","ChapterTopicVideoID":21427,"CourseChapterTopicPlaylistID":99473,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.800","Text":"Hello. In this question,"},{"Start":"00:01.800 ","End":"00:05.820","Text":"we\u0027re being asked to calculate the electric field"},{"Start":"00:05.820 ","End":"00:10.530","Text":"along the axis of symmetry so here we can say that this is"},{"Start":"00:10.530 ","End":"00:15.660","Text":"the z axes of a ring of"},{"Start":"00:15.660 ","End":"00:21.540","Text":"radius r and charge density per unit length of Lambda."},{"Start":"00:21.540 ","End":"00:28.770","Text":"What we\u0027re trying to find is our electric field in the z direction."},{"Start":"00:28.770 ","End":"00:30.915","Text":"This is what we\u0027re trying to find."},{"Start":"00:30.915 ","End":"00:35.830","Text":"If we choose some arbitrary point along our z axis,"},{"Start":"00:35.830 ","End":"00:40.640","Text":"we say that it is at some height z."},{"Start":"00:40.640 ","End":"00:47.645","Text":"Then what we can do is we can split up our ring"},{"Start":"00:47.645 ","End":"00:56.625","Text":"into small pieces so that each piece can be regarded as a point charge."},{"Start":"00:56.625 ","End":"01:01.395","Text":"Each pieces of length dl which is very small length,"},{"Start":"01:01.395 ","End":"01:08.810","Text":"and then we can say that the distance from dl"},{"Start":"01:08.810 ","End":"01:18.280","Text":"to some arbitrary point on the z-axis is given by our r vector."},{"Start":"01:18.280 ","End":"01:24.620","Text":"We can say that also the radius of the ring is"},{"Start":"01:24.620 ","End":"01:31.680","Text":"given as R and that the angle between r and the x-axis is Theta."},{"Start":"01:32.590 ","End":"01:35.510","Text":"Let us begin."},{"Start":"01:35.510 ","End":"01:40.175","Text":"The first thing that we\u0027re going to do is we\u0027re going to say"},{"Start":"01:40.175 ","End":"01:44.720","Text":"that our piece dl has charge dq and dq,"},{"Start":"01:44.720 ","End":"01:47.510","Text":"as we know, is equal to charge density per"},{"Start":"01:47.510 ","End":"01:52.675","Text":"unit length multiplied by the length of the piece dl."},{"Start":"01:52.675 ","End":"01:56.660","Text":"Now, because we\u0027re dealing with a circle,"},{"Start":"01:56.660 ","End":"02:04.520","Text":"so instead of using dx or dy because we have both x and y components in the cycle,"},{"Start":"02:04.520 ","End":"02:07.655","Text":"we\u0027ll use polar coordinates."},{"Start":"02:07.655 ","End":"02:13.580","Text":"Because the radius of our circle is constant, it\u0027s never changing."},{"Start":"02:13.580 ","End":"02:16.520","Text":"You can write this as Lambda i,"},{"Start":"02:16.520 ","End":"02:21.940","Text":"the radius of the circle multiplied by d Theta,"},{"Start":"02:21.940 ","End":"02:24.360","Text":"so the change in angle."},{"Start":"02:26.150 ","End":"02:29.985","Text":"We\u0027re going to use Coulomb\u0027s law."},{"Start":"02:29.985 ","End":"02:32.780","Text":"The first thing that we\u0027re going to do is we\u0027re going to find"},{"Start":"02:32.780 ","End":"02:38.700","Text":"the magnitude of our electric field, dE."},{"Start":"02:39.650 ","End":"02:45.949","Text":"That means that we\u0027re going to have some electric field in this direction."},{"Start":"02:45.949 ","End":"02:51.215","Text":"This is going to be dE and we want to calculate what this is."},{"Start":"02:51.215 ","End":"02:53.540","Text":"Now, first we\u0027re going to do just the magnitude"},{"Start":"02:53.540 ","End":"02:56.570","Text":"and then afterwards we\u0027re going to deal with the direction."},{"Start":"02:56.570 ","End":"03:05.060","Text":"The electric field is equal to k multiplied by the charge so here it\u0027s dq divided by"},{"Start":"03:05.060 ","End":"03:09.275","Text":"the distance between our piece and our point"},{"Start":"03:09.275 ","End":"03:15.690","Text":"along the axis of symmetry so here we have r^2."},{"Start":"03:16.280 ","End":"03:21.450","Text":"Now let\u0027s see what our r is equal to."},{"Start":"03:21.450 ","End":"03:25.595","Text":"Our r is going to be equal to via Pythagoras."},{"Start":"03:25.595 ","End":"03:34.275","Text":"We can see that here we have a 90 degree angle so this is equal to R^2 plus z^2."},{"Start":"03:34.275 ","End":"03:41.160","Text":"We\u0027re going up z and then the square root of all of that."},{"Start":"03:41.160 ","End":"03:47.130","Text":"Remember that the square root can be denoted by this power symbol."},{"Start":"03:49.070 ","End":"03:58.070","Text":"What we want to do is we want to prepare what we need in order to sum up"},{"Start":"03:58.070 ","End":"04:07.500","Text":"all of our dl\u0027s in order to find the total force along our z axis."},{"Start":"04:07.500 ","End":"04:11.870","Text":"What we\u0027re going to notice is that through symmetry,"},{"Start":"04:11.870 ","End":"04:18.140","Text":"we can see that our electric field if our Lambda is positive,"},{"Start":"04:18.140 ","End":"04:21.830","Text":"it\u0027s only going in the positive z direction and if our Lambda is negative,"},{"Start":"04:21.830 ","End":"04:24.460","Text":"it\u0027s going in the negative direction."},{"Start":"04:24.460 ","End":"04:29.610","Text":"There\u0027s no components in the y or in the x directions."},{"Start":"04:29.950 ","End":"04:34.160","Text":"We can see that as shown in previous videos,"},{"Start":"04:34.160 ","End":"04:37.730","Text":"if we draw out the x or"},{"Start":"04:37.730 ","End":"04:42.415","Text":"y component and then the z component we\u0027ll see that the piece over here,"},{"Start":"04:42.415 ","End":"04:47.600","Text":"so the x or y component will cancel out with another piece opposite."},{"Start":"04:47.600 ","End":"04:55.175","Text":"We only have a z component so let\u0027s then write that our dE in"},{"Start":"04:55.175 ","End":"05:05.300","Text":"the z direction is going to be equal to the magnitude of our dE multiplied by,"},{"Start":"05:05.300 ","End":"05:07.805","Text":"so let\u0027s see what it\u0027s going to be multiplied by."},{"Start":"05:07.805 ","End":"05:16.660","Text":"If we say that this angle between dE and our z axis as Alpha,"},{"Start":"05:16.660 ","End":"05:21.350","Text":"then we can see that this angle will also be Alpha."},{"Start":"05:21.350 ","End":"05:30.805","Text":"Then we can see that our z component of our electric field is going to be given by,"},{"Start":"05:30.805 ","End":"05:37.590","Text":"so it\u0027s the adjacent side to our Alpha so it\u0027s going to be given by cosine of Alpha."},{"Start":"05:37.590 ","End":"05:40.610","Text":"I\u0027m going to leave this here as a positive because I\u0027m"},{"Start":"05:40.610 ","End":"05:43.790","Text":"assuming that Lambda over here is positive."},{"Start":"05:43.790 ","End":"05:45.200","Text":"If however it\u0027s negative,"},{"Start":"05:45.200 ","End":"05:50.030","Text":"then when you substitute in your value for Lambda over here for your dq,"},{"Start":"05:50.030 ","End":"05:52.640","Text":"you\u0027ll get a negative number and then it will"},{"Start":"05:52.640 ","End":"05:55.850","Text":"point to your dE and the negative z direction."},{"Start":"05:55.850 ","End":"05:58.590","Text":"It doesn\u0027t really matter at this point."},{"Start":"05:58.610 ","End":"06:01.545","Text":"Let\u0027s start substituting in everything."},{"Start":"06:01.545 ","End":"06:04.770","Text":"Our dE is equal to kdq,"},{"Start":"06:04.770 ","End":"06:14.865","Text":"where dq is equal to Lambda Rd Theta divided by r^2 multiplied by cosine of Alpha."},{"Start":"06:14.865 ","End":"06:20.910","Text":"Cosine of Alpha is equal to adjacent over hypotenuse so"},{"Start":"06:20.910 ","End":"06:29.495","Text":"our adjacent side is this arbitrary length z divided by the hypotenuse."},{"Start":"06:29.495 ","End":"06:31.250","Text":"What\u0027s our hypotenuse?"},{"Start":"06:31.250 ","End":"06:35.030","Text":"This length r, this red arrow."},{"Start":"06:35.030 ","End":"06:38.000","Text":"Of course, we know what our length r is."},{"Start":"06:38.000 ","End":"06:43.770","Text":"It\u0027s given over here so now let\u0027s substitute in all of that."},{"Start":"06:43.770 ","End":"06:53.550","Text":"Therefore, so we will have k Lambda Rd Theta multiplied by z."},{"Start":"06:53.550 ","End":"06:56.250","Text":"Let\u0027s put the z over here and then we have r^3."},{"Start":"06:56.250 ","End":"07:06.152","Text":"We\u0027ll have (R^2+z^2)^3/2,"},{"Start":"07:06.152 ","End":"07:10.520","Text":"from this 1/2 over here."},{"Start":"07:10.680 ","End":"07:18.430","Text":"Now we have an electric field in the z-direction from one piece,"},{"Start":"07:18.430 ","End":"07:19.630","Text":"but we want to find"},{"Start":"07:19.630 ","End":"07:25.030","Text":"the total electric field from all of the pieces in the ring put together."},{"Start":"07:25.030 ","End":"07:29.395","Text":"Then we can say that E_z electric field in the z direction,"},{"Start":"07:29.395 ","End":"07:34.880","Text":"we only have a z component is going to be equal to the integral of dE_z."},{"Start":"07:35.250 ","End":"07:40.945","Text":"This is going to be equal to the integral of"},{"Start":"07:40.945 ","End":"07:47.830","Text":"k lambda Rzd Theta"},{"Start":"07:47.830 ","End":"07:57.205","Text":"divided by (R^2+z^2)^3/2,"},{"Start":"07:57.205 ","End":"07:59.811","Text":"where our bounds are going to be,"},{"Start":"07:59.811 ","End":"08:04.000","Text":"we\u0027re going to start from zero and end at 2Pi."},{"Start":"08:04.000 ","End":"08:06.790","Text":"We\u0027ll start at Theta is equal to 0,"},{"Start":"08:06.790 ","End":"08:10.585","Text":"which means that we\u0027re going along our x-axis."},{"Start":"08:10.585 ","End":"08:12.415","Text":"As r Theta increases,"},{"Start":"08:12.415 ","End":"08:16.240","Text":"we\u0027re going to go around an entire circle."},{"Start":"08:16.240 ","End":"08:19.820","Text":"As we know, that\u0027s given by 2Pi radians,"},{"Start":"08:19.820 ","End":"08:22.780","Text":"an entire circle is 2Pi radians."},{"Start":"08:23.090 ","End":"08:29.420","Text":"We\u0027re integrating with respect to Theta."},{"Start":"08:29.420 ","End":"08:39.085","Text":"Once we complete this integration will get that this is equal to k lambda r multiplied by"},{"Start":"08:39.085 ","End":"08:43.990","Text":"2Pi z divided by"},{"Start":"08:43.990 ","End":"08:51.680","Text":"(R^2+z^2)^3/2."},{"Start":"08:52.950 ","End":"08:59.635","Text":"This is our electric field in the z direction, and of course,"},{"Start":"08:59.635 ","End":"09:01.810","Text":"our electric field in the x-direction,"},{"Start":"09:01.810 ","End":"09:03.805","Text":"and our electric field in the y-direction,"},{"Start":"09:03.805 ","End":"09:08.719","Text":"as we saw before, is equal to 0 due to the symmetry."},{"Start":"09:09.270 ","End":"09:16.555","Text":"Then we can just write that this is going to be in the z-direction."},{"Start":"09:16.555 ","End":"09:20.875","Text":"If our z it\u0027s bigger than 0, this over here."},{"Start":"09:20.875 ","End":"09:27.350","Text":"Then the negative z direction if our z is smaller than 0."},{"Start":"09:28.980 ","End":"09:32.815","Text":"These are the answers to question Number 1."},{"Start":"09:32.815 ","End":"09:34.885","Text":"Now let\u0027s look at question Number 2."},{"Start":"09:34.885 ","End":"09:38.155","Text":"Now, question Number 2 is asking us to do the exact same thing,"},{"Start":"09:38.155 ","End":"09:41.350","Text":"calculate the electric field along the axis of symmetry."},{"Start":"09:41.350 ","End":"09:49.780","Text":"The z axis of a disk of radius r and charge density per unit surface area Sigma."},{"Start":"09:49.780 ","End":"09:56.125","Text":"All we\u0027re doing this time is we\u0027re just saying that instead of a ring,"},{"Start":"09:56.125 ","End":"09:57.400","Text":"so one dimension,"},{"Start":"09:57.400 ","End":"09:59.380","Text":"we\u0027re dealing with two dimensions,"},{"Start":"09:59.380 ","End":"10:03.380","Text":"it means that this disk is full."},{"Start":"10:04.020 ","End":"10:07.345","Text":"Let\u0027s begin and see what we need to change."},{"Start":"10:07.345 ","End":"10:10.220","Text":"I\u0027m going to write everything in green."},{"Start":"10:11.220 ","End":"10:18.985","Text":"Let\u0027s begin here for question Number 2 by defining what our dq is."},{"Start":"10:18.985 ","End":"10:23.635","Text":"When we were dealing with a one-dimensional circle,"},{"Start":"10:23.635 ","End":"10:27.820","Text":"we could choose just a piece along the circle itself."},{"Start":"10:27.820 ","End":"10:30.865","Text":"However, now we\u0027re dealing with an area."},{"Start":"10:30.865 ","End":"10:33.580","Text":"In order to make this more general,"},{"Start":"10:33.580 ","End":"10:40.150","Text":"we\u0027re going to choose some arbitrary area inside somewhere,"},{"Start":"10:40.150 ","End":"10:43.195","Text":"the area of our disk."},{"Start":"10:43.195 ","End":"10:47.125","Text":"Now let\u0027s see what the charge is on this area."},{"Start":"10:47.125 ","End":"10:49.630","Text":"Our charge is going to be Sigma."},{"Start":"10:49.630 ","End":"10:53.875","Text":"Because that\u0027s the charge density per unit of surface area."},{"Start":"10:53.875 ","End":"10:59.095","Text":"Then it\u0027s going to be multiplied by this area over here."},{"Start":"10:59.095 ","End":"11:04.090","Text":"This area isn\u0027t exactly a square because it has some curvature."},{"Start":"11:04.090 ","End":"11:05.800","Text":"What does this equal to?"},{"Start":"11:05.800 ","End":"11:11.230","Text":"It\u0027s going to be equal to rdrd Theta."},{"Start":"11:11.230 ","End":"11:19.520","Text":"Because now r distance is changing and let\u0027s call this r tag."},{"Start":"11:19.560 ","End":"11:23.605","Text":"We have this distance now,"},{"Start":"11:23.605 ","End":"11:25.795","Text":"is going to be our r tag."},{"Start":"11:25.795 ","End":"11:30.920","Text":"We can now delete this because it\u0027s not relevant."},{"Start":"11:31.260 ","End":"11:34.900","Text":"We have our r tag,"},{"Start":"11:34.900 ","End":"11:39.760","Text":"which is some distance from the center of"},{"Start":"11:39.760 ","End":"11:46.940","Text":"the disk until our area rdrd Theta because we\u0027re dealing with the area."},{"Start":"11:48.420 ","End":"11:55.825","Text":"Now we can add in our r vector from this point over here."},{"Start":"11:55.825 ","End":"12:04.215","Text":"Now, this is our r vector from here. Let\u0027s carry on."},{"Start":"12:04.215 ","End":"12:09.339","Text":"Now we can find the magnitude of our electric field,"},{"Start":"12:09.339 ","End":"12:15.910","Text":"which is being applied from this area over here onto our z axes."},{"Start":"12:15.910 ","End":"12:18.655","Text":"Let\u0027s see what that\u0027s going to be equal to."},{"Start":"12:18.655 ","End":"12:26.455","Text":"That\u0027s going to be equal to kdq divided by our r^2,"},{"Start":"12:26.455 ","End":"12:28.795","Text":"which is this red arrow over here."},{"Start":"12:28.795 ","End":"12:34.030","Text":"Such that our r is going to be this time equal to,"},{"Start":"12:34.030 ","End":"12:35.350","Text":"instead of R^2,"},{"Start":"12:35.350 ","End":"12:37.465","Text":"the total radius of the circle,"},{"Start":"12:37.465 ","End":"12:41.875","Text":"it\u0027s going to be equal to small so r tag squared,"},{"Start":"12:41.875 ","End":"12:48.670","Text":"which is from the center until the point that we\u0027re at."},{"Start":"12:48.670 ","End":"12:55.210","Text":"That we\u0027re finding our dq hat plus our z^2."},{"Start":"12:55.210 ","End":"12:58.675","Text":"Then the square root of all of that."},{"Start":"12:58.675 ","End":"13:06.595","Text":"All I\u0027ve done is I\u0027ve changed this R into the r tag because that\u0027s a variable."},{"Start":"13:06.595 ","End":"13:09.370","Text":"That is our magnitude."},{"Start":"13:09.370 ","End":"13:15.265","Text":"Now we can see that just like with the ring."},{"Start":"13:15.265 ","End":"13:17.020","Text":"The same with the desk."},{"Start":"13:17.020 ","End":"13:23.530","Text":"We\u0027re only going to have an electric field in the positive or negative z axes."},{"Start":"13:23.530 ","End":"13:26.380","Text":"No electric field components in X or Y."},{"Start":"13:26.380 ","End":"13:29.150","Text":"That\u0027s due to symmetry."},{"Start":"13:30.360 ","End":"13:37.360","Text":"We can say therefore that our dEz is simply going to be equal"},{"Start":"13:37.360 ","End":"13:45.460","Text":"to the magnitude of our dE multiplied by,"},{"Start":"13:45.460 ","End":"13:50.275","Text":"we can just define this as a different Alpha,"},{"Start":"13:50.275 ","End":"13:56.530","Text":"multiplied by again cosine of Alpha."},{"Start":"13:56.530 ","End":"14:00.715","Text":"Then we can say that that is equal to our dE,"},{"Start":"14:00.715 ","End":"14:02.545","Text":"which is equal to kdq,"},{"Start":"14:02.545 ","End":"14:12.925","Text":"which is Sigma r tag dr tag d Theta divided by r^2."},{"Start":"14:12.925 ","End":"14:16.535","Text":"That\u0027s going to be, let\u0027s write it in the meantime, r^2."},{"Start":"14:16.535 ","End":"14:21.305","Text":"Then cosine of Alpha is going to be the same thing that it was over here."},{"Start":"14:21.305 ","End":"14:24.395","Text":"It\u0027s adjacent over hypotenuse."},{"Start":"14:24.395 ","End":"14:30.610","Text":"That\u0027s going to be z divided by r. Again,"},{"Start":"14:30.610 ","End":"14:32.500","Text":"we\u0027re going to be left with k"},{"Start":"14:32.500 ","End":"14:42.625","Text":"Sigma r tag dr tag d Theta multiplied by z divided by r^3,"},{"Start":"14:42.625 ","End":"14:53.110","Text":"where r^3 is r tag squared plus z^2 to the power of 3/2."},{"Start":"14:53.110 ","End":"15:01.060","Text":"Now, we\u0027re ready to integrate along the entire desk."},{"Start":"15:01.060 ","End":"15:07.075","Text":"Then we\u0027ll say that our E_z is = the integral of dE_z,"},{"Start":"15:07.075 ","End":"15:10.030","Text":"which is going to be equal to."},{"Start":"15:10.030 ","End":"15:12.430","Text":"now we\u0027re integrating along area."},{"Start":"15:12.430 ","End":"15:15.445","Text":"There\u0027s 2 integration signs."},{"Start":"15:15.445 ","End":"15:22.690","Text":"Then we\u0027re going to have k z Sigma r tag,"},{"Start":"15:22.690 ","End":"15:28.105","Text":"dr d theta divided by"},{"Start":"15:28.105 ","End":"15:34.735","Text":"r tag^2 plus z ^2^3 over 2."},{"Start":"15:34.735 ","End":"15:38.545","Text":"Then our bounds for our r tag,"},{"Start":"15:38.545 ","End":"15:45.669","Text":"we\u0027re integrating from radius of 0 until our maximum disk radius,"},{"Start":"15:45.669 ","End":"15:48.580","Text":"which is r. For r Theta,"},{"Start":"15:48.580 ","End":"15:51.475","Text":"where again starting at Theta\u0027s = 0,"},{"Start":"15:51.475 ","End":"15:55.495","Text":"and then we\u0027re going around a full circle."},{"Start":"15:55.495 ","End":"15:58.400","Text":"Up until 2Pi."},{"Start":"15:58.710 ","End":"16:02.050","Text":"Now when we\u0027re integrating along the theta,"},{"Start":"16:02.050 ","End":"16:04.480","Text":"we can see that Theta is in a variable."},{"Start":"16:04.480 ","End":"16:07.180","Text":"what we could have done is we could have gone"},{"Start":"16:07.180 ","End":"16:12.010","Text":"straight to what we got in question number 1."},{"Start":"16:12.010 ","End":"16:17.949","Text":"Then substituting in instead of Lambda,"},{"Start":"16:17.949 ","End":"16:24.880","Text":"we\u0027d have put in a Sigma and then we would just have to do"},{"Start":"16:24.880 ","End":"16:29.350","Text":"another integration on this answer and then"},{"Start":"16:29.350 ","End":"16:34.810","Text":"another integration from 0 to R and our r tag, dr tag."},{"Start":"16:34.810 ","End":"16:40.750","Text":"We could have just skipped this step and just gone straight to integrating on this."},{"Start":"16:40.750 ","End":"16:43.210","Text":"However, to understand it a little bit more,"},{"Start":"16:43.210 ","End":"16:45.505","Text":"Let\u0027s do it like so."},{"Start":"16:45.505 ","End":"16:47.890","Text":"I\u0027m going to begin over here."},{"Start":"16:47.890 ","End":"16:53.290","Text":"We have that our E_z is therefore going to be = 0 to"},{"Start":"16:53.290 ","End":"17:01.720","Text":"R and it\u0027s going to be 2Pi k Sigma r tag,"},{"Start":"17:01.720 ","End":"17:05.680","Text":"dr tag divided by"},{"Start":"17:05.680 ","End":"17:12.805","Text":"r tag ^2 plus z^2 to the power of 3/2."},{"Start":"17:12.805 ","End":"17:14.980","Text":"Can you see how similar this is?"},{"Start":"17:14.980 ","End":"17:19.120","Text":"We\u0027ve just substituted lambda for a Sigma and"},{"Start":"17:19.120 ","End":"17:24.250","Text":"then we\u0027ve added in our rdr and then instead of having the capital R,"},{"Start":"17:24.250 ","End":"17:27.080","Text":"we have our r tag over here."},{"Start":"17:27.080 ","End":"17:29.760","Text":"Now let\u0027s integrate."},{"Start":"17:29.760 ","End":"17:36.295","Text":"Let\u0027s make a substitution."},{"Start":"17:36.295 ","End":"17:40.930","Text":"Let\u0027s say that u is equal to what\u0027s inside the brackets over here."},{"Start":"17:40.930 ","End":"17:42.820","Text":"Now, this is a very common integral,"},{"Start":"17:42.820 ","End":"17:44.995","Text":"so it\u0027s useful to remember this technique."},{"Start":"17:44.995 ","End":"17:48.415","Text":"r tag ^2 plus z^2."},{"Start":"17:48.415 ","End":"17:57.170","Text":"Then we can say that our du is going to be equal = 2 r tag dr tag."},{"Start":"17:57.900 ","End":"18:04.930","Text":"Therefore, we\u0027re going to get 2Pi k Sigma,"},{"Start":"18:04.930 ","End":"18:08.440","Text":"are constants so we can move them out of the integration sign."},{"Start":"18:08.440 ","End":"18:14.680","Text":"Then we\u0027re integrating from 0 to R on du divided"},{"Start":"18:14.680 ","End":"18:21.745","Text":"by 2, divided by u^3/2."},{"Start":"18:21.745 ","End":"18:25.195","Text":"We\u0027re integrating via substitution."},{"Start":"18:25.195 ","End":"18:30.940","Text":"Then we can say that this is simply going to be"},{"Start":"18:30.940 ","End":"18:38.680","Text":"equal to Pi k. The 2 over here cancels out with a 2 over here."},{"Start":"18:38.680 ","End":"18:47.000","Text":"Integral from 0 till r of t to the negative 3/2."},{"Start":"18:48.510 ","End":"18:51.250","Text":"U^3 over 2."},{"Start":"18:51.250 ","End":"18:54.220","Text":"If we have 1 divided by u^3/2,"},{"Start":"18:54.220 ","End":"19:02.120","Text":"it\u0027s the same as writing u to the power negative 3/2 du."},{"Start":"19:02.280 ","End":"19:06.565","Text":"Then we can solve this."},{"Start":"19:06.565 ","End":"19:12.920","Text":"We\u0027ll get that this is equal to negatve 2Pi k"},{"Start":"19:16.140 ","End":"19:23.380","Text":"u to the power negative 1/2 and then we can substitute in from"},{"Start":"19:23.380 ","End":"19:30.250","Text":"0 to R and just remember that we have to work back what our u is equal to."},{"Start":"19:30.250 ","End":"19:39.960","Text":"Then we\u0027ll have that this is equal to negative 2Pi k Sigma multiplied by"},{"Start":"19:39.960 ","End":"19:49.560","Text":"r tag^2+z^2"},{"Start":"19:49.560 ","End":"19:52.120","Text":"to the power of 1/2."},{"Start":"19:52.410 ","End":"20:00.940","Text":"Again, putting in our bounds 0-R. Then once we substitute that in,"},{"Start":"20:00.940 ","End":"20:10.125","Text":"we\u0027ll get that this is= 2Pi k multiplied by 1 divided by z"},{"Start":"20:10.125 ","End":"20:13.740","Text":"minus 1 divided by"},{"Start":"20:13.740 ","End":"20:20.620","Text":"r^2 plus z^2 and then the square root"},{"Start":"20:20.620 ","End":"20:25.090","Text":"of that to the power of half."},{"Start":"20:25.090 ","End":"20:31.150","Text":"This is our electric field in the z direction for a disk and of course,"},{"Start":"20:31.150 ","End":"20:36.730","Text":"our electric field in the x-direction and our electric field in the y-direction = 0."},{"Start":"20:36.730 ","End":"20:39.370","Text":"They cancel out because of symmetry."},{"Start":"20:39.370 ","End":"20:43.150","Text":"What\u0027s important to note is that all that we"},{"Start":"20:43.150 ","End":"20:49.465","Text":"changed from the beginning was simply what our dq is equal to."},{"Start":"20:49.465 ","End":"20:55.285","Text":"Because now our dq is a portion of the area."},{"Start":"20:55.285 ","End":"20:58.780","Text":"A piece of area rather than a piece of length."},{"Start":"20:58.780 ","End":"21:02.590","Text":"We had to change that around a little bit and"},{"Start":"21:02.590 ","End":"21:07.420","Text":"then our magnitude of our electric field is going to be the exact same thing."},{"Start":"21:07.420 ","End":"21:13.615","Text":"Our distance, the only thing that\u0027s changing is that our radius is changing."},{"Start":"21:13.615 ","End":"21:17.485","Text":"We add in another variable over here because we\u0027re integrating"},{"Start":"21:17.485 ","End":"21:22.555","Text":"all the way out and not just along 1 constant value."},{"Start":"21:22.555 ","End":"21:29.800","Text":"Then we just do the integration just like we would have been used to in 1 dimension,"},{"Start":"21:29.800 ","End":"21:34.120","Text":"but we just do this in 2 dimensions."},{"Start":"21:34.120 ","End":"21:36.550","Text":"Because it was in 2 dimensions,"},{"Start":"21:36.550 ","End":"21:40.150","Text":"so it\u0027s slightly more complicated and we had another variable,"},{"Start":"21:40.150 ","End":"21:45.760","Text":"of r tag over here and our variable was also in the denominator,"},{"Start":"21:45.760 ","End":"21:50.635","Text":"which meant that in order to do this integration and this is a very useful trick."},{"Start":"21:50.635 ","End":"21:54.760","Text":"You define a new variable u and that\u0027s equal to"},{"Start":"21:54.760 ","End":"21:59.455","Text":"what\u0027s inside the brackets and the denominator and then your du."},{"Start":"21:59.455 ","End":"22:04.180","Text":"You differentiate with regarding to your variable."},{"Start":"22:04.180 ","End":"22:07.750","Text":"Then, you assembly, substitute that in and you do integration"},{"Start":"22:07.750 ","End":"22:12.235","Text":"by substitution and then you get your final answer."},{"Start":"22:12.235 ","End":"22:15.230","Text":"That\u0027s the end of that lesson."}],"ID":22261},{"Watched":false,"Name":"Exercise 3","Duration":"30m 43s","ChapterTopicVideoID":21428,"CourseChapterTopicPlaylistID":99473,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.414","Text":"Hello. In this question,"},{"Start":"00:02.414 ","End":"00:04.589","Text":"we\u0027re being told that we have a cylinder,"},{"Start":"00:04.589 ","End":"00:11.580","Text":"and that the height of the cylinder is given and it is equal to H. We\u0027re"},{"Start":"00:11.580 ","End":"00:14.880","Text":"being told that the radius of the cylinder is"},{"Start":"00:14.880 ","End":"00:18.895","Text":"equal to r. That\u0027s a constant radius throughout,"},{"Start":"00:18.895 ","End":"00:24.435","Text":"and that we have charge density per unit volume Rho,"},{"Start":"00:24.435 ","End":"00:26.894","Text":"which means that our cylinder is full."},{"Start":"00:26.894 ","End":"00:29.654","Text":"Now, what we\u0027re being asked to find is,"},{"Start":"00:29.654 ","End":"00:35.444","Text":"what is the electric field at some point along the axis of symmetry,"},{"Start":"00:35.444 ","End":"00:38.800","Text":"which here is the z-axis?"},{"Start":"00:38.800 ","End":"00:43.335","Text":"If I say that this point is at some height z,"},{"Start":"00:43.335 ","End":"00:49.490","Text":"my question is, what is my electric field at height z?"},{"Start":"00:49.490 ","End":"00:53.509","Text":"Right now for this question,"},{"Start":"00:53.509 ","End":"00:57.080","Text":"let\u0027s imagine that Rho is some constant value,"},{"Start":"00:57.080 ","End":"01:01.130","Text":"and later we\u0027ll discuss what happens if Rho is changing."},{"Start":"01:01.130 ","End":"01:03.994","Text":"Let\u0027s say that this is our x-axis,"},{"Start":"01:03.994 ","End":"01:06.110","Text":"this is our y-axis."},{"Start":"01:06.110 ","End":"01:11.420","Text":"Now what we\u0027re going to do is I\u0027m going to choose some point,"},{"Start":"01:11.420 ","End":"01:18.109","Text":"arbitrary piece of area inside my cylinder."},{"Start":"01:18.109 ","End":"01:21.410","Text":"We can say that this point is at some height,"},{"Start":"01:21.410 ","End":"01:24.515","Text":"z tag, a general height,"},{"Start":"01:24.515 ","End":"01:34.040","Text":"and it is a distance of r from the axis of symmetry to my area."},{"Start":"01:34.040 ","End":"01:38.584","Text":"The total radius, the furthest I can come out of the cylinder is"},{"Start":"01:38.584 ","End":"01:45.800","Text":"R. But I\u0027m right now located at some distance from the center of r. Now,"},{"Start":"01:45.800 ","End":"01:54.200","Text":"of course, if I draw the projection on the x-y plane for this r,"},{"Start":"01:54.200 ","End":"01:56.165","Text":"so it\u0027s going to be up until,"},{"Start":"01:56.165 ","End":"01:59.583","Text":"let\u0027s say somewhere here,"},{"Start":"01:59.583 ","End":"02:03.710","Text":"then I can say that the angle between"},{"Start":"02:03.710 ","End":"02:08.470","Text":"my x-axis and this projection of r on my x-y plane,"},{"Start":"02:08.470 ","End":"02:13.465","Text":"so I can say that this angle over here is Theta."},{"Start":"02:13.465 ","End":"02:18.110","Text":"Now we have our diagram and we want to"},{"Start":"02:18.110 ","End":"02:22.430","Text":"work out what our electric field is at this arbitrary point z."},{"Start":"02:22.430 ","End":"02:27.169","Text":"The first thing that we have to do is we have to find out what our charge is,"},{"Start":"02:27.169 ","End":"02:33.920","Text":"dq, on this small piece of area or piece of volume."},{"Start":"02:33.920 ","End":"02:39.409","Text":"We know that our dq can be 1 of 3 things."},{"Start":"02:39.409 ","End":"02:43.348","Text":"We either have Lambda dl for unit length,"},{"Start":"02:43.348 ","End":"02:47.659","Text":"we have Sigma ds for unit area,"},{"Start":"02:47.659 ","End":"02:51.439","Text":"and we have Rho dv for unit volume."},{"Start":"02:51.439 ","End":"02:54.004","Text":"Because we\u0027re dealing with a solid cylinder,"},{"Start":"02:54.004 ","End":"02:57.875","Text":"we\u0027re going to be using our Rho dv."},{"Start":"02:57.875 ","End":"03:02.700","Text":"Therefore, our dq is going to be equal to Rho."},{"Start":"03:02.700 ","End":"03:04.344","Text":"Now, what is our dv?"},{"Start":"03:04.344 ","End":"03:09.349","Text":"We have to use cylindrical coordinates because we\u0027re dealing with a cylinder,"},{"Start":"03:09.349 ","End":"03:11.458","Text":"and it\u0027s a full cylinder,"},{"Start":"03:11.458 ","End":"03:13.355","Text":"so we have to remember our Jacobian,"},{"Start":"03:13.355 ","End":"03:18.990","Text":"so it\u0027s going to be equal to rdr dTheta and then dz."},{"Start":"03:18.990 ","End":"03:22.635","Text":"However, here it\u0027s going to be z tag because"},{"Start":"03:22.635 ","End":"03:27.280","Text":"z is our arbitrary point where we\u0027re measuring our electric field."},{"Start":"03:27.280 ","End":"03:33.479","Text":"Our z tag is what is changing as we move up and down our cylinder itself."},{"Start":"03:34.730 ","End":"03:38.809","Text":"Now let\u0027s find the electric field,"},{"Start":"03:38.809 ","End":"03:46.430","Text":"which is applied by this point charge dq on our arbitrary point dz."},{"Start":"03:46.430 ","End":"03:49.565","Text":"We have our dE,"},{"Start":"03:49.565 ","End":"03:52.825","Text":"which is going to be equal to via Coulomb\u0027s law,"},{"Start":"03:52.825 ","End":"03:57.764","Text":"kdq divided by r Tilde squared."},{"Start":"03:57.764 ","End":"04:01.275","Text":"What is this r Tilde?"},{"Start":"04:01.275 ","End":"04:06.154","Text":"I\u0027m not writing r because r is the distance from the axis of symmetry"},{"Start":"04:06.154 ","End":"04:11.150","Text":"until our small volume unit over here."},{"Start":"04:11.150 ","End":"04:12.625","Text":"Now, r Tilde,"},{"Start":"04:12.625 ","End":"04:15.500","Text":"when we\u0027re dealing with this in Coulomb\u0027s law,"},{"Start":"04:15.500 ","End":"04:20.014","Text":"we\u0027re dealing with the r that goes from"},{"Start":"04:20.014 ","End":"04:27.094","Text":"our point charge until our area where we\u0027re testing the electric field."},{"Start":"04:27.094 ","End":"04:33.660","Text":"This is our r Tilde and it\u0027s also a vector."},{"Start":"04:35.420 ","End":"04:41.175","Text":"Great. What is our r Tilde equal to?"},{"Start":"04:41.175 ","End":"04:43.169","Text":"Let\u0027s take a look."},{"Start":"04:43.169 ","End":"04:48.069","Text":"Our r Tilde is going to be equal to the square root of,"},{"Start":"04:48.069 ","End":"04:49.921","Text":"so using Pythagoras,"},{"Start":"04:49.921 ","End":"04:53.034","Text":"we\u0027re going to see that it\u0027s the square root of r squared,"},{"Start":"04:53.034 ","End":"04:59.409","Text":"where r is from the axis of symmetry until our unit volume or our point charge over here."},{"Start":"04:59.409 ","End":"05:04.074","Text":"Then we have to go up this distance."},{"Start":"05:04.074 ","End":"05:09.639","Text":"What exactly is this distance over here?"},{"Start":"05:09.639 ","End":"05:11.755","Text":"This is the length that we want."},{"Start":"05:11.755 ","End":"05:18.385","Text":"This is going to be equal to our point z over here where we\u0027re testing,"},{"Start":"05:18.385 ","End":"05:22.419","Text":"minus the height that we\u0027re measuring over here,"},{"Start":"05:22.419 ","End":"05:25.555","Text":"so minus z tag."},{"Start":"05:25.555 ","End":"05:28.146","Text":"As our z tag goes lower,"},{"Start":"05:28.146 ","End":"05:31.055","Text":"so our green arrow is going to get longer."},{"Start":"05:31.055 ","End":"05:32.629","Text":"Again, as our z tag goes higher,"},{"Start":"05:32.629 ","End":"05:35.555","Text":"then our green line will be less,"},{"Start":"05:35.555 ","End":"05:39.590","Text":"so it\u0027s going to be r Tilde is equal to the square root of r squared"},{"Start":"05:39.590 ","End":"05:45.149","Text":"plus z minus z tag squared."},{"Start":"05:47.510 ","End":"05:52.880","Text":"What we\u0027ve figured out here is the magnitude of our dE."},{"Start":"05:52.880 ","End":"05:55.039","Text":"We figured out what our r Tilde is."},{"Start":"05:55.039 ","End":"06:01.805","Text":"Now what do we want to do is we want to find our total electric field."},{"Start":"06:01.805 ","End":"06:03.679","Text":"We know that it\u0027s a vector quantity,"},{"Start":"06:03.679 ","End":"06:09.169","Text":"so it\u0027s going to be pointing in specific directions or have specific components."},{"Start":"06:09.169 ","End":"06:13.535","Text":"Now, we can see that from symmetry, our dE,"},{"Start":"06:13.535 ","End":"06:19.420","Text":"from this point over here is going to be pointing in this direction."},{"Start":"06:19.420 ","End":"06:21.980","Text":"We can see from symmetry,"},{"Start":"06:21.980 ","End":"06:27.079","Text":"that the x and y components of the electric fields are going to cancel out."},{"Start":"06:27.079 ","End":"06:30.175","Text":"Because we can see that if we take"},{"Start":"06:30.175 ","End":"06:36.559","Text":"some unit volume from the opposite side of our cylinder,"},{"Start":"06:36.559 ","End":"06:41.450","Text":"it\u0027s going to apply an electric field in this direction dE."},{"Start":"06:41.450 ","End":"06:45.559","Text":"Then we can see from symmetry that the x and y components cancel out and we\u0027re"},{"Start":"06:45.559 ","End":"06:50.620","Text":"just going to have 2 times the z component."},{"Start":"06:51.290 ","End":"06:56.540","Text":"All we have to do now is find the electric field"},{"Start":"06:56.540 ","End":"07:02.309","Text":"from all of these little pieces only in the z direction."},{"Start":"07:03.260 ","End":"07:09.035","Text":"Now we\u0027re trying to find our dE in the z direction."},{"Start":"07:09.035 ","End":"07:15.290","Text":"We know that this is going to be equal to the magnitude of our electric field,"},{"Start":"07:15.290 ","End":"07:19.265","Text":"and then we have to find its z component."},{"Start":"07:19.265 ","End":"07:25.819","Text":"If we say that this angle over here between our dE and our z axes,"},{"Start":"07:25.819 ","End":"07:27.845","Text":"so let\u0027s say that this angle is Alpha,"},{"Start":"07:27.845 ","End":"07:32.440","Text":"which means that this angle over here between our r Tilde,"},{"Start":"07:32.440 ","End":"07:35.434","Text":"and our z axis is also going to be Alpha,"},{"Start":"07:35.434 ","End":"07:40.950","Text":"it\u0027s a bit small, but you can see between the z axis and our r Tilde vector."},{"Start":"07:40.950 ","End":"07:43.020","Text":"We have our Alpha."},{"Start":"07:43.510 ","End":"07:49.355","Text":"Now we can see that"},{"Start":"07:49.355 ","End":"07:56.360","Text":"our z component of the electric field is going to be on the adjacent side,"},{"Start":"07:56.360 ","End":"07:59.764","Text":"so the side adjacent to our Alpha angle."},{"Start":"07:59.764 ","End":"08:08.364","Text":"We\u0027re going to multiply our electric field magnitude by cosine of that angle,"},{"Start":"08:08.364 ","End":"08:09.874","Text":"so cosine Alpha,"},{"Start":"08:09.874 ","End":"08:15.209","Text":"because we\u0027re finding the components on the adjacent side to our angle."},{"Start":"08:15.620 ","End":"08:19.099","Text":"Now, what is our cosine Alpha?"},{"Start":"08:19.099 ","End":"08:22.265","Text":"We know from our trig identities,"},{"Start":"08:22.265 ","End":"08:30.379","Text":"that cosine of an angle is equal to the adjacent side divided by the hypotenuse."},{"Start":"08:30.379 ","End":"08:35.060","Text":"That means that we\u0027re going to have the magnitude of our electric field multiplied by"},{"Start":"08:35.060 ","End":"08:41.014","Text":"the adjacent side is this length with our green arrow over here."},{"Start":"08:41.014 ","End":"08:44.614","Text":"That\u0027s our weird length of z minus z tag,"},{"Start":"08:44.614 ","End":"08:47.045","Text":"divided by our hypotenuse,"},{"Start":"08:47.045 ","End":"08:49.675","Text":"which is this r Tilde."},{"Start":"08:49.675 ","End":"08:55.984","Text":"We saw that our r Tilde is equal to the square root of r squared"},{"Start":"08:55.984 ","End":"09:03.700","Text":"plus z minus z tags squared to the power of 1/2."},{"Start":"09:07.200 ","End":"09:13.690","Text":"Now let\u0027s write out our dE_z in full."},{"Start":"09:13.690 ","End":"09:19.345","Text":"We have the magnitude of our electric field is going to be equal to kdq,"},{"Start":"09:19.345 ","End":"09:23.830","Text":"so k, and then our dq is Rho,"},{"Start":"09:23.830 ","End":"09:28.870","Text":"rdr, dTheta dz tag,"},{"Start":"09:28.870 ","End":"09:33.440","Text":"divided by r tilde squared,"},{"Start":"09:34.380 ","End":"09:38.020","Text":"and then multiplied by this over here."},{"Start":"09:38.020 ","End":"09:40.255","Text":"This is also r tilde,"},{"Start":"09:40.255 ","End":"09:43.195","Text":"so we\u0027re going to have r tilde cubed."},{"Start":"09:43.195 ","End":"09:52.870","Text":"We\u0027ll have in the denominator r^2 plus z minus z tag squared to the power of 3/2."},{"Start":"09:52.870 ","End":"09:57.549","Text":"I\u0027ve just multiplied the denominator of my dE with the denominator of this,"},{"Start":"09:57.549 ","End":"10:00.460","Text":"what we got for our cosine Alpha over here."},{"Start":"10:00.460 ","End":"10:05.904","Text":"Then this is going to be multiplied by z minus z tag."},{"Start":"10:05.904 ","End":"10:12.085","Text":"If we want to find the full electric field in the z direction,"},{"Start":"10:12.085 ","End":"10:15.115","Text":"we\u0027re going to have to integrate along this."},{"Start":"10:15.115 ","End":"10:17.665","Text":"What does that mean? That means,"},{"Start":"10:17.665 ","End":"10:19.150","Text":"because we\u0027re dealing with volume,"},{"Start":"10:19.150 ","End":"10:20.785","Text":"because we have a solid cylinder,"},{"Start":"10:20.785 ","End":"10:25.085","Text":"we\u0027re going to have to do a triple integral."},{"Start":"10:25.085 ","End":"10:27.670","Text":"Let\u0027s see this over here."},{"Start":"10:29.070 ","End":"10:34.479","Text":"Now we\u0027re going to integrate along our r. We\u0027re going from a radius of"},{"Start":"10:34.479 ","End":"10:40.075","Text":"0 from the axis of symmetry until our final radius,"},{"Start":"10:40.075 ","End":"10:44.770","Text":"which is R. Because if we go larger than R,"},{"Start":"10:44.770 ","End":"10:46.990","Text":"we\u0027re outside of our cylinder and then we"},{"Start":"10:46.990 ","End":"10:50.740","Text":"have charged density per unit volume equal to 0."},{"Start":"10:50.740 ","End":"10:55.780","Text":"We\u0027re going from 0 to r. Then let\u0027s talk about what we\u0027re doing with Theta."},{"Start":"10:55.780 ","End":"11:01.210","Text":"What we want to do is we want to go from here where Theta is equal to 0,"},{"Start":"11:01.210 ","End":"11:05.275","Text":"and then we want to complete a full circle."},{"Start":"11:05.275 ","End":"11:08.530","Text":"We\u0027re going from 0 until 2Pi."},{"Start":"11:08.530 ","End":"11:11.020","Text":"2Pi is a full circle."},{"Start":"11:11.020 ","End":"11:16.990","Text":"Now we\u0027ve integrated throughout the depth of our cylinder,"},{"Start":"11:16.990 ","End":"11:20.485","Text":"in this direction, along the radial direction,"},{"Start":"11:20.485 ","End":"11:22.344","Text":"and in a whole circle."},{"Start":"11:22.344 ","End":"11:25.494","Text":"Now what we want to do is we want to sum that up"},{"Start":"11:25.494 ","End":"11:29.199","Text":"throughout the total height of our cylinder,"},{"Start":"11:29.199 ","End":"11:34.869","Text":"so we\u0027re going from height is equal to 0 into height is equal to H. From 0 until"},{"Start":"11:34.869 ","End":"11:42.100","Text":"H. Now what we\u0027re going to do is we\u0027re going to solve this integration."},{"Start":"11:42.100 ","End":"11:43.525","Text":"It\u0027s a lot of algebra."},{"Start":"11:43.525 ","End":"11:46.045","Text":"I advise, if you\u0027re short with time,"},{"Start":"11:46.045 ","End":"11:48.730","Text":"to just skip ahead to the final answer,"},{"Start":"11:48.730 ","End":"11:52.599","Text":"however, you want to see how we do this integration,"},{"Start":"11:52.599 ","End":"11:55.339","Text":"you can carry on watching."},{"Start":"11:55.860 ","End":"12:04.270","Text":"Let\u0027s begin. We can see that when we\u0027re integrating with respect to Theta,"},{"Start":"12:04.270 ","End":"12:09.235","Text":"we can see that we have no variables of Theta in here."},{"Start":"12:09.235 ","End":"12:15.609","Text":"That means that we can just write 2Pi multiplied by everything."},{"Start":"12:15.609 ","End":"12:18.690","Text":"Now let\u0027s take out all of the constants,"},{"Start":"12:18.690 ","End":"12:23.815","Text":"so we\u0027re going to have 2Pi from integrating with respect to Theta,"},{"Start":"12:23.815 ","End":"12:26.395","Text":"multiplied by k, which is a constant,"},{"Start":"12:26.395 ","End":"12:30.220","Text":"multiplied by Rho in the rest of variables."},{"Start":"12:30.220 ","End":"12:41.245","Text":"Then we\u0027re going to have our integral from 0 to H and our integral from 0 to R of rdrd,"},{"Start":"12:41.245 ","End":"12:46.435","Text":"and then z minus z tag,"},{"Start":"12:46.435 ","End":"12:53.305","Text":"dz tag divided by r^2 plus z"},{"Start":"12:53.305 ","End":"13:01.150","Text":"minus z tag squared to the power of 3/2."},{"Start":"13:01.150 ","End":"13:08.860","Text":"Now let\u0027s take a look at integrating with respect to our z tag rather."},{"Start":"13:08.860 ","End":"13:17.980","Text":"What we\u0027re going to do is we\u0027re going to define a new variable."},{"Start":"13:17.980 ","End":"13:20.485","Text":"We\u0027re going to say that we have our u,"},{"Start":"13:20.485 ","End":"13:25.209","Text":"which is going to be equal to what is here inside the brackets,"},{"Start":"13:25.209 ","End":"13:32.510","Text":"so u is going to be equal to r^2 plus z minus z tag squared."},{"Start":"13:32.510 ","End":"13:38.520","Text":"Then our du is going to be simply equal to 2 multiplied"},{"Start":"13:38.520 ","End":"13:44.294","Text":"by z minus z tag and then multiplied by the anti-derivative,"},{"Start":"13:44.294 ","End":"13:49.300","Text":"which is negative 1, dz tag."},{"Start":"13:49.950 ","End":"13:52.780","Text":"Now let\u0027s substitute this in."},{"Start":"13:52.780 ","End":"13:55.580","Text":"We have 2PikRho."},{"Start":"13:57.570 ","End":"14:05.410","Text":"Then we\u0027re going to have the integral from 0 to H. Then we have rdr,"},{"Start":"14:05.410 ","End":"14:07.750","Text":"and then, let\u0027s change this."},{"Start":"14:07.750 ","End":"14:16.720","Text":"We can see that our dz tag is going to be equal to du divided by negative 2z minus z tag."},{"Start":"14:16.720 ","End":"14:23.559","Text":"That means that we can substitute in here a negative 1/2"},{"Start":"14:23.559 ","End":"14:30.190","Text":"multiplied by du,"},{"Start":"14:30.190 ","End":"14:31.794","Text":"we called our variable u,"},{"Start":"14:31.794 ","End":"14:35.990","Text":"so u divided by u^3/2."},{"Start":"14:39.660 ","End":"14:44.049","Text":"Then our negative 1/2 can cancel out with our 2 over here."},{"Start":"14:44.049 ","End":"14:54.130","Text":"We\u0027ll get negative Pi k Rho integral from 0 to H. Then I forgot to also add this"},{"Start":"14:54.130 ","End":"14:56.840","Text":"in 0 to R."},{"Start":"14:57.780 ","End":"15:04.630","Text":"Now notice that I\u0027m just going to cross out these borders."},{"Start":"15:04.630 ","End":"15:09.259","Text":"Because we\u0027re substituting in this u variable."},{"Start":"15:09.510 ","End":"15:11.964","Text":"Our bounds are going to change."},{"Start":"15:11.964 ","End":"15:13.630","Text":"In the meantime, I\u0027ll leave this boundless,"},{"Start":"15:13.630 ","End":"15:16.420","Text":"and then when we go back to our z tag variable,"},{"Start":"15:16.420 ","End":"15:25.240","Text":"then we can substitute back in our balance of 0 to R. Then we have rdr and then"},{"Start":"15:25.240 ","End":"15:27.380","Text":"we have du,"},{"Start":"15:27.960 ","End":"15:37.270","Text":"divided by u^3/2."},{"Start":"15:37.270 ","End":"15:39.250","Text":"Now, let\u0027s integrate."},{"Start":"15:39.250 ","End":"15:41.154","Text":"Scroll a little bit down."},{"Start":"15:41.154 ","End":"15:46.765","Text":"This is going to be equal to negative Pi k Rho,"},{"Start":"15:46.765 ","End":"15:51.835","Text":"0 to H, rdr."},{"Start":"15:51.835 ","End":"15:54.159","Text":"Now let\u0027s integrate."},{"Start":"15:54.159 ","End":"16:01.285","Text":"Now we have u to the power of negative 3/2du."},{"Start":"16:01.285 ","End":"16:03.969","Text":"Let\u0027s see what this is equal to."},{"Start":"16:03.969 ","End":"16:09.819","Text":"This is simply going to be equal to negative"},{"Start":"16:09.819 ","End":"16:16.825","Text":"2u to the power of negative 1/2."},{"Start":"16:16.825 ","End":"16:20.034","Text":"Then we have to place in our bounds."},{"Start":"16:20.034 ","End":"16:27.055","Text":"But first, let\u0027s substitute back in over here what our u was equal to."},{"Start":"16:27.055 ","End":"16:31.409","Text":"This is going to then be equal to negative 2,"},{"Start":"16:31.409 ","End":"16:36.339","Text":"so u is r squared plus z"},{"Start":"16:36.339 ","End":"16:43.240","Text":"minus z tag squared to the power of negative 1/2."},{"Start":"16:43.240 ","End":"16:52.450","Text":"Now, we can substitute back in our bounds of 0 until R. A quick note."},{"Start":"16:52.450 ","End":"16:56.470","Text":"I got confused between our integration signs."},{"Start":"16:56.470 ","End":"16:58.780","Text":"We have 0 to R not 0 to H,"},{"Start":"16:58.780 ","End":"17:00.505","Text":"which was written here."},{"Start":"17:00.505 ","End":"17:03.925","Text":"Then here again, also 0 to R, 0 to R,"},{"Start":"17:03.925 ","End":"17:11.769","Text":"and our 0 to H is what we\u0027ve changed u to r integrating by substitution."},{"Start":"17:11.769 ","End":"17:15.639","Text":"There was a bit of a mix-up. Bear with me."},{"Start":"17:15.639 ","End":"17:17.905","Text":"Now let\u0027s rewrite this."},{"Start":"17:17.905 ","End":"17:20.335","Text":"Let\u0027s go all the way here."},{"Start":"17:20.335 ","End":"17:22.210","Text":"We\u0027re beginning here."},{"Start":"17:22.210 ","End":"17:30.175","Text":"Negative 2 multiplied by negative Pi k Rho is going to be 2Pi k Rho."},{"Start":"17:30.175 ","End":"17:36.784","Text":"Then 0 to R(r) multiplied by this over here,"},{"Start":"17:36.784 ","End":"17:40.884","Text":"so r^2 plus z minus"},{"Start":"17:40.884 ","End":"17:47.035","Text":"z tag squared to the power of negative 1/2."},{"Start":"17:47.035 ","End":"17:50.510","Text":"When we\u0027re substituting in here,"},{"Start":"17:53.760 ","End":"18:04.135","Text":"our bounds of 0 until H. Now let\u0027s substitute everything in."},{"Start":"18:04.135 ","End":"18:10.400","Text":"We\u0027ll have 2Pi k Rho integral from 0 to R(rdr),"},{"Start":"18:11.460 ","End":"18:15.054","Text":"which is going to multiply out."},{"Start":"18:15.054 ","End":"18:18.019","Text":"Now we\u0027re going to substitute in our bounds."},{"Start":"18:18.019 ","End":"18:22.024","Text":"We\u0027re going to have r^2 plus z"},{"Start":"18:22.024 ","End":"18:29.780","Text":"minus H^2 to the power of negative 1/2,"},{"Start":"18:29.780 ","End":"18:38.085","Text":"minus r^2 plus z^2,"},{"Start":"18:38.085 ","End":"18:40.374","Text":"so when we\u0027re substituting in 0,"},{"Start":"18:40.374 ","End":"18:48.655","Text":"and this is also going to be to the power of negative 1/2."},{"Start":"18:48.655 ","End":"18:52.690","Text":"Now we\u0027re going to need to integrate with respect"},{"Start":"18:52.690 ","End":"18:56.140","Text":"to r. Now notice that because this is to the power of negative"},{"Start":"18:56.140 ","End":"19:02.079","Text":"1/2 that just means that we have rdr divided by this to the power of"},{"Start":"19:02.079 ","End":"19:12.250","Text":"1/2 plus rdr divided by this or minus rdr divided by this to the power of 1/2."},{"Start":"19:12.250 ","End":"19:15.909","Text":"We have the exact same integration format"},{"Start":"19:15.909 ","End":"19:19.660","Text":"that we had over here when we were integrating with respect to z."},{"Start":"19:19.660 ","End":"19:23.140","Text":"What we did was integration via substitution,"},{"Start":"19:23.140 ","End":"19:25.719","Text":"and there we said that we\u0027re going to use"},{"Start":"19:25.719 ","End":"19:29.545","Text":"our substitution of u which is going to equal to this."},{"Start":"19:29.545 ","End":"19:33.980","Text":"Let\u0027s do the exact same thing."},{"Start":"19:34.710 ","End":"19:45.970","Text":"We can see that we can split this up into 2Pik Rho and then we\u0027re integrating"},{"Start":"19:45.970 ","End":"19:56.739","Text":"from 0 to R of rdr divided by r"},{"Start":"19:56.739 ","End":"20:06.969","Text":"squared plus z minus h squared to the power of 1.5 minus"},{"Start":"20:06.969 ","End":"20:12.639","Text":"the integration from 0 to r of rdr divided"},{"Start":"20:12.639 ","End":"20:20.140","Text":"by r squared plus z squared to the power of 1/2."},{"Start":"20:20.140 ","End":"20:26.090","Text":"We can split up this 1 integral into 2 integrals."},{"Start":"20:26.100 ","End":"20:34.330","Text":"Now over here, let\u0027s set up our substitutions."},{"Start":"20:34.330 ","End":"20:36.280","Text":"Let\u0027s deal with this 1."},{"Start":"20:36.280 ","End":"20:40.825","Text":"Let\u0027s say that we have our t which is going to be equal to u just like what we did here."},{"Start":"20:40.825 ","End":"20:44.320","Text":"It\u0027s going to be equal to what\u0027s inside this bracket over here,"},{"Start":"20:44.320 ","End":"20:47.754","Text":"so that\u0027s going to be equal to r squared plus z"},{"Start":"20:47.754 ","End":"20:52.524","Text":"minus h squared and then we\u0027ll have that rdt"},{"Start":"20:52.524 ","End":"21:02.059","Text":"is going to be equal to 2 rdr and that\u0027s it."},{"Start":"21:02.220 ","End":"21:07.870","Text":"Now, let\u0027s take a look at our substitution for this integral,"},{"Start":"21:07.870 ","End":"21:15.730","Text":"so here we have something different going on inside our bracket over here."},{"Start":"21:15.730 ","End":"21:22.210","Text":"Let\u0027s say that we have y which is equal to r squared plus z squared,"},{"Start":"21:22.210 ","End":"21:28.910","Text":"so then we\u0027ll have that our dy is going to also be equal to 2 rdr."},{"Start":"21:29.400 ","End":"21:33.039","Text":"Perfect. Let\u0027s carry on with our integration,"},{"Start":"21:33.039 ","End":"21:38.155","Text":"so we have 2Pik rho multiplied by."},{"Start":"21:38.155 ","End":"21:40.120","Text":"We have our integration."},{"Start":"21:40.120 ","End":"21:45.010","Text":"Now I\u0027m going to leave out the bounds because now we\u0027re using"},{"Start":"21:45.010 ","End":"21:47.860","Text":"this substitution which means that the bounds are changing and"},{"Start":"21:47.860 ","End":"21:51.760","Text":"we\u0027ll put them back in afterwards."},{"Start":"21:51.760 ","End":"21:54.759","Text":"Our rdr is simply going to"},{"Start":"21:54.759 ","End":"21:59.710","Text":"be=1/2 dt divided"},{"Start":"21:59.710 ","End":"22:05.679","Text":"by t^1.5,"},{"Start":"22:05.679 ","End":"22:10.943","Text":"and we\u0027ll have minus the integral."},{"Start":"22:10.943 ","End":"22:13.660","Text":"Again, I\u0027m leaving out the bounds because we\u0027re integrating via"},{"Start":"22:13.660 ","End":"22:17.305","Text":"substitution and again we\u0027re going to have 1/2,"},{"Start":"22:17.305 ","End":"22:27.310","Text":"but this time dy because it\u0027s a little bit different divided by y^1.5."},{"Start":"22:27.310 ","End":"22:33.025","Text":"All of this inside the square brackets is being multiplied by this 2Pik Rho."},{"Start":"22:33.025 ","End":"22:38.275","Text":"Let\u0027s make this a little bit clearer and I\u0027m going to just move to a different color,"},{"Start":"22:38.275 ","End":"22:43.480","Text":"so let\u0027s see this integral right over here."},{"Start":"22:43.480 ","End":"22:46.112","Text":"This is going to be equal to."},{"Start":"22:46.112 ","End":"22:50.785","Text":"We\u0027re going to have 1/2 and then we\u0027re integrating."},{"Start":"22:50.785 ","End":"22:57.159","Text":"We can write this also as t^negative 1/2,"},{"Start":"22:57.159 ","End":"23:05.060","Text":"so we have this 1/2 multiplied by 2."},{"Start":"23:05.820 ","End":"23:14.926","Text":"Then this is going to be multiplied by t^1.5,"},{"Start":"23:14.926 ","End":"23:20.760","Text":"and then we\u0027re going to substitute in our borders or a bound soon."},{"Start":"23:20.760 ","End":"23:24.315","Text":"Now let\u0027s deal with this integral."},{"Start":"23:24.315 ","End":"23:29.664","Text":"Again, it\u0027s going to be equal to the exact same thing at this stage."},{"Start":"23:29.664 ","End":"23:33.524","Text":"Just instead of t we\u0027re substituting in y instead and then"},{"Start":"23:33.524 ","End":"23:38.120","Text":"soon we\u0027ll move back to our original values for t and y,"},{"Start":"23:38.120 ","End":"23:40.989","Text":"so this we have a negative over here."},{"Start":"23:40.989 ","End":"23:50.275","Text":"It\u0027s going to be negative 1.5 times 2 multiplied by y^1/2,"},{"Start":"23:50.275 ","End":"23:54.520","Text":"and of course substituting in our bounds soon."},{"Start":"23:54.520 ","End":"23:57.609","Text":"Now, let\u0027s see."},{"Start":"23:57.609 ","End":"24:01.510","Text":"Let\u0027s substitute in what our t is equal to,"},{"Start":"24:01.510 ","End":"24:04.868","Text":"so 1/2 and our 2 cancel out from both places."},{"Start":"24:04.868 ","End":"24:07.413","Text":"We\u0027re going to have t^1/2,"},{"Start":"24:07.413 ","End":"24:12.790","Text":"so we\u0027re going to have r^2 plus z minus"},{"Start":"24:12.790 ","End":"24:21.040","Text":"h^2^1/2 because that\u0027s what our t is equal to,"},{"Start":"24:21.040 ","End":"24:27.174","Text":"and then of course we\u0027ll substitute in our borders."},{"Start":"24:27.174 ","End":"24:31.614","Text":"Then we\u0027ll have a negative and now let\u0027s substitute in our y."},{"Start":"24:31.614 ","End":"24:36.564","Text":"Our y is equal to r^2 plus z^2^1/2"},{"Start":"24:36.564 ","End":"24:42.609","Text":"and also substitute in our borders over here."},{"Start":"24:42.609 ","End":"24:51.280","Text":"Let\u0027s scroll down a little bit more and let\u0027s write this out nice and neatly."},{"Start":"24:51.280 ","End":"24:59.170","Text":"We\u0027re going to have 2Pik Rho multiplied by,"},{"Start":"24:59.170 ","End":"25:09.895","Text":"and then we\u0027re going to have r^2 plus z minus h^2^1/2."},{"Start":"25:09.895 ","End":"25:16.233","Text":"Then our borders are as we saw before between 0 and r,"},{"Start":"25:16.233 ","End":"25:26.233","Text":"and then we\u0027re going to have minus r^2 plus z^2^1/2,"},{"Start":"25:26.233 ","End":"25:31.430","Text":"and again our borders are 0-r."},{"Start":"25:34.830 ","End":"25:38.709","Text":"I just wrote out this integration again."},{"Start":"25:38.709 ","End":"25:43.884","Text":"Once I integrated by using substitution and then I"},{"Start":"25:43.884 ","End":"25:49.539","Text":"went back to my original variables and so that means that I can put my bounds in."},{"Start":"25:49.539 ","End":"25:52.509","Text":"To be honest I could have even put them over here,"},{"Start":"25:52.509 ","End":"26:00.160","Text":"so from 0-r. Now let\u0027s substitute in our bounds."},{"Start":"26:00.160 ","End":"26:08.890","Text":"We can write in that we have 2Pik Rho multiplied by."},{"Start":"26:08.890 ","End":"26:12.670","Text":"We have R squared."},{"Start":"26:12.670 ","End":"26:17.530","Text":"Sorry, this is capital R squared of course,"},{"Start":"26:17.530 ","End":"26:27.380","Text":"plus z minus h^2^1/2 minus then we substitute in our 0,"},{"Start":"26:27.380 ","End":"26:33.084","Text":"so minus 0 plus."},{"Start":"26:33.084 ","End":"26:38.283","Text":"It\u0027s just going to be z minus h^2 and"},{"Start":"26:38.283 ","End":"26:44.736","Text":"then this is going to be to the power of 1/2 so we can already see that these cross off,"},{"Start":"26:44.736 ","End":"26:49.224","Text":"and then we have minus and now this,"},{"Start":"26:49.224 ","End":"26:53.364","Text":"so we\u0027re going to have again minus r^2"},{"Start":"26:53.364 ","End":"27:00.205","Text":"plus z^2^2 minus minus."},{"Start":"27:00.205 ","End":"27:01.644","Text":"It\u0027s going to be a plus,"},{"Start":"27:01.644 ","End":"27:03.745","Text":"and now we\u0027re substituting in now is 0."},{"Start":"27:03.745 ","End":"27:07.894","Text":"Now it\u0027s simply going to be z^2^1/2,"},{"Start":"27:07.894 ","End":"27:11.845","Text":"and that\u0027s going to cross off as well."},{"Start":"27:11.845 ","End":"27:13.843","Text":"This is very exciting,"},{"Start":"27:13.843 ","End":"27:17.440","Text":"we\u0027re finishing this long integration."},{"Start":"27:17.440 ","End":"27:24.474","Text":"We have 2Pik Rho and now we\u0027ve crossed off all of these powers,"},{"Start":"27:24.474 ","End":"27:26.859","Text":"and so let\u0027s write this out."},{"Start":"27:26.859 ","End":"27:32.904","Text":"We have r^2 plus z minus"},{"Start":"27:32.904 ","End":"27:41.890","Text":"h^2^1/2 minus z,"},{"Start":"27:41.890 ","End":"27:44.020","Text":"minus minus h, so plus h,"},{"Start":"27:44.020 ","End":"27:51.355","Text":"and then we have minus r^2 plus z^2^1/2,"},{"Start":"27:51.355 ","End":"27:54.700","Text":"and then plus z."},{"Start":"27:54.700 ","End":"27:58.135","Text":"We have negative z plus z;"},{"Start":"27:58.135 ","End":"28:00.129","Text":"so those can cancel out,"},{"Start":"28:00.129 ","End":"28:10.285","Text":"and we\u0027re going to be left with in the end 2Pik Rho multiplied by r^2 plus"},{"Start":"28:10.285 ","End":"28:20.890","Text":"z minus h^2^1/2 minus r squared plus"},{"Start":"28:20.890 ","End":"28:25.840","Text":"z^2^1/2 plus"},{"Start":"28:25.840 ","End":"28:32.080","Text":"h. This is very exciting."},{"Start":"28:32.080 ","End":"28:42.025","Text":"This is our final value for our electric field in the z-direction from a solid cylinder."},{"Start":"28:42.025 ","End":"28:46.960","Text":"Now what we can do is we can check something with limits."},{"Start":"28:46.960 ","End":"28:55.254","Text":"Let\u0027s say our radius is much larger than our point z which has some arbitrary point"},{"Start":"28:55.254 ","End":"28:58.180","Text":"above our cylinder and much larger than"},{"Start":"28:58.180 ","End":"29:03.760","Text":"h. The radius of the cylinder is much larger than also the height of the cylinder,"},{"Start":"29:03.760 ","End":"29:08.010","Text":"so it looks like a giant disk."},{"Start":"29:08.010 ","End":"29:11.789","Text":"If we\u0027re looking at that,"},{"Start":"29:11.789 ","End":"29:17.760","Text":"that would mean that this z minus h will be insignificant compared to our r^2"},{"Start":"29:17.760 ","End":"29:23.979","Text":"and z^2 over here would also be insignificant to our r^2,"},{"Start":"29:23.979 ","End":"29:29.432","Text":"and that would mean that our electric field in the z-direction will be equal to."},{"Start":"29:29.432 ","End":"29:33.415","Text":"We\u0027ll have 2Pik Rho,"},{"Start":"29:33.415 ","End":"29:38.519","Text":"and then we have r^2^1/2 minus r^2^1/2 plus h,"},{"Start":"29:38.519 ","End":"29:42.849","Text":"so multiplied by h. What we"},{"Start":"29:42.849 ","End":"29:47.890","Text":"can see is that this is the electric field of an infinite plane."},{"Start":"29:47.890 ","End":"29:51.564","Text":"Just to remember the electric field of an infinite plane and"},{"Start":"29:51.564 ","End":"29:57.355","Text":"the z-direction is equal to 2Pi Sigma k,"},{"Start":"29:57.355 ","End":"30:04.750","Text":"and we can see that our Sigma over here is simply=Rho H. Now,"},{"Start":"30:04.750 ","End":"30:07.180","Text":"if you look at the units of Rho;"},{"Start":"30:07.180 ","End":"30:12.460","Text":"so that\u0027s charged density per unit volume multiplied by H which has height,"},{"Start":"30:12.460 ","End":"30:18.655","Text":"we will get the same units of our Sigma which is the charge density per unit area."},{"Start":"30:18.655 ","End":"30:22.450","Text":"We can really see that when we set limits where"},{"Start":"30:22.450 ","End":"30:27.790","Text":"the radius of our cylinder is much larger than our arbitrary point"},{"Start":"30:27.790 ","End":"30:31.179","Text":"above the z-axis and much larger than the height of"},{"Start":"30:31.179 ","End":"30:36.490","Text":"the cylinder we really get the value for an electric field of an infinite plane,"},{"Start":"30:36.490 ","End":"30:39.880","Text":"and that\u0027s how we can see that this is correct."},{"Start":"30:39.880 ","End":"30:42.890","Text":"That\u0027s the end of this lesson."}],"ID":22262},{"Watched":false,"Name":"Exercise 4","Duration":"33m 38s","ChapterTopicVideoID":21429,"CourseChapterTopicPlaylistID":99473,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.235","Text":"Hello. In this question,"},{"Start":"00:02.235 ","End":"00:06.825","Text":"we\u0027re given a ring of radius R. We\u0027re being told that 1/2"},{"Start":"00:06.825 ","End":"00:12.224","Text":"of the ring has charge density per unit length Lambda,"},{"Start":"00:12.224 ","End":"00:19.365","Text":"and that the second 1/2 has charge density per unit length of negative Lambda."},{"Start":"00:19.365 ","End":"00:25.104","Text":"We\u0027re being asked to find the electric field along the rings axis of symmetry."},{"Start":"00:25.104 ","End":"00:31.144","Text":"Let\u0027s call this axis of symmetry the z-axis."},{"Start":"00:31.144 ","End":"00:33.545","Text":"What we\u0027ve done up until now,"},{"Start":"00:33.545 ","End":"00:41.354","Text":"is that we\u0027ve split up our ring into tiny little pieces where each piece is of length dl."},{"Start":"00:41.354 ","End":"00:45.349","Text":"Then we found some arbitrary point along our z-axis,"},{"Start":"00:45.349 ","End":"00:48.890","Text":"and we\u0027ve worked out what the electric field is going"},{"Start":"00:48.890 ","End":"00:52.649","Text":"to be on the z-axis due to this little piece."},{"Start":"00:52.649 ","End":"00:55.205","Text":"Then we sum along around the circle."},{"Start":"00:55.205 ","End":"00:58.339","Text":"How we solve these questions is that we first worked"},{"Start":"00:58.339 ","End":"01:02.300","Text":"out the magnitude of the electric field on this section,"},{"Start":"01:02.300 ","End":"01:07.220","Text":"and then afterwards we worked out the direction."},{"Start":"01:07.220 ","End":"01:10.520","Text":"This time we\u0027re going to do this slightly"},{"Start":"01:10.520 ","End":"01:14.540","Text":"differently and we\u0027re going to be working out the electric field on"},{"Start":"01:14.540 ","End":"01:22.640","Text":"this arbitrary point via working out the vector field on this point straight away."},{"Start":"01:22.640 ","End":"01:26.090","Text":"So we\u0027re going to have size and direction and then we\u0027re going to"},{"Start":"01:26.090 ","End":"01:30.455","Text":"integrate along this vector quantity."},{"Start":"01:30.455 ","End":"01:34.620","Text":"Let\u0027s see how we\u0027re going to do this."},{"Start":"01:34.720 ","End":"01:43.219","Text":"Up until now, we said that the vector pointing from our section,"},{"Start":"01:43.219 ","End":"01:49.925","Text":"dl, up until the arbitrary point on our z-axis was defined as so."},{"Start":"01:49.925 ","End":"01:53.360","Text":"We said that we would always take the magnitude of it."},{"Start":"01:53.360 ","End":"01:56.879","Text":"So let\u0027s say that the vectors I tilde,"},{"Start":"01:56.879 ","End":"01:59.480","Text":"so we\u0027re looking at this purple line."},{"Start":"01:59.480 ","End":"02:02.495","Text":"The magnitude of r tilde, we always said,"},{"Start":"02:02.495 ","End":"02:06.440","Text":"was equal to the total radius of the circle,"},{"Start":"02:06.440 ","End":"02:11.420","Text":"so the distance from the center until the edge of the ring."},{"Start":"02:11.420 ","End":"02:15.755","Text":"Here it\u0027s I^2 because here it\u0027s a constant value,"},{"Start":"02:15.755 ","End":"02:20.584","Text":"plus some length from here,"},{"Start":"02:20.584 ","End":"02:25.853","Text":"the origin, and up to our arbitrary heights along the z-axis."},{"Start":"02:25.853 ","End":"02:28.870","Text":"We said that that arbitrary height was at z,"},{"Start":"02:28.870 ","End":"02:33.410","Text":"so z^2, and then we would take the square root of all of that."},{"Start":"02:33.410 ","End":"02:36.830","Text":"When we\u0027re dealing with vector quantities,"},{"Start":"02:36.830 ","End":"02:39.650","Text":"because now we want to work out the electric field as"},{"Start":"02:39.650 ","End":"02:42.410","Text":"a vector quantity straight away rather than"},{"Start":"02:42.410 ","End":"02:45.718","Text":"first working out the magnitude and then the direction."},{"Start":"02:45.718 ","End":"02:51.000","Text":"So what we say is that our r tilde vector,"},{"Start":"02:51.000 ","End":"02:52.670","Text":"this purple line,"},{"Start":"02:52.670 ","End":"02:56.779","Text":"is equal to our r vector,"},{"Start":"02:56.779 ","End":"03:01.239","Text":"which is the vector pointing along the axes of symmetry,"},{"Start":"03:01.239 ","End":"03:06.119","Text":"minus our r tag vector,"},{"Start":"03:06.119 ","End":"03:14.240","Text":"where r tag vector is the vector pointing from the origin to our point dl."},{"Start":"03:14.240 ","End":"03:18.530","Text":"This is a common way of writing this,"},{"Start":"03:18.530 ","End":"03:20.430","Text":"so it\u0027s useful to learn this."},{"Start":"03:20.430 ","End":"03:24.028","Text":"As we can see, that is exactly what our r tilde vector is."},{"Start":"03:24.028 ","End":"03:25.675","Text":"Our purple vector,"},{"Start":"03:25.675 ","End":"03:27.050","Text":"if we\u0027re beginning from here,"},{"Start":"03:27.050 ","End":"03:32.030","Text":"if we go back, r tag vector and up r vector,"},{"Start":"03:32.030 ","End":"03:41.389","Text":"we\u0027ll get this purple vector over here for r tilde."},{"Start":"03:41.389 ","End":"03:46.084","Text":"We\u0027ve gone to this method,"},{"Start":"03:46.084 ","End":"03:47.735","Text":"so now let\u0027s see what else."},{"Start":"03:47.735 ","End":"03:49.024","Text":"Now up until now,"},{"Start":"03:49.024 ","End":"03:56.300","Text":"we first worked out the magnitude of our electric field and we said that was"},{"Start":"03:56.300 ","End":"04:04.047","Text":"equal to kdq divided by our r^2,"},{"Start":"04:04.047 ","End":"04:14.359","Text":"where this is also equal to r. Another way of writing this,"},{"Start":"04:14.359 ","End":"04:19.954","Text":"that now we\u0027re going to be using this trick over here,"},{"Start":"04:19.954 ","End":"04:29.585","Text":"is that we\u0027re going to write that our dE vector is going to be equal to kdq as per usual."},{"Start":"04:29.585 ","End":"04:40.444","Text":"Then we\u0027re going to write that this is divided by the magnitude of our I tilde squared."},{"Start":"04:40.444 ","End":"04:47.494","Text":"This we always said was in some r direction, the unit vector."},{"Start":"04:47.494 ","End":"04:58.759","Text":"Another way of writing the unit vector is by writing our r vector divided by r. Then"},{"Start":"04:58.759 ","End":"05:04.399","Text":"we\u0027re going to instead write our r tilde divided"},{"Start":"05:04.399 ","End":"05:12.679","Text":"by the magnitude of r tilde,"},{"Start":"05:12.679 ","End":"05:16.865","Text":"because this small r is the magnitude."},{"Start":"05:16.865 ","End":"05:22.910","Text":"Then what we get is kdq divided by"},{"Start":"05:22.910 ","End":"05:32.729","Text":"the magnitude of our I tilde vector cubed in the r tilde vector direction."},{"Start":"05:32.770 ","End":"05:41.194","Text":"We can rewrite this by substituting in what our I tilde vector is equal to."},{"Start":"05:41.194 ","End":"05:46.530","Text":"We have k multiplied by dq,"},{"Start":"05:47.420 ","End":"05:51.404","Text":"because that\u0027s the charge of our dl section,"},{"Start":"05:51.404 ","End":"05:55.593","Text":"divided by our I tilde vector,"},{"Start":"05:55.593 ","End":"05:59.569","Text":"so we have the size which is equal to r vector,"},{"Start":"05:59.569 ","End":"06:02.240","Text":"which is this green arrow here,"},{"Start":"06:02.240 ","End":"06:05.614","Text":"minus r tag vector,"},{"Start":"06:05.614 ","End":"06:09.425","Text":"which is this green arrow here, cubed."},{"Start":"06:09.425 ","End":"06:13.104","Text":"Then in the direction of r tilde vector,"},{"Start":"06:13.104 ","End":"06:21.855","Text":"which is given by r vector minus r tag vector."},{"Start":"06:21.855 ","End":"06:27.964","Text":"When we integrate, we\u0027re going to be integrating along our r tag vector,"},{"Start":"06:27.964 ","End":"06:34.557","Text":"because that is what is changing its direction as we rotate around the circle,"},{"Start":"06:34.557 ","End":"06:40.489","Text":"whereas our r vector is always going to be pointing to our arbitrary point that we chose."},{"Start":"06:40.489 ","End":"06:44.390","Text":"This arbitrary point is always going to be the same point,"},{"Start":"06:44.390 ","End":"06:47.434","Text":"the same position along the z-axis,"},{"Start":"06:47.434 ","End":"06:53.104","Text":"but our r tag vector is moving around in order to sum up all of our dls."},{"Start":"06:53.104 ","End":"06:57.929","Text":"That\u0027s why we\u0027re going to integrate along this variable."},{"Start":"06:58.160 ","End":"07:04.079","Text":"Let\u0027s rub out some things to make a little bit more room."},{"Start":"07:04.270 ","End":"07:09.739","Text":"Really quickly, before we begin the rest of"},{"Start":"07:09.739 ","End":"07:12.200","Text":"the process which is defining what each of"},{"Start":"07:12.200 ","End":"07:15.095","Text":"these vectors is equal to and then doing the integration,"},{"Start":"07:15.095 ","End":"07:19.290","Text":"let\u0027s just see what our dq is equal to."},{"Start":"07:20.270 ","End":"07:23.460","Text":"It\u0027s going to be Lambda,"},{"Start":"07:23.460 ","End":"07:29.050","Text":"be it the positive Lambda or the negative Lambda multiplied by dl,"},{"Start":"07:29.050 ","End":"07:31.305","Text":"the length of each piece."},{"Start":"07:31.305 ","End":"07:35.320","Text":"What is our dl therefore going to be equal to?"},{"Start":"07:35.320 ","End":"07:38.470","Text":"Because we\u0027re working along a circle,"},{"Start":"07:38.470 ","End":"07:40.630","Text":"so we\u0027re going to be using polar coordinates."},{"Start":"07:40.630 ","End":"07:44.000","Text":"Because our radius is constant,"},{"Start":"07:44.000 ","End":"07:45.960","Text":"we just have 1 line,"},{"Start":"07:45.960 ","End":"07:50.740","Text":"so it\u0027s going to be rd Theta."},{"Start":"07:51.230 ","End":"07:56.289","Text":"Let\u0027s begin. First of all,"},{"Start":"07:56.289 ","End":"07:58.209","Text":"what we want to find out is,"},{"Start":"07:58.209 ","End":"08:02.193","Text":"what is our r vector?"},{"Start":"08:02.193 ","End":"08:05.984","Text":"Once we find our r vector and our r tag vector,"},{"Start":"08:05.984 ","End":"08:09.870","Text":"we\u0027ll know what our r tilde vector is which is this over here,"},{"Start":"08:09.870 ","End":"08:13.505","Text":"and then we can work this out."},{"Start":"08:13.505 ","End":"08:14.600","Text":"Our r vector, as we can see,"},{"Start":"08:14.600 ","End":"08:19.446","Text":"we\u0027re going from the origin and up the z axis,"},{"Start":"08:19.446 ","End":"08:24.620","Text":"so we can see that our r vector has 0,0 in"},{"Start":"08:24.620 ","End":"08:30.739","Text":"the x and y direction and then some z value in the z direction."},{"Start":"08:30.739 ","End":"08:37.831","Text":"You\u0027ll remember from previous lessons that we used to define this arbitrary length as z."},{"Start":"08:37.831 ","End":"08:41.839","Text":"Here we\u0027ve just defined it as a vector quantity,"},{"Start":"08:41.839 ","End":"08:46.670","Text":"but we can see that it only has z in the z direction,"},{"Start":"08:46.670 ","End":"08:51.560","Text":"which means the exact same thing as what we\u0027ve done in the previous lessons."},{"Start":"08:51.560 ","End":"08:59.270","Text":"Now, let\u0027s see what our r tag vector is going to be equal to."},{"Start":"08:59.270 ","End":"09:04.039","Text":"We can see that our r tag vector is moving along a circle,"},{"Start":"09:04.039 ","End":"09:10.134","Text":"which means that it\u0027s always going to have some x and some y components."},{"Start":"09:10.134 ","End":"09:12.694","Text":"When we\u0027re dealing with polar coordinates,"},{"Start":"09:12.694 ","End":"09:16.244","Text":"our x-coordinate is always going to be"},{"Start":"09:16.244 ","End":"09:24.234","Text":"R cosine of the angle between the x-axis and r vector, so cosine Theta."},{"Start":"09:24.234 ","End":"09:31.095","Text":"Then our y component is always going to be R sine of Theta."},{"Start":"09:31.095 ","End":"09:37.559","Text":"Then we can see that this r tag vector has no z component, so 0."},{"Start":"09:37.559 ","End":"09:39.950","Text":"Here we have R\u0027s because,"},{"Start":"09:39.950 ","End":"09:44.179","Text":"again, our radius of the circle is constant."},{"Start":"09:44.179 ","End":"09:49.835","Text":"Therefore, we can write that our r tilde vector,"},{"Start":"09:49.835 ","End":"09:55.370","Text":"which is equal to r vector minus r tag vector is"},{"Start":"09:55.370 ","End":"10:01.245","Text":"going to be equal to 0 minus R cosine Theta,"},{"Start":"10:01.245 ","End":"10:04.829","Text":"will be negative R cosine of Theta,"},{"Start":"10:04.829 ","End":"10:07.410","Text":"0 minus R sine Theta,"},{"Start":"10:07.410 ","End":"10:11.115","Text":"so negative R sine of Theta,"},{"Start":"10:11.115 ","End":"10:15.925","Text":"and z minus 0 is z."},{"Start":"10:15.925 ","End":"10:23.705","Text":"Now, we want to know because we can see that we have the absolute value"},{"Start":"10:23.705 ","End":"10:27.890","Text":"of our r minus r tag vectors here to the power"},{"Start":"10:27.890 ","End":"10:32.180","Text":"of 3 but the important thing right now is that we\u0027re taking the absolute value."},{"Start":"10:32.180 ","End":"10:36.070","Text":"That means that the absolute value of our r tilde,"},{"Start":"10:36.070 ","End":"10:38.080","Text":"which is the magnitude of r tilde,"},{"Start":"10:38.080 ","End":"10:43.324","Text":"means the magnitude of our r minus r tag vectors."},{"Start":"10:43.324 ","End":"10:45.605","Text":"What is that going to be equal to?"},{"Start":"10:45.605 ","End":"10:49.850","Text":"That\u0027s going to be equal to the square root of our x component squared,"},{"Start":"10:49.850 ","End":"10:51.919","Text":"so a negative squared is going to be a positive,"},{"Start":"10:51.919 ","End":"10:59.765","Text":"so we\u0027re going to have R^2 cosine^2 Theta plus our y component squared."},{"Start":"10:59.765 ","End":"11:01.889","Text":"Again, the negative cancels out,"},{"Start":"11:01.889 ","End":"11:06.725","Text":"so we\u0027ll have R^2 sine squared Theta."},{"Start":"11:06.725 ","End":"11:09.079","Text":"Plus here, we have a z component as well,"},{"Start":"11:09.079 ","End":"11:12.049","Text":"so plus our z component squared,"},{"Start":"11:12.049 ","End":"11:14.539","Text":"and the square root of all of that."},{"Start":"11:14.539 ","End":"11:17.854","Text":"This is going to be equal to, as we know,"},{"Start":"11:17.854 ","End":"11:22.100","Text":"cosine^2 Theta plus sine squared Theta is equal to 1."},{"Start":"11:22.100 ","End":"11:26.089","Text":"It\u0027s going to be the square root of R^2,"},{"Start":"11:26.089 ","End":"11:28.865","Text":"and then in brackets,"},{"Start":"11:28.865 ","End":"11:39.285","Text":"cosine^2 Theta plus sine squared Theta plus z^2."},{"Start":"11:39.285 ","End":"11:44.785","Text":"Then we know that this is equal to 1,"},{"Start":"11:44.785 ","End":"11:49.375","Text":"so we\u0027ll get that the magnitude of this,"},{"Start":"11:49.375 ","End":"11:52.165","Text":"which is in the denominator of our dE,"},{"Start":"11:52.165 ","End":"11:59.161","Text":"is going to be equal to r^2 plus z^2."},{"Start":"11:59.161 ","End":"12:02.535","Text":"Then if you remember to our previous lessons,"},{"Start":"12:02.535 ","End":"12:06.975","Text":"this is also the value that we got in the denominator."},{"Start":"12:06.975 ","End":"12:09.484","Text":"Then of course we wrote it"},{"Start":"12:09.484 ","End":"12:13.720","Text":"as what\u0027s inside the square root sign and then to the power of 1/2,"},{"Start":"12:13.720 ","End":"12:15.070","Text":"which means the same thing."},{"Start":"12:15.070 ","End":"12:17.919","Text":"Then when we write that to the power of 3,"},{"Start":"12:17.919 ","End":"12:20.690","Text":"it\u0027s going to be to the power of 3/2."},{"Start":"12:21.540 ","End":"12:27.700","Text":"Therefore, let\u0027s write out what our dE vector."},{"Start":"12:27.700 ","End":"12:29.169","Text":"We\u0027re working with a vector,"},{"Start":"12:29.169 ","End":"12:31.240","Text":"so we have magnitude and direction now."},{"Start":"12:31.240 ","End":"12:34.375","Text":"That\u0027s going to be equal to kd_q,"},{"Start":"12:34.375 ","End":"12:43.959","Text":"so that\u0027s k Lambda Rd Theta divided by the magnitude of this,"},{"Start":"12:43.959 ","End":"12:45.190","Text":"which we just worked out,"},{"Start":"12:45.190 ","End":"12:53.800","Text":"so we have r^2 plus z^2 to the power of 1/2 to the power of 3, so 3/2."},{"Start":"12:53.800 ","End":"13:00.745","Text":"Then in brackets, we have in our r tilde vector direction,"},{"Start":"13:00.745 ","End":"13:03.744","Text":"which is equal to this. How can we write that?"},{"Start":"13:03.744 ","End":"13:05.860","Text":"We can just write out the coordinates,"},{"Start":"13:05.860 ","End":"13:14.439","Text":"so that\u0027s going to be going in the negative R cosine Theta direction in the x direction,"},{"Start":"13:14.439 ","End":"13:19.555","Text":"negative R sine Theta in the y-direction,"},{"Start":"13:19.555 ","End":"13:21.970","Text":"and z in the z-direction."},{"Start":"13:21.970 ","End":"13:25.989","Text":"Notice, I just took this from this over here,"},{"Start":"13:25.989 ","End":"13:28.009","Text":"what we worked out."},{"Start":"13:28.680 ","End":"13:32.290","Text":"Now, we\u0027re ready to integrate."},{"Start":"13:32.290 ","End":"13:40.270","Text":"We can say that our E field is going to be the integral on our dE."},{"Start":"13:40.270 ","End":"13:47.439","Text":"That is going to be the integral on k Lambda Rd Theta"},{"Start":"13:47.439 ","End":"13:56.154","Text":"divided by r^2 plus z^2 to the power of 3/2."},{"Start":"13:56.154 ","End":"14:00.745","Text":"Then we have it in the negative r cosine Theta,"},{"Start":"14:00.745 ","End":"14:05.844","Text":"negative R sine of Theta, and z."},{"Start":"14:05.844 ","End":"14:07.944","Text":"Now, another way to write this,"},{"Start":"14:07.944 ","End":"14:10.029","Text":"which is the exact same thing as this,"},{"Start":"14:10.029 ","End":"14:13.644","Text":"is if we integrate along our x-axis,"},{"Start":"14:13.644 ","End":"14:16.285","Text":"our y-axis, and our z-axis separately."},{"Start":"14:16.285 ","End":"14:20.189","Text":"Then we can split up our integration like so."},{"Start":"14:20.189 ","End":"14:25.810","Text":"I\u0027ll have k Lambda Rd Theta divided"},{"Start":"14:25.810 ","End":"14:31.840","Text":"by R^2 plus z^2 to the power of 3/2."},{"Start":"14:31.840 ","End":"14:34.810","Text":"Then we\u0027ll have multiplied by this,"},{"Start":"14:34.810 ","End":"14:38.380","Text":"denoting the x-component in the x-direction,"},{"Start":"14:38.380 ","End":"14:42.417","Text":"so negative r cosine of Theta,"},{"Start":"14:42.417 ","End":"14:44.860","Text":"and this is in the x-direction."},{"Start":"14:44.860 ","End":"14:52.299","Text":"Then we\u0027ll have plus the integration of k Lambda Rd Theta divided"},{"Start":"14:52.299 ","End":"15:02.169","Text":"by R^2 plus z^2 to the power of 3/2 multiplied by the component in the y-direction."},{"Start":"15:02.169 ","End":"15:06.904","Text":"So negative R sine of Theta and this is in the y-direction,"},{"Start":"15:06.904 ","End":"15:15.789","Text":"plus the integration of k Lambda Rd Theta divided by R^2"},{"Start":"15:15.789 ","End":"15:17.829","Text":"plus z^2 to the power of"},{"Start":"15:17.829 ","End":"15:23.290","Text":"3/2 multiplied"},{"Start":"15:23.290 ","End":"15:28.675","Text":"by the z-component in the z-direction."},{"Start":"15:28.675 ","End":"15:35.030","Text":"What we\u0027ve written here is the exact same thing as what we\u0027ve written here."},{"Start":"15:36.540 ","End":"15:39.955","Text":"Now, let\u0027s solve the integral."},{"Start":"15:39.955 ","End":"15:45.475","Text":"First, what I want to do is I want to solve the integral in the y-direction,"},{"Start":"15:45.475 ","End":"15:50.215","Text":"and then afterwards we\u0027ll look at the x and the z direction."},{"Start":"15:50.215 ","End":"15:53.095","Text":"Let\u0027s write this over here."},{"Start":"15:53.095 ","End":"16:01.225","Text":"Let\u0027s write that our E field in the y-direction is equal to the integral of"},{"Start":"16:01.225 ","End":"16:09.399","Text":"k Lambda Rd Theta divided by r^2 plus z^2 to the power of"},{"Start":"16:09.399 ","End":"16:19.674","Text":"3/2 in the negative R sine Theta y-direction."},{"Start":"16:19.674 ","End":"16:23.530","Text":"Now, I don\u0027t have to write y-direction right now"},{"Start":"16:23.530 ","End":"16:29.305","Text":"because we know that my E field in the y-direction is obviously in the y-direction."},{"Start":"16:29.305 ","End":"16:35.440","Text":"Now, what I\u0027m noticing is that in the question I\u0027m being told that half of the circle"},{"Start":"16:35.440 ","End":"16:43.235","Text":"has charge density Lambda and the other half has charged density of negative Lambda."},{"Start":"16:43.235 ","End":"16:49.959","Text":"That means that I have to split up this integration into two integrals."},{"Start":"16:49.959 ","End":"16:55.629","Text":"One, along the section of the circle that has charge density Lambda,"},{"Start":"16:55.629 ","End":"16:58.694","Text":"and then one further integration along"},{"Start":"16:58.694 ","End":"17:03.580","Text":"the section of the circle that has charge density negative Lambda."},{"Start":"17:04.590 ","End":"17:08.785","Text":"Here is our x-axis,"},{"Start":"17:08.785 ","End":"17:13.030","Text":"which means that when our R tag vector is along here,"},{"Start":"17:13.030 ","End":"17:15.730","Text":"our Theta is equal to 0."},{"Start":"17:15.730 ","End":"17:17.586","Text":"We\u0027re going to go from 0,"},{"Start":"17:17.586 ","End":"17:21.895","Text":"and then we\u0027re going to go all the way because we\u0027re being told it\u0027s half a circle,"},{"Start":"17:21.895 ","End":"17:24.625","Text":"so we\u0027re going to go half a circle around."},{"Start":"17:24.625 ","End":"17:26.574","Text":"That\u0027s Pi radians."},{"Start":"17:26.574 ","End":"17:30.470","Text":"We\u0027re going to write this again."},{"Start":"17:30.780 ","End":"17:37.405","Text":"Remember this equation of by heart by the time we\u0027re done with this."},{"Start":"17:37.405 ","End":"17:43.795","Text":"This is going to be negative R sine Theta."},{"Start":"17:43.795 ","End":"17:47.965","Text":"That is for our positive Lambda direction."},{"Start":"17:47.965 ","End":"17:50.875","Text":"Then we have to add in our second integral,"},{"Start":"17:50.875 ","End":"17:53.289","Text":"which is going to be from here,"},{"Start":"17:53.289 ","End":"17:54.670","Text":"halfway along the circle,"},{"Start":"17:54.670 ","End":"17:59.559","Text":"so that\u0027s from Pi radians until we complete a full circle."},{"Start":"17:59.559 ","End":"18:02.785","Text":"So that\u0027s up until 2Pi radians."},{"Start":"18:02.785 ","End":"18:04.870","Text":"Then again k,"},{"Start":"18:04.870 ","End":"18:07.720","Text":"now this time, instead of Lambda,"},{"Start":"18:07.720 ","End":"18:13.570","Text":"we\u0027re substituting in negative Lambda because here we have our negative charge density,"},{"Start":"18:13.570 ","End":"18:18.355","Text":"so negative Lambda Rd Theta"},{"Start":"18:18.355 ","End":"18:25.480","Text":"divided by r^2 plus z^2 to the power of 3/2."},{"Start":"18:25.480 ","End":"18:30.440","Text":"Again, negative R sine Theta."},{"Start":"18:32.340 ","End":"18:36.804","Text":"Now, in order to solve this slightly quicker,"},{"Start":"18:36.804 ","End":"18:38.710","Text":"I\u0027m just going to say what our answer is."},{"Start":"18:38.710 ","End":"18:46.569","Text":"We\u0027re going to get that our E_y is equal to negative 4 k"},{"Start":"18:46.569 ","End":"18:50.919","Text":"lambda r^2 squared divided by r^2"},{"Start":"18:50.919 ","End":"18:58.250","Text":"plus z^2 to the power of 3/2."},{"Start":"18:59.820 ","End":"19:02.499","Text":"This is our answer for this."},{"Start":"19:02.499 ","End":"19:08.395","Text":"Now, let\u0027s look at our electric field in the x direction."},{"Start":"19:08.395 ","End":"19:13.660","Text":"Here we can see that it\u0027s going to be equal to 0. Why is this?"},{"Start":"19:13.660 ","End":"19:16.989","Text":"From symmetry, we can see that here we\u0027re going to have"},{"Start":"19:16.989 ","End":"19:21.040","Text":"our electric field pointing in one x direction,"},{"Start":"19:21.040 ","End":"19:23.800","Text":"and due to this positive charge density,"},{"Start":"19:23.800 ","End":"19:26.019","Text":"the electric field is going to be pointing in"},{"Start":"19:26.019 ","End":"19:29.965","Text":"the opposite x direction and then they\u0027ll cancel out."},{"Start":"19:29.965 ","End":"19:31.600","Text":"You can see that from symmetry."},{"Start":"19:31.600 ","End":"19:34.270","Text":"If however, you can\u0027t see that from symmetry,"},{"Start":"19:34.270 ","End":"19:36.774","Text":"all you have to do is copy out"},{"Start":"19:36.774 ","End":"19:43.600","Text":"this integration like so and then set it out like we did for our E_Y field."},{"Start":"19:43.600 ","End":"19:48.429","Text":"With the two integrations from 0 to Pi for the positive charge density,"},{"Start":"19:48.429 ","End":"19:52.090","Text":"and then from Pi to 2Pi for the negative charge density."},{"Start":"19:52.090 ","End":"19:56.890","Text":"You can even just literally write out these two lines, again,"},{"Start":"19:56.890 ","End":"20:03.715","Text":"just replacing R sine Theta with cosine Theta for our x direction."},{"Start":"20:03.715 ","End":"20:05.950","Text":"Instead of sine Theta, cosine Theta,"},{"Start":"20:05.950 ","End":"20:08.350","Text":"and then when you do the integration,"},{"Start":"20:08.350 ","End":"20:11.600","Text":"you\u0027ll see that you will get 0."},{"Start":"20:12.300 ","End":"20:18.625","Text":"Now, let\u0027s talk about our E field in the z direction."},{"Start":"20:18.625 ","End":"20:20.439","Text":"Again, through symmetry,"},{"Start":"20:20.439 ","End":"20:22.510","Text":"we\u0027re also going to get 0."},{"Start":"20:22.510 ","End":"20:27.219","Text":"Why is that? From our semicircle with a negative charge density,"},{"Start":"20:27.219 ","End":"20:33.189","Text":"it will have an electric field in this horizontal or diagonal direction."},{"Start":"20:33.189 ","End":"20:37.179","Text":"From the positive charge density direction,"},{"Start":"20:37.179 ","End":"20:40.750","Text":"we\u0027re going to have an electric field in this direction."},{"Start":"20:40.750 ","End":"20:44.290","Text":"Then we can see if we split this up into the separate components,"},{"Start":"20:44.290 ","End":"20:50.440","Text":"so we\u0027ll have a z component in this direction and the z component in this direction."},{"Start":"20:50.440 ","End":"20:54.355","Text":"Therefore, the two z components will cancel each other out."},{"Start":"20:54.355 ","End":"20:59.620","Text":"From symmetry, our E field in the z direction is also going to be equal to 0."},{"Start":"20:59.620 ","End":"21:03.415","Text":"Alternatively, again, if you cannot see the symmetry,"},{"Start":"21:03.415 ","End":"21:06.515","Text":"then we can do what we did for E_y,"},{"Start":"21:06.515 ","End":"21:09.010","Text":"exactly the same for our E_z."},{"Start":"21:09.010 ","End":"21:10.450","Text":"We write it out,"},{"Start":"21:10.450 ","End":"21:14.560","Text":"and then instead of writing negative r sine Theta, we\u0027re just going to write z."},{"Start":"21:14.560 ","End":"21:18.369","Text":"Then again, we split it up into two integrals because we have"},{"Start":"21:18.369 ","End":"21:23.935","Text":"half a circle with this charge density and half a circle with a negative charge density."},{"Start":"21:23.935 ","End":"21:27.580","Text":"Then, because we\u0027re integrating along d Theta,"},{"Start":"21:27.580 ","End":"21:30.549","Text":"but we don\u0027t have any variable"},{"Start":"21:30.549 ","End":"21:33.534","Text":"with Theta here anymore because we\u0027re going to be dealing just with z,"},{"Start":"21:33.534 ","End":"21:35.394","Text":"there\u0027s no other Theta here."},{"Start":"21:35.394 ","End":"21:38.710","Text":"So we\u0027re integrating along a constant."},{"Start":"21:38.710 ","End":"21:42.610","Text":"Then when we substitute in our bounds,"},{"Start":"21:42.610 ","End":"21:45.974","Text":"we\u0027ll have Pi times this,"},{"Start":"21:45.974 ","End":"21:47.729","Text":"where this is z over here."},{"Start":"21:47.729 ","End":"21:54.555","Text":"So Pi times this minus and then here when we substitute in these bounds as well,"},{"Start":"21:54.555 ","End":"21:57.840","Text":"2Pi minus Pi will also have Pi times this,"},{"Start":"21:57.840 ","End":"22:02.275","Text":"but this time a negative because of the negative charge density."},{"Start":"22:02.275 ","End":"22:07.284","Text":"We\u0027ll have Pi times some constant minus Pi times the same constant,"},{"Start":"22:07.284 ","End":"22:11.150","Text":"which is again going to be equal to 0."},{"Start":"22:12.180 ","End":"22:17.875","Text":"Now, we saw that we worked out the E field in"},{"Start":"22:17.875 ","End":"22:23.650","Text":"all the different directions via working straight away with a vector quantity."},{"Start":"22:23.650 ","End":"22:28.150","Text":"Now, what I\u0027m going to do is I\u0027m going to solve the exact same question,"},{"Start":"22:28.150 ","End":"22:30.130","Text":"but in the original method that we learned,"},{"Start":"22:30.130 ","End":"22:34.180","Text":"which is first by working out the magnitude of the E field and then"},{"Start":"22:34.180 ","End":"22:39.579","Text":"sorting out the direction afterwards."},{"Start":"22:39.579 ","End":"22:45.100","Text":"Now we\u0027re repeating the question but in the previous methods that we\u0027ve learned,"},{"Start":"22:45.100 ","End":"22:48.340","Text":"where we first work out the magnitude and then the direction."},{"Start":"22:48.340 ","End":"22:51.354","Text":"Now here are the answers that we got from doing it"},{"Start":"22:51.354 ","End":"22:57.159","Text":"straight integration along vector quantities. Here are our answers."},{"Start":"22:57.159 ","End":"22:59.380","Text":"Let\u0027s see if we get the same thing."},{"Start":"22:59.380 ","End":"23:04.780","Text":"The first thing that we want to do is we want to find out what our dq is equal to,"},{"Start":"23:04.780 ","End":"23:07.045","Text":"so the charge of this length dl."},{"Start":"23:07.045 ","End":"23:10.899","Text":"As we know, it\u0027s going to be Lambda positive or negative."},{"Start":"23:10.899 ","End":"23:17.410","Text":"We\u0027ll soon discuss this along the length dl."},{"Start":"23:17.410 ","End":"23:20.260","Text":"That\u0027s going to be equal 2 because we\u0027re working in"},{"Start":"23:20.260 ","End":"23:23.140","Text":"polar coordinates and our radius is constant."},{"Start":"23:23.140 ","End":"23:27.010","Text":"It\u0027s going to be Lambda Rd Theta."},{"Start":"23:27.010 ","End":"23:32.185","Text":"Now, let\u0027s work out the magnitude of the electric field."},{"Start":"23:32.185 ","End":"23:40.945","Text":"We know that this is going to be equal to kdq divided by this purple line^2."},{"Start":"23:40.945 ","End":"23:45.280","Text":"That\u0027s going to be our r tilde^2."},{"Start":"23:45.280 ","End":"23:48.730","Text":"The purple arrow is our r tilde vector,"},{"Start":"23:48.730 ","End":"23:53.154","Text":"and where our r tilde is equal to"},{"Start":"23:53.154 ","End":"24:00.534","Text":"the magnitude of our r tilde vector,"},{"Start":"24:00.534 ","End":"24:03.556","Text":"which is going to be the square root of."},{"Start":"24:03.556 ","End":"24:07.495","Text":"So we see we have here R,"},{"Start":"24:07.495 ","End":"24:09.057","Text":"the radius of the circle."},{"Start":"24:09.057 ","End":"24:17.155","Text":"So it\u0027s going to be R^2 plus this arbitrary distance over here till this point,"},{"Start":"24:17.155 ","End":"24:21.265","Text":"let\u0027s call this z, so plus z^2."},{"Start":"24:21.265 ","End":"24:27.865","Text":"The square root of that, so R plus z and then the squared."},{"Start":"24:27.865 ","End":"24:33.370","Text":"Now let\u0027s find our electric field in the y-direction first."},{"Start":"24:33.370 ","End":"24:35.890","Text":"Let\u0027s have our dEy."},{"Start":"24:35.890 ","End":"24:40.675","Text":"That\u0027s equal to the magnitude of our electric field,"},{"Start":"24:40.675 ","End":"24:43.795","Text":"so the total magnitude, which is this."},{"Start":"24:43.795 ","End":"24:48.560","Text":"Then we have to choose the direction."},{"Start":"24:49.170 ","End":"24:56.860","Text":"Now, finding our direction here is going to be slightly more difficult. Why is that?"},{"Start":"24:56.860 ","End":"24:58.930","Text":"Because usually we were dealing with"},{"Start":"24:58.930 ","End":"25:02.365","Text":"a constant charge distribution of some kind of Lambda,"},{"Start":"25:02.365 ","End":"25:06.415","Text":"but here we have half positive Lambda and half negative Lambda,"},{"Start":"25:06.415 ","End":"25:09.235","Text":"which means that due to symmetry,"},{"Start":"25:09.235 ","End":"25:15.760","Text":"we\u0027re going to have all components of our electric field canceling out with each other."},{"Start":"25:15.760 ","End":"25:18.864","Text":"What we did before is we said that"},{"Start":"25:18.864 ","End":"25:26.245","Text":"our dE was in this direction and we said that this angle is Alpha,"},{"Start":"25:26.245 ","End":"25:27.895","Text":"which means that this angle is Alpha."},{"Start":"25:27.895 ","End":"25:30.085","Text":"Then we multiplied this by"},{"Start":"25:30.085 ","End":"25:35.755","Text":"cosine Alpha or whatever it would be or sine Alpha depending which direction."},{"Start":"25:35.755 ","End":"25:40.285","Text":"Now, here, because we can see that we have the positive and negative Lambdas,"},{"Start":"25:40.285 ","End":"25:45.970","Text":"so we know that the E field in the z direction is going to cancel out."},{"Start":"25:45.970 ","End":"25:54.100","Text":"So in order to find in which direction our E field is acting in,"},{"Start":"25:54.100 ","End":"25:58.790","Text":"we\u0027re going to have to do this in 2 steps."},{"Start":"25:59.340 ","End":"26:09.760","Text":"Before we would say that here we would take the projection of our dE in the z direction,"},{"Start":"26:09.760 ","End":"26:14.995","Text":"so this would be our dE_z, but now,"},{"Start":"26:14.995 ","End":"26:17.124","Text":"because we don\u0027t have any z components,"},{"Start":"26:17.124 ","End":"26:24.950","Text":"we want to find the projection of this on our x,y plane."},{"Start":"26:25.620 ","End":"26:31.030","Text":"We can see that if we take sine of Alpha,"},{"Start":"26:31.030 ","End":"26:33.400","Text":"so what we\u0027re doing,"},{"Start":"26:33.400 ","End":"26:40.990","Text":"this red arrow is the same as this red arrow over here because they\u0027re both parallel,"},{"Start":"26:40.990 ","End":"26:43.030","Text":"and this red arrow,"},{"Start":"26:43.030 ","End":"26:46.465","Text":"it\u0027s the side opposite to our angle Alpha."},{"Start":"26:46.465 ","End":"26:48.130","Text":"From our trig identities,"},{"Start":"26:48.130 ","End":"26:50.680","Text":"we know that we\u0027re going to be using sine of Alpha."},{"Start":"26:50.680 ","End":"26:57.305","Text":"The magnitude of the electric field multiplied by sine of Alpha,"},{"Start":"26:57.305 ","End":"27:05.410","Text":"so all of this is going to be our electric field in the x,y plane."},{"Start":"27:05.660 ","End":"27:17.030","Text":"What we\u0027ve done is we\u0027ve taken our projection of our electric field on the x,y plane."},{"Start":"27:17.310 ","End":"27:21.760","Text":"We can see that our r vector is pointing to this point,"},{"Start":"27:21.760 ","End":"27:28.735","Text":"so let\u0027s copy our red arrow down to the x,y plane."},{"Start":"27:28.735 ","End":"27:32.799","Text":"We can see that both arrows are parallel."},{"Start":"27:32.799 ","End":"27:36.130","Text":"Just the red arrow is pointing in the opposite direction."},{"Start":"27:36.130 ","End":"27:40.885","Text":"Now specifically here, because we\u0027re dealing with the electric field in the y direction,"},{"Start":"27:40.885 ","End":"27:44.020","Text":"we\u0027re only going to want this y component."},{"Start":"27:44.020 ","End":"27:51.224","Text":"Imagine that all the red arrows are pointing in the same direction."},{"Start":"27:51.224 ","End":"27:55.420","Text":"Imagine they\u0027re all slightly at a diagonal."},{"Start":"27:55.610 ","End":"28:01.784","Text":"Now we can see that in order to get"},{"Start":"28:01.784 ","End":"28:10.315","Text":"our r tag vector to point in the direction of our red vector,"},{"Start":"28:10.315 ","End":"28:17.020","Text":"so that\u0027s the projection of our electric field dE on the x, y plane."},{"Start":"28:17.020 ","End":"28:22.555","Text":"We can see that between the x axis and our r tag vector,"},{"Start":"28:22.555 ","End":"28:26.305","Text":"we have an angle of Theta but then,"},{"Start":"28:26.305 ","End":"28:29.140","Text":"because we\u0027re pointing in the opposite direction,"},{"Start":"28:29.140 ","End":"28:32.005","Text":"in order to spin around to this opposite direction,"},{"Start":"28:32.005 ","End":"28:35.664","Text":"we\u0027re going to have to add in half a circle,"},{"Start":"28:35.664 ","End":"28:37.765","Text":"or 180 degrees,"},{"Start":"28:37.765 ","End":"28:40.135","Text":"or Pi radians,"},{"Start":"28:40.135 ","End":"28:43.795","Text":"because if you have a vector pointing in this direction,"},{"Start":"28:43.795 ","End":"28:46.104","Text":"in order to get it to point in this direction,"},{"Start":"28:46.104 ","End":"28:48.895","Text":"you have to spin it 180 degrees,"},{"Start":"28:48.895 ","End":"28:52.100","Text":"which is equal to Pi radians."},{"Start":"28:52.410 ","End":"28:56.600","Text":"You spin it round half a circle."},{"Start":"28:58.560 ","End":"29:05.815","Text":"Now what we want is we want our y component of that angle."},{"Start":"29:05.815 ","End":"29:09.850","Text":"Our total angle is Theta plus Pi."},{"Start":"29:09.850 ","End":"29:14.200","Text":"This is our angle and we want our y component."},{"Start":"29:14.200 ","End":"29:17.125","Text":"This is our total angle."},{"Start":"29:17.125 ","End":"29:21.025","Text":"Now, we\u0027re going to take our y component."},{"Start":"29:21.025 ","End":"29:26.830","Text":"As we know, because our Theta is defined relative to our x axis,"},{"Start":"29:26.830 ","End":"29:30.264","Text":"that means that a y component is always going to be"},{"Start":"29:30.264 ","End":"29:36.025","Text":"sine of the angle between the x axis and our vector."},{"Start":"29:36.025 ","End":"29:41.545","Text":"That means it\u0027s going to be sine of Theta plus Pi."},{"Start":"29:41.545 ","End":"29:49.750","Text":"What we did is we found our electric field projected onto our x,"},{"Start":"29:49.750 ","End":"29:52.090","Text":"y plane, which is what we can see here."},{"Start":"29:52.090 ","End":"29:57.520","Text":"Then we found the y components of this projection on the x,"},{"Start":"29:57.520 ","End":"30:01.929","Text":"y plane via multiplying the projection by sine of"},{"Start":"30:01.929 ","End":"30:07.970","Text":"the angle between the x axis and our r tag vector."},{"Start":"30:09.000 ","End":"30:15.130","Text":"We can say that this is equal to our magnitude of dE, which is kdq,"},{"Start":"30:15.130 ","End":"30:22.645","Text":"which is k Lambda Rd Theta divided by our r tilde^2."},{"Start":"30:22.645 ","End":"30:24.415","Text":"Then our sine Alpha,"},{"Start":"30:24.415 ","End":"30:25.839","Text":"just like we did before,"},{"Start":"30:25.839 ","End":"30:33.805","Text":"we can see that sine of Alpha is going to be our adjacent divided by our hypotenuse."},{"Start":"30:33.805 ","End":"30:37.900","Text":"It\u0027s going to be our adjacent side, which is z,"},{"Start":"30:37.900 ","End":"30:40.990","Text":"divided by our hypotenuse,"},{"Start":"30:40.990 ","End":"30:44.659","Text":"the magnitude, which is going to be our r tilde."},{"Start":"30:44.659 ","End":"30:50.390","Text":"Then multiplied by sine of Theta plus Pi."},{"Start":"30:50.790 ","End":"30:54.010","Text":"This is going to be equal 2,"},{"Start":"30:54.010 ","End":"30:57.745","Text":"now substituting in what our r tilde is which is equal to this."},{"Start":"30:57.745 ","End":"31:00.595","Text":"Then we have r tilde^3 in the denominator."},{"Start":"31:00.595 ","End":"31:08.980","Text":"We\u0027re going to have k Lambda Rd Theta z divided by r tilde^3."},{"Start":"31:08.980 ","End":"31:17.740","Text":"That\u0027s going to be R^2 plus z^2^3/2 from this,"},{"Start":"31:17.740 ","End":"31:23.390","Text":"and then in the direction sine Theta plus Pi."},{"Start":"31:25.620 ","End":"31:28.000","Text":"Now let\u0027s integrate."},{"Start":"31:28.000 ","End":"31:32.080","Text":"Let\u0027s find our electric field in the y direction."},{"Start":"31:32.080 ","End":"31:36.070","Text":"This is the integral of dEy."},{"Start":"31:36.070 ","End":"31:39.055","Text":"Then again, we\u0027re going to split them,"},{"Start":"31:39.055 ","End":"31:41.170","Text":"this integral into 2 integrals,"},{"Start":"31:41.170 ","End":"31:46.269","Text":"1 along the area where we have positive charge density and another integral along"},{"Start":"31:46.269 ","End":"31:51.880","Text":"the length here where we have negative charge density."},{"Start":"31:51.880 ","End":"31:54.459","Text":"We\u0027re going to have again from 0 till"},{"Start":"31:54.459 ","End":"32:04.570","Text":"Pi k Lambda zRd Theta divided by"},{"Start":"32:04.570 ","End":"32:13.224","Text":"R^2 plus z^2^3/2 sine Theta"},{"Start":"32:13.224 ","End":"32:20.560","Text":"plus Pi plus our second integral along the area of the negative charge density."},{"Start":"32:20.560 ","End":"32:23.799","Text":"We\u0027re going to have from Pi till"},{"Start":"32:23.799 ","End":"32:31.315","Text":"2Pi of k negative Lambda zRd Theta"},{"Start":"32:31.315 ","End":"32:40.570","Text":"divided by R^2 plus z^2^3/2 sine"},{"Start":"32:40.570 ","End":"32:42.804","Text":"of Theta plus Pi."},{"Start":"32:42.804 ","End":"32:50.065","Text":"Then when we do this integration and we substitute in our bounds,"},{"Start":"32:50.065 ","End":"32:55.659","Text":"we\u0027ll see that we will get negative 4k"},{"Start":"32:55.659 ","End":"33:00.579","Text":"Lambda R^2 divided by"},{"Start":"33:00.579 ","End":"33:07.870","Text":"R^2 plus z^2^3/2 in the y direction,"},{"Start":"33:07.870 ","End":"33:12.619","Text":"which is exactly what we got over here as well."},{"Start":"33:12.740 ","End":"33:22.015","Text":"Then we can say that our E_x and E_z are also equal to 0,"},{"Start":"33:22.015 ","End":"33:24.744","Text":"our electric fields, due to symmetry,"},{"Start":"33:24.744 ","End":"33:28.705","Text":"which was explained in the first half of the video."},{"Start":"33:28.705 ","End":"33:31.884","Text":"Either you can see it due to symmetry or you can"},{"Start":"33:31.884 ","End":"33:36.280","Text":"just integrate along them and you\u0027ll see that they\u0027ll cancel out."},{"Start":"33:36.280 ","End":"33:39.350","Text":"That\u0027s the end of this lesson."}],"ID":22263},{"Watched":false,"Name":"Exercise 5","Duration":"33m 36s","ChapterTopicVideoID":21430,"CourseChapterTopicPlaylistID":99473,"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":"00:04.499","Text":"we\u0027re going to be answering the following question."},{"Start":"00:04.499 ","End":"00:10.454","Text":"A ring of radius R has a non-uniform charge density,"},{"Start":"00:10.454 ","End":"00:13.800","Text":"which is dependent on the angle between the x-axis,"},{"Start":"00:13.800 ","End":"00:17.070","Text":"according to the equation that Lambda, as a function of Theta,"},{"Start":"00:17.070 ","End":"00:23.820","Text":"is equal to Lambda_0 multiplied by sine of Theta."},{"Start":"00:23.820 ","End":"00:25.514","Text":"Question Number 1 is,"},{"Start":"00:25.514 ","End":"00:29.260","Text":"what is the total charge of the ring?"},{"Start":"00:30.110 ","End":"00:36.900","Text":"Let\u0027s say that we take a small piece of this ring."},{"Start":"00:36.900 ","End":"00:46.245","Text":"We know that this piece over here has a charge of dq."},{"Start":"00:46.245 ","End":"00:51.769","Text":"So let\u0027s see, it\u0027s going to be"},{"Start":"00:51.769 ","End":"00:58.020","Text":"equal to the charge density multiplied by the length, dl."},{"Start":"00:59.150 ","End":"01:05.915","Text":"First of all, we know that our Lambda over here is dependent on Theta,"},{"Start":"01:05.915 ","End":"01:08.860","Text":"so we can substitute that in,"},{"Start":"01:08.860 ","End":"01:15.870","Text":"so it\u0027s Lambda_0 multiplied by sine of Theta, then our dl."},{"Start":"01:15.870 ","End":"01:18.800","Text":"So what is this? We can see that we\u0027re dealing with a ring,"},{"Start":"01:18.800 ","End":"01:20.780","Text":"which is a circle,"},{"Start":"01:20.780 ","End":"01:22.535","Text":"and it\u0027s in 1 dimension."},{"Start":"01:22.535 ","End":"01:26.786","Text":"So that means that we\u0027re going to be dealing with polar coordinates,"},{"Start":"01:26.786 ","End":"01:30.095","Text":"and because our radius is constant,"},{"Start":"01:30.095 ","End":"01:40.080","Text":"so this is simply going to be dl is equal to Rd Theta."},{"Start":"01:40.080 ","End":"01:43.940","Text":"Now, in order to find the total charge of the ring Q,"},{"Start":"01:43.940 ","End":"01:48.640","Text":"what we\u0027re going to do is we\u0027re going to integrate along dq."},{"Start":"01:48.640 ","End":"01:51.510","Text":"So we\u0027re going to integrate like so,"},{"Start":"01:51.510 ","End":"01:56.475","Text":"and our bounds are from 0 degrees or 0 radians,"},{"Start":"01:56.475 ","End":"01:59.870","Text":"and we\u0027re going along a full circle,"},{"Start":"01:59.870 ","End":"02:01.530","Text":"the full ring,"},{"Start":"02:01.530 ","End":"02:08.070","Text":"so that\u0027s 360 degrees or 2 Pi radians."},{"Start":"02:09.830 ","End":"02:13.815","Text":"Because we\u0027re going from 0 until 2 Pi,"},{"Start":"02:13.815 ","End":"02:18.094","Text":"when we integrate along sine of Theta and plug in these values,"},{"Start":"02:18.094 ","End":"02:24.429","Text":"we\u0027re going to get a total charge of 0."},{"Start":"02:25.250 ","End":"02:30.950","Text":"Let\u0027s see why the integral of sine Theta and also, by the way,"},{"Start":"02:30.950 ","End":"02:35.090","Text":"the integral of cosine Theta is equal to 0 when we\u0027re"},{"Start":"02:35.090 ","End":"02:39.935","Text":"going along an entire period or a whole oscillation."},{"Start":"02:39.935 ","End":"02:46.134","Text":"So if we draw over here a graph in sine Theta,"},{"Start":"02:46.134 ","End":"02:51.800","Text":"so 1 oscillation or 1 period of sine Theta looks like this,"},{"Start":"02:51.800 ","End":"02:55.045","Text":"where this is 0, and this over here is 2 Pi."},{"Start":"02:55.045 ","End":"02:57.239","Text":"So this is a full cycle."},{"Start":"02:57.239 ","End":"02:59.754","Text":"So what are we doing when we\u0027re integrating?"},{"Start":"02:59.754 ","End":"03:02.364","Text":"We\u0027re finding the area under the graph,"},{"Start":"03:02.364 ","End":"03:05.749","Text":"so we add up this area over here,"},{"Start":"03:05.749 ","End":"03:08.514","Text":"and we add up this area over here,"},{"Start":"03:08.514 ","End":"03:13.909","Text":"and we can see that this area is the same size as this area."},{"Start":"03:13.909 ","End":"03:16.962","Text":"However, their signs are opposite."},{"Start":"03:16.962 ","End":"03:19.959","Text":"This is positive, and this is negative."},{"Start":"03:19.959 ","End":"03:23.895","Text":"All in all, it equals to 0."},{"Start":"03:23.895 ","End":"03:27.140","Text":"The average area under the graph is equal to 0,"},{"Start":"03:27.140 ","End":"03:28.370","Text":"and the same, of course,"},{"Start":"03:28.370 ","End":"03:29.885","Text":"with cosine of Theta,"},{"Start":"03:29.885 ","End":"03:39.702","Text":"if we do 1 whole oscillation like so."},{"Start":"03:39.702 ","End":"03:44.224","Text":"So then this area under here"},{"Start":"03:44.224 ","End":"03:49.819","Text":"plus this area which is in a minus plus this area will give us a total of 0,"},{"Start":"03:49.819 ","End":"03:52.860","Text":"where this is 0, and this is 2 Pi."},{"Start":"03:54.170 ","End":"03:57.495","Text":"This is the answer to Question 1."},{"Start":"03:57.495 ","End":"04:01.344","Text":"The total charge of the ring is equal to 0."},{"Start":"04:01.344 ","End":"04:04.760","Text":"Now let\u0027s take a look at Question Number 2,"},{"Start":"04:04.760 ","End":"04:10.399","Text":"which is to find the electric field at each point on the ring\u0027s axis of symmetry."},{"Start":"04:10.399 ","End":"04:12.575","Text":"This is the axis of symmetry,"},{"Start":"04:12.575 ","End":"04:17.495","Text":"and it also goes down into the negative direction."},{"Start":"04:17.495 ","End":"04:20.975","Text":"So let\u0027s take this arbitrary point,"},{"Start":"04:20.975 ","End":"04:24.470","Text":"and let\u0027s say that it\u0027s at a position, z,"},{"Start":"04:24.470 ","End":"04:28.009","Text":"above the center of the ring."},{"Start":"04:28.009 ","End":"04:33.814","Text":"What we want to do is to find the electric field at this arbitrary point."},{"Start":"04:33.814 ","End":"04:35.494","Text":"How are we going to do this?"},{"Start":"04:35.494 ","End":"04:40.984","Text":"We\u0027re going to use the idea of superposition or the superposition principle."},{"Start":"04:40.984 ","End":"04:43.054","Text":"So what are we going to do?"},{"Start":"04:43.054 ","End":"04:47.780","Text":"Again, we\u0027re going to split the ring up into little pieces,"},{"Start":"04:47.780 ","End":"04:50.104","Text":"each with this charge, dq,"},{"Start":"04:50.104 ","End":"04:55.634","Text":"and then we\u0027re going to find the electric field at this point, z,"},{"Start":"04:55.634 ","End":"04:57.665","Text":"due to this piece, dq,"},{"Start":"04:57.665 ","End":"05:04.319","Text":"and then we\u0027re going to sum up along all of the points in the ring."},{"Start":"05:05.450 ","End":"05:10.744","Text":"Let\u0027s start with this piece over here, dq,"},{"Start":"05:10.744 ","End":"05:16.899","Text":"where we saw that dq is equal to this here."},{"Start":"05:16.899 ","End":"05:19.170","Text":"So let\u0027s find dE,"},{"Start":"05:19.170 ","End":"05:26.010","Text":"the small electric field at this point over here, z."},{"Start":"05:26.010 ","End":"05:32.570","Text":"From Coulomb\u0027s law, we know that this is equal to kdq divided by r^2,"},{"Start":"05:32.570 ","End":"05:39.339","Text":"where r^2 is the vector pointing from dq until this point, z."},{"Start":"05:40.640 ","End":"05:45.030","Text":"Of course, dE is a vector quantity,"},{"Start":"05:45.030 ","End":"05:48.530","Text":"so what we have over here is the magnitude of dE,"},{"Start":"05:48.530 ","End":"05:51.560","Text":"so the size of the electric field, and of course,"},{"Start":"05:51.560 ","End":"05:58.190","Text":"the direction of the electric field is in the direction of our r vector."},{"Start":"05:58.190 ","End":"06:05.490","Text":"So the electric field is going to carry on in this line over here."},{"Start":"06:07.850 ","End":"06:14.670","Text":"So what is the magnitude of our r vector?"},{"Start":"06:14.670 ","End":"06:18.325","Text":"What is the size? We\u0027re just using Pythagoras."},{"Start":"06:18.325 ","End":"06:21.190","Text":"We can see that the r vector is the hypotenuse,"},{"Start":"06:21.190 ","End":"06:26.005","Text":"and then we have 1 of the sides of this triangle,"},{"Start":"06:26.005 ","End":"06:27.640","Text":"which is of length r,"},{"Start":"06:27.640 ","End":"06:30.774","Text":"and the other side is of length z."},{"Start":"06:30.774 ","End":"06:38.320","Text":"From Pythagoras, we know that the magnitude of r is simply R^2 plus Z^2,"},{"Start":"06:38.320 ","End":"06:45.534","Text":"and then we take the square root of all of that."},{"Start":"06:45.534 ","End":"06:51.669","Text":"Of course, we can just write this as r. Now we can substitute all of this,"},{"Start":"06:51.669 ","End":"06:52.819","Text":"we know what our dq is,"},{"Start":"06:52.819 ","End":"06:55.565","Text":"and we know what our r is, into our equation."},{"Start":"06:55.565 ","End":"07:01.204","Text":"So now we have the magnitude of this small E-field."},{"Start":"07:01.204 ","End":"07:03.290","Text":"But of course, the E-field has a vector,"},{"Start":"07:03.290 ","End":"07:06.300","Text":"and we also want to know the direction."},{"Start":"07:07.160 ","End":"07:10.880","Text":"In order to find the direction,"},{"Start":"07:10.880 ","End":"07:14.945","Text":"the first thing that we\u0027re going to have to do is we\u0027re going to have to split up"},{"Start":"07:14.945 ","End":"07:20.314","Text":"our dE electric field into the different components,"},{"Start":"07:20.314 ","End":"07:24.644","Text":"so the z component and the xy components,"},{"Start":"07:24.644 ","End":"07:27.709","Text":"because we can see that every piece is going"},{"Start":"07:27.709 ","End":"07:32.405","Text":"to apply an electric field over here in a slightly different direction."},{"Start":"07:32.405 ","End":"07:36.480","Text":"We can see this piece over here applies an electric field in this direction,"},{"Start":"07:36.480 ","End":"07:38.900","Text":"but the piece over here on this side will apply"},{"Start":"07:38.900 ","End":"07:43.090","Text":"an electric field in this direction over here."},{"Start":"07:43.090 ","End":"07:50.969","Text":"Let\u0027s first find the small electric field, z component."},{"Start":"07:51.230 ","End":"07:53.820","Text":"In order to find the z component,"},{"Start":"07:53.820 ","End":"07:59.815","Text":"let\u0027s define this angle over here as the angle Phi."},{"Start":"07:59.815 ","End":"08:01.815","Text":"Then, in that case,"},{"Start":"08:01.815 ","End":"08:06.570","Text":"dE_z is equal to the size of the field,"},{"Start":"08:06.570 ","End":"08:14.600","Text":"so the magnitude of dE multiplied by the projection of this in the z direction."},{"Start":"08:14.600 ","End":"08:17.540","Text":"So if this angle over here is Phi,"},{"Start":"08:17.540 ","End":"08:21.610","Text":"and we know that this side over here is the hypotenuse,"},{"Start":"08:21.610 ","End":"08:23.805","Text":"and this is the z side,"},{"Start":"08:23.805 ","End":"08:28.790","Text":"the z-axis, and we want the projection on the z-axis which we"},{"Start":"08:28.790 ","End":"08:34.390","Text":"can see is the adjacent side to our angle Phi."},{"Start":"08:34.390 ","End":"08:37.129","Text":"From a trig identities, SOH-CAH-TOA,"},{"Start":"08:37.129 ","End":"08:39.900","Text":"sine is opposite over hypotenuse,"},{"Start":"08:39.900 ","End":"08:42.680","Text":"cosine is adjacent over hypotenuse,"},{"Start":"08:42.680 ","End":"08:46.745","Text":"and tangent is opposite over adjacent."},{"Start":"08:46.745 ","End":"08:54.733","Text":"Specifically, here we have adjacent and hypotenuse which corresponds to cosine,"},{"Start":"08:54.733 ","End":"08:58.955","Text":"so this is going to be multiplied by cosine of Phi."},{"Start":"08:58.955 ","End":"09:01.925","Text":"That\u0027s all great. However,"},{"Start":"09:01.925 ","End":"09:05.291","Text":"we aren\u0027t given in the question what Phi is."},{"Start":"09:05.291 ","End":"09:10.085","Text":"We\u0027re just given what Lambda_0 is and what r is equal to."},{"Start":"09:10.085 ","End":"09:14.240","Text":"What do we have to do is now we have to find out what Phi is in"},{"Start":"09:14.240 ","End":"09:19.260","Text":"terms of the values that we are given in the question."},{"Start":"09:19.460 ","End":"09:23.435","Text":"As we can see, if this angle over here is Phi,"},{"Start":"09:23.435 ","End":"09:28.645","Text":"so that means that this angle over here is also Phi."},{"Start":"09:28.645 ","End":"09:32.179","Text":"Then in that case, if this is Phi,"},{"Start":"09:32.179 ","End":"09:37.730","Text":"then we can see that the adjacent angle to Phi is z,"},{"Start":"09:37.730 ","End":"09:42.544","Text":"and the hypotenuse is this, our r vector."},{"Start":"09:42.544 ","End":"09:51.129","Text":"So we can say that cosine of Phi is equal to z divided by r,"},{"Start":"09:51.129 ","End":"09:54.350","Text":"again, because of SOH-CAH-TOA."},{"Start":"09:54.350 ","End":"09:58.640","Text":"Now we can integrate along dE_z,"},{"Start":"09:58.640 ","End":"10:08.070","Text":"in order to find the total electric field in the z direction, so therefore,"},{"Start":"10:08.070 ","End":"10:11.104","Text":"we\u0027ll get that E_z is equal to,"},{"Start":"10:11.104 ","End":"10:15.500","Text":"so it\u0027s the integral of the magnitude of dE,"},{"Start":"10:15.500 ","End":"10:18.305","Text":"which is equal to k, and then dq,"},{"Start":"10:18.305 ","End":"10:25.910","Text":"which is Lambda_0 multiplied by sine of Theta multiplied by Rd Theta."},{"Start":"10:25.910 ","End":"10:30.274","Text":"Then we\u0027re multiplying this by cosine of Phi,"},{"Start":"10:30.274 ","End":"10:37.940","Text":"which is z divided by r. I just substituted in dE,"},{"Start":"10:37.940 ","End":"10:44.565","Text":"and dq, and cosine of Phi."},{"Start":"10:44.565 ","End":"10:50.405","Text":"Of course, we\u0027re integrating along the entire ring,"},{"Start":"10:50.405 ","End":"10:53.320","Text":"that\u0027s from 0 up till 2 Pi."},{"Start":"10:53.320 ","End":"10:57.880","Text":"Again, we\u0027re integrating along sine of Theta from 0 until 2 Pi,"},{"Start":"10:57.880 ","End":"11:01.459","Text":"so for 1 whole oscillation,"},{"Start":"11:01.459 ","End":"11:06.760","Text":"so therefore E_z is going to be equal to 0."},{"Start":"11:07.100 ","End":"11:12.771","Text":"Of course, I forgot to put in the r squared over here."},{"Start":"11:12.771 ","End":"11:17.669","Text":"If we join all of this,"},{"Start":"11:17.669 ","End":"11:20.550","Text":"so we have R^2 multiplied by r,"},{"Start":"11:20.550 ","End":"11:22.955","Text":"so that\u0027s r^3,"},{"Start":"11:22.955 ","End":"11:24.390","Text":"so if we just cube this,"},{"Start":"11:24.390 ","End":"11:28.860","Text":"(R^2 plus Z^2) 3/2."},{"Start":"11:28.860 ","End":"11:31.095","Text":"But either way, because of the sine Theta,"},{"Start":"11:31.095 ","End":"11:34.287","Text":"it ends up as 0."},{"Start":"11:34.287 ","End":"11:37.945","Text":"Why does this make sense,"},{"Start":"11:37.945 ","End":"11:40.809","Text":"aside from knowing that sine Theta,"},{"Start":"11:40.809 ","End":"11:43.299","Text":"if we integrate along its entire period,"},{"Start":"11:43.299 ","End":"11:44.635","Text":"is equal to 0?"},{"Start":"11:44.635 ","End":"11:50.020","Text":"We can see from this equation for Lambda as a function of Theta that,"},{"Start":"11:50.020 ","End":"11:54.142","Text":"between the angle of 0 and Pi,"},{"Start":"11:54.142 ","End":"11:55.839","Text":"so for half a circle,"},{"Start":"11:55.839 ","End":"11:59.784","Text":"we\u0027re going to have a positive charge density."},{"Start":"11:59.784 ","End":"12:05.605","Text":"Then from Pi and all the way on this half and 2 Pi,"},{"Start":"12:05.605 ","End":"12:08.619","Text":"we\u0027re going to have a negative charge density,"},{"Start":"12:08.619 ","End":"12:14.319","Text":"exactly like what we saw before with sine Theta."},{"Start":"12:14.319 ","End":"12:17.897","Text":"Sp along half a circle we\u0027re in the positive range over here,"},{"Start":"12:17.897 ","End":"12:20.057","Text":"so we\u0027ll have a positive charge density."},{"Start":"12:20.057 ","End":"12:21.970","Text":"Then from Pi until 2 Pi,"},{"Start":"12:21.970 ","End":"12:23.440","Text":"we\u0027re in negative range over here,"},{"Start":"12:23.440 ","End":"12:26.080","Text":"so we\u0027ll have a negative charge density."},{"Start":"12:26.080 ","End":"12:28.750","Text":"That means that if we take this piece over here,"},{"Start":"12:28.750 ","End":"12:34.224","Text":"dq, we saw that its electric field is going in this way."},{"Start":"12:34.224 ","End":"12:39.110","Text":"If we choose the piece over here which is right opposite this,"},{"Start":"12:39.110 ","End":"12:42.158","Text":"let\u0027s draw it in red,"},{"Start":"12:42.158 ","End":"12:48.000","Text":"so we can see that it\u0027s going to have a negative value for dq,"},{"Start":"12:48.000 ","End":"12:56.155","Text":"so that means that it\u0027s going to be in this direction for this piece over here."},{"Start":"12:56.155 ","End":"12:58.270","Text":"So what we can see is that,"},{"Start":"12:58.270 ","End":"13:02.409","Text":"for the component in the z-direction, for the blue,"},{"Start":"13:02.409 ","End":"13:06.284","Text":"we have a component going in the positive z direction,"},{"Start":"13:06.284 ","End":"13:10.450","Text":"and for the red, we have a component going in the negative z direction."},{"Start":"13:10.450 ","End":"13:12.415","Text":"They\u0027re both of equal magnitude,"},{"Start":"13:12.415 ","End":"13:14.712","Text":"and so therefore they cancel each other out."},{"Start":"13:14.712 ","End":"13:17.664","Text":"That\u0027s why that in the z direction,"},{"Start":"13:17.664 ","End":"13:20.664","Text":"with this changing charge density,"},{"Start":"13:20.664 ","End":"13:26.779","Text":"we get a value of 0 for the E-field in the z direction."},{"Start":"13:28.350 ","End":"13:35.349","Text":"Now we want to find the electric field in the y direction."},{"Start":"13:35.349 ","End":"13:42.415","Text":"We know that the electric field in the y direction is going to be the integral on dE_y,"},{"Start":"13:42.415 ","End":"13:50.844","Text":"which is going to be the integral on kdq divided by r^2."},{"Start":"13:50.844 ","End":"13:57.849","Text":"Then we have to multiply it by its direction or its projection on the y axis,"},{"Start":"13:57.849 ","End":"14:00.890","Text":"or specifically, the xy-plane."},{"Start":"14:03.470 ","End":"14:09.464","Text":"Here we\u0027re going to draw the projection line for the xy-plane,"},{"Start":"14:09.464 ","End":"14:14.460","Text":"and then we take down this line which is perpendicular to the xy-plane."},{"Start":"14:14.460 ","End":"14:16.655","Text":"This is 90 degrees."},{"Start":"14:16.655 ","End":"14:23.569","Text":"Of course, this angle over here is our fee from before."},{"Start":"14:24.150 ","End":"14:31.044","Text":"The projection over here is the same as if we would project it up over here,"},{"Start":"14:31.044 ","End":"14:36.174","Text":"so we can see it\u0027s the opposite side to our angle Phi."},{"Start":"14:36.174 ","End":"14:39.670","Text":"Therefore, from our SOH-CAH-TOA,"},{"Start":"14:39.670 ","End":"14:42.910","Text":"we know that we have to multiply all of this to get the projection."},{"Start":"14:42.910 ","End":"14:47.930","Text":"We have to multiply all of this by sine of Phi."},{"Start":"14:48.510 ","End":"14:55.900","Text":"This angle over here dealing with Phi is the projection on the xy-plane."},{"Start":"14:55.900 ","End":"14:58.630","Text":"Let\u0027s write this xy-plane."},{"Start":"14:58.630 ","End":"15:04.254","Text":"But we\u0027re specifically speaking about the E-field along the y-axis."},{"Start":"15:04.254 ","End":"15:09.684","Text":"That means that we have to take this now projection on the xy-plane,"},{"Start":"15:09.684 ","End":"15:16.525","Text":"and we have to find the projection on the y-axis, specifically."},{"Start":"15:16.525 ","End":"15:19.450","Text":"This line over here,"},{"Start":"15:19.450 ","End":"15:21.219","Text":"the projection on the xy-plane,"},{"Start":"15:21.219 ","End":"15:26.359","Text":"is the exact same as this line over here in black."},{"Start":"15:26.670 ","End":"15:31.059","Text":"In fact, let\u0027s draw it in red."},{"Start":"15:31.059 ","End":"15:34.585","Text":"This line over here is the same as this line over here,"},{"Start":"15:34.585 ","End":"15:38.185","Text":"which is the exact same as this line over here."},{"Start":"15:38.185 ","End":"15:43.794","Text":"Now if we draw the extension of the y-axis over here,"},{"Start":"15:43.794 ","End":"15:47.080","Text":"so we can see that if this angle is Theta,"},{"Start":"15:47.080 ","End":"15:50.169","Text":"and the angle between the x and y axes"},{"Start":"15:50.169 ","End":"15:53.320","Text":"is of course 90 degrees because they are perpendicular,"},{"Start":"15:53.320 ","End":"16:02.244","Text":"then that means that this angle over here is equal to 90 minus Theta."},{"Start":"16:02.244 ","End":"16:09.114","Text":"Then, therefore, that means that this angle over here between this red line,"},{"Start":"16:09.114 ","End":"16:12.173","Text":"so the projection on the xy-plane,"},{"Start":"16:12.173 ","End":"16:17.470","Text":"and the y-axis is also going to be equal to 90"},{"Start":"16:17.470 ","End":"16:24.380","Text":"minus Theta because they\u0027re both meeting at a point these 2 angles."},{"Start":"16:24.510 ","End":"16:30.970","Text":"Then we can see that we want the projection along the y-axis,"},{"Start":"16:30.970 ","End":"16:36.475","Text":"which is the side adjacent to the angle."},{"Start":"16:36.475 ","End":"16:38.934","Text":"So the side adjacent,"},{"Start":"16:38.934 ","End":"16:41.470","Text":"therefore, we have to use cosine."},{"Start":"16:41.470 ","End":"16:45.804","Text":"So this is going to be multiplied by cosine of the angle,"},{"Start":"16:45.804 ","End":"16:51.280","Text":"which is equal to 90 minus Theta."},{"Start":"16:51.280 ","End":"16:55.570","Text":"Something that\u0027s useful to remember is that cosine"},{"Start":"16:55.570 ","End":"17:01.014","Text":"of 90 minus Theta is the same as sine of Theta."},{"Start":"17:01.014 ","End":"17:05.754","Text":"This is an identity that\u0027s very useful to remember."},{"Start":"17:05.754 ","End":"17:13.809","Text":"Now let\u0027s start plugging in all of our variables so let\u0027s write this over here."},{"Start":"17:13.809 ","End":"17:19.719","Text":"E_y is going to be equal to the integral of k,"},{"Start":"17:19.719 ","End":"17:21.909","Text":"and then dq,"},{"Start":"17:21.909 ","End":"17:23.200","Text":"we saw over here,"},{"Start":"17:23.200 ","End":"17:31.849","Text":"was equal to Lambda_0 multiplied by sine of Theta multiplied by Rd Theta."},{"Start":"17:32.280 ","End":"17:35.979","Text":"This is divided by r^2."},{"Start":"17:35.979 ","End":"17:39.069","Text":"Soon we\u0027ll substitute in what r^2 is equal to."},{"Start":"17:39.069 ","End":"17:46.380","Text":"Then we\u0027re multiplying this by sine of Phi."},{"Start":"17:46.380 ","End":"17:51.400","Text":"As we saw, if this angle over here is Phi,"},{"Start":"17:51.400 ","End":"17:55.608","Text":"then this angle over here is Phi."},{"Start":"17:55.608 ","End":"18:00.985","Text":"Then we can see that we want to be on the xy-plane,"},{"Start":"18:00.985 ","End":"18:04.600","Text":"so then we\u0027re using this side over here,"},{"Start":"18:04.600 ","End":"18:08.030","Text":"capital R, the radius of the ring,"},{"Start":"18:08.030 ","End":"18:12.425","Text":"and that is the opposite side to the angle,"},{"Start":"18:12.425 ","End":"18:14.769","Text":"and we have our radius over here,"},{"Start":"18:14.769 ","End":"18:17.600","Text":"which is the hypotenuse."},{"Start":"18:17.880 ","End":"18:22.435","Text":"Therefore, we can say that sine of Phi,"},{"Start":"18:22.435 ","End":"18:26.263","Text":"which is equal to opposite over hypotenuse,"},{"Start":"18:26.263 ","End":"18:34.165","Text":"so that is simply going to be equal to R divided by R. Now sine of Phi,"},{"Start":"18:34.165 ","End":"18:42.085","Text":"which is R divided by r. Then we\u0027re multiplying by cosine of 90 minus Theta,"},{"Start":"18:42.085 ","End":"18:46.779","Text":"which is equal to sine of Theta."},{"Start":"18:46.779 ","End":"18:50.710","Text":"Of course, our bounds is along the entire ring,"},{"Start":"18:50.710 ","End":"18:53.750","Text":"so from 0 until 2 Pi."},{"Start":"18:54.930 ","End":"18:57.865","Text":"Let\u0027s simplify this out."},{"Start":"18:57.865 ","End":"18:59.725","Text":"Let\u0027s take out all the constants."},{"Start":"18:59.725 ","End":"19:03.220","Text":"We have k Lambda_0,"},{"Start":"19:03.220 ","End":"19:06.835","Text":"and then we have R multiplied by R,"},{"Start":"19:06.835 ","End":"19:14.020","Text":"so R^2, and then all of this is divided by R^3."},{"Start":"19:14.020 ","End":"19:23.140","Text":"So that is going to be capital R^2 plus z^2 to the power 3/2."},{"Start":"19:23.140 ","End":"19:30.609","Text":"Then all of this is multiplying by the integral of sine Theta times sine Theta,"},{"Start":"19:30.609 ","End":"19:34.914","Text":"so sine^2 Theta d Theta,"},{"Start":"19:34.914 ","End":"19:38.240","Text":"from 0 until 2 Pi."},{"Start":"19:38.790 ","End":"19:46.434","Text":"Now, something very important to notice is that we can see that the projection"},{"Start":"19:46.434 ","End":"19:54.144","Text":"is going to be along the y-axis but in the negative y direction."},{"Start":"19:54.144 ","End":"20:00.440","Text":"So just remember to add in the negative over here."},{"Start":"20:02.010 ","End":"20:06.910","Text":"Now we have sine squared of Theta which, already,"},{"Start":"20:06.910 ","End":"20:09.369","Text":"when we integrate between 0 and 2 Pi,"},{"Start":"20:09.369 ","End":"20:12.214","Text":"will not be equal to 0."},{"Start":"20:12.214 ","End":"20:14.625","Text":"That\u0027s the first thing that we have to remember."},{"Start":"20:14.625 ","End":"20:16.755","Text":"In order to do this integral,"},{"Start":"20:16.755 ","End":"20:21.369","Text":"we\u0027re going to use a trig identity for sine^2 Theta."},{"Start":"20:21.810 ","End":"20:28.809","Text":"There\u0027s an identity that says that cosine of 2 Theta is equal"},{"Start":"20:28.809 ","End":"20:35.605","Text":"to 1 minus 2 sine^2 Theta."},{"Start":"20:35.605 ","End":"20:38.814","Text":"Therefore, if we isolate out the sine squared Theta,"},{"Start":"20:38.814 ","End":"20:45.640","Text":"we\u0027ll get that 1 minus cosine of 2 Theta"},{"Start":"20:45.640 ","End":"20:53.185","Text":"divided by 2 is equal to sine^2 Theta."},{"Start":"20:53.185 ","End":"20:58.849","Text":"So this is what we\u0027re going to substitute in to our equation."},{"Start":"20:59.640 ","End":"21:06.250","Text":"We have negative k Lambda_0 R^2 divided"},{"Start":"21:06.250 ","End":"21:12.415","Text":"by R^2 plus z^2 to the power of 3/2."},{"Start":"21:12.415 ","End":"21:17.786","Text":"Then we\u0027re going to be integrating between 0 and 2 Pi of"},{"Start":"21:17.786 ","End":"21:25.239","Text":"1 minus cosine of 2 Theta divided by 2 d Theta."},{"Start":"21:25.239 ","End":"21:29.049","Text":"Let\u0027s carry that on over here."},{"Start":"21:29.049 ","End":"21:35.364","Text":"First of all, when we integrate along cosine of 2 Theta from 0 until 2 Pi,"},{"Start":"21:35.364 ","End":"21:38.398","Text":"or for 1 whole oscillation,"},{"Start":"21:38.398 ","End":"21:44.109","Text":"so just like cosine of Theta and just like sine of Theta, it is equal to 0."},{"Start":"21:44.109 ","End":"21:49.429","Text":"So that means that we\u0027re just going to be integrating along 1/2."},{"Start":"21:50.450 ","End":"21:54.164","Text":"If we integrate 1/2 d Theta,"},{"Start":"21:54.164 ","End":"21:56.910","Text":"so it\u0027s going to be equal to 1/2 of Theta."},{"Start":"21:56.910 ","End":"21:59.804","Text":"When we substitute in 2 Pi and 0,"},{"Start":"21:59.804 ","End":"22:05.598","Text":"so we\u0027re going to have 1/2 of 2 Pi minus 1/2 of 0 which is simply equal to Pi,"},{"Start":"22:05.598 ","End":"22:11.330","Text":"so all we\u0027re going to do is we\u0027re going to multiply all of these constants by Pi."},{"Start":"22:11.330 ","End":"22:18.340","Text":"Therefore we\u0027ll get that our electric field in the y direction is equal to"},{"Start":"22:18.340 ","End":"22:25.869","Text":"negative k Pi Lambda_0 R^2 divided"},{"Start":"22:25.869 ","End":"22:36.033","Text":"by R^2 plus z^2 to the power of 3/2."},{"Start":"22:36.033 ","End":"22:39.700","Text":"So we got that the E-field in the z direction is equal to 0,"},{"Start":"22:39.700 ","End":"22:43.278","Text":"the E-field in the y direction is equal to this,"},{"Start":"22:43.278 ","End":"22:48.519","Text":"and the E-field in the x direction is also equal to 0."},{"Start":"22:48.519 ","End":"22:50.455","Text":"So how did I get this?"},{"Start":"22:50.455 ","End":"22:53.410","Text":"I\u0027m going to write this out in blue."},{"Start":"22:53.410 ","End":"22:59.427","Text":"In order to get along the x-axis,"},{"Start":"22:59.427 ","End":"23:03.699","Text":"instead of having the projection over here,"},{"Start":"23:03.699 ","End":"23:06.970","Text":"so this represents the prediction on the xy-plane."},{"Start":"23:06.970 ","End":"23:09.189","Text":"Then in order to get the x-axis,"},{"Start":"23:09.189 ","End":"23:12.970","Text":"instead of having cosine of 90 minus Theta,"},{"Start":"23:12.970 ","End":"23:17.709","Text":"we\u0027ll have here sine of 90 minus"},{"Start":"23:17.709 ","End":"23:24.519","Text":"Theta which is equal to simply cosine of Theta."},{"Start":"23:24.519 ","End":"23:30.470","Text":"Then what we\u0027ll get over here,"},{"Start":"23:30.930 ","End":"23:34.585","Text":"instead of sine^2 Theta,"},{"Start":"23:34.585 ","End":"23:41.544","Text":"we\u0027ll have sine Theta multiplied by cosine of Theta which,"},{"Start":"23:41.544 ","End":"23:43.270","Text":"according to the trig identity,"},{"Start":"23:43.270 ","End":"23:49.480","Text":"is equal to sine of 2 Theta divided by 2."},{"Start":"23:49.480 ","End":"23:58.285","Text":"So then we\u0027ll plug in over here sine of 2 Theta divided by 2."},{"Start":"23:58.285 ","End":"24:02.515","Text":"Then again, we\u0027re integrating along sine of 2 Theta,"},{"Start":"24:02.515 ","End":"24:06.793","Text":"which just like cosine of 2 Theta and sine of Theta and cosine of Theta,"},{"Start":"24:06.793 ","End":"24:10.465","Text":"if we\u0027re integrating along 1 whole period,"},{"Start":"24:10.465 ","End":"24:13.090","Text":"it\u0027s going to be equal to 0."},{"Start":"24:13.090 ","End":"24:15.919","Text":"We won\u0027t have this 1/2 over here,"},{"Start":"24:15.919 ","End":"24:19.285","Text":"so it will just be sine of 2 Theta divided by 2,"},{"Start":"24:19.285 ","End":"24:24.439","Text":"and therefore, our whole integral will be equal to 0."},{"Start":"24:25.110 ","End":"24:30.020","Text":"Now let\u0027s go on to answer Question 3."},{"Start":"24:31.470 ","End":"24:33.970","Text":"Here\u0027s the answer for Question 1,"},{"Start":"24:33.970 ","End":"24:38.320","Text":"and here\u0027s the answer to Question 2. Now Question 3."},{"Start":"24:38.320 ","End":"24:42.775","Text":"What is the electric field when z,"},{"Start":"24:42.775 ","End":"24:46.000","Text":"so our value along the z-axis,"},{"Start":"24:46.000 ","End":"24:47.455","Text":"or the axis of symmetry,"},{"Start":"24:47.455 ","End":"24:52.674","Text":"is much larger than the radius of the ring?"},{"Start":"24:52.674 ","End":"24:55.855","Text":"What type of electric field is this?"},{"Start":"24:55.855 ","End":"24:59.379","Text":"Is it similar to the electric field of a point charge,"},{"Start":"24:59.379 ","End":"25:02.349","Text":"of an infinite plane, of a rod?"},{"Start":"25:02.349 ","End":"25:05.750","Text":"So that\u0027s what this means."},{"Start":"25:05.900 ","End":"25:15.189","Text":"So z is much larger than the radius of the ring."},{"Start":"25:15.189 ","End":"25:20.500","Text":"In that case, I can do some approximations."},{"Start":"25:20.500 ","End":"25:23.904","Text":"For our E-field in the y direction,"},{"Start":"25:23.904 ","End":"25:32.934","Text":"I can write that negative k Pi Lambda_0 R^2 divided by,"},{"Start":"25:32.934 ","End":"25:40.990","Text":"so here I have z^2 which is going to be a very big number plus R^2,"},{"Start":"25:40.990 ","End":"25:43.570","Text":"which is going to be a very insignificant."},{"Start":"25:43.570 ","End":"25:49.750","Text":"It\u0027s the equivalent of me saying 1 million plus 1."},{"Start":"25:49.750 ","End":"25:56.365","Text":"So the difference between having my denominator being a million and 1^3/2,"},{"Start":"25:56.365 ","End":"26:00.470","Text":"or 1,000,000^3/2 is very small,"},{"Start":"26:00.470 ","End":"26:02.785","Text":"and so therefore, we can do that approximation."},{"Start":"26:02.785 ","End":"26:05.304","Text":"So I can just take out the R^2,"},{"Start":"26:05.304 ","End":"26:11.320","Text":"and I can just say that this is divided by z^2 to the power of 3/2,"},{"Start":"26:11.320 ","End":"26:20.929","Text":"which is equal to negative k Pi Lambda_0 R^2 divided by z^3."},{"Start":"26:21.390 ","End":"26:27.369","Text":"What\u0027s important to see here is that our field is diminishing at a rate"},{"Start":"26:27.369 ","End":"26:33.130","Text":"of 1 divided by z^3 or 1 divided by R^3."},{"Start":"26:33.130 ","End":"26:37.569","Text":"What electric field also diminishes at this rate?"},{"Start":"26:37.569 ","End":"26:43.119","Text":"It\u0027s the electric field of a dipole or of an electric dipole."},{"Start":"26:43.119 ","End":"26:49.389","Text":"The electric field of a dipole is equal to,"},{"Start":"26:49.389 ","End":"26:51.129","Text":"this is a really horrible equation,"},{"Start":"26:51.129 ","End":"27:00.869","Text":"k multiplied by 3p dot r in the r direction minus p,"},{"Start":"27:00.869 ","End":"27:05.159","Text":"and all of this is divided by r^3."},{"Start":"27:05.159 ","End":"27:11.365","Text":"We can see that this electric field in the y direction,"},{"Start":"27:11.365 ","End":"27:14.260","Text":"when we\u0027re located very far away from this ring,"},{"Start":"27:14.260 ","End":"27:18.429","Text":"is behaving in a way similar to the electric field of a dipole."},{"Start":"27:18.429 ","End":"27:21.954","Text":"We\u0027ve already seen that the electric field for"},{"Start":"27:21.954 ","End":"27:28.494","Text":"a point charge is equal to kQ divided by r^2."},{"Start":"27:28.494 ","End":"27:30.025","Text":"So a point charge,"},{"Start":"27:30.025 ","End":"27:33.984","Text":"the electric field diminishes at a rate of 1 divided by r^2."},{"Start":"27:33.984 ","End":"27:36.580","Text":"But here we\u0027re at 1 divided by r^3."},{"Start":"27:36.580 ","End":"27:37.930","Text":"So whenever you see that,"},{"Start":"27:37.930 ","End":"27:44.119","Text":"you should know that this is a field similar to that of an electric dipole."},{"Start":"27:44.640 ","End":"27:49.720","Text":"Another interesting thing to note over here is that we saw in Question"},{"Start":"27:49.720 ","End":"27:54.670","Text":"1 that the total charge of the ring is equal to 0."},{"Start":"27:54.670 ","End":"27:56.875","Text":"We saw that Q is equal to 0."},{"Start":"27:56.875 ","End":"27:58.975","Text":"Therefore, in this case over here,"},{"Start":"27:58.975 ","End":"28:03.475","Text":"if we were to look at our ring from very far away,"},{"Start":"28:03.475 ","End":"28:08.469","Text":"then we wouldn\u0027t be able to look at it as the electric field of"},{"Start":"28:08.469 ","End":"28:13.330","Text":"a point because the electric field at this size,"},{"Start":"28:13.330 ","End":"28:17.770","Text":"when it\u0027s 1 divided by r^2 is going to be equal to 0."},{"Start":"28:17.770 ","End":"28:22.074","Text":"That\u0027s why it doesn\u0027t have the electric field of a point charge."},{"Start":"28:22.074 ","End":"28:27.699","Text":"But of course we do know that there is charge density over here."},{"Start":"28:27.699 ","End":"28:30.790","Text":"It\u0027s non-uniform, but there is a charge density,"},{"Start":"28:30.790 ","End":"28:35.285","Text":"and we know that every little piece along this ring does have some charge,"},{"Start":"28:35.285 ","End":"28:39.580","Text":"so to say that there\u0027s no electric field is ridiculous."},{"Start":"28:39.580 ","End":"28:42.984","Text":"We know that there is. We also worked it out in the previous question."},{"Start":"28:42.984 ","End":"28:45.715","Text":"Therefore, we have to go to"},{"Start":"28:45.715 ","End":"28:52.360","Text":"the next biggest possible option for an electric field which is that of a dipole,"},{"Start":"28:52.360 ","End":"28:56.070","Text":"and that is 1 divided by r^3,"},{"Start":"28:56.070 ","End":"29:00.909","Text":"so it\u0027s significantly smaller than the electric field for a point charge."},{"Start":"29:00.909 ","End":"29:04.944","Text":"However, we can see that when we\u0027re dealing with the electric field of a dipole,"},{"Start":"29:04.944 ","End":"29:09.910","Text":"the electric field doesn\u0027t suddenly become 0 and non-existent."},{"Start":"29:09.910 ","End":"29:12.505","Text":"So we can see that we do have an electric field."},{"Start":"29:12.505 ","End":"29:17.724","Text":"Therefore, we can see that our ring is acting like a dipole,"},{"Start":"29:17.724 ","End":"29:21.530","Text":"especially when we\u0027re located far away."},{"Start":"29:21.540 ","End":"29:26.590","Text":"The only reason we\u0027re getting this dipole electric field is"},{"Start":"29:26.590 ","End":"29:32.600","Text":"because the total charge on the ring is equal to 0."},{"Start":"29:32.910 ","End":"29:35.500","Text":"If it wasn\u0027t equal to 0,"},{"Start":"29:35.500 ","End":"29:39.680","Text":"we would be able to consider this ring as a point charge."},{"Start":"29:41.490 ","End":"29:44.980","Text":"So we see that if we have some charge,"},{"Start":"29:44.980 ","End":"29:47.515","Text":"let\u0027s say it\u0027s a positive over here,"},{"Start":"29:47.515 ","End":"29:53.860","Text":"then if we consider this ring as a positive charge very far away,"},{"Start":"29:53.860 ","End":"29:56.544","Text":"we can see it as a point charge,"},{"Start":"29:56.544 ","End":"30:00.789","Text":"and then it will have an electric field like so,"},{"Start":"30:00.789 ","End":"30:04.766","Text":"but we\u0027re dealing with something that is similar to a dipole,"},{"Start":"30:04.766 ","End":"30:08.064","Text":"so we\u0027re dealing with something that looks like so."},{"Start":"30:08.064 ","End":"30:10.509","Text":"Then if it\u0027s like this,"},{"Start":"30:10.509 ","End":"30:16.420","Text":"then we have the next electric field down the line after that of a point charge,"},{"Start":"30:16.420 ","End":"30:18.550","Text":"which is that of a dipole."},{"Start":"30:18.550 ","End":"30:22.030","Text":"Then we can say that we have the electric field of a dipole."},{"Start":"30:22.030 ","End":"30:28.040","Text":"However, what happens if we have an electric field like so?"},{"Start":"30:30.240 ","End":"30:33.475","Text":"Now we have 2 dipoles."},{"Start":"30:33.475 ","End":"30:37.190","Text":"Here we have an electric field of a dipole,"},{"Start":"30:38.310 ","End":"30:40.750","Text":"here we have the electric field of a dipole, but we can see that"},{"Start":"30:40.750 ","End":"30:41.820","Text":"the charges don\u0027t match."},{"Start":"30:41.820 ","End":"30:43.569","Text":"Here we have a plus and a minus,"},{"Start":"30:43.569 ","End":"30:45.865","Text":"and here we have a minus and a plus."},{"Start":"30:45.865 ","End":"30:52.944","Text":"So overall, the electric field of these 2 dipoles together would be equal to 0."},{"Start":"30:52.944 ","End":"30:58.254","Text":"In this case, E_dipole would also be equal to 0."},{"Start":"30:58.254 ","End":"31:03.130","Text":"In this case where we have 2 dipoles in this type of formation,"},{"Start":"31:03.130 ","End":"31:07.525","Text":"then we call this the electric field of a quadrupole."},{"Start":"31:07.525 ","End":"31:11.545","Text":"This is the next electric field down the line."},{"Start":"31:11.545 ","End":"31:18.880","Text":"The electric field here is proportional to 1 divided by r^4."},{"Start":"31:18.880 ","End":"31:21.460","Text":"So we can see that it\u0027s much weaker than"},{"Start":"31:21.460 ","End":"31:25.809","Text":"the electric field of a dipole k. It\u0027s 10 times as weak,"},{"Start":"31:25.809 ","End":"31:33.319","Text":"and it\u0027s 100 times as weak as the electric field for a point charge."},{"Start":"31:33.660 ","End":"31:38.845","Text":"Of course, if we add in another 2 dipoles in here,"},{"Start":"31:38.845 ","End":"31:43.104","Text":"then this electric field of the quadrupole will also be equal to 0,"},{"Start":"31:43.104 ","End":"31:49.539","Text":"and then we\u0027ll go on to an electric field proportional to 1 divided by r^5,"},{"Start":"31:49.539 ","End":"31:52.340","Text":"and so on and so forth."},{"Start":"31:53.040 ","End":"31:57.129","Text":"Of course, we work this out using Coulomb\u0027s law."},{"Start":"31:57.129 ","End":"32:03.094","Text":"But we can also use the equation for a dipole,"},{"Start":"32:03.094 ","End":"32:06.555","Text":"where in this equation, we need to know the value of"},{"Start":"32:06.555 ","End":"32:11.129","Text":"p. Now the value of p refers to the dipole moment."},{"Start":"32:11.129 ","End":"32:17.664","Text":"So if we\u0027re dealing with a standard dipole like so,"},{"Start":"32:17.664 ","End":"32:20.545","Text":"p will be equal to qd,"},{"Start":"32:20.545 ","End":"32:23.680","Text":"where d is the distance between the 2 dipoles."},{"Start":"32:23.680 ","End":"32:27.145","Text":"However, when we\u0027re dealing with something like"},{"Start":"32:27.145 ","End":"32:32.275","Text":"this where we have a charge distributed all along a ring,"},{"Start":"32:32.275 ","End":"32:33.970","Text":"and it\u0027s continuous,"},{"Start":"32:33.970 ","End":"32:41.214","Text":"then we would have to sort this out into the different components for x and y."},{"Start":"32:41.214 ","End":"32:43.059","Text":"Then we would find, let\u0027s say,"},{"Start":"32:43.059 ","End":"32:48.440","Text":"our x component as the integral of x dq."},{"Start":"32:48.540 ","End":"32:55.105","Text":"Then of course, the y component will be the integral on y dq."},{"Start":"32:55.105 ","End":"33:00.229","Text":"This is spoken about in the chapter on dipoles."},{"Start":"33:00.360 ","End":"33:05.289","Text":"We could have used this equation over here to solve this question,"},{"Start":"33:05.289 ","End":"33:08.161","Text":"where are r hats over here?"},{"Start":"33:08.161 ","End":"33:15.069","Text":"Our r vector is the unit vector pointing from the origin until our points."},{"Start":"33:15.069 ","End":"33:17.439","Text":"So this is the r hat vector."},{"Start":"33:17.439 ","End":"33:24.295","Text":"In actual fact, the r hat vector in this case is in the z direction."},{"Start":"33:24.295 ","End":"33:27.684","Text":"Then we could have just plugged all of this in,"},{"Start":"33:27.684 ","End":"33:29.904","Text":"and solved it like so,"},{"Start":"33:29.904 ","End":"33:33.820","Text":"and then we would have gotten this answer as well."},{"Start":"33:33.820 ","End":"33:37.429","Text":"That is the end of the lesson."}],"ID":22264},{"Watched":false,"Name":"Exercise 6","Duration":"20m 42s","ChapterTopicVideoID":21431,"CourseChapterTopicPlaylistID":99473,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.950","Text":"Hello. In this question,"},{"Start":"00:01.950 ","End":"00:06.060","Text":"we\u0027re being told that we have a wire of length L that"},{"Start":"00:06.060 ","End":"00:10.500","Text":"has a charge Q uniformly distributed along its length."},{"Start":"00:10.500 ","End":"00:14.190","Text":"We\u0027re being asked to calculate the electric field along"},{"Start":"00:14.190 ","End":"00:20.355","Text":"the axis perpendicular to the wire located at its center."},{"Start":"00:20.355 ","End":"00:25.510","Text":"We\u0027re working out the electric field along the y-axis over here."},{"Start":"00:25.910 ","End":"00:29.010","Text":"How are we going to solve this question?"},{"Start":"00:29.010 ","End":"00:32.925","Text":"What we\u0027re going to do is we\u0027re going to consider our wire"},{"Start":"00:32.925 ","End":"00:37.710","Text":"is made out of many, many point charges."},{"Start":"00:37.710 ","End":"00:43.790","Text":"As we know, the electric field due to a point charge is given by Coulomb\u0027s law,"},{"Start":"00:43.790 ","End":"00:52.240","Text":"which is equal to kq divided by r squared in the radial direction."},{"Start":"00:52.370 ","End":"00:58.414","Text":"What we\u0027re going to do is we\u0027re going to split up our wire into lots of little pieces."},{"Start":"00:58.414 ","End":"01:01.175","Text":"Here we have 1 piece."},{"Start":"01:01.175 ","End":"01:07.565","Text":"Let\u0027s say that the distance to this piece is this distance x."},{"Start":"01:07.565 ","End":"01:12.200","Text":"Then we\u0027re going to work out the electric field due to this piece of"},{"Start":"01:12.200 ","End":"01:17.780","Text":"a wire with this point charge at this random points along the y-axis over here."},{"Start":"01:17.780 ","End":"01:20.060","Text":"Once we work it out for this piece,"},{"Start":"01:20.060 ","End":"01:23.750","Text":"then we\u0027re just going to sum along the entire wire,"},{"Start":"01:23.750 ","End":"01:26.220","Text":"and then we\u0027ll have our answer."},{"Start":"01:27.380 ","End":"01:35.455","Text":"We can say that this little point charge over here has a charge of dq."},{"Start":"01:35.455 ","End":"01:39.240","Text":"That means that here we\u0027ll add in our dq."},{"Start":"01:39.240 ","End":"01:46.735","Text":"That means that our electric field due to this point charge dq is going to be dE."},{"Start":"01:46.735 ","End":"01:50.360","Text":"That\u0027s great. Now what I have to do is I have to find out"},{"Start":"01:50.360 ","End":"01:53.765","Text":"what this charge dq is equal to."},{"Start":"01:53.765 ","End":"01:56.195","Text":"Dq, as we know,"},{"Start":"01:56.195 ","End":"02:01.280","Text":"is equal to the unit charge per unit length,"},{"Start":"02:01.280 ","End":"02:02.535","Text":"which is Lambda,"},{"Start":"02:02.535 ","End":"02:05.250","Text":"multiplied by the length."},{"Start":"02:05.250 ","End":"02:10.260","Text":"Here we\u0027re dealing with some small length, dl."},{"Start":"02:10.260 ","End":"02:12.930","Text":"What is our Lambda?"},{"Start":"02:12.930 ","End":"02:15.350","Text":"It\u0027s our charge per unit length."},{"Start":"02:15.350 ","End":"02:18.950","Text":"We\u0027re given in the question that we have a charge Q"},{"Start":"02:18.950 ","End":"02:22.940","Text":"uniformly distributed along the length L. That"},{"Start":"02:22.940 ","End":"02:26.345","Text":"means that our Lambda is going to be equal to"},{"Start":"02:26.345 ","End":"02:30.755","Text":"the total charge divided by the total length,"},{"Start":"02:30.755 ","End":"02:34.310","Text":"Q divided by L. Now what is our dl?"},{"Start":"02:34.310 ","End":"02:37.295","Text":"Our dl is this length over here."},{"Start":"02:37.295 ","End":"02:40.595","Text":"Now because we\u0027re dealing with our x-axis over here,"},{"Start":"02:40.595 ","End":"02:45.515","Text":"so we can say that dl is equal to dx."},{"Start":"02:45.515 ","End":"02:48.845","Text":"Here we can write dx."},{"Start":"02:48.845 ","End":"02:51.205","Text":"That is our dq."},{"Start":"02:51.205 ","End":"02:55.190","Text":"Now what do we have to do is we have to find what our r is."},{"Start":"02:55.190 ","End":"03:00.125","Text":"Our r is the vector connecting"},{"Start":"03:00.125 ","End":"03:06.635","Text":"this point charge over here to our point along the y-axis,"},{"Start":"03:06.635 ","End":"03:09.865","Text":"where we want to measure our electric field."},{"Start":"03:09.865 ","End":"03:13.160","Text":"This is our vector r pointing from"},{"Start":"03:13.160 ","End":"03:18.215","Text":"a point charge until a point where we\u0027re measuring the electric field."},{"Start":"03:18.215 ","End":"03:21.695","Text":"Now in the equation for our electric field,"},{"Start":"03:21.695 ","End":"03:26.275","Text":"we can see that we\u0027re using the magnitude of our r vector."},{"Start":"03:26.275 ","End":"03:32.100","Text":"That means that we\u0027re just taking the size of this arrow."},{"Start":"03:33.130 ","End":"03:37.625","Text":"Let\u0027s now work out the magnitude of"},{"Start":"03:37.625 ","End":"03:45.770","Text":"our vector r. We can say that we\u0027re picking some arbitrary point along our y-axis."},{"Start":"03:45.770 ","End":"03:47.923","Text":"Let\u0027s say we pick this point,"},{"Start":"03:47.923 ","End":"03:55.716","Text":"and that this point is at a distance y away from the origin."},{"Start":"03:55.716 ","End":"03:58.145","Text":"Now by using Pythagoras,"},{"Start":"03:58.145 ","End":"04:09.029","Text":"we can get that r is going to be equal to the square root of x squared plus y squared."},{"Start":"04:10.610 ","End":"04:14.720","Text":"This is just Pythagoras and that will give us this length."},{"Start":"04:14.720 ","End":"04:19.190","Text":"Now we can also work out our E field by"},{"Start":"04:19.190 ","End":"04:23.885","Text":"plugging in both the size and direction of our r vector."},{"Start":"04:23.885 ","End":"04:28.790","Text":"However, we\u0027re going to do it first in the way where we work out the magnitude,"},{"Start":"04:28.790 ","End":"04:34.530","Text":"and then afterwards we will figure out the direction that it is pointing in."},{"Start":"04:36.100 ","End":"04:39.650","Text":"Now we have what our dq is equal to,"},{"Start":"04:39.650 ","End":"04:41.885","Text":"and we have what our r is equal to."},{"Start":"04:41.885 ","End":"04:45.500","Text":"Now we can work out the magnitude of"},{"Start":"04:45.500 ","End":"04:50.595","Text":"our electric field due to this point charge over here."},{"Start":"04:50.595 ","End":"04:58.230","Text":"That is simply going to be equal to kdq divided by r squared,"},{"Start":"04:58.230 ","End":"05:03.765","Text":"where we plug in our r over here and our dq over here."},{"Start":"05:03.765 ","End":"05:05.990","Text":"Now we have the magnitude."},{"Start":"05:05.990 ","End":"05:10.680","Text":"Now let\u0027s figure out the direction of the electric field."},{"Start":"05:10.750 ","End":"05:16.130","Text":"Now the direction of the E field is very, very important."},{"Start":"05:16.130 ","End":"05:20.075","Text":"We have to figure out in which direction the E field will be"},{"Start":"05:20.075 ","End":"05:26.120","Text":"before we can sum up the E field due to each point charge."},{"Start":"05:26.120 ","End":"05:28.550","Text":"Let\u0027s do this in green."},{"Start":"05:28.550 ","End":"05:32.400","Text":"I\u0027m going to carry on the line of my r vector."},{"Start":"05:32.500 ","End":"05:42.770","Text":"This is the direction of my E field due to this point charge over here in green."},{"Start":"05:42.770 ","End":"05:49.200","Text":"It carries on parallel to my red r vector arrow."},{"Start":"05:49.910 ","End":"05:53.690","Text":"Now as we know with vectors,"},{"Start":"05:53.690 ","End":"05:57.709","Text":"we always have to split them up into the different components."},{"Start":"05:57.709 ","End":"06:03.445","Text":"Here we have to split up this vector into the x and the y component."},{"Start":"06:03.445 ","End":"06:06.590","Text":"That\u0027s the only way that we can sum up vectors."},{"Start":"06:06.590 ","End":"06:10.235","Text":"If we say that we have some kind of angle over here,"},{"Start":"06:10.235 ","End":"06:12.610","Text":"let\u0027s call it Alpha,"},{"Start":"06:12.610 ","End":"06:23.150","Text":"we can say that our y component of the E field due to this point charge will be equal to"},{"Start":"06:23.150 ","End":"06:28.770","Text":"the size of our E field multiplied by"},{"Start":"06:28.770 ","End":"06:36.170","Text":"cosine of our angle Alpha over here."},{"Start":"06:36.170 ","End":"06:41.290","Text":"dE_y or the y component,"},{"Start":"06:41.480 ","End":"06:44.540","Text":"that\u0027s going to be this vector over here."},{"Start":"06:44.540 ","End":"06:46.325","Text":"This is dE_y."},{"Start":"06:46.325 ","End":"06:51.770","Text":"The y component of our electric field due to this point charge over here is"},{"Start":"06:51.770 ","End":"06:57.790","Text":"going to be the projection of our green vector on the y-axis."},{"Start":"06:57.790 ","End":"07:00.860","Text":"That\u0027s simply going to be the magnitude of"},{"Start":"07:00.860 ","End":"07:06.630","Text":"our E field multiplied by cosine of this angle over here."},{"Start":"07:07.790 ","End":"07:12.980","Text":"Now we can see that there\u0027s this cosine Alpha over here."},{"Start":"07:12.980 ","End":"07:16.805","Text":"Now, we just defined what Alpha is."},{"Start":"07:16.805 ","End":"07:20.585","Text":"We just said that this angle over here is Alpha."},{"Start":"07:20.585 ","End":"07:24.555","Text":"However, what is this angle Alpha actually?"},{"Start":"07:24.555 ","End":"07:28.400","Text":"As we know, if this angle over here is Alpha,"},{"Start":"07:28.400 ","End":"07:34.235","Text":"that means that this angle over here is also Alpha."},{"Start":"07:34.235 ","End":"07:38.750","Text":"Now, if we remember from simple trigonometry,"},{"Start":"07:38.750 ","End":"07:48.480","Text":"we know that cosine of an angle is equal to the adjacent side,"},{"Start":"07:48.480 ","End":"07:51.850","Text":"so the adjacent side here is y,"},{"Start":"07:51.850 ","End":"07:55.370","Text":"divided by the hypotenuse side,"},{"Start":"07:55.370 ","End":"08:00.120","Text":"which over here is the magnitude of our r vector."},{"Start":"08:01.190 ","End":"08:04.555","Text":"Now we can plug all of that in."},{"Start":"08:04.555 ","End":"08:07.480","Text":"The magnitude of our E field is going to be"},{"Start":"08:07.480 ","End":"08:14.835","Text":"kdq divided by r squared multiplied by cosine Alpha,"},{"Start":"08:14.835 ","End":"08:22.409","Text":"so multiplied by y divided by r. Now let\u0027s plug in everything."},{"Start":"08:22.409 ","End":"08:26.805","Text":"We have k multiplied by dq."},{"Start":"08:26.805 ","End":"08:32.295","Text":"Our dq is equal to Q divided by L,"},{"Start":"08:32.295 ","End":"08:40.035","Text":"dx, multiplied by y divided by r cubed."},{"Start":"08:40.035 ","End":"08:45.650","Text":"Our r if we raise it to the power of 3,"},{"Start":"08:45.650 ","End":"08:49.120","Text":"then we also have to raise this side to the power of 3."},{"Start":"08:49.120 ","End":"08:54.620","Text":"Then we\u0027ll get that r cubed is equal to x squared plus y"},{"Start":"08:54.620 ","End":"09:01.910","Text":"squared to the power of 3 divided by 2."},{"Start":"09:01.910 ","End":"09:08.160","Text":"Now let\u0027s just rewrite this where the dx and the y\u0027s switch sides."},{"Start":"09:08.380 ","End":"09:10.535","Text":"Now we have it."},{"Start":"09:10.535 ","End":"09:14.165","Text":"This is our y component of"},{"Start":"09:14.165 ","End":"09:21.060","Text":"our E field due to this point charge over here."},{"Start":"09:21.060 ","End":"09:26.840","Text":"Now in order to find the total electric field in the y-direction,"},{"Start":"09:26.840 ","End":"09:32.350","Text":"we\u0027re going to have to sum up on all of these point charges along our wire."},{"Start":"09:32.350 ","End":"09:35.524","Text":"That means that we\u0027re going to have to integrate."},{"Start":"09:35.524 ","End":"09:41.630","Text":"Our E field in the y direction or the y component of our E field is"},{"Start":"09:41.630 ","End":"09:47.878","Text":"going to be equal to the integral on our dE_y."},{"Start":"09:47.878 ","End":"09:50.920","Text":"What does that mean?"},{"Start":"09:50.920 ","End":"09:53.095","Text":"This is going to be equal to,"},{"Start":"09:53.095 ","End":"09:57.895","Text":"and now we can take all of the constants outside of the integration sign."},{"Start":"09:57.895 ","End":"10:01.360","Text":"Notice that we\u0027re integrating with respect to x,"},{"Start":"10:01.360 ","End":"10:02.740","Text":"which means that y,"},{"Start":"10:02.740 ","End":"10:07.090","Text":"both here and in the denominator are constants."},{"Start":"10:07.090 ","End":"10:09.700","Text":"Let\u0027s take out all of our constants."},{"Start":"10:09.700 ","End":"10:18.415","Text":"We\u0027re going to have that this is equal to kyQ divided by L. Then"},{"Start":"10:18.415 ","End":"10:24.000","Text":"integrating with respect to x on"},{"Start":"10:24.000 ","End":"10:31.660","Text":"x^2 plus y^2 to the power of 3 over 2."},{"Start":"10:31.660 ","End":"10:36.670","Text":"Before we integrate, let\u0027s work out our integration bounds."},{"Start":"10:36.670 ","End":"10:42.325","Text":"In the question, we\u0027re being told that the axis is perpendicular to the wire and located"},{"Start":"10:42.325 ","End":"10:48.535","Text":"at its center which means that this is our point 0,"},{"Start":"10:48.535 ","End":"10:53.320","Text":"but the total length of our wire is equal to L. First,"},{"Start":"10:53.320 ","End":"10:57.130","Text":"we\u0027re going to start integrating from this point over here."},{"Start":"10:57.130 ","End":"11:03.470","Text":"This point is located at negative L divided by 2."},{"Start":"11:05.160 ","End":"11:11.095","Text":"Then we\u0027re summing up past 0 and up until this point over here,"},{"Start":"11:11.095 ","End":"11:17.990","Text":"and this point over here is equal to positive L divided by 2."},{"Start":"11:18.570 ","End":"11:22.825","Text":"Now, what we can see is that we have this integral."},{"Start":"11:22.825 ","End":"11:25.810","Text":"Now, this format for integration is"},{"Start":"11:25.810 ","End":"11:29.365","Text":"very common and it will come up a lot in these types of questions."},{"Start":"11:29.365 ","End":"11:32.095","Text":"Instead of starting to work out this integral,"},{"Start":"11:32.095 ","End":"11:37.375","Text":"it\u0027s easy to just know the format of what its solution looks like."},{"Start":"11:37.375 ","End":"11:40.885","Text":"Let\u0027s just write out the solution to this integral first."},{"Start":"11:40.885 ","End":"11:47.305","Text":"We have kyQ divided by L multiplied by."},{"Start":"11:47.305 ","End":"11:51.865","Text":"What we have here is we\u0027re integrating with respect to x,"},{"Start":"11:51.865 ","End":"11:56.320","Text":"which means that this y^2 over here is a constant."},{"Start":"11:56.320 ","End":"11:58.390","Text":"It isn\u0027t a variable."},{"Start":"11:58.390 ","End":"12:06.265","Text":"In these cases, we always add up or put in to the numerator."},{"Start":"12:06.265 ","End":"12:09.880","Text":"The square root of our variable,"},{"Start":"12:09.880 ","End":"12:15.025","Text":"which here is x, divided by our constant,"},{"Start":"12:15.025 ","End":"12:18.715","Text":"which here is y^2, y^2 is a constant number,"},{"Start":"12:18.715 ","End":"12:23.890","Text":"multiplied by this whole expression inside the brackets,"},{"Start":"12:23.890 ","End":"12:27.280","Text":"so x^2 plus y^2."},{"Start":"12:27.280 ","End":"12:34.120","Text":"Then we raise it to the power of a half."},{"Start":"12:34.120 ","End":"12:36.730","Text":"Then of course we add in our bounds,"},{"Start":"12:36.730 ","End":"12:44.470","Text":"which here specifically is between negative L divided by 2 until positive L divided by 2."},{"Start":"12:44.470 ","End":"12:47.260","Text":"Now let\u0027s substitute in our bounds."},{"Start":"12:47.260 ","End":"12:52.105","Text":"I wrote over here the general solution for this type of integral."},{"Start":"12:52.105 ","End":"12:54.640","Text":"Now remember if there was no bounds like here,"},{"Start":"12:54.640 ","End":"12:58.540","Text":"you always have to add in an integration constant,"},{"Start":"12:58.540 ","End":"13:01.570","Text":"so plus c. Here we had bounds,"},{"Start":"13:01.570 ","End":"13:03.130","Text":"so I don\u0027t have to add in my plus"},{"Start":"13:03.130 ","End":"13:07.435","Text":"c. It\u0027s useful to write this out in your equation sheet."},{"Start":"13:07.435 ","End":"13:12.080","Text":"Let\u0027s carry on. Let\u0027s substitute in our bounds over here."},{"Start":"13:12.600 ","End":"13:16.060","Text":"First of all, we can see that we have a y^2 here."},{"Start":"13:16.060 ","End":"13:19.105","Text":"This y cancels out with one of these ys."},{"Start":"13:19.105 ","End":"13:25.640","Text":"Then we\u0027ll have kQ divided by Ly."},{"Start":"13:26.220 ","End":"13:30.895","Text":"Now, we substitute L divided by 2 divided by,"},{"Start":"13:30.895 ","End":"13:33.760","Text":"so this y has already been written over here,"},{"Start":"13:33.760 ","End":"13:36.160","Text":"and then we have L over"},{"Start":"13:36.160 ","End":"13:45.100","Text":"2^2 plus y^2 to the power of 1/2 minus,"},{"Start":"13:45.100 ","End":"13:48.130","Text":"and now we substitute in our negative L over 2,"},{"Start":"13:48.130 ","End":"13:52.495","Text":"so minus negative L over 2 divided by,"},{"Start":"13:52.495 ","End":"13:54.820","Text":"and then we\u0027ll get the same denominator because"},{"Start":"13:54.820 ","End":"13:58.000","Text":"L over 2 is the same as negative L over 2."},{"Start":"13:58.000 ","End":"14:06.640","Text":"Here again, L over 2^2 plus y^2 to the power of 1/2."},{"Start":"14:06.640 ","End":"14:09.205","Text":"Now because we have a minus and a minus,"},{"Start":"14:09.205 ","End":"14:11.155","Text":"this becomes a plus,"},{"Start":"14:11.155 ","End":"14:14.050","Text":"and because both of the denominators are the same,"},{"Start":"14:14.050 ","End":"14:17.950","Text":"so we can add up these fractions very easily."},{"Start":"14:17.950 ","End":"14:21.730","Text":"Then we\u0027ll have L over 2 plus L over 2,"},{"Start":"14:21.730 ","End":"14:30.515","Text":"which will simply become L. Then this L will cancel out with L outside of the brackets."},{"Start":"14:30.515 ","End":"14:33.720","Text":"Let\u0027s quickly write out the final answer."},{"Start":"14:33.720 ","End":"14:38.295","Text":"What we\u0027ll have is we\u0027ll have our kQ over here,"},{"Start":"14:38.295 ","End":"14:41.955","Text":"divided by our y."},{"Start":"14:41.955 ","End":"14:46.050","Text":"Then our L\u0027s canceled out,"},{"Start":"14:46.050 ","End":"14:52.615","Text":"and so now we\u0027re just multiplying by 1 divided by L"},{"Start":"14:52.615 ","End":"15:00.470","Text":"over 2^2 plus y^2 to the power of 1/2."},{"Start":"15:00.470 ","End":"15:05.660","Text":"This is the y component of our electric field."},{"Start":"15:05.660 ","End":"15:08.600","Text":"This will be pointing in this direction."},{"Start":"15:08.600 ","End":"15:11.440","Text":"Now, what about the x component?"},{"Start":"15:11.440 ","End":"15:16.195","Text":"We can see from symmetry that our x component will cancel out."},{"Start":"15:16.195 ","End":"15:18.265","Text":"Let\u0027s just demonstrate this."},{"Start":"15:18.265 ","End":"15:22.660","Text":"If we split this up into the x and y components, as we said,"},{"Start":"15:22.660 ","End":"15:25.885","Text":"this blue arrow is our y component,"},{"Start":"15:25.885 ","End":"15:34.195","Text":"and then this blue arrow over here will be our x component."},{"Start":"15:34.195 ","End":"15:38.560","Text":"Now, from the other side,"},{"Start":"15:38.560 ","End":"15:41.755","Text":"if we go the same distance x,"},{"Start":"15:41.755 ","End":"15:43.990","Text":"but in the negative x-direction,"},{"Start":"15:43.990 ","End":"15:46.735","Text":"and we take this point charge over here."},{"Start":"15:46.735 ","End":"15:55.450","Text":"We can see that it will produce an electric field in this direction."},{"Start":"15:55.450 ","End":"16:00.685","Text":"Then we can see if we use again our blue arrow,"},{"Start":"16:00.685 ","End":"16:04.960","Text":"we can split up this vector into x and y components."},{"Start":"16:04.960 ","End":"16:07.495","Text":"Again, we\u0027ll have our y component,"},{"Start":"16:07.495 ","End":"16:10.240","Text":"and so the 2 y components will add up and there\u0027ll be"},{"Start":"16:10.240 ","End":"16:13.135","Text":"pointing in this positive y-direction,"},{"Start":"16:13.135 ","End":"16:17.830","Text":"and the x component will be this over here."},{"Start":"16:17.830 ","End":"16:23.440","Text":"As we can see, is going to be equal and opposite to"},{"Start":"16:23.440 ","End":"16:29.695","Text":"the x component of the E field produced from this point charge on the opposite side."},{"Start":"16:29.695 ","End":"16:34.120","Text":"As we can see, the x component from symmetry will cancel out,"},{"Start":"16:34.120 ","End":"16:39.670","Text":"and we\u0027re only going to be left with a y component of our E field."},{"Start":"16:39.670 ","End":"16:43.915","Text":"That\u0027s an explanation and now I\u0027m just going to show it"},{"Start":"16:43.915 ","End":"16:48.400","Text":"algebraically so that it\u0027s a bit more clear."},{"Start":"16:48.400 ","End":"16:53.065","Text":"Now just to show what we would do with our x component,"},{"Start":"16:53.065 ","End":"16:55.480","Text":"also with this example, if you work it out,"},{"Start":"16:55.480 ","End":"16:58.360","Text":"you\u0027ll see that we get to 0 in the x-direction,"},{"Start":"16:58.360 ","End":"17:03.220","Text":"which shows that the E field will cancel out in the x-direction but also in general so"},{"Start":"17:03.220 ","End":"17:08.245","Text":"that you know the method of what to do just in case the x component doesn\u0027t cancel out."},{"Start":"17:08.245 ","End":"17:12.534","Text":"An example of when it wouldn\u0027t cancel out is if we have the same wire,"},{"Start":"17:12.534 ","End":"17:15.010","Text":"however, we\u0027re told that the origin,"},{"Start":"17:15.010 ","End":"17:17.095","Text":"instead of being at the center of the wire,"},{"Start":"17:17.095 ","End":"17:21.295","Text":"is located at one of the extremum of the wire,"},{"Start":"17:21.295 ","End":"17:23.965","Text":"so then the x component wouldn\u0027t cancel out."},{"Start":"17:23.965 ","End":"17:26.905","Text":"Let\u0027s see how we would work it out."},{"Start":"17:26.905 ","End":"17:30.175","Text":"Our x component of our E field,"},{"Start":"17:30.175 ","End":"17:33.325","Text":"similarly to how we worked out the y component,"},{"Start":"17:33.325 ","End":"17:36.760","Text":"is going to be equal to the magnitude"},{"Start":"17:36.760 ","End":"17:41.005","Text":"of our electric field due to one of the point charges,"},{"Start":"17:41.005 ","End":"17:45.265","Text":"and now we\u0027re taking this side over here,"},{"Start":"17:45.265 ","End":"17:47.080","Text":"or this over here,"},{"Start":"17:47.080 ","End":"17:50.455","Text":"the opposite side in order to be in the x-direction."},{"Start":"17:50.455 ","End":"17:54.070","Text":"That means instead of cosine of this angle Alpha,"},{"Start":"17:54.070 ","End":"17:57.880","Text":"we\u0027re going to be using sin of this angle Alpha."},{"Start":"17:57.880 ","End":"18:02.125","Text":"Now, of course, Alpha is something that we just made up, so what does it mean?"},{"Start":"18:02.125 ","End":"18:05.260","Text":"Again, this is our Alpha angle over here."},{"Start":"18:05.260 ","End":"18:08.455","Text":"As we know from our basic trig identities,"},{"Start":"18:08.455 ","End":"18:15.625","Text":"sin of an angle is equal to the opposite side divided by the hypotenuse."},{"Start":"18:15.625 ","End":"18:22.780","Text":"Here, the opposite side is our x and the hypotenuse is the magnitude of our vector."},{"Start":"18:22.780 ","End":"18:25.825","Text":"Sin of Alpha is in this example,"},{"Start":"18:25.825 ","End":"18:33.444","Text":"x divided by r. Now we\u0027re going to substitute in what our magnitude is."},{"Start":"18:33.444 ","End":"18:40.285","Text":"The magnitude of our dE multiplied by here sin of Alpha."},{"Start":"18:40.285 ","End":"18:42.445","Text":"We\u0027re just going to rewrite this."},{"Start":"18:42.445 ","End":"18:51.130","Text":"We\u0027ll get that this is equal to kdQ divided by r^2 multiplied by"},{"Start":"18:51.130 ","End":"18:55.360","Text":"x divided by r. Now I\u0027m going"},{"Start":"18:55.360 ","End":"18:59.845","Text":"to straight away put this into an integral form where we substitute in our dq,"},{"Start":"18:59.845 ","End":"19:02.335","Text":"which is the same dq as we got before,"},{"Start":"19:02.335 ","End":"19:03.865","Text":"and substitute in our r,"},{"Start":"19:03.865 ","End":"19:06.550","Text":"which is the same r that we got before."},{"Start":"19:06.550 ","End":"19:13.380","Text":"This is going to be equal to kQ divided by L, and again,"},{"Start":"19:13.380 ","End":"19:21.170","Text":"we\u0027re integrating from negative L divided by 2 until L divided by 2."},{"Start":"19:21.810 ","End":"19:26.035","Text":"Our integral this time because we have an x over here,"},{"Start":"19:26.035 ","End":"19:30.630","Text":"it\u0027s going to be xdx divided"},{"Start":"19:30.630 ","End":"19:38.675","Text":"by x^2 plus y^2 to the power of 3 over 2."},{"Start":"19:38.675 ","End":"19:42.650","Text":"Again, we\u0027re integrating with just respect to x,"},{"Start":"19:42.650 ","End":"19:47.330","Text":"which means that our y over here is considered a constant."},{"Start":"19:47.330 ","End":"19:48.950","Text":"Then in order to do this,"},{"Start":"19:48.950 ","End":"19:52.095","Text":"we have to use substitution."},{"Start":"19:52.095 ","End":"19:55.480","Text":"We can define some new variable, let\u0027s call it t,"},{"Start":"19:55.480 ","End":"20:00.855","Text":"and say that this is equal to x^2 plus y^2."},{"Start":"20:00.855 ","End":"20:08.120","Text":"Then we can therefore say that dt is simply going to be equal to 2x dx."},{"Start":"20:08.120 ","End":"20:11.030","Text":"The y\u0027s canceled out because it\u0027s a constant."},{"Start":"20:11.030 ","End":"20:17.615","Text":"Then we simply substitute this all in over here and we\u0027ll get an answer."},{"Start":"20:17.615 ","End":"20:20.210","Text":"If you want to carry on this calculation,"},{"Start":"20:20.210 ","End":"20:24.140","Text":"you\u0027ll see that we really do get 0 for"},{"Start":"20:24.140 ","End":"20:29.345","Text":"the x component of our E field in this example over here, and of course,"},{"Start":"20:29.345 ","End":"20:32.739","Text":"if we\u0027re already doing this integration over here,"},{"Start":"20:32.739 ","End":"20:40.385","Text":"what we would have over here is our total E field in the x-direction."},{"Start":"20:40.385 ","End":"20:43.020","Text":"That\u0027s the end of this lesson."}],"ID":22265},{"Watched":false,"Name":"Exercise 7","Duration":"16m 35s","ChapterTopicVideoID":21290,"CourseChapterTopicPlaylistID":99473,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.235","Text":"Hello. In this lesson,"},{"Start":"00:02.235 ","End":"00:04.785","Text":"we\u0027re going to be answering the following question."},{"Start":"00:04.785 ","End":"00:12.315","Text":"So a ring of radius R has a charge density of Lambda 1 on its right hand side,"},{"Start":"00:12.315 ","End":"00:17.220","Text":"and a charge density of Lambda 2 on its left hand side."},{"Start":"00:17.220 ","End":"00:24.115","Text":"The right hand side has a piece missing formed by this angle of 2 Alpha."},{"Start":"00:24.115 ","End":"00:31.280","Text":"We\u0027re being asked, what is the electric field at the center of the ring?"},{"Start":"00:31.280 ","End":"00:34.280","Text":"The first thing that we\u0027re going to do is"},{"Start":"00:34.280 ","End":"00:37.610","Text":"we\u0027re going to say that the center is the origin,"},{"Start":"00:37.610 ","End":"00:40.280","Text":"and this is where we\u0027re looking for the electric field,"},{"Start":"00:40.280 ","End":"00:44.720","Text":"and that going like so right through the center."},{"Start":"00:44.720 ","End":"00:49.520","Text":"Over here, we\u0027ll have our x-axis,"},{"Start":"00:49.520 ","End":"00:52.895","Text":"and then going up like so,"},{"Start":"00:52.895 ","End":"00:56.970","Text":"we\u0027ll have our y-axis."},{"Start":"00:57.800 ","End":"01:07.260","Text":"It looks like so, and now I\u0027m going to choose a random piece of the ring over here."},{"Start":"01:08.150 ","End":"01:10.745","Text":"This piece over here,"},{"Start":"01:10.745 ","End":"01:14.645","Text":"so we have this radius r over here,"},{"Start":"01:14.645 ","End":"01:17.750","Text":"pointing from the origin until this piece,"},{"Start":"01:17.750 ","End":"01:21.530","Text":"and then we have this angle that is formed between"},{"Start":"01:21.530 ","End":"01:25.785","Text":"the x-axis and the radius pointing to the piece over here,"},{"Start":"01:25.785 ","End":"01:28.350","Text":"and let\u0027s call this angle Theta."},{"Start":"01:28.350 ","End":"01:30.570","Text":"Of course, Theta is a variable,"},{"Start":"01:30.570 ","End":"01:33.500","Text":"it\u0027s constantly changing as this piece,"},{"Start":"01:33.500 ","End":"01:36.710","Text":"which is an arbitrary piece of the ring that I chose."},{"Start":"01:36.710 ","End":"01:41.960","Text":"As we rotate around and sum up along the whole ring, so obviously,"},{"Start":"01:41.960 ","End":"01:48.150","Text":"the piece is moving, and therefore angle Theta is going to be changing as well."},{"Start":"01:49.400 ","End":"01:55.565","Text":"Of course, this piece of ring has some charge."},{"Start":"01:55.565 ","End":"01:57.560","Text":"It has some charge,"},{"Start":"01:57.560 ","End":"01:59.950","Text":"dq. What is dq?"},{"Start":"01:59.950 ","End":"02:04.595","Text":"We know that dq is always equal to Lambda dl because we\u0027re dealing with a ring,"},{"Start":"02:04.595 ","End":"02:08.060","Text":"so it\u0027s 1 dimension. What is dl?"},{"Start":"02:08.060 ","End":"02:12.060","Text":"We can see that we\u0027re dealing with polar coordinates,"},{"Start":"02:12.060 ","End":"02:16.070","Text":"so we\u0027re dealing with r and Theta because we\u0027re going around in a circle."},{"Start":"02:16.070 ","End":"02:19.145","Text":"First of all, we know that our R is going to be constant"},{"Start":"02:19.145 ","End":"02:23.150","Text":"because we\u0027re going along a ring, not a disc,"},{"Start":"02:23.150 ","End":"02:27.410","Text":"so the r is always going to be equal or constant,"},{"Start":"02:27.410 ","End":"02:33.615","Text":"so dl in polar coordinates will be Rd Theta."},{"Start":"02:33.615 ","End":"02:37.700","Text":"Well, I just substituted in this R because I know that it\u0027s constant,"},{"Start":"02:37.700 ","End":"02:40.145","Text":"and d Theta, as we just said,"},{"Start":"02:40.145 ","End":"02:44.240","Text":"Theta\u0027s changing, so we\u0027re going to integrate along that soon,"},{"Start":"02:44.240 ","End":"02:48.515","Text":"and I know that there\u0027s different Lambda values."},{"Start":"02:48.515 ","End":"02:50.720","Text":"Soon we\u0027re going to see how we deal with that,"},{"Start":"02:50.720 ","End":"02:53.970","Text":"this is just the general equation."},{"Start":"02:54.560 ","End":"02:59.675","Text":"What we\u0027re trying to find is the electric field over here at the center of the ring."},{"Start":"02:59.675 ","End":"03:03.830","Text":"The electric field formed by this arbitrary point on"},{"Start":"03:03.830 ","End":"03:08.570","Text":"the ring over here is first of all going to be going in this direction,"},{"Start":"03:08.570 ","End":"03:11.630","Text":"so from the peace until the center,"},{"Start":"03:11.630 ","End":"03:14.510","Text":"so it\u0027s going in the opposite direction to"},{"Start":"03:14.510 ","End":"03:18.470","Text":"its position vector so we can draw it over here,"},{"Start":"03:18.470 ","End":"03:20.390","Text":"but it\u0027s going to be parallel,"},{"Start":"03:20.390 ","End":"03:22.400","Text":"so in the same line."},{"Start":"03:22.400 ","End":"03:25.370","Text":"Of course, this is dE,"},{"Start":"03:25.370 ","End":"03:30.410","Text":"so it\u0027s the small electric field caused at the center from"},{"Start":"03:30.410 ","End":"03:36.686","Text":"this single piece of the ring over here."},{"Start":"03:36.686 ","End":"03:41.180","Text":"Now let\u0027s find the size of this dE."},{"Start":"03:41.180 ","End":"03:47.170","Text":"The size of this small electric field is going to be from Coulomb\u0027s law,"},{"Start":"03:47.170 ","End":"03:52.449","Text":"k dq divided by r^2."},{"Start":"03:52.449 ","End":"03:57.640","Text":"Well, we can just substitute in the radius of the ring because it\u0027s not changing,"},{"Start":"03:57.640 ","End":"04:02.395","Text":"so it\u0027s k dq divided by capital R^2,"},{"Start":"04:02.395 ","End":"04:06.620","Text":"where soon we\u0027ll substitute in our dq."},{"Start":"04:06.950 ","End":"04:11.770","Text":"The next thing that we have to do is we have to"},{"Start":"04:11.770 ","End":"04:16.135","Text":"split up this dE into its different components,"},{"Start":"04:16.135 ","End":"04:18.905","Text":"so it\u0027s x, y-components."},{"Start":"04:18.905 ","End":"04:24.830","Text":"What we can do is we can carry on our x-axis over here,"},{"Start":"04:24.830 ","End":"04:28.130","Text":"so this is the negative section of the x-axis,"},{"Start":"04:28.130 ","End":"04:32.890","Text":"and here we have the negative section of the y-axis."},{"Start":"04:32.890 ","End":"04:38.570","Text":"First of all, we know that if this angle over here is Theta,"},{"Start":"04:38.570 ","End":"04:42.895","Text":"then that means that this angle over here is Theta."},{"Start":"04:42.895 ","End":"04:50.105","Text":"Now we\u0027re trying to find the projection of our dE in the x-direction,"},{"Start":"04:50.105 ","End":"04:53.510","Text":"so the x-component of our de."},{"Start":"04:53.510 ","End":"04:56.480","Text":"Let\u0027s write that over here, so dE_x,"},{"Start":"04:56.480 ","End":"05:02.795","Text":"the projection in the x-direction is simply equal to the size of the vector,"},{"Start":"05:02.795 ","End":"05:04.585","Text":"so the size of dE,"},{"Start":"05:04.585 ","End":"05:07.715","Text":"and then we\u0027re going to multiply it by,"},{"Start":"05:07.715 ","End":"05:10.850","Text":"so we can see that it\u0027s on the x-axis,"},{"Start":"05:10.850 ","End":"05:13.940","Text":"so here it\u0027s going to be cosine of Theta."},{"Start":"05:13.940 ","End":"05:18.440","Text":"However, we can see that the x component is going to"},{"Start":"05:18.440 ","End":"05:23.555","Text":"be projected something like so in this direction,"},{"Start":"05:23.555 ","End":"05:27.245","Text":"so we can see it\u0027s going in the negative x-direction."},{"Start":"05:27.245 ","End":"05:30.215","Text":"We have to add in here the negative,"},{"Start":"05:30.215 ","End":"05:32.840","Text":"so negative cosine of Theta."},{"Start":"05:32.840 ","End":"05:38.570","Text":"Now, another way that you could have written it is found this whole angle over here,"},{"Start":"05:38.570 ","End":"05:44.900","Text":"so we would have had 90 plus 90,"},{"Start":"05:44.900 ","End":"05:48.550","Text":"so that\u0027s 180, plus this Theta."},{"Start":"05:48.550 ","End":"05:57.770","Text":"You could have also said that this is equal to dE multiplied by cosine of,"},{"Start":"05:57.770 ","End":"06:01.880","Text":"and then the angle would have been 180 plus Theta"},{"Start":"06:01.880 ","End":"06:07.560","Text":"or just Theta plus 180 degrees, it\u0027s the same thing."},{"Start":"06:09.500 ","End":"06:17.720","Text":"Now, what we want to do is we want to find the total electric field in the x-direction."},{"Start":"06:17.720 ","End":"06:24.495","Text":"The electric field from all the pieces along the ring in the x-direction,"},{"Start":"06:24.495 ","End":"06:29.640","Text":"so that means that we want to find E_x."},{"Start":"06:29.640 ","End":"06:35.840","Text":"Of course, that means that we\u0027re going to be integrating along dE_x."},{"Start":"06:35.840 ","End":"06:39.425","Text":"That is going to be equal to the integral of,"},{"Start":"06:39.425 ","End":"06:43.030","Text":"so it\u0027s going to be over here,"},{"Start":"06:43.030 ","End":"06:49.005","Text":"so dE is equal to k dq."},{"Start":"06:49.005 ","End":"06:51.740","Text":"Let\u0027s substitute in already our dq,"},{"Start":"06:51.740 ","End":"06:57.080","Text":"so our dq is Lambda Rd Theta."},{"Start":"06:57.080 ","End":"07:02.800","Text":"So k dq divided by R^2,"},{"Start":"07:02.800 ","End":"07:06.040","Text":"so this is our dE,"},{"Start":"07:06.040 ","End":"07:09.230","Text":"and then this is going to be multiplied by,"},{"Start":"07:09.230 ","End":"07:11.530","Text":"so I\u0027m going to choose this over here,"},{"Start":"07:11.530 ","End":"07:15.890","Text":"so multiplied by negative cosine of Theta."},{"Start":"07:15.890 ","End":"07:22.050","Text":"But you could have also chosen this version of multiplied by cosine of Theta plus 180."},{"Start":"07:22.740 ","End":"07:29.740","Text":"Now, what we want to do is we want to take into account that in this side of the ring,"},{"Start":"07:29.740 ","End":"07:32.470","Text":"we have charged density of Lambda 1,"},{"Start":"07:32.470 ","End":"07:35.920","Text":"and on this side we have charged density of Lambda 2."},{"Start":"07:35.920 ","End":"07:39.115","Text":"This is where we\u0027re going to substitute this in."},{"Start":"07:39.115 ","End":"07:43.975","Text":"What we\u0027re going to do is we\u0027re going to split this up into 2 integrals."},{"Start":"07:43.975 ","End":"07:47.560","Text":"Where one is where integrating along the section of"},{"Start":"07:47.560 ","End":"07:51.950","Text":"Lambda 1 and the other one along the section of Lambda 2."},{"Start":"07:52.500 ","End":"07:55.510","Text":"Let\u0027s write this a fresh,"},{"Start":"07:55.510 ","End":"07:58.360","Text":"Our E_x, so let\u0027s split this up."},{"Start":"07:58.360 ","End":"08:00.985","Text":"Now we\u0027re going to have to split this up into 3 sections."},{"Start":"08:00.985 ","End":"08:05.680","Text":"We have this section over here for Lambda 1,"},{"Start":"08:05.680 ","End":"08:09.436","Text":"then we have this half-a-ring section over here for Lambda 2,"},{"Start":"08:09.436 ","End":"08:13.555","Text":"and then we go back to Lambda 1 over here."},{"Start":"08:13.555 ","End":"08:18.655","Text":"We have to split this section for Lambda 1 into 2 integrals because here,"},{"Start":"08:18.655 ","End":"08:21.110","Text":"we have empty space."},{"Start":"08:21.510 ","End":"08:24.290","Text":"First of all,"},{"Start":"08:24.290 ","End":"08:27.580","Text":"we can cancel out one of the Rs over here,"},{"Start":"08:27.580 ","End":"08:29.695","Text":"and then we can take out the constants."},{"Start":"08:29.695 ","End":"08:33.780","Text":"The constants are k divided by R,"},{"Start":"08:33.780 ","End":"08:40.300","Text":"and Lambda is changing so we\u0027re going to put that inside the brackets over here."},{"Start":"08:41.110 ","End":"08:49.240","Text":"Our first integral is we\u0027re integrating along this Lambda 1. Let\u0027s plug it in."},{"Start":"08:49.240 ","End":"08:59.455","Text":"We have Lambda 1 multiplied by negative cosine of Theta d Theta."},{"Start":"08:59.455 ","End":"09:01.285","Text":"What are our bounds?"},{"Start":"09:01.285 ","End":"09:06.400","Text":"We can see that this angle over here is Alpha,"},{"Start":"09:06.400 ","End":"09:08.260","Text":"and this angle over here is Alpha."},{"Start":"09:08.260 ","End":"09:11.270","Text":"Together, we have this angle 2 Alpha."},{"Start":"09:12.060 ","End":"09:14.560","Text":"All of this space is empty,"},{"Start":"09:14.560 ","End":"09:19.480","Text":"which means that the first point where we have some charged density is at"},{"Start":"09:19.480 ","End":"09:26.035","Text":"this place over here which is located at the angle Alpha."},{"Start":"09:26.035 ","End":"09:32.080","Text":"Then this carries on up until we get to this angle over here,"},{"Start":"09:32.080 ","End":"09:34.465","Text":"which we can see is at 90 degrees."},{"Start":"09:34.465 ","End":"09:36.070","Text":"We could say 90 degrees,"},{"Start":"09:36.070 ","End":"09:37.900","Text":"or we can do this in radians,"},{"Start":"09:37.900 ","End":"09:40.735","Text":"Pi divided by 2."},{"Start":"09:40.735 ","End":"09:44.950","Text":"That\u0027s the first integral and then the next integral."},{"Start":"09:44.950 ","End":"09:47.395","Text":"Now we\u0027re going along a Lambda 2."},{"Start":"09:47.395 ","End":"09:55.175","Text":"We have Lambda 2 multiplied by negative cosine of Theta d Theta."},{"Start":"09:55.175 ","End":"09:57.975","Text":"Then our bounds are from over here,"},{"Start":"09:57.975 ","End":"10:02.125","Text":"which is at a 90-degree angle or at an angle of Pi over 2,"},{"Start":"10:02.125 ","End":"10:04.495","Text":"and then we go all the way over here."},{"Start":"10:04.495 ","End":"10:09.505","Text":"We\u0027re going a 180 degrees or Pi degrees."},{"Start":"10:09.505 ","End":"10:16.330","Text":"We\u0027re going from Pi over 2 until 3 Pi over 2, a full semicircle."},{"Start":"10:16.330 ","End":"10:18.700","Text":"Then our final integral."},{"Start":"10:18.700 ","End":"10:22.840","Text":"We\u0027re back to this section with charge density of Lambda 1."},{"Start":"10:22.840 ","End":"10:29.920","Text":"Lambda 1 multiplied by negative cosine of Theta d Theta,"},{"Start":"10:29.920 ","End":"10:33.790","Text":"and our bounds are this time from over here,"},{"Start":"10:33.790 ","End":"10:42.430","Text":"which is at 3 Pi over 2 until we get to this section over here,"},{"Start":"10:42.430 ","End":"10:44.140","Text":"this is our last piece."},{"Start":"10:44.140 ","End":"10:46.885","Text":"A full circle is 2 Pi."},{"Start":"10:46.885 ","End":"10:51.559","Text":"Starting from here all the way up until the end is 2 Pi,"},{"Start":"10:51.559 ","End":"10:59.455","Text":"but what 2 Pi and then we go down an angle of Alpha."},{"Start":"10:59.455 ","End":"11:04.640","Text":"We can write down 2 Pi minus Alpha."},{"Start":"11:08.430 ","End":"11:13.255","Text":"Now, let\u0027s write out what this is equal to."},{"Start":"11:13.255 ","End":"11:20.710","Text":"This is equal to k divided by R. Then this will be equal to,"},{"Start":"11:20.710 ","End":"11:23.170","Text":"so we\u0027ll have Lambda 1,"},{"Start":"11:23.170 ","End":"11:26.830","Text":"and then the integral of negative cosine of Theta d Theta is"},{"Start":"11:26.830 ","End":"11:31.540","Text":"simply equal to negative sine of Theta"},{"Start":"11:31.540 ","End":"11:34.630","Text":"where we\u0027re substituting in the bounds of"},{"Start":"11:34.630 ","End":"11:40.060","Text":"Alpha into Pi divided by 2 plus and then we\u0027ll have Lambda 2."},{"Start":"11:40.060 ","End":"11:44.650","Text":"Then again, negative sine Theta because it\u0027s the same integral."},{"Start":"11:44.650 ","End":"11:50.440","Text":"The bounds are from Pi over 2 until 3 Pi over 2,"},{"Start":"11:50.440 ","End":"11:56.665","Text":"and then plus Lambda 1 multiplied by negative sine of Theta and"},{"Start":"11:56.665 ","End":"12:04.580","Text":"also between 3 Pi over 2 until 2 Pi minus Alpha."},{"Start":"12:04.830 ","End":"12:07.240","Text":"What does this equal to?"},{"Start":"12:07.240 ","End":"12:14.785","Text":"Again, k divided by R. Let\u0027s put in our Lambda 1s into the same place."},{"Start":"12:14.785 ","End":"12:19.255","Text":"We\u0027ll have Lambda 1, and then this is going to be multiplied by."},{"Start":"12:19.255 ","End":"12:27.590","Text":"First of all, we\u0027re going to have negative sine of Pi over 2,"},{"Start":"12:28.320 ","End":"12:37.315","Text":"negative sine of Pi over 2 is equal to negative 1."},{"Start":"12:37.315 ","End":"12:43.375","Text":"Then we\u0027re going to have minus negative sine of Alpha."},{"Start":"12:43.375 ","End":"12:49.690","Text":"That\u0027s going to be plus sine of Alpha because the negative and negative will cancel out,"},{"Start":"12:49.690 ","End":"12:53.320","Text":"and then we have plus Lambda 1 of,"},{"Start":"12:53.320 ","End":"13:00.655","Text":"so then now we have negative sine of 2 Pi minus Alpha."},{"Start":"13:00.655 ","End":"13:02.680","Text":"Let\u0027s put that in."},{"Start":"13:02.680 ","End":"13:08.575","Text":"Negative sine of 2 Pi minus Alpha."},{"Start":"13:08.575 ","End":"13:13.855","Text":"This is the same as being equal to,"},{"Start":"13:13.855 ","End":"13:18.610","Text":"just from here, from the sine not taking into account the negative over here."},{"Start":"13:18.610 ","End":"13:21.640","Text":"Then we\u0027ll get a negative and a negative and they\u0027ll cancel out,"},{"Start":"13:21.640 ","End":"13:25.345","Text":"but this is equal to sine of Alpha."},{"Start":"13:25.345 ","End":"13:31.315","Text":"Sine of 2 Pi minus Alpha is the same as negative sine of Alpha."},{"Start":"13:31.315 ","End":"13:39.445","Text":"Then we\u0027re subtracting negative sine of 3 Pi over 2,"},{"Start":"13:39.445 ","End":"13:43.700","Text":"which is equal to 1."},{"Start":"13:45.960 ","End":"13:52.180","Text":"We get that negative sine of 3 Pi over 2 is equal to 1,"},{"Start":"13:52.180 ","End":"13:54.460","Text":"but then we have to add in a minus because it\u0027s"},{"Start":"13:54.460 ","End":"13:57.505","Text":"the bottom bound and we always subtract that."},{"Start":"13:57.505 ","End":"13:59.710","Text":"That is for Lambda 1."},{"Start":"13:59.710 ","End":"14:03.220","Text":"Now let\u0027s see for Lambda 2."},{"Start":"14:03.220 ","End":"14:10.135","Text":"We have Lambda 2 multiplied by negative sine of 3 Pi over 2,"},{"Start":"14:10.135 ","End":"14:13.250","Text":"that is equal to 1."},{"Start":"14:14.070 ","End":"14:19.225","Text":"Then we\u0027re subtracting negative sine of Pi over 2,"},{"Start":"14:19.225 ","End":"14:20.643","Text":"which is negative 1,"},{"Start":"14:20.643 ","End":"14:24.740","Text":"so we\u0027re subtracting 1."},{"Start":"14:26.640 ","End":"14:30.985","Text":"Now let\u0027s just write this neatly."},{"Start":"14:30.985 ","End":"14:39.730","Text":"We\u0027ll get that E_x is equal to k divided by R multiplied by Lambda 1."},{"Start":"14:39.730 ","End":"14:45.445","Text":"Then here we have sine Alpha minus minus sine of Alpha,"},{"Start":"14:45.445 ","End":"14:48.825","Text":"so we have 2 sine of Alpha,"},{"Start":"14:48.825 ","End":"14:51.420","Text":"and then we also have negative 1,"},{"Start":"14:51.420 ","End":"14:55.270","Text":"negative 1, so negative 2."},{"Start":"14:55.270 ","End":"14:58.930","Text":"Then plus Lambda 2,"},{"Start":"14:58.930 ","End":"15:02.320","Text":"and then we have 1 minus minus 1, which is 2."},{"Start":"15:02.320 ","End":"15:06.410","Text":"Plus 2 times Lambda 2."},{"Start":"15:07.950 ","End":"15:13.975","Text":"This is the final answer for the x-component of our electric field."},{"Start":"15:13.975 ","End":"15:16.450","Text":"Now for the y-component,"},{"Start":"15:16.450 ","End":"15:19.645","Text":"let\u0027s just scroll up over here."},{"Start":"15:19.645 ","End":"15:24.115","Text":"We saw that dE_x was equal to this."},{"Start":"15:24.115 ","End":"15:32.105","Text":"Our dE_y is simply going to be equal to the size of our small electric field."},{"Start":"15:32.105 ","End":"15:37.220","Text":"Then it\u0027s going to be multiplied by instead of negative cosine of Theta,"},{"Start":"15:37.220 ","End":"15:41.565","Text":"it will be negative sine of Theta."},{"Start":"15:41.565 ","End":"15:45.410","Text":"This is for dE_y, and then all we\u0027ll"},{"Start":"15:45.410 ","End":"15:49.310","Text":"do is the exact same thing that we did for the x-component."},{"Start":"15:49.310 ","End":"15:54.680","Text":"We\u0027ll integrate along, taking into account all of the bounds."},{"Start":"15:54.680 ","End":"16:01.025","Text":"But what we\u0027ll see at the end is that our electric field in the y-direction cancels out,"},{"Start":"16:01.025 ","End":"16:02.810","Text":"it\u0027s equal to 0."},{"Start":"16:02.810 ","End":"16:06.230","Text":"This is because all along the y-axis,"},{"Start":"16:06.230 ","End":"16:11.590","Text":"there\u0027s an equal and opposite electric force working on"},{"Start":"16:11.590 ","End":"16:17.315","Text":"it or electric field rather nor the electric force,"},{"Start":"16:17.315 ","End":"16:19.520","Text":"an equal and opposite electric field."},{"Start":"16:19.520 ","End":"16:21.995","Text":"That is why in the y direction,"},{"Start":"16:21.995 ","End":"16:25.070","Text":"the electric field will cancel out."},{"Start":"16:25.070 ","End":"16:27.215","Text":"But if you want to check yourself,"},{"Start":"16:27.215 ","End":"16:33.215","Text":"then feel free to do this calculation and integrate along the y-axis."},{"Start":"16:33.215 ","End":"16:36.090","Text":"That\u0027s the end of this lesson."}],"ID":21370},{"Watched":false,"Name":"Exercise 8","Duration":"14m 43s","ChapterTopicVideoID":21432,"CourseChapterTopicPlaylistID":99473,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.740","Text":"Hello. In this lesson,"},{"Start":"00:01.740 ","End":"00:04.185","Text":"we\u0027re going to be answering the following question."},{"Start":"00:04.185 ","End":"00:08.820","Text":"Two rods, each of length L and charge density Lambda,"},{"Start":"00:08.820 ","End":"00:12.900","Text":"are placed along the x-axis as shown in the diagram."},{"Start":"00:12.900 ","End":"00:17.830","Text":"What force are the rods applying to one another?"},{"Start":"00:19.640 ","End":"00:24.630","Text":"We can see that every point on this rod, let\u0027s call this rod,"},{"Start":"00:24.630 ","End":"00:29.580","Text":"rod number 1 has some charge,"},{"Start":"00:29.580 ","End":"00:33.885","Text":"dq, and every point on this rod,"},{"Start":"00:33.885 ","End":"00:40.170","Text":"let\u0027s call this rod number 2 also has some charge, dq."},{"Start":"00:40.170 ","End":"00:45.680","Text":"What we can see is that every little piece on this rod is going"},{"Start":"00:45.680 ","End":"00:51.460","Text":"to be applying a force on every little piece on this rod."},{"Start":"00:51.460 ","End":"01:00.630","Text":"We\u0027re going to have to sum up the forces on both rods at every single point."},{"Start":"01:02.450 ","End":"01:08.515","Text":"This over here, let\u0027s say that this is located at position"},{"Start":"01:08.515 ","End":"01:17.950","Text":"x′ is applying an electric field which has felt by this point over here,"},{"Start":"01:17.950 ","End":"01:21.740","Text":"so here we are feeling an E-field."},{"Start":"01:21.740 ","End":"01:27.455","Text":"Let\u0027s call this point over here that it\u0027s position along the x-axis as x,"},{"Start":"01:27.455 ","End":"01:30.870","Text":"so we have x and x tag."},{"Start":"01:31.370 ","End":"01:37.790","Text":"Now what I want to do is I want to find what E-field this point on the rod"},{"Start":"01:37.790 ","End":"01:44.135","Text":"over here is inducing over here at this point on the rod."},{"Start":"01:44.135 ","End":"01:49.530","Text":"This is a very small E-field because it comes just from 1 point."},{"Start":"01:49.530 ","End":"01:52.630","Text":"We\u0027re going to call this dE."},{"Start":"01:55.070 ","End":"01:57.450","Text":"Let\u0027s start calculating,"},{"Start":"01:57.450 ","End":"02:01.620","Text":"so we have our dE and first of all,"},{"Start":"02:01.620 ","End":"02:06.795","Text":"we can see that the E-field will be in the x-direction so I\u0027ll write that out soon."},{"Start":"02:06.795 ","End":"02:10.620","Text":"We\u0027re considering this as a point charge,"},{"Start":"02:10.620 ","End":"02:15.520","Text":"so from Coulomb\u0027s law the electric field due to a point charge is equal"},{"Start":"02:15.520 ","End":"02:25.690","Text":"to kdq divided by r^2 and of course it\u0027s in the x-direction."},{"Start":"02:26.240 ","End":"02:28.980","Text":"Let\u0027s just call this over here,"},{"Start":"02:28.980 ","End":"02:39.010","Text":"dq tag because dq tag is the charge at point x tag."},{"Start":"02:39.650 ","End":"02:43.845","Text":"What is this dq tag equal to?"},{"Start":"02:43.845 ","End":"02:50.120","Text":"So dq tag is equal to the charge density per unit length,"},{"Start":"02:50.120 ","End":"02:54.635","Text":"which is Lambda multiplied by this small length dl,"},{"Start":"02:54.635 ","End":"02:56.995","Text":"or let\u0027s call it dl tag."},{"Start":"02:56.995 ","End":"02:58.995","Text":"What is this dl tag?"},{"Start":"02:58.995 ","End":"03:01.800","Text":"Because it\u0027s in the x-direction,"},{"Start":"03:01.800 ","End":"03:07.210","Text":"we can just write this as Lambda dx tag."},{"Start":"03:08.120 ","End":"03:18.810","Text":"So dx tag is the length of this point over here on the x tag region,"},{"Start":"03:18.810 ","End":"03:22.820","Text":"and dx will be the length of this point over here,"},{"Start":"03:22.820 ","End":"03:25.830","Text":"at this point x."},{"Start":"03:26.920 ","End":"03:32.935","Text":"What is r? r is the distance from over here,"},{"Start":"03:32.935 ","End":"03:37.700","Text":"from a point x tag until a point x."},{"Start":"03:37.700 ","End":"03:43.979","Text":"Enjoying that length right now and it goes like this,"},{"Start":"03:43.979 ","End":"03:45.255","Text":"it\u0027s also a vector,"},{"Start":"03:45.255 ","End":"03:48.005","Text":"so this is our r vector."},{"Start":"03:48.005 ","End":"03:51.160","Text":"Now, if we want to find the size of r,"},{"Start":"03:51.160 ","End":"03:56.455","Text":"it\u0027s simply equal to this length from the origin up until here,"},{"Start":"03:56.455 ","End":"04:03.420","Text":"which is x minus the origin up until here, which is x tag."},{"Start":"04:03.420 ","End":"04:08.275","Text":"This whole length is x minus x tag,"},{"Start":"04:08.275 ","End":"04:11.330","Text":"and then we\u0027re left with this length over here."},{"Start":"04:11.330 ","End":"04:16.745","Text":"Now I have my dq tag and my r vector."},{"Start":"04:16.745 ","End":"04:21.395","Text":"I can now plug this in to this equation over here,"},{"Start":"04:21.395 ","End":"04:29.825","Text":"so I\u0027ll get that dE is equal to k multiplied by dq tag,"},{"Start":"04:29.825 ","End":"04:35.150","Text":"which is Lambda dx tag and all of this is in the x-direction"},{"Start":"04:35.150 ","End":"04:42.165","Text":"divided by r which is x minus x tag, and it\u0027s r^2."},{"Start":"04:42.165 ","End":"04:53.300","Text":"Then if I want to find the total electric field due to this rod at this point,"},{"Start":"04:53.300 ","End":"04:55.910","Text":"so right now we\u0027re ignoring the fact that over here we also have a charge,"},{"Start":"04:55.910 ","End":"04:58.820","Text":"we\u0027re just finding the electric field at"},{"Start":"04:58.820 ","End":"05:02.870","Text":"this point due to all the point charges on this rod over here."},{"Start":"05:02.870 ","End":"05:05.420","Text":"What we\u0027re going to do is we\u0027re just going to"},{"Start":"05:05.420 ","End":"05:11.175","Text":"integrate for d along all of the points on this rod."},{"Start":"05:11.175 ","End":"05:15.800","Text":"What\u0027s important to note over here in our integral is that"},{"Start":"05:15.800 ","End":"05:21.380","Text":"our point x over here during this integral is a constant."},{"Start":"05:21.380 ","End":"05:26.860","Text":"First of all, from the integral we can see that we\u0027re integrating along dx tag,"},{"Start":"05:26.860 ","End":"05:31.354","Text":"so x tag is what\u0027s changing and x is a constant."},{"Start":"05:31.354 ","End":"05:33.155","Text":"But also, as we just said,"},{"Start":"05:33.155 ","End":"05:36.560","Text":"we\u0027re finding the electric field at this point x,"},{"Start":"05:36.560 ","End":"05:40.400","Text":"which means intuitively that this x point over here for"},{"Start":"05:40.400 ","End":"05:44.510","Text":"now has to be a constant and what we\u0027re doing"},{"Start":"05:44.510 ","End":"05:48.500","Text":"is we\u0027re trying to find the electric field"},{"Start":"05:48.500 ","End":"05:52.655","Text":"at this point due to a point charge over here,"},{"Start":"05:52.655 ","End":"05:54.290","Text":"and here, and here, and here,"},{"Start":"05:54.290 ","End":"05:57.665","Text":"and so on and so forth until the end of this rod."},{"Start":"05:57.665 ","End":"06:00.620","Text":"What we can see is that our x tag is moving."},{"Start":"06:00.620 ","End":"06:03.905","Text":"First our x tag will represent a point right over here at"},{"Start":"06:03.905 ","End":"06:07.580","Text":"0 and then it will move along the rod until we get"},{"Start":"06:07.580 ","End":"06:11.967","Text":"to this point over here L. Our"},{"Start":"06:11.967 ","End":"06:19.735","Text":"bound because we\u0027re integrating along this rod is from 0 up until L,"},{"Start":"06:19.735 ","End":"06:21.285","Text":"so those are bounds."},{"Start":"06:21.285 ","End":"06:22.620","Text":"Now let\u0027s integrate."},{"Start":"06:22.620 ","End":"06:26.690","Text":"This is an easy integral k and Lambda are constants,"},{"Start":"06:26.690 ","End":"06:30.805","Text":"so we can write them over here on the side."},{"Start":"06:30.805 ","End":"06:39.814","Text":"Then we have dx, sorry this is meant to be squared, dx divided by x minus x′^2."},{"Start":"06:39.814 ","End":"06:49.545","Text":"If we say that t is equal to x minus x tag,"},{"Start":"06:49.545 ","End":"06:52.260","Text":"we didn\u0027t have to put in."},{"Start":"06:52.260 ","End":"06:59.820","Text":"Therefore, dt= -dx tag."},{"Start":"06:59.820 ","End":"07:09.310","Text":"Therefore,"},{"Start":"07:09.310 ","End":"07:14.038","Text":"if we have dx tag"},{"Start":"07:14.038 ","End":"07:23.170","Text":"which is equals to simply negative dt divided by t^2."},{"Start":"07:23.170 ","End":"07:28.255","Text":"This we know is the same as having an integral"},{"Start":"07:28.255 ","End":"07:34.810","Text":"of t to the negative 2 dt and then the negative over here."},{"Start":"07:34.810 ","End":"07:40.870","Text":"Then that is going to be equal to negative t to the negative 1,"},{"Start":"07:40.870 ","End":"07:46.630","Text":"then we have to multiply by negative 1."},{"Start":"07:46.630 ","End":"07:51.530","Text":"That\u0027s just going to be equal to 1 divided by t,"},{"Start":"07:51.530 ","End":"07:57.655","Text":"where of course this negative 1 comes from the inner derivative."},{"Start":"07:57.655 ","End":"08:05.125","Text":"1 divided by t is simply 1 divided by x minus x tag."},{"Start":"08:05.125 ","End":"08:11.870","Text":"This is going to be multiplied by 1 divided by x minus x tag."},{"Start":"08:14.040 ","End":"08:18.685","Text":"Then of course we have to substitute in our bounds."},{"Start":"08:18.685 ","End":"08:25.210","Text":"That is going to be equal to k Lambda multiplied by 1 divided"},{"Start":"08:25.210 ","End":"08:32.200","Text":"by x minus L minus 1 divided by x minus 0,"},{"Start":"08:32.200 ","End":"08:34.730","Text":"so that\u0027s just x."},{"Start":"08:34.950 ","End":"08:44.810","Text":"This is the total electric field that is acting on this point over here due to this rod."},{"Start":"08:45.510 ","End":"08:50.710","Text":"Now our next step is to find force at this point."},{"Start":"08:50.710 ","End":"08:52.720","Text":"We found the electric field at this point,"},{"Start":"08:52.720 ","End":"08:57.410","Text":"and now we want to find the force."},{"Start":"08:58.050 ","End":"09:02.650","Text":"The force at this point is going to be"},{"Start":"09:02.650 ","End":"09:10.495","Text":"dF and this is equal to the charge at this point, dq."},{"Start":"09:10.495 ","End":"09:11.560","Text":"I\u0027m reminding you,"},{"Start":"09:11.560 ","End":"09:17.380","Text":"the charge at this point x tag over here was dq tag and the charge at this point x is"},{"Start":"09:17.380 ","End":"09:24.325","Text":"dq and then of course we multiply it by the field at that point x."},{"Start":"09:24.325 ","End":"09:31.330","Text":"What is dq? So dq is going to be equal to as we saw,"},{"Start":"09:31.330 ","End":"09:35.980","Text":"Lambda dl without a tag because we\u0027re speaking about this point over here,"},{"Start":"09:35.980 ","End":"09:43.130","Text":"so that\u0027s going to be Lambda dx because we\u0027re moving along the x-axis."},{"Start":"09:43.410 ","End":"09:48.085","Text":"Now we can plug in our dq into this equation."},{"Start":"09:48.085 ","End":"09:54.130","Text":"We\u0027ll get that dF is equal to dq,"},{"Start":"09:54.130 ","End":"09:58.420","Text":"which is equal to Lambda dx multiplied by"},{"Start":"09:58.420 ","End":"10:04.035","Text":"the electric field at this point x which we found over here is equal to k Lambda,"},{"Start":"10:04.035 ","End":"10:07.840","Text":"so we\u0027ll get something with Lambda^2 and then all of"},{"Start":"10:07.840 ","End":"10:12.070","Text":"this is multiplied by 1 divided by x minus l,"},{"Start":"10:12.070 ","End":"10:15.715","Text":"minus 1 divided by x."},{"Start":"10:15.715 ","End":"10:23.395","Text":"Now what we want to do is we want to find the total force on this entire rod over here."},{"Start":"10:23.395 ","End":"10:28.435","Text":"Not just on this point but on the whole rod over here."},{"Start":"10:28.435 ","End":"10:31.870","Text":"Now we can see that our x is a variable,"},{"Start":"10:31.870 ","End":"10:37.075","Text":"our x is moving between the point 2L to the point 3L."},{"Start":"10:37.075 ","End":"10:39.790","Text":"That we know intuitively and also we can see over"},{"Start":"10:39.790 ","End":"10:44.480","Text":"here we have dx which means that we\u0027re integrating along the x."},{"Start":"10:45.330 ","End":"10:54.130","Text":"Then we can say that the total force acting on the second rod due to the first rod is"},{"Start":"10:54.130 ","End":"11:03.170","Text":"the integral in dF and then our bounds are of course between 2L and 3L."},{"Start":"11:03.600 ","End":"11:08.589","Text":"Now let\u0027s do this integration."},{"Start":"11:08.589 ","End":"11:10.420","Text":"What we\u0027re going to get,"},{"Start":"11:10.420 ","End":"11:12.100","Text":"let\u0027s take out the constants."},{"Start":"11:12.100 ","End":"11:15.609","Text":"Our constants are k and Lambda times Lambda,"},{"Start":"11:15.609 ","End":"11:23.140","Text":"k Lambda^2, and then we\u0027re integrating along this dx."},{"Start":"11:23.140 ","End":"11:28.720","Text":"If we integrate 1 divided by x minus L,"},{"Start":"11:28.720 ","End":"11:35.530","Text":"we know that this is going to be equal to ln of x minus L and"},{"Start":"11:35.530 ","End":"11:43.585","Text":"then we have minus the integral of 1 divided by x is just ln of x."},{"Start":"11:43.585 ","End":"11:47.500","Text":"Then we\u0027re going to plug in our bounds,"},{"Start":"11:47.500 ","End":"11:52.880","Text":"our bounds are of course 2L until 3L."},{"Start":"11:53.400 ","End":"11:56.590","Text":"Before we substitute in our bounds,"},{"Start":"11:56.590 ","End":"11:59.965","Text":"we know that we can play a trick with the lns over here,"},{"Start":"11:59.965 ","End":"12:05.410","Text":"k Lambda^2 and another way of writing ln x minus L minus"},{"Start":"12:05.410 ","End":"12:11.010","Text":"ln of x is to simply write ln,"},{"Start":"12:11.010 ","End":"12:16.804","Text":"so x minus L divided by x,"},{"Start":"12:16.804 ","End":"12:23.360","Text":"and then we substitute in afterwards 2L and 3L."},{"Start":"12:23.730 ","End":"12:26.590","Text":"This is something that you can always use."},{"Start":"12:26.590 ","End":"12:34.570","Text":"Just remember the number in the numerator is always the one with a positive sign and"},{"Start":"12:34.570 ","End":"12:37.405","Text":"the number in the denominator"},{"Start":"12:37.405 ","End":"12:43.580","Text":"is the number in here where the coefficient of ln is negative."},{"Start":"12:43.890 ","End":"12:47.935","Text":"Now let\u0027s carry this onto here,"},{"Start":"12:47.935 ","End":"12:52.795","Text":"so we have k Lambda^2 and now let\u0027s substitute in."},{"Start":"12:52.795 ","End":"12:55.975","Text":"First we\u0027re just substituting in 3L,"},{"Start":"12:55.975 ","End":"13:01.420","Text":"so we\u0027re going to have ln of 3L minus L divided"},{"Start":"13:01.420 ","End":"13:08.230","Text":"by 3L and then minus and now we substituting in 2L,"},{"Start":"13:08.230 ","End":"13:16.010","Text":"so minus ln of 2L minus L divided by 2L."},{"Start":"13:18.540 ","End":"13:21.640","Text":"We have k Lambda^2."},{"Start":"13:21.640 ","End":"13:26.260","Text":"Then we have 3L minus L is 2L divided by 3L,"},{"Start":"13:26.260 ","End":"13:30.355","Text":"so we\u0027re going to have ln of 2 divided by"},{"Start":"13:30.355 ","End":"13:36.910","Text":"3 minus ln of 2L minus L is L divided by 2L,"},{"Start":"13:36.910 ","End":"13:39.445","Text":"that\u0027s going to be a 1/2,"},{"Start":"13:39.445 ","End":"13:41.005","Text":"the Ls cancel out."},{"Start":"13:41.005 ","End":"13:43.510","Text":"Then again we can do our trick."},{"Start":"13:43.510 ","End":"13:49.765","Text":"We\u0027ll have k Lambda^2 multiplied by ln of,"},{"Start":"13:49.765 ","End":"13:55.495","Text":"so here we\u0027ll have 2/3 divided by 1/2."},{"Start":"13:55.495 ","End":"13:59.020","Text":"Then we know how fractions work,"},{"Start":"13:59.020 ","End":"14:00.310","Text":"so at the end,"},{"Start":"14:00.310 ","End":"14:03.710","Text":"let\u0027s write it over here."},{"Start":"14:04.050 ","End":"14:11.260","Text":"We\u0027ll get that the force along the rod is equal to k Lambda^2"},{"Start":"14:11.260 ","End":"14:18.680","Text":"multiplied by ln 4/3."},{"Start":"14:19.020 ","End":"14:21.205","Text":"This is the final answer."},{"Start":"14:21.205 ","End":"14:24.795","Text":"Of course, it\u0027s always in the positive x direction."},{"Start":"14:24.795 ","End":"14:29.130","Text":"We saw that from the diagram as well."},{"Start":"14:29.130 ","End":"14:31.750","Text":"But I also could have written it in."},{"Start":"14:31.750 ","End":"14:37.400","Text":"You can just imagine that here we have it written in."},{"Start":"14:38.220 ","End":"14:43.040","Text":"That is the end of this lesson."}],"ID":22266},{"Watched":false,"Name":"Exercise 9","Duration":"13m 40s","ChapterTopicVideoID":21433,"CourseChapterTopicPlaylistID":99473,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.995","Text":"Hello. In this lesson,"},{"Start":"00:01.995 ","End":"00:04.830","Text":"we\u0027re going to be answering the following question."},{"Start":"00:04.830 ","End":"00:12.225","Text":"A solid full disk of radius R has a charge density per unit area of Sigma."},{"Start":"00:12.225 ","End":"00:17.670","Text":"Then a rod of length L is placed along the disk\u0027s axis"},{"Start":"00:17.670 ","End":"00:23.220","Text":"of symmetry at a height d above the disk center."},{"Start":"00:23.220 ","End":"00:29.625","Text":"The rod has a charge density per unit length of Lambda."},{"Start":"00:29.625 ","End":"00:34.725","Text":"What force does the rod exert on the disk?"},{"Start":"00:34.725 ","End":"00:40.115","Text":"First of all, we know from Newton\u0027s third law that the force"},{"Start":"00:40.115 ","End":"00:46.955","Text":"exerted by the rod on the disk is equal to the force exerted by the disk on the rod."},{"Start":"00:46.955 ","End":"00:51.635","Text":"The force exerted by rod on"},{"Start":"00:51.635 ","End":"00:58.895","Text":"disk is equal to the force exerted by disk on rod."},{"Start":"00:58.895 ","End":"01:02.880","Text":"This is Newton\u0027s third law."},{"Start":"01:03.310 ","End":"01:12.385","Text":"What we\u0027re going to do is we\u0027re going to find the force that the disk applies to the rod."},{"Start":"01:12.385 ","End":"01:17.900","Text":"First of all, what we\u0027re going to do is we\u0027re going to cut up the rod into lots of"},{"Start":"01:17.900 ","End":"01:23.355","Text":"little lengths where each length is a very small length dl,"},{"Start":"01:23.355 ","End":"01:29.049","Text":"and we can take it as that each length like this is just like a point."},{"Start":"01:29.049 ","End":"01:33.875","Text":"We\u0027re splitting up this rod into lots of little point charges,"},{"Start":"01:33.875 ","End":"01:36.560","Text":"and then what we\u0027re going to do is we\u0027re going to"},{"Start":"01:36.560 ","End":"01:39.845","Text":"use the equation that we already know for"},{"Start":"01:39.845 ","End":"01:48.450","Text":"the force that is applied by a disk onto its axis of symmetry."},{"Start":"01:49.520 ","End":"01:57.515","Text":"Here is an equation for the electric field of a solid disk along its axis of symmetry."},{"Start":"01:57.515 ","End":"02:02.645","Text":"It\u0027s equal to 2 Pi k Sigma,"},{"Start":"02:02.645 ","End":"02:04.609","Text":"which is the charge density,"},{"Start":"02:04.609 ","End":"02:06.320","Text":"multiplied by z,"},{"Start":"02:06.320 ","End":"02:07.865","Text":"the direction that it\u0027s going in."},{"Start":"02:07.865 ","End":"02:15.260","Text":"This is the z-axis multiplied by 1 divided by z minus 1"},{"Start":"02:15.260 ","End":"02:18.380","Text":"divided by the square root of the radius of"},{"Start":"02:18.380 ","End":"02:24.315","Text":"the disc squared plus the height in the z direction squared."},{"Start":"02:24.315 ","End":"02:29.900","Text":"Then what we\u0027re going to do is we\u0027re going to work out the E-field along every point,"},{"Start":"02:29.900 ","End":"02:35.920","Text":"so then we\u0027ll get E-field like so."},{"Start":"02:36.050 ","End":"02:41.459","Text":"Then we\u0027ll find the force at this point."},{"Start":"02:41.459 ","End":"02:43.984","Text":"This is the E-field,"},{"Start":"02:43.984 ","End":"02:47.330","Text":"but we\u0027ll find the force at this point due to knowing"},{"Start":"02:47.330 ","End":"02:52.325","Text":"the E-field at this exact point by using the equation,"},{"Start":"02:52.325 ","End":"03:01.020","Text":"that force is equal to the charge at this point multiplied by the E-field."},{"Start":"03:01.420 ","End":"03:09.275","Text":"Now of course, this charge is a tiny point charge that we\u0027ve split this rod into."},{"Start":"03:09.275 ","End":"03:12.050","Text":"It\u0027s going to be dq,"},{"Start":"03:12.050 ","End":"03:17.970","Text":"and then that means that this is going to be a tiny infinitesimal force,"},{"Start":"03:17.970 ","End":"03:21.590","Text":"so dF, and of course,"},{"Start":"03:21.590 ","End":"03:26.855","Text":"the direction of the force will be in the direction of the E-field."},{"Start":"03:26.855 ","End":"03:29.570","Text":"What we\u0027re going to do is we\u0027re going to find"},{"Start":"03:29.570 ","End":"03:33.410","Text":"the force at some arbitrary point along the rod,"},{"Start":"03:33.410 ","End":"03:39.110","Text":"and then we\u0027re going to integrate or sub up along the whole length of the rod,"},{"Start":"03:39.110 ","End":"03:45.185","Text":"and then we\u0027ll have the answer for the force that the disk applies to the rod,"},{"Start":"03:45.185 ","End":"03:49.970","Text":"which of course is equal to the force that the rod applies to the disk,"},{"Start":"03:49.970 ","End":"03:52.800","Text":"and that is what we\u0027re trying to find."},{"Start":"03:53.540 ","End":"03:57.590","Text":"The reason I can\u0027t just straight away find"},{"Start":"03:57.590 ","End":"04:02.645","Text":"the force by multiplying by this E-field is because the force,"},{"Start":"04:02.645 ","End":"04:07.010","Text":"or the E-field rather is dependent on the height that I am at z."},{"Start":"04:07.010 ","End":"04:10.640","Text":"The force being applied to"},{"Start":"04:10.640 ","End":"04:15.275","Text":"this point over here is different to the force being applied to this point over here,"},{"Start":"04:15.275 ","End":"04:16.760","Text":"because the z values,"},{"Start":"04:16.760 ","End":"04:20.840","Text":"the height along the z-axis of these 2 points is different and"},{"Start":"04:20.840 ","End":"04:26.320","Text":"therefore the force at these 2 points will be different."},{"Start":"04:27.140 ","End":"04:34.575","Text":"What is dq? What is this charge over here dq equal to?"},{"Start":"04:34.575 ","End":"04:40.580","Text":"So dq is equal to the charge density per unit length,"},{"Start":"04:40.580 ","End":"04:44.585","Text":"which has Lambda multiplied by this unit length,"},{"Start":"04:44.585 ","End":"04:48.040","Text":"which is some small length dl."},{"Start":"04:48.040 ","End":"04:53.370","Text":"What is dl because L is located where the rod is located"},{"Start":"04:53.370 ","End":"04:59.670","Text":"along the z-axis so we can say that this is equal to Lambda dz."},{"Start":"05:00.370 ","End":"05:07.085","Text":"Now we can say that dF is equal to dq,"},{"Start":"05:07.085 ","End":"05:12.035","Text":"which is Lambda dz multiplied by the E-field,"},{"Start":"05:12.035 ","End":"05:14.885","Text":"which is this over here."},{"Start":"05:14.885 ","End":"05:24.300","Text":"This is equal to 2 Pi k Sigma z multiplied by 1 divided by z"},{"Start":"05:24.300 ","End":"05:29.296","Text":"minus 1 divided by the square root of R^2 plus"},{"Start":"05:29.296 ","End":"05:38.020","Text":"z^2 and all of this is of course in the z direction."},{"Start":"05:38.020 ","End":"05:42.140","Text":"Then in order to find the total force,"},{"Start":"05:42.140 ","End":"05:45.415","Text":"this is what I\u0027m trying to calculate,"},{"Start":"05:45.415 ","End":"05:50.250","Text":"all I\u0027m going to do is integrate along everything."},{"Start":"05:51.290 ","End":"05:56.345","Text":"Now, let\u0027s just talk about our limits."},{"Start":"05:56.345 ","End":"06:00.905","Text":"We\u0027re going from this height over here,"},{"Start":"06:00.905 ","End":"06:03.080","Text":"which is where the rod begins,"},{"Start":"06:03.080 ","End":"06:07.055","Text":"which we\u0027re being told is that a height of d above its center."},{"Start":"06:07.055 ","End":"06:13.970","Text":"We\u0027re integrating from d and then up until this height over here."},{"Start":"06:13.970 ","End":"06:20.140","Text":"The whole length of the rod is L plus this height over here,"},{"Start":"06:20.140 ","End":"06:25.940","Text":"d. This total over here is just L plus d. So L plus d,"},{"Start":"06:25.940 ","End":"06:36.635","Text":"L plus d. Let\u0027s scroll down so we have a bit more space."},{"Start":"06:36.635 ","End":"06:44.220","Text":"Let\u0027s begin. F is going to be equal 2."},{"Start":"06:44.220 ","End":"06:47.190","Text":"We can take out our constants already,"},{"Start":"06:47.190 ","End":"06:54.250","Text":"so have 2 Pi k Sigma, Lambda."},{"Start":"06:54.800 ","End":"06:58.440","Text":"Then we\u0027re integrating along."},{"Start":"06:58.440 ","End":"07:00.435","Text":"We have this z over here."},{"Start":"07:00.435 ","End":"07:02.840","Text":"We\u0027re going to open up these brackets."},{"Start":"07:02.840 ","End":"07:04.850","Text":"We have z divided by z,"},{"Start":"07:04.850 ","End":"07:08.125","Text":"which is going to be equal to 1."},{"Start":"07:08.125 ","End":"07:13.490","Text":"Minus z divided by"},{"Start":"07:13.490 ","End":"07:18.950","Text":"the square root of R^2 plus z^2."},{"Start":"07:18.950 ","End":"07:23.610","Text":"The square root can be written like so,"},{"Start":"07:23.610 ","End":"07:27.195","Text":"and all of this is of course the z,"},{"Start":"07:27.195 ","End":"07:30.760","Text":"and this is of course all in the z direction."},{"Start":"07:31.130 ","End":"07:38.450","Text":"This is going from d to L Plus d. We can integrate along with this 1."},{"Start":"07:38.450 ","End":"07:39.755","Text":"It\u0027s going to be very easy."},{"Start":"07:39.755 ","End":"07:44.110","Text":"We have 2 Pi k Sigma,"},{"Start":"07:44.110 ","End":"07:49.670","Text":"Lambda in the z direction and all of this is multiplied by,"},{"Start":"07:49.670 ","End":"07:58.155","Text":"so here we have z along the bounds of d to L plus d,"},{"Start":"07:58.155 ","End":"08:02.675","Text":"and then minus, and now we have to integrate along this."},{"Start":"08:02.675 ","End":"08:06.920","Text":"Let\u0027s use integration by substitution."},{"Start":"08:06.920 ","End":"08:12.935","Text":"Let\u0027s say that t is equal to what we have here in the brackets."},{"Start":"08:12.935 ","End":"08:16.925","Text":"So I^2 plus z^2."},{"Start":"08:16.925 ","End":"08:24.030","Text":"Therefore, dt is going to be equal to 2z, dz."},{"Start":"08:24.430 ","End":"08:27.840","Text":"Here we have z, dz."},{"Start":"08:29.680 ","End":"08:33.010","Text":"We have half of our dt."},{"Start":"08:33.010 ","End":"08:36.135","Text":"We can write dt,"},{"Start":"08:36.135 ","End":"08:45.300","Text":"and then we have all of this is multiplied by 1/2 and then divided by t^(1/2)."},{"Start":"08:45.610 ","End":"08:48.860","Text":"We\u0027re going to integrate along with that,"},{"Start":"08:48.860 ","End":"08:52.415","Text":"and then we\u0027ll substitute back for our z values,"},{"Start":"08:52.415 ","End":"08:56.340","Text":"and then we\u0027ll plug in our limits."},{"Start":"08:57.250 ","End":"09:00.395","Text":"Let\u0027s carry this on."},{"Start":"09:00.395 ","End":"09:08.990","Text":"We\u0027re going to have 2 Pi k Sigma Lambda and the z direction multiplied by."},{"Start":"09:08.990 ","End":"09:15.525","Text":"Now, let\u0027s substitute z in with its limits, with its bounds."},{"Start":"09:15.525 ","End":"09:23.810","Text":"We have L plus d minus d. This is being multiplied by L and then"},{"Start":"09:23.810 ","End":"09:32.600","Text":"minus the integral of 1/2t 2 because this t^(1/2) is in the denominator."},{"Start":"09:32.600 ","End":"09:37.400","Text":"We can write it as t to the negative 1/2 dt."},{"Start":"09:37.400 ","End":"09:46.745","Text":"This is going to be equal to 2 Pi k Sigma Lambda in the z direction,"},{"Start":"09:46.745 ","End":"09:53.675","Text":"multiplied by L minus so then t^(1/2)."},{"Start":"09:53.675 ","End":"09:54.980","Text":"If we integrate that,"},{"Start":"09:54.980 ","End":"10:00.950","Text":"that\u0027s going to be t^(1/2) k not minus 1/2,"},{"Start":"10:00.950 ","End":"10:06.655","Text":"and then we\u0027re going to have to divide it by 1/2."},{"Start":"10:06.655 ","End":"10:13.415","Text":"What we\u0027re going to have is 1/2 multiplied by t^(1/2),"},{"Start":"10:13.415 ","End":"10:15.425","Text":"because we\u0027ve integrated this,"},{"Start":"10:15.425 ","End":"10:21.570","Text":"and then multiplied by 1 divided by 1/2."},{"Start":"10:21.710 ","End":"10:27.245","Text":"Then what we\u0027re just going to get is that these 2 cross out."},{"Start":"10:27.245 ","End":"10:32.030","Text":"In the end, what we\u0027re going to get is that this is equal to"},{"Start":"10:32.030 ","End":"10:36.430","Text":"2 Pi k Sigma Lambda and"},{"Start":"10:36.430 ","End":"10:44.070","Text":"the z direction multiplied by L minus t^(1/2)."},{"Start":"10:45.550 ","End":"10:51.785","Text":"Now let\u0027s substitute back in what our t is in terms of z."},{"Start":"10:51.785 ","End":"10:53.570","Text":"We\u0027re going to have that this is equal to"},{"Start":"10:53.570 ","End":"11:00.305","Text":"2 Pi k Sigma Lambda in the z direction multiplied by"},{"Start":"11:00.305 ","End":"11:10.030","Text":"L minus the square root of t. That is going to be R^2 plus z^2 is t,"},{"Start":"11:10.030 ","End":"11:11.855","Text":"and then we take the square root,"},{"Start":"11:11.855 ","End":"11:14.120","Text":"and then we can plug in the bounds,"},{"Start":"11:14.120 ","End":"11:21.695","Text":"which is from d to L plus d. Now let\u0027s plug in our bounds."},{"Start":"11:21.695 ","End":"11:23.705","Text":"This is our last stage."},{"Start":"11:23.705 ","End":"11:32.010","Text":"This is going to equal to 2 Pi k Sigma Lambda L minus."},{"Start":"11:32.010 ","End":"11:34.275","Text":"Now let\u0027s plug this in."},{"Start":"11:34.275 ","End":"11:39.095","Text":"We have R^2 plus z^2,"},{"Start":"11:39.095 ","End":"11:44.775","Text":"where z^2 is now L plus d^2,"},{"Start":"11:44.775 ","End":"11:53.150","Text":"and then we take the square root of all of this."},{"Start":"11:53.150 ","End":"11:58.850","Text":"Minus over here, we have R^2 plus z^2,"},{"Start":"11:58.850 ","End":"12:02.885","Text":"where z is now a d plus d^2,"},{"Start":"12:02.885 ","End":"12:06.840","Text":"and we take the square root of all of this."},{"Start":"12:09.110 ","End":"12:14.105","Text":"We\u0027ve just plugged in the bounds over here,"},{"Start":"12:14.105 ","End":"12:18.080","Text":"and then we can close these brackets over here."},{"Start":"12:18.080 ","End":"12:24.600","Text":"This closes with the L and all of this is in the z direction."},{"Start":"12:25.880 ","End":"12:31.650","Text":"This is the force of the disk on the rod."},{"Start":"12:31.650 ","End":"12:34.875","Text":"This isn\u0027t exactly our final answer."},{"Start":"12:34.875 ","End":"12:39.540","Text":"Because we\u0027re dealing with the directions over here,"},{"Start":"12:39.650 ","End":"12:49.955","Text":"so the force of the rod on the disk is going to be the exact magnitude,"},{"Start":"12:49.955 ","End":"12:51.755","Text":"but in the opposite direction,"},{"Start":"12:51.755 ","End":"12:54.480","Text":"in the negative z direction."},{"Start":"12:54.760 ","End":"12:58.205","Text":"Therefore, our answer,"},{"Start":"12:58.205 ","End":"13:03.935","Text":"F rod on disk because that is what we were being asked in the question is equal to"},{"Start":"13:03.935 ","End":"13:12.040","Text":"negative 2 Pi k Sigma Lambda L minus"},{"Start":"13:12.040 ","End":"13:20.985","Text":"R^2 plus L plus d^2,"},{"Start":"13:20.985 ","End":"13:29.645","Text":"and then the square root of this minus R^2 plus d^2 square root,"},{"Start":"13:29.645 ","End":"13:34.530","Text":"and then in the z direction."},{"Start":"13:34.530 ","End":"13:40.290","Text":"This is onset and this is the end of the lesson."}],"ID":22267},{"Watched":false,"Name":"Superposition Principle","Duration":"6m 13s","ChapterTopicVideoID":21434,"CourseChapterTopicPlaylistID":99473,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.295","Text":"Hello. In this lesson,"},{"Start":"00:02.295 ","End":"00:05.894","Text":"we\u0027re going to be speaking about the superposition principle."},{"Start":"00:05.894 ","End":"00:07.885","Text":"Now what is this principle,"},{"Start":"00:07.885 ","End":"00:09.705","Text":"like in the previous lesson,"},{"Start":"00:09.705 ","End":"00:11.820","Text":"where we had our q_ 1 and our q_ 2,"},{"Start":"00:11.820 ","End":"00:17.580","Text":"and we wanted to find out what force there was between both of the charges."},{"Start":"00:17.580 ","End":"00:20.160","Text":"We found out that that force was F_ 1,"},{"Start":"00:20.160 ","End":"00:22.590","Text":"given by Coulomb\u0027s Law."},{"Start":"00:22.590 ","End":"00:26.570","Text":"Now, here when we\u0027re dealing with the superposition principle,"},{"Start":"00:26.570 ","End":"00:30.455","Text":"is that we have also our q_ 1 acting on our q_ 2,"},{"Start":"00:30.455 ","End":"00:34.580","Text":"but we have at least 1 other charged particle."},{"Start":"00:34.580 ","End":"00:36.875","Text":"For instance here, a q_ 3,"},{"Start":"00:36.875 ","End":"00:39.555","Text":"also acting on our q_ 2."},{"Start":"00:39.555 ","End":"00:41.195","Text":"What we want to know,"},{"Start":"00:41.195 ","End":"00:48.120","Text":"is what is the total force being applied on our q_ 2?"},{"Start":"00:48.560 ","End":"00:53.510","Text":"We\u0027re not interested what\u0027s happening between our q_ 1 and our q_ 3,"},{"Start":"00:53.510 ","End":"00:56.495","Text":"or if there\u0027s other particles and what\u0027s happening between them."},{"Start":"00:56.495 ","End":"00:58.475","Text":"We\u0027re specifically interested,"},{"Start":"00:58.475 ","End":"01:01.835","Text":"in what is the total force being applied on q_ 2,"},{"Start":"01:01.835 ","End":"01:06.960","Text":"due to the interaction with q_ 1 and q_ 3."},{"Start":"01:06.960 ","End":"01:10.250","Text":"The superposition principle simply tells us,"},{"Start":"01:10.250 ","End":"01:11.810","Text":"that the total force,"},{"Start":"01:11.810 ","End":"01:20.135","Text":"our F_ T is going to be equal to the force F_ 1 from our first charged particle,"},{"Start":"01:20.135 ","End":"01:29.070","Text":"plus here it will be F_ 3 from our q_3,"},{"Start":"01:29.150 ","End":"01:36.240","Text":"plus which however many more charged particles we have in our system."},{"Start":"01:36.760 ","End":"01:40.535","Text":"Now, just as we learned in mechanics,"},{"Start":"01:40.535 ","End":"01:43.355","Text":"when we were summing up forces over there,"},{"Start":"01:43.355 ","End":"01:45.470","Text":"we have to do the same over here."},{"Start":"01:45.470 ","End":"01:47.780","Text":"That is written over here in this note."},{"Start":"01:47.780 ","End":"01:50.720","Text":"In order to add the forces or fields,"},{"Start":"01:50.720 ","End":"01:52.085","Text":"we\u0027ll see that in a second."},{"Start":"01:52.085 ","End":"01:55.000","Text":"We must first separate out the components."},{"Start":"01:55.000 ","End":"01:58.910","Text":"That means, when we figured out what our F_ 1 is,"},{"Start":"01:58.910 ","End":"02:02.090","Text":"our force between our q_ 1 and our q_ 2."},{"Start":"02:02.090 ","End":"02:05.540","Text":"Then we separate that out into the force in"},{"Start":"02:05.540 ","End":"02:09.320","Text":"the x-direction and the force in the y-direction."},{"Start":"02:09.320 ","End":"02:12.380","Text":"Then we do the same thing for our F_ 3."},{"Start":"02:12.380 ","End":"02:17.405","Text":"We find out what components of F_ 3 is in the x-direction,"},{"Start":"02:17.405 ","End":"02:19.340","Text":"and which is in the y-direction."},{"Start":"02:19.340 ","End":"02:24.295","Text":"Then we sum up our F_ 1 and F_ 3 x\u0027s with x\u0027s and y\u0027s with y\u0027s."},{"Start":"02:24.295 ","End":"02:29.000","Text":"Then we will get our F total."},{"Start":"02:29.330 ","End":"02:34.910","Text":"Now, let\u0027s look at our electric field superposition principle."},{"Start":"02:34.910 ","End":"02:37.190","Text":"Here we have the exact same diagram."},{"Start":"02:37.190 ","End":"02:41.060","Text":"Just remember with our electric field, because it\u0027s theoretical."},{"Start":"02:41.060 ","End":"02:47.015","Text":"It\u0027s predicting what force we\u0027re going to have if we put some charge over here."},{"Start":"02:47.015 ","End":"02:48.965","Text":"When we\u0027re dealing with our E field,"},{"Start":"02:48.965 ","End":"02:53.310","Text":"we don\u0027t actually have a charge over here, it\u0027s imaginary."},{"Start":"02:53.690 ","End":"02:58.460","Text":"What we do is we work out what our E field will be due to our Charge"},{"Start":"02:58.460 ","End":"03:03.365","Text":"1 on some test charged particle over here."},{"Start":"03:03.365 ","End":"03:07.335","Text":"Then we find out what our E field over here"},{"Start":"03:07.335 ","End":"03:12.055","Text":"is going to be due to a charged particle Number 3."},{"Start":"03:12.055 ","End":"03:18.335","Text":"Then we\u0027re not going to work out the E fields of the interaction between q_ 1 and q_ 3,"},{"Start":"03:18.335 ","End":"03:21.215","Text":"we simply want to know what the E field is,"},{"Start":"03:21.215 ","End":"03:23.300","Text":"just due to our q_ 3."},{"Start":"03:23.300 ","End":"03:26.900","Text":"If there wasn\u0027t a q_ 1 and the same with our E field,"},{"Start":"03:26.900 ","End":"03:29.635","Text":"if there was just q_ 1 and no q_ 3."},{"Start":"03:29.635 ","End":"03:35.900","Text":"Then in order to find the total electric field in our area and our vicinity,"},{"Start":"03:35.900 ","End":"03:39.755","Text":"then we again use the superposition principle exactly"},{"Start":"03:39.755 ","End":"03:44.115","Text":"like we did with our forces just this time with our E field."},{"Start":"03:44.115 ","End":"03:47.300","Text":"Our total E field is going to be equal to here,"},{"Start":"03:47.300 ","End":"03:51.230","Text":"specifically our E_ 1 plus our E_ 3."},{"Start":"03:51.230 ","End":"03:53.690","Text":"Then of course, if we have other charged particles,"},{"Start":"03:53.690 ","End":"03:55.850","Text":"we\u0027ll just add on E_ 4,"},{"Start":"03:55.850 ","End":"03:57.710","Text":"E_ 5, and so on and so forth."},{"Start":"03:57.710 ","End":"04:01.715","Text":"Again over here, we have to split up"},{"Start":"04:01.715 ","End":"04:06.440","Text":"our E fields for E_ 3 and E_ 1 into their x and y components."},{"Start":"04:06.440 ","End":"04:14.880","Text":"Then add in our x\u0027s and x\u0027s and y\u0027s and y\u0027s in order to find out our total E field."},{"Start":"04:16.520 ","End":"04:22.280","Text":"Now what I quickly want to demonstrate is why our force for instance,"},{"Start":"04:22.280 ","End":"04:25.755","Text":"and it\u0027s going to be the exact same idea for our field."},{"Start":"04:25.755 ","End":"04:27.440","Text":"Let\u0027s just speak about the force."},{"Start":"04:27.440 ","End":"04:30.410","Text":"Why we\u0027re just going to have a component,"},{"Start":"04:30.410 ","End":"04:33.805","Text":"for instance, in this example in the x-direction."},{"Start":"04:33.805 ","End":"04:35.795","Text":"If we look at our q_ 1,"},{"Start":"04:35.795 ","End":"04:38.719","Text":"it\u0027s applying a force on q_ 2 in this direction,"},{"Start":"04:38.719 ","End":"04:44.015","Text":"which means that it has its x-component going like so,"},{"Start":"04:44.015 ","End":"04:47.245","Text":"and its y component going like so."},{"Start":"04:47.245 ","End":"04:50.155","Text":"Now let\u0027s talk about our q_ 3."},{"Start":"04:50.155 ","End":"04:53.824","Text":"It has force F_ 3 going in this direction."},{"Start":"04:53.824 ","End":"04:59.205","Text":"Remember, the force is following on from the direction of our vector,"},{"Start":"04:59.205 ","End":"05:09.365","Text":"which means that it has a component going in the x-direction as well,"},{"Start":"05:09.365 ","End":"05:13.595","Text":"and a component going in the negative y-direction."},{"Start":"05:13.595 ","End":"05:17.630","Text":"Now assuming that the size of F_ 1 and F_ 3 are the same,"},{"Start":"05:17.630 ","End":"05:20.630","Text":"so assuming that our charges and,"},{"Start":"05:20.630 ","End":"05:27.240","Text":"let\u0027s assume that our y components are the same length."},{"Start":"05:27.240 ","End":"05:29.475","Text":"This isn\u0027t drawn to scale."},{"Start":"05:29.475 ","End":"05:34.165","Text":"We can see that we have our blue arrow going up."},{"Start":"05:34.165 ","End":"05:37.620","Text":"Let\u0027s say it\u0027s a bit higher, like so."},{"Start":"05:37.620 ","End":"05:42.845","Text":"We have our green arrow going down and we can see that they\u0027ll cancel each other out."},{"Start":"05:42.845 ","End":"05:46.235","Text":"However, in the positive x-direction,"},{"Start":"05:46.235 ","End":"05:49.475","Text":"we have our blue arrow and a green arrow."},{"Start":"05:49.475 ","End":"05:51.920","Text":"We can add up those 2 arrows,"},{"Start":"05:51.920 ","End":"05:56.970","Text":"because they\u0027re both going in the same direction. They add up."},{"Start":"05:56.970 ","End":"06:02.120","Text":"Then we can see that we\u0027ll only have a force in this x-direction."},{"Start":"06:02.120 ","End":"06:10.470","Text":"That\u0027s to give you a little bit of an intuition of why the y components here cancel out."},{"Start":"06:10.470 ","End":"06:13.450","Text":"That\u0027s the end of this lesson."}],"ID":22268},{"Watched":false,"Name":"Exercise 10","Duration":"14m 17s","ChapterTopicVideoID":21291,"CourseChapterTopicPlaylistID":99473,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.670","Text":"Hello. In this question,"},{"Start":"00:02.670 ","End":"00:05.550","Text":"we\u0027re being asked to calculate the force acting on"},{"Start":"00:05.550 ","End":"00:09.420","Text":"a charge located at the bottom right-hand corner of a square."},{"Start":"00:09.420 ","End":"00:12.000","Text":"q and a are given."},{"Start":"00:12.000 ","End":"00:16.845","Text":"The bottom right-hand corner of the square is our charge q_4."},{"Start":"00:16.845 ","End":"00:21.570","Text":"Now, we\u0027re being told that there are charges on"},{"Start":"00:21.570 ","End":"00:27.270","Text":"every single corner of the square and that the side of the square is equal to a."},{"Start":"00:27.270 ","End":"00:29.130","Text":"We have charge q_1,"},{"Start":"00:29.130 ","End":"00:31.470","Text":"which is equal to negative 2q,"},{"Start":"00:31.470 ","End":"00:33.390","Text":"q_2, which is equal to q,"},{"Start":"00:33.390 ","End":"00:35.370","Text":"q_3, which is equal to 2q,"},{"Start":"00:35.370 ","End":"00:40.160","Text":"and q_4, which is equal to negative q."},{"Start":"00:40.160 ","End":"00:47.200","Text":"Now we\u0027re trying to find the total force on this q_4 over here."},{"Start":"00:47.200 ","End":"00:48.980","Text":"How are we going to do this?"},{"Start":"00:48.980 ","End":"00:53.120","Text":"We\u0027re going to use the superposition principle in order to find out"},{"Start":"00:53.120 ","End":"00:57.940","Text":"the force which each charge is exerting on q_4."},{"Start":"00:57.940 ","End":"01:01.275","Text":"Let\u0027s begin by looking at our q_3."},{"Start":"01:01.275 ","End":"01:05.900","Text":"Now we can see that q_3 has some positive charge,"},{"Start":"01:05.900 ","End":"01:09.393","Text":"whereas q_4 has some negative charge,"},{"Start":"01:09.393 ","End":"01:12.560","Text":"which means that between these 2 charges,"},{"Start":"01:12.560 ","End":"01:15.955","Text":"there\u0027s going to be an attractive force."},{"Start":"01:15.955 ","End":"01:22.400","Text":"As we know, our force goes from negative to positive."},{"Start":"01:22.400 ","End":"01:25.520","Text":"We can draw an arrow over here,"},{"Start":"01:25.520 ","End":"01:29.980","Text":"which is, let\u0027s call it our F_3."},{"Start":"01:29.980 ","End":"01:37.790","Text":"It\u0027s a force which is being applied on the line joining up our q_4 and our q_3."},{"Start":"01:37.790 ","End":"01:43.200","Text":"It\u0027s in the direction from q_4 to q_3."},{"Start":"01:45.170 ","End":"01:50.195","Text":"Now what we have to do is we want to find out the force between these 2."},{"Start":"01:50.195 ","End":"01:56.329","Text":"First, let\u0027s decide which are positive x-direction, and y-direction."},{"Start":"01:56.329 ","End":"02:00.385","Text":"Let\u0027s say that this is our positive x direction and"},{"Start":"02:00.385 ","End":"02:06.100","Text":"that this is our positive y-direction."},{"Start":"02:06.140 ","End":"02:09.350","Text":"When we\u0027re trying to find the force,"},{"Start":"02:09.350 ","End":"02:11.470","Text":"our force F_3,"},{"Start":"02:11.470 ","End":"02:15.740","Text":"we know that we\u0027re going to have to be using Coulomb\u0027s law."},{"Start":"02:15.740 ","End":"02:20.600","Text":"We know that that\u0027s going to be equal to k multiplied"},{"Start":"02:20.600 ","End":"02:25.965","Text":"by our charge q_3 multiplied by our charge q_4."},{"Start":"02:25.965 ","End":"02:32.660","Text":"q_3, q_4 divided by the distance between the 2."},{"Start":"02:32.660 ","End":"02:38.600","Text":"We know that each side of the square is a distance a or a length a,"},{"Start":"02:38.600 ","End":"02:43.850","Text":"so divided by distance squared, so a squared."},{"Start":"02:43.850 ","End":"02:49.415","Text":"And then we can see that our force is being applied in this line,"},{"Start":"02:49.415 ","End":"02:54.810","Text":"which is going in the negative x-direction."},{"Start":"02:56.030 ","End":"02:59.820","Text":"This is in the negative x-direction."},{"Start":"02:59.820 ","End":"03:05.535","Text":"Now, let\u0027s put in all of our numbers."},{"Start":"03:05.535 ","End":"03:08.930","Text":"We\u0027re going to have negative k. Now,"},{"Start":"03:08.930 ","End":"03:11.320","Text":"q_3 is 2q,"},{"Start":"03:11.320 ","End":"03:13.785","Text":"q_4 is negative q,"},{"Start":"03:13.785 ","End":"03:16.820","Text":"but I\u0027m just going to write q and not a negative."},{"Start":"03:16.820 ","End":"03:19.880","Text":"Now the reason that I\u0027m not doing that is because I already"},{"Start":"03:19.880 ","End":"03:24.710","Text":"said that my force is being applied in the negative x-direction."},{"Start":"03:24.710 ","End":"03:30.715","Text":"Now, how do I know that my arrow is pointing in the negative x direction?"},{"Start":"03:30.715 ","End":"03:34.760","Text":"That\u0027s because right at the beginning before we began the calculation,"},{"Start":"03:34.760 ","End":"03:37.520","Text":"we already said that we saw that"},{"Start":"03:37.520 ","End":"03:41.150","Text":"q_3 has a positive charge and the q_4 has a negative charge."},{"Start":"03:41.150 ","End":"03:44.300","Text":"Which means that there\u0027s going to be an attractive force between the 2."},{"Start":"03:44.300 ","End":"03:48.835","Text":"We already said that we know that it\u0027s going to go from negative to positive."},{"Start":"03:48.835 ","End":"03:51.650","Text":"Then we already said that that is going to be here in"},{"Start":"03:51.650 ","End":"03:55.940","Text":"this specific diagram in the negative x direction."},{"Start":"03:55.940 ","End":"04:01.730","Text":"There\u0027s no need now to put in another minus over here,"},{"Start":"04:01.730 ","End":"04:08.110","Text":"because we already substituted that in just by drawing in our diagram."},{"Start":"04:08.110 ","End":"04:12.890","Text":"That\u0027s then divided by a squared and then the x-direction."},{"Start":"04:12.890 ","End":"04:17.434","Text":"In total, we\u0027re going to have negative 2kq"},{"Start":"04:17.434 ","End":"04:25.139","Text":"squared divided by a squared in the x direction."},{"Start":"04:25.960 ","End":"04:32.780","Text":"Now let\u0027s do the same for our force between q_4 and q_2."},{"Start":"04:32.780 ","End":"04:36.530","Text":"The first thing that we can see is that q_4 still has"},{"Start":"04:36.530 ","End":"04:42.240","Text":"some negative charge and that q_2 has a positive charge."},{"Start":"04:42.240 ","End":"04:44.120","Text":"As we saw before,"},{"Start":"04:44.120 ","End":"04:46.925","Text":"between a negative charge and a positive charge,"},{"Start":"04:46.925 ","End":"04:52.884","Text":"we\u0027re going to have an attractive force and it\u0027s going to go from negative to positive."},{"Start":"04:52.884 ","End":"04:55.275","Text":"Let\u0027s draw this arrow."},{"Start":"04:55.275 ","End":"05:03.050","Text":"This is going to be F_2 and it\u0027s going in this direction from negative to positive,"},{"Start":"05:03.050 ","End":"05:07.180","Text":"which as we can see, is the positive y direction."},{"Start":"05:07.180 ","End":"05:10.680","Text":"Now again, let\u0027s use Coulomb\u0027s law."},{"Start":"05:10.680 ","End":"05:13.950","Text":"We have F_2 is equal 2."},{"Start":"05:13.950 ","End":"05:18.260","Text":"Now, because we can see that our F_2 is going"},{"Start":"05:18.260 ","End":"05:21.860","Text":"from our negative charge to our positive charge,"},{"Start":"05:21.860 ","End":"05:24.785","Text":"we can see that specifically how we chose our axis,"},{"Start":"05:24.785 ","End":"05:27.439","Text":"is pointing in the positive y direction."},{"Start":"05:27.439 ","End":"05:30.635","Text":"This time, I\u0027m not going to put a minus over here."},{"Start":"05:30.635 ","End":"05:34.070","Text":"I\u0027m just going to write k multiplied by"},{"Start":"05:34.070 ","End":"05:41.950","Text":"q_2 multiplied by q_4 divided by the distance between the charges squared."},{"Start":"05:41.950 ","End":"05:47.045","Text":"Again, because we\u0027re dealing with a square shape and where each side is of length a,"},{"Start":"05:47.045 ","End":"05:49.220","Text":"so divided by a squared."},{"Start":"05:49.220 ","End":"05:52.040","Text":"This is in the y-direction,"},{"Start":"05:52.040 ","End":"05:55.085","Text":"specifically the positive y-direction."},{"Start":"05:55.085 ","End":"05:57.770","Text":"Now we can substitute in our numbers."},{"Start":"05:57.770 ","End":"06:02.390","Text":"We have k, then q_2 is equal to q,"},{"Start":"06:02.390 ","End":"06:05.195","Text":"q_4 is equal to negative q,"},{"Start":"06:05.195 ","End":"06:08.330","Text":"except we\u0027re just going to substitute in q because,"},{"Start":"06:08.330 ","End":"06:11.330","Text":"why aren\u0027t we putting in the negative here for the same reason?"},{"Start":"06:11.330 ","End":"06:15.380","Text":"Because we see that our force is going from a negative charge"},{"Start":"06:15.380 ","End":"06:19.840","Text":"to our positive charge and it\u0027s pointing specifically in the positive y direction."},{"Start":"06:19.840 ","End":"06:25.444","Text":"Then divided by a squared in the positive y direction."},{"Start":"06:25.444 ","End":"06:33.060","Text":"Now we have kq squared divided by a squared in the y-direction."},{"Start":"06:34.550 ","End":"06:40.400","Text":"Now, 1 of the last steps is that we have to"},{"Start":"06:40.400 ","End":"06:48.270","Text":"find what our force from Charge number 1 is going to be on our Charge number 4."},{"Start":"06:49.100 ","End":"06:51.470","Text":"Let\u0027s take a look. First of all,"},{"Start":"06:51.470 ","End":"06:57.715","Text":"we can see that q_1 has a negative charge and q_4 has a negative charge."},{"Start":"06:57.715 ","End":"07:01.550","Text":"We\u0027re going to have repulsive forces between the 2."},{"Start":"07:01.550 ","End":"07:02.779","Text":"But in the meantime,"},{"Start":"07:02.779 ","End":"07:05.011","Text":"let\u0027s not deal with that,"},{"Start":"07:05.011 ","End":"07:08.420","Text":"and what we really want to do right now, therefore,"},{"Start":"07:08.420 ","End":"07:10.580","Text":"so that we don\u0027t have to deal with our minuses and"},{"Start":"07:10.580 ","End":"07:13.844","Text":"our signs and our direction and which direction everything\u0027s pointing."},{"Start":"07:13.844 ","End":"07:18.120","Text":"Let\u0027s find the size of F_1."},{"Start":"07:18.120 ","End":"07:27.545","Text":"F_1 is going to be this force over here. Let\u0027s work it out."},{"Start":"07:27.545 ","End":"07:35.514","Text":"Now, we can see that our force is going to be applied along this line, the diagonal."},{"Start":"07:35.514 ","End":"07:41.725","Text":"Now we know that because this side is of length a and this side is of length a,"},{"Start":"07:41.725 ","End":"07:48.150","Text":"that means that the length of this horizontal is going to be root 2a."},{"Start":"07:49.700 ","End":"07:55.410","Text":"Now let\u0027s just write down the size of our F1 without dealing with the direction"},{"Start":"07:55.410 ","End":"08:00.410","Text":"that it\u0027s pointing be it in this downwards diagonal or an upwards diagonal."},{"Start":"08:00.410 ","End":"08:03.900","Text":"We know that it\u0027s going to be equal to k multiplied by"},{"Start":"08:03.900 ","End":"08:10.515","Text":"q_1 multiplied by q_4 divided by the distance between the 2 points,"},{"Start":"08:10.515 ","End":"08:14.930","Text":"which is root 2 multiplied by a squared."},{"Start":"08:14.930 ","End":"08:17.015","Text":"Now let\u0027s substitute in our numbers."},{"Start":"08:17.015 ","End":"08:20.810","Text":"It\u0027s going to be k multiplied by q_1."},{"Start":"08:20.810 ","End":"08:23.745","Text":"Again, we\u0027re not dealing with the minus sign yet."},{"Start":"08:23.745 ","End":"08:27.790","Text":"K multiplied by 2q multiplied by q_4,"},{"Start":"08:27.790 ","End":"08:32.665","Text":"which is q divided by 2a squared."},{"Start":"08:32.665 ","End":"08:40.220","Text":"That\u0027s going to be 2kq squared over 2a squared."},{"Start":"08:40.710 ","End":"08:47.375","Text":"Now what we want to do is we want to find not just the magnitude of this vector,"},{"Start":"08:47.375 ","End":"08:52.057","Text":"but also its direction."},{"Start":"08:52.057 ","End":"08:54.510","Text":"How are we going to write this vector?"},{"Start":"08:54.510 ","End":"08:57.595","Text":"We\u0027re going to raise it, the magnitude"},{"Start":"08:57.595 ","End":"09:01.205","Text":"multiplied by the different directions that it\u0027s pointing in."},{"Start":"09:01.205 ","End":"09:02.435","Text":"Let\u0027s write the magnitude."},{"Start":"09:02.435 ","End":"09:08.740","Text":"We have 2kq squared over 2a squared."},{"Start":"09:08.750 ","End":"09:10.945","Text":"You can cross these out,"},{"Start":"09:10.945 ","End":"09:13.190","Text":"to be honest, these 2 twos."},{"Start":"09:13.190 ","End":"09:19.364","Text":"Now, what I can see is that the direction that it\u0027s pointing."},{"Start":"09:19.364 ","End":"09:21.050","Text":"Let\u0027s draw this in blue."},{"Start":"09:21.050 ","End":"09:25.315","Text":"We can see that we have a negative y component is going in"},{"Start":"09:25.315 ","End":"09:30.370","Text":"this direction and a positive X component."},{"Start":"09:31.280 ","End":"09:37.330","Text":"What does that mean? That means that I have this magnitude multiplied"},{"Start":"09:37.330 ","End":"09:44.980","Text":"by the x component is going to be cosine of my angle Theta."},{"Start":"09:44.980 ","End":"09:46.745","Text":"Soon we\u0027ll see what Theta is."},{"Start":"09:46.745 ","End":"09:51.070","Text":"Then my y component is negative because we\u0027re going in"},{"Start":"09:51.070 ","End":"09:58.670","Text":"the negative y direction multiplied by sine of the angle Theta."},{"Start":"10:01.400 ","End":"10:06.910","Text":"Of course, don\u0027t forget to write in your x-hat and y-hat because the"},{"Start":"10:06.910 ","End":"10:09.250","Text":"cosine Theta\u0027s obviously in the x direction and"},{"Start":"10:09.250 ","End":"10:12.410","Text":"the negative sign Theta\u0027s obviously in the y-direction."},{"Start":"10:12.410 ","End":"10:15.625","Text":"Now, what is our angle Theta?"},{"Start":"10:15.625 ","End":"10:18.970","Text":"It\u0027s going to be this angle over here."},{"Start":"10:18.970 ","End":"10:24.524","Text":"As we can see, because we\u0027re dealing with half the angle of 1 of the squares corners,"},{"Start":"10:24.524 ","End":"10:27.738","Text":"so it\u0027s going to be 45 degrees,"},{"Start":"10:27.738 ","End":"10:29.800","Text":"this angle over here."},{"Start":"10:31.350 ","End":"10:34.095","Text":"What does this equal?"},{"Start":"10:34.095 ","End":"10:42.130","Text":"This is equal to kq squared over a squared multiplied by cosine of"},{"Start":"10:42.130 ","End":"10:51.100","Text":"45 degrees is equal to 1 over root 2 in the x-direction."},{"Start":"10:51.100 ","End":"10:54.229","Text":"Then sine of 45 degrees is also 1 over root 2,"},{"Start":"10:54.229 ","End":"11:02.710","Text":"so negative sine 45 degrees is negative 1 over root 2 in the y-direction."},{"Start":"11:02.710 ","End":"11:04.930","Text":"Then another way of writing this,"},{"Start":"11:04.930 ","End":"11:06.150","Text":"if we just want to shorten this,"},{"Start":"11:06.150 ","End":"11:11.030","Text":"we can write kq squared divided by a squared."},{"Start":"11:11.030 ","End":"11:18.195","Text":"Then we have 1 over root 2 negative 1 over root 2."},{"Start":"11:18.195 ","End":"11:24.260","Text":"Written like this means the exact same thing as this over here."},{"Start":"11:25.230 ","End":"11:29.143","Text":"Now, we have the final stop,"},{"Start":"11:29.143 ","End":"11:30.905","Text":"the final thing that we have to do."},{"Start":"11:30.905 ","End":"11:34.180","Text":"We\u0027re going to use the superposition principle and we\u0027re going to find"},{"Start":"11:34.180 ","End":"11:37.595","Text":"out the total force acting on q_4."},{"Start":"11:37.595 ","End":"11:42.490","Text":"Now, because we can see that we have different components,"},{"Start":"11:42.490 ","End":"11:45.610","Text":"we have forces acting only in the x-direction,"},{"Start":"11:45.610 ","End":"11:46.865","Text":"only in the y direction,"},{"Start":"11:46.865 ","End":"11:48.605","Text":"or a combination of the 2."},{"Start":"11:48.605 ","End":"11:52.570","Text":"We\u0027re going to find the total force is the sum"},{"Start":"11:52.570 ","End":"11:56.435","Text":"of all of the forces in each direction separately."},{"Start":"11:56.435 ","End":"11:58.955","Text":"Let\u0027s start with the x-direction."},{"Start":"11:58.955 ","End":"12:00.740","Text":"We have F_3,"},{"Start":"12:00.740 ","End":"12:03.945","Text":"which is acting in the x-direction,"},{"Start":"12:03.945 ","End":"12:08.245","Text":"and F_2 is acting in the y direction."},{"Start":"12:08.245 ","End":"12:10.930","Text":"Then we also have F_1,"},{"Start":"12:11.850 ","End":"12:14.495","Text":"which has an x-component,"},{"Start":"12:14.495 ","End":"12:16.405","Text":"which is what we\u0027re going to add."},{"Start":"12:16.405 ","End":"12:18.935","Text":"Let\u0027s see what this is equal to."},{"Start":"12:18.935 ","End":"12:27.855","Text":"F_3 is equal to negative 2kq squared divided by a squared."},{"Start":"12:27.855 ","End":"12:34.530","Text":"Then plus the x component of F_1 is going to be kq"},{"Start":"12:34.530 ","End":"12:42.185","Text":"squared over a squared multiplied by 1 divided by root 2."},{"Start":"12:42.185 ","End":"12:48.550","Text":"When you solve that, we\u0027re simply going to get kq squared divided by a"},{"Start":"12:48.550 ","End":"12:55.730","Text":"squared multiplied by 1 over root 2 minus 2."},{"Start":"12:58.380 ","End":"13:03.120","Text":"Now here, we don\u0027t have to write that this is in the x-direction because we\u0027re"},{"Start":"13:03.120 ","End":"13:07.910","Text":"summing all of the forces in the x-direction so of course,"},{"Start":"13:07.910 ","End":"13:11.450","Text":"this force is going to be acting in the x-direction."},{"Start":"13:11.450 ","End":"13:14.785","Text":"Now, we\u0027re going to do the same thing for"},{"Start":"13:14.785 ","End":"13:19.075","Text":"the sum of the forces acting on q_4 in the y direction."},{"Start":"13:19.075 ","End":"13:30.010","Text":"Here we know that we have F_2 acting in the y-direction plus our y component of F_1."},{"Start":"13:30.010 ","End":"13:36.620","Text":"Our F_2 is equal to kq squared divided by a squared."},{"Start":"13:36.620 ","End":"13:42.865","Text":"Then our y component of F_1 is going to be, so plus."},{"Start":"13:42.865 ","End":"13:48.070","Text":"Then we have kq squared divided by a"},{"Start":"13:48.070 ","End":"13:54.515","Text":"squared multiplied by negative 1 over root 2."},{"Start":"13:54.515 ","End":"13:56.935","Text":"Then, when we sum all of that,"},{"Start":"13:56.935 ","End":"14:01.120","Text":"we\u0027ll get kq squared divided by a"},{"Start":"14:01.120 ","End":"14:07.790","Text":"squared multiplied by 1 minus 1 over root 2."},{"Start":"14:07.790 ","End":"14:14.560","Text":"That\u0027s our answer for the sum of all of the forces on q_4 acting in the y-direction."},{"Start":"14:14.560 ","End":"14:17.950","Text":"That is the end of this lesson."}],"ID":21371},{"Watched":false,"Name":"Exercise 11","Duration":"15m 59s","ChapterTopicVideoID":21435,"CourseChapterTopicPlaylistID":99473,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.355","Text":"Hello. In this question,"},{"Start":"00:02.355 ","End":"00:08.475","Text":"we have 3 identical charges which are located at the corners of an equilateral triangle."},{"Start":"00:08.475 ","End":"00:14.755","Text":"The magnitude of each charge is going to be equal to 2 microcoulombs."},{"Start":"00:14.755 ","End":"00:19.445","Text":"The length of each edge of the triangle is equal to 4 meters."},{"Start":"00:19.445 ","End":"00:22.595","Text":"Find the force that each charge experiences"},{"Start":"00:22.595 ","End":"00:26.530","Text":"as a result of the other charges in the system."},{"Start":"00:26.530 ","End":"00:29.505","Text":"How am I going to do that?"},{"Start":"00:29.505 ","End":"00:33.590","Text":"Now first of all, I\u0027m going to use Coulomb\u0027s law."},{"Start":"00:33.590 ","End":"00:39.560","Text":"Now, Coulomb\u0027s law defines the force experienced between 2 charges."},{"Start":"00:39.560 ","End":"00:46.100","Text":"This is equal to kq_1 multiplied by"},{"Start":"00:46.100 ","End":"00:49.610","Text":"q_2 divided by the distance"},{"Start":"00:49.610 ","End":"00:54.244","Text":"between the 2 charges and because our force is a vector quantity,"},{"Start":"00:54.244 ","End":"00:59.180","Text":"it\u0027s going to be pointing in the direction of the line,"},{"Start":"00:59.180 ","End":"01:03.390","Text":"which is joining up our q_1 and our q_2."},{"Start":"01:03.500 ","End":"01:13.580","Text":"Now, our k is sometimes also given by 1 divided by 4 Pi Epsilon naught."},{"Start":"01:13.580 ","End":"01:16.100","Text":"Instead of writing k, we can also write this expression."},{"Start":"01:16.100 ","End":"01:24.525","Text":"It\u0027s the same thing, and this is equal to approximately 9 times 10^9,"},{"Start":"01:24.525 ","End":"01:35.580","Text":"and the units for our k value is in meters^2 divided by Coulombs^2 multiplied by newtons."},{"Start":"01:35.800 ","End":"01:37.925","Text":"Now, in our question,"},{"Start":"01:37.925 ","End":"01:42.305","Text":"we\u0027re being told that our q\u0027s are all identical."},{"Start":"01:42.305 ","End":"01:46.295","Text":"Let\u0027s say that this is q_1,"},{"Start":"01:46.295 ","End":"01:50.000","Text":"this is q_2 and this is q_3."},{"Start":"01:50.000 ","End":"01:53.515","Text":"Now let\u0027s see."},{"Start":"01:53.515 ","End":"01:56.840","Text":"q_1 is equal to q_2,"},{"Start":"01:56.840 ","End":"01:59.225","Text":"which is equal to q_3,"},{"Start":"01:59.225 ","End":"02:03.940","Text":"and this is equal to 2 microcoulombs."},{"Start":"02:03.940 ","End":"02:09.515","Text":"What\u0027s the problem with our charge being given in microcoulombs?"},{"Start":"02:09.515 ","End":"02:14.975","Text":"When we substitute in this number into our force equation,"},{"Start":"02:14.975 ","End":"02:20.120","Text":"we\u0027re going to have to start battling around with our units for k,"},{"Start":"02:20.120 ","End":"02:25.445","Text":"because our k is being given with a factor of coulombs and not microcoulomb,"},{"Start":"02:25.445 ","End":"02:28.370","Text":"then it\u0027s very easy to get confused with"},{"Start":"02:28.370 ","End":"02:32.555","Text":"the units and then maybe lose some marks in your test."},{"Start":"02:32.555 ","End":"02:39.545","Text":"What we\u0027re going to do is we\u0027re going to convert our microcoulombs into coulombs."},{"Start":"02:39.545 ","End":"02:44.600","Text":"Because then our charge is going to be given in coulombs,"},{"Start":"02:44.600 ","End":"02:49.760","Text":"and then our distance between the 2 charges is going to be given in meters."},{"Start":"02:49.760 ","End":"02:57.170","Text":"Therefore, our force is going to then be given in newtons because"},{"Start":"02:57.170 ","End":"02:59.900","Text":"everything\u0027s going to cancel out due to the units of our k"},{"Start":"02:59.900 ","End":"03:05.020","Text":"and force in newtons is exactly what we want to get."},{"Start":"03:05.020 ","End":"03:08.615","Text":"What we\u0027re going to do is we\u0027re going to convert"},{"Start":"03:08.615 ","End":"03:14.225","Text":"2 microcoulombs into some value of just coulombs."},{"Start":"03:14.225 ","End":"03:18.065","Text":"That\u0027s going to be 2 times, now microcoulombs,"},{"Start":"03:18.065 ","End":"03:24.815","Text":"micro is 10^negative 6 coulombs."},{"Start":"03:24.815 ","End":"03:30.410","Text":"Now let\u0027s find our forces which"},{"Start":"03:30.410 ","End":"03:36.325","Text":"are being applied between our charges and our charge number 3."},{"Start":"03:36.325 ","End":"03:42.815","Text":"The first thing that we have to do is we\u0027re going to have to divide our axes,"},{"Start":"03:42.815 ","End":"03:45.425","Text":"decide which direction is which."},{"Start":"03:45.425 ","End":"03:47.510","Text":"Let\u0027s say that this is"},{"Start":"03:47.510 ","End":"03:54.000","Text":"the positive y direction and that this is the positive x direction."},{"Start":"03:54.640 ","End":"04:00.350","Text":"The force between q_1 and q_3 is going"},{"Start":"04:00.350 ","End":"04:05.150","Text":"to be pointing in this direction along the same line joining them."},{"Start":"04:05.150 ","End":"04:07.715","Text":"Let\u0027s call this F_1."},{"Start":"04:07.715 ","End":"04:14.880","Text":"Then the force pointing between our q_2 and our q_3 is going to be in this direction,"},{"Start":"04:14.880 ","End":"04:17.980","Text":"and let\u0027s call this F_2."},{"Start":"04:18.380 ","End":"04:23.795","Text":"Let\u0027s begin. Let\u0027s try and find out what our F_1 is equal to."},{"Start":"04:23.795 ","End":"04:27.484","Text":"Our F_1, which is a vector quantity,"},{"Start":"04:27.484 ","End":"04:30.395","Text":"is going to be equal to our k,"},{"Start":"04:30.395 ","End":"04:38.475","Text":"which is 9 times 10^9 multiplied by q_1."},{"Start":"04:38.475 ","End":"04:48.275","Text":"Our q_1 is going to be 2 times 10^negative 6 multiplied by q_2."},{"Start":"04:48.275 ","End":"04:51.040","Text":"Here specifically it\u0027s our q_3,"},{"Start":"04:51.040 ","End":"04:56.200","Text":"that\u0027s also 2 times 10^negative 6,"},{"Start":"04:56.200 ","End":"05:00.565","Text":"divided by the distance between the 2."},{"Start":"05:00.565 ","End":"05:04.275","Text":"We know that the distance is 4 meters."},{"Start":"05:04.275 ","End":"05:07.665","Text":"The distance^2, 4^2."},{"Start":"05:07.665 ","End":"05:11.735","Text":"Then we can see that the units our m^2"},{"Start":"05:11.735 ","End":"05:15.890","Text":"at the top is going to divide by our m^2 down here."},{"Start":"05:15.890 ","End":"05:20.490","Text":"Our C^2 at the bottom is going to be canceled out by"},{"Start":"05:20.490 ","End":"05:22.650","Text":"our C^2 due to multiplying"},{"Start":"05:22.650 ","End":"05:27.950","Text":"the 2 charges and we\u0027re simply going to be left with the Newton."},{"Start":"05:28.880 ","End":"05:35.135","Text":"Now you can either work this out in your head or use a calculator."},{"Start":"05:35.135 ","End":"05:41.090","Text":"Our powers to 9 and negative 6 will cancel out some of them and we\u0027ll be left"},{"Start":"05:41.090 ","End":"05:48.740","Text":"with 9 divided by 4 multiplied by 10^negative 3 newtons,"},{"Start":"05:48.740 ","End":"05:56.690","Text":"which is also equal to 2.25 times 10^negative 3 newtons."},{"Start":"05:56.690 ","End":"05:59.135","Text":"Now we have our force in newtons."},{"Start":"05:59.135 ","End":"06:02.525","Text":"Now notice we haven\u0027t considered the direction."},{"Start":"06:02.525 ","End":"06:06.200","Text":"Let\u0027s say that this is the magnitude."},{"Start":"06:06.200 ","End":"06:11.510","Text":"However, if we want to write our force F_1 with the direction,"},{"Start":"06:11.510 ","End":"06:20.650","Text":"we can say that this is equal to 2.25 times 10^negative 3,"},{"Start":"06:20.710 ","End":"06:25.055","Text":"we can see it\u0027s going in the positive x direction,"},{"Start":"06:25.055 ","End":"06:28.410","Text":"so in the x direction."},{"Start":"06:29.540 ","End":"06:36.090","Text":"Now let\u0027s take a look at what our value for F_2 is going to be."},{"Start":"06:36.090 ","End":"06:38.905","Text":"Because we\u0027re dealing with an equilateral triangle,"},{"Start":"06:38.905 ","End":"06:43.240","Text":"that means that the distance between q_1 and q_3 is equal to"},{"Start":"06:43.240 ","End":"06:49.010","Text":"the distance between the q_2 and q_3 and all of our charges are also identical."},{"Start":"06:49.010 ","End":"06:58.135","Text":"That means that the magnitude of F_2 is equal to the magnitude of F_1,"},{"Start":"06:58.135 ","End":"07:07.070","Text":"which is equal to 2.25 times 10^negative 3 newtons."},{"Start":"07:07.070 ","End":"07:10.920","Text":"Now, in order to find our direction,"},{"Start":"07:10.920 ","End":"07:12.980","Text":"we\u0027re going to look back at our diagram."},{"Start":"07:12.980 ","End":"07:14.420","Text":"Now as we can see,"},{"Start":"07:14.420 ","End":"07:19.260","Text":"our F_2 is in some diagonal direction."},{"Start":"07:19.780 ","End":"07:24.830","Text":"Because we know that this is an equilateral triangle,"},{"Start":"07:24.830 ","End":"07:30.740","Text":"that means that every angle in the triangle is going to be equal to 60 degrees."},{"Start":"07:30.740 ","End":"07:35.130","Text":"Now because these are 2 opposite angles at a point,"},{"Start":"07:35.130 ","End":"07:40.115","Text":"and we know that this angle is going to be the same as this angle,"},{"Start":"07:40.115 ","End":"07:43.920","Text":"so this angle here is also 60 degrees."},{"Start":"07:45.390 ","End":"07:53.875","Text":"Now we know that using our SOHCAHTOA or trig identities that"},{"Start":"07:53.875 ","End":"08:03.925","Text":"our component in the x-direction for F_2 is therefore,"},{"Start":"08:03.925 ","End":"08:06.115","Text":"let\u0027s write it over here."},{"Start":"08:06.115 ","End":"08:09.415","Text":"Our F_2 vector is going to be equal to"},{"Start":"08:09.415 ","End":"08:16.915","Text":"the magnitude of F_2 multiplied by its x-direction."},{"Start":"08:16.915 ","End":"08:26.710","Text":"The x-direction is the magnitude multiplied by cosine of this angle,"},{"Start":"08:26.710 ","End":"08:29.030","Text":"so multiplied by cosine of 60."},{"Start":"08:29.640 ","End":"08:39.550","Text":"Then our F_2 vector in the y-direction is going to be the magnitude of the force,"},{"Start":"08:39.550 ","End":"08:45.070","Text":"F_2 multiplied by sine of the angle,"},{"Start":"08:45.070 ","End":"08:47.455","Text":"so here it\u0027s sine of 60."},{"Start":"08:47.455 ","End":"08:49.540","Text":"Let\u0027s substitute this in."},{"Start":"08:49.540 ","End":"08:56.020","Text":"This is going to be equal to 2.25 times 10 to the negative 3."},{"Start":"08:56.020 ","End":"08:59.800","Text":"This is the magnitude of F_2 and then cosine 60"},{"Start":"08:59.800 ","End":"09:06.475","Text":"is equal to a 1/2 and this is going to be in the x-direction."},{"Start":"09:06.475 ","End":"09:09.670","Text":"Now, in the y-direction we have the magnitude of F_2 which is"},{"Start":"09:09.670 ","End":"09:15.740","Text":"2.25 times 10 to the negative 3."},{"Start":"09:16.620 ","End":"09:21.490","Text":"This is going to be multiplied by root"},{"Start":"09:21.490 ","End":"09:28.150","Text":"3/2 which is sine 60 and of course our units are newtons."},{"Start":"09:28.150 ","End":"09:31.300","Text":"Now, something that\u0027s important to note is that we can see that"},{"Start":"09:31.300 ","End":"09:35.500","Text":"our y-component is going in the negative y-direction,"},{"Start":"09:35.500 ","End":"09:36.970","Text":"so let\u0027s just draw this."},{"Start":"09:36.970 ","End":"09:40.450","Text":"Along our diagonal we have the y-component"},{"Start":"09:40.450 ","End":"09:43.840","Text":"going like so which is the negative y-direction as we defined"},{"Start":"09:43.840 ","End":"09:46.810","Text":"it here and the x-component going like"},{"Start":"09:46.810 ","End":"09:50.830","Text":"so which is the positive x-direction as defined here."},{"Start":"09:50.830 ","End":"09:56.125","Text":"That means that we have to put a negative over here and a negative over here."},{"Start":"09:56.125 ","End":"10:00.280","Text":"If we put this into our calculator to find out"},{"Start":"10:00.280 ","End":"10:04.360","Text":"exactly what this is we\u0027ll get that our force for F_2 in"},{"Start":"10:04.360 ","End":"10:09.130","Text":"the x-direction is equal to 1.125 times"},{"Start":"10:09.130 ","End":"10:14.290","Text":"10 to the negative 3 newtons and that our force F_2 in"},{"Start":"10:14.290 ","End":"10:17.590","Text":"the y-direction is going to be equal to negative"},{"Start":"10:17.590 ","End":"10:26.150","Text":"1.949 times 10 to the negative 3 newtons."},{"Start":"10:27.600 ","End":"10:34.090","Text":"Now what I want to do is I want to use my superposition principle in order to"},{"Start":"10:34.090 ","End":"10:41.360","Text":"find the total or the sum of all of the forces on my q_3."},{"Start":"10:41.820 ","End":"10:45.430","Text":"Because I have my force is acting in"},{"Start":"10:45.430 ","End":"10:48.220","Text":"different directions I have x-components and y-components,"},{"Start":"10:48.220 ","End":"10:51.730","Text":"so I\u0027m going to have to split up the sum of all of the forces."},{"Start":"10:51.730 ","End":"10:55.945","Text":"I\u0027ll have the sum of all of the forces in my x-direction and I\u0027ll have"},{"Start":"10:55.945 ","End":"11:01.075","Text":"the sum of all of the forces on my q_3 in the y-direction."},{"Start":"11:01.075 ","End":"11:06.610","Text":"In the x-direction I\u0027m going to have F_1 plus"},{"Start":"11:06.610 ","End":"11:13.030","Text":"F_2 and then the y-direction we\u0027ll deal with in a second."},{"Start":"11:13.030 ","End":"11:20.710","Text":"F_1 is only acting in the x-direction which is equal to 2.25 times"},{"Start":"11:20.710 ","End":"11:29.020","Text":"10 to the negative 3 plus my F_2 in the x-direction which is this over here,"},{"Start":"11:29.020 ","End":"11:37.090","Text":"so plus 1.125 times 10 to the negative 3."},{"Start":"11:37.090 ","End":"11:47.350","Text":"This in total is going to be equal to 3.375 times 10 to the negative 3 newtons."},{"Start":"11:47.350 ","End":"11:53.420","Text":"That\u0027s the total force acting on my charge to q_3 in the x-direction."},{"Start":"11:53.940 ","End":"12:01.945","Text":"In the y-direction I only have my y-component of F_2 acting."},{"Start":"12:01.945 ","End":"12:12.925","Text":"That\u0027s going to be equal to negative 1.949 times 10 to the negative 3 newtons."},{"Start":"12:12.925 ","End":"12:21.450","Text":"Now, if we wanted to find the total magnitude of the force being experienced by our q_3,"},{"Start":"12:21.450 ","End":"12:25.350","Text":"so not taking into account in which direction the forces are pointing but"},{"Start":"12:25.350 ","End":"12:30.100","Text":"just what our q_3 feels with the size of the total force,"},{"Start":"12:30.100 ","End":"12:34.285","Text":"so that means that the magnitude of the force experienced"},{"Start":"12:34.285 ","End":"12:39.070","Text":"is going to be equal to the square root of"},{"Start":"12:39.070 ","End":"12:44.650","Text":"the sum of all of our forces in the x-direction squared"},{"Start":"12:44.650 ","End":"12:51.800","Text":"plus the sum of all of our forces in the y-direction squared."},{"Start":"12:53.040 ","End":"12:58.900","Text":"Then we\u0027re just going to sub in 3.375 times 10 to the 3 squared"},{"Start":"12:58.900 ","End":"13:04.180","Text":"plus negative 1.949 times 10 to the negative 3 squared."},{"Start":"13:04.180 ","End":"13:08.725","Text":"Then when you put this in your calculator we\u0027re going to get"},{"Start":"13:08.725 ","End":"13:14.830","Text":"3.897 times 10 to"},{"Start":"13:14.830 ","End":"13:19.432","Text":"the negative 3 newtons."},{"Start":"13:19.432 ","End":"13:24.910","Text":"That is the total force just taking into account the magnitude"},{"Start":"13:24.910 ","End":"13:32.840","Text":"experienced by our charge q_3 due to the forces being applied to it by q_2 and q_1."},{"Start":"13:33.180 ","End":"13:37.165","Text":"Now, if we want to see"},{"Start":"13:37.165 ","End":"13:43.000","Text":"what vector direction this magnitude of the total force is pointing in,"},{"Start":"13:43.000 ","End":"13:49.330","Text":"so let\u0027s draw that so we can see that due to the symmetry in the problem"},{"Start":"13:49.330 ","End":"13:52.675","Text":"that our q_1 and q_2 are equal and"},{"Start":"13:52.675 ","End":"13:56.500","Text":"the distance between each one of them to our q_3 is equal."},{"Start":"13:56.500 ","End":"14:04.630","Text":"We can see that from symmetry our vector direction is going to be going like so"},{"Start":"14:04.630 ","End":"14:07.660","Text":"such that each angle over here is going to be"},{"Start":"14:07.660 ","End":"14:13.525","Text":"30 degrees because it\u0027s bisecting the 60 degree angle over here."},{"Start":"14:13.525 ","End":"14:18.775","Text":"Now, if we weren\u0027t using symmetry and we wanted to work it out mathematically,"},{"Start":"14:18.775 ","End":"14:20.545","Text":"how would we do that?"},{"Start":"14:20.545 ","End":"14:24.535","Text":"To find the angle that our force is pointing in."},{"Start":"14:24.535 ","End":"14:30.985","Text":"We can say that tan of the angle is going to be equal to"},{"Start":"14:30.985 ","End":"14:35.020","Text":"the sum of all the forces in the y-direction divided"},{"Start":"14:35.020 ","End":"14:39.535","Text":"by the sum of all of the forces in the x-direction."},{"Start":"14:39.535 ","End":"14:46.900","Text":"Then when you work it out so you take this divided by this and then"},{"Start":"14:46.900 ","End":"14:54.980","Text":"you arctan both sides so then we\u0027ll get that our Theta is really equal to 30 degrees."},{"Start":"14:55.980 ","End":"14:59.710","Text":"Our force vector, the sum of all of the forces is"},{"Start":"14:59.710 ","End":"15:02.845","Text":"going to be going along this gray arrow."},{"Start":"15:02.845 ","End":"15:06.430","Text":"It\u0027s magnitude is Sigma F which we worked out"},{"Start":"15:06.430 ","End":"15:10.870","Text":"here and its angle with regards to the x-axis is just"},{"Start":"15:10.870 ","End":"15:14.770","Text":"going to be 30 degrees and here we can see that it\u0027s going to"},{"Start":"15:14.770 ","End":"15:19.510","Text":"be below the x-axis just by looking at the problem."},{"Start":"15:19.510 ","End":"15:23.980","Text":"Then in a very similar direction we don\u0027t have to work it out."},{"Start":"15:23.980 ","End":"15:25.420","Text":"But because the symmetry, again,"},{"Start":"15:25.420 ","End":"15:29.900","Text":"it\u0027s an equilateral triangle and all the charges are identical we"},{"Start":"15:29.900 ","End":"15:34.550","Text":"can say that we\u0027re going to have a Sigma F force pointing in"},{"Start":"15:34.550 ","End":"15:39.860","Text":"this direction for q_2 due to the forces applied by q_1 and q_3"},{"Start":"15:39.860 ","End":"15:46.180","Text":"and an exact mirror image of what\u0027s happening on this charge,"},{"Start":"15:46.180 ","End":"15:49.610","Text":"so we\u0027ll have Sigma F pointing in this direction for"},{"Start":"15:49.610 ","End":"15:56.045","Text":"q_1 due to the forces acting on it by q_2 and q_3."},{"Start":"15:56.045 ","End":"15:59.100","Text":"That\u0027s the end of this lesson."}],"ID":22269},{"Watched":false,"Name":"Exercise 12","Duration":"9m 50s","ChapterTopicVideoID":21436,"CourseChapterTopicPlaylistID":99473,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.475","Text":"Hello. In this question,"},{"Start":"00:02.475 ","End":"00:06.570","Text":"we\u0027re being told that 2 balls of mass m and"},{"Start":"00:06.570 ","End":"00:11.925","Text":"identical charge are hung from the ceiling via strings of length L,"},{"Start":"00:11.925 ","End":"00:16.710","Text":"so each string over here is of length L. We\u0027re told that the angle between the strings is"},{"Start":"00:16.710 ","End":"00:22.530","Text":"30 degrees and we\u0027re being asked what is the charge on each ball?"},{"Start":"00:22.530 ","End":"00:26.415","Text":"What we\u0027re trying to find is this."},{"Start":"00:26.415 ","End":"00:29.890","Text":"Each ball has this charge q."},{"Start":"00:30.530 ","End":"00:33.390","Text":"How are we going to do this?"},{"Start":"00:33.390 ","End":"00:36.825","Text":"Now, we\u0027re going to use the little trick from mechanics."},{"Start":"00:36.825 ","End":"00:41.950","Text":"We know that this system is now stationary."},{"Start":"00:42.290 ","End":"00:44.845","Text":"If the system is stationary,"},{"Start":"00:44.845 ","End":"00:50.885","Text":"that means that the net force on each ball is going to be 0."},{"Start":"00:50.885 ","End":"00:57.774","Text":"Before we begin, we are going to have to draw a 3 body diagram."},{"Start":"00:57.774 ","End":"01:00.700","Text":"The first force that we have is T,"},{"Start":"01:00.700 ","End":"01:03.070","Text":"which is the tension from the string."},{"Start":"01:03.070 ","End":"01:08.140","Text":"Then we have our weight mg pointing downwards and, of course,"},{"Start":"01:08.140 ","End":"01:11.155","Text":"we have our force that comes from Coulomb\u0027s law,"},{"Start":"01:11.155 ","End":"01:18.145","Text":"which is as a product from this charged ball q exerting a force on this charge ball q."},{"Start":"01:18.145 ","End":"01:22.650","Text":"Because they both have the same charge, so they\u0027re repelling."},{"Start":"01:22.650 ","End":"01:24.270","Text":"The force is repulsion,"},{"Start":"01:24.270 ","End":"01:27.790","Text":"so this force is going to go in this direction and then"},{"Start":"01:27.790 ","End":"01:32.990","Text":"an equal and opposite force will point in this direction for Coulomb\u0027s law."},{"Start":"01:34.440 ","End":"01:41.080","Text":"Before we begin, let\u0027s define our y-direction to be in"},{"Start":"01:41.080 ","End":"01:47.520","Text":"this direction and our x-direction to be in this direction."},{"Start":"01:47.520 ","End":"01:56.000","Text":"Now, we can say that this force F is going to be equal to."},{"Start":"01:56.000 ","End":"02:01.765","Text":"We know from Coulomb\u0027s law that it\u0027s going to be kq multiplied by q,"},{"Start":"02:01.765 ","End":"02:07.585","Text":"so q^2, divided by the distance between the 2 balls."},{"Start":"02:07.585 ","End":"02:10.890","Text":"Let\u0027s call this distance r,"},{"Start":"02:10.890 ","End":"02:16.865","Text":"so divided by r^2 and we can see that the force is pointing in this direction,"},{"Start":"02:16.865 ","End":"02:20.880","Text":"which is the positive x-direction."},{"Start":"02:20.990 ","End":"02:25.040","Text":"We don\u0027t exactly know what our value for r is,"},{"Start":"02:25.040 ","End":"02:28.700","Text":"what exactly is this distance and we can solve"},{"Start":"02:28.700 ","End":"02:32.795","Text":"this easily and this is by using the cosine law."},{"Start":"02:32.795 ","End":"02:35.540","Text":"The cosine law states that r^2,"},{"Start":"02:35.540 ","End":"02:37.550","Text":"this distance squared,"},{"Start":"02:37.550 ","End":"02:44.460","Text":"is going to be equal to the length of this side of the triangle squared,"},{"Start":"02:44.460 ","End":"02:49.670","Text":"so that\u0027s L^2 plus the length of this side of the triangle squared,"},{"Start":"02:49.670 ","End":"02:52.520","Text":"which is plus L^2."},{"Start":"02:52.520 ","End":"03:00.020","Text":"Both the strings are of length L. Then minus 2 times the length of this side,"},{"Start":"03:00.020 ","End":"03:01.846","Text":"times the length of this side,"},{"Start":"03:01.846 ","End":"03:10.475","Text":"so that\u0027s L^2 multiplied by cosine of the angle between these 2 sides,"},{"Start":"03:10.475 ","End":"03:13.570","Text":"which is 30 degrees."},{"Start":"03:13.570 ","End":"03:16.920","Text":"That is going to be equal to"},{"Start":"03:16.920 ","End":"03:26.890","Text":"2L^2 minus 2L^2 and then cosine 30 is equal to root 3/2."},{"Start":"03:26.890 ","End":"03:31.532","Text":"Then in total, this is going to be equal to L^2,"},{"Start":"03:31.532 ","End":"03:34.980","Text":"2 minus root 3."},{"Start":"03:36.010 ","End":"03:41.555","Text":"Now that we have our value for our distance between charges,"},{"Start":"03:41.555 ","End":"03:46.370","Text":"let\u0027s work out what are other forces acting on?"},{"Start":"03:46.370 ","End":"03:48.560","Text":"What we can do is,"},{"Start":"03:48.560 ","End":"03:50.425","Text":"let\u0027s write this over here."},{"Start":"03:50.425 ","End":"03:52.925","Text":"The sum of the forces,"},{"Start":"03:52.925 ","End":"03:55.340","Text":"let\u0027s begin on the y-axis."},{"Start":"03:55.340 ","End":"03:59.029","Text":"As we know,"},{"Start":"03:59.029 ","End":"04:03.017","Text":"our y-axis is going like so, so this is our y."},{"Start":"04:03.017 ","End":"04:06.080","Text":"Let\u0027s see what this is equal to."},{"Start":"04:06.080 ","End":"04:12.675","Text":"What we can do is we can draw a line going down here."},{"Start":"04:12.675 ","End":"04:15.935","Text":"This line is parallel to our y-axis,"},{"Start":"04:15.935 ","End":"04:21.620","Text":"which means that this angle over here is going to be 15 degrees."},{"Start":"04:21.620 ","End":"04:24.320","Text":"Now from alternate angles,"},{"Start":"04:24.320 ","End":"04:27.800","Text":"we know that if this angle is 15 degrees,"},{"Start":"04:27.800 ","End":"04:33.420","Text":"then we know that this angle is also going to be 15 degrees."},{"Start":"04:33.430 ","End":"04:39.575","Text":"That means that the sum of all of our forces in our y-direction are going to be equal 2."},{"Start":"04:39.575 ","End":"04:42.635","Text":"We have our tension in our positive y-direction"},{"Start":"04:42.635 ","End":"04:46.685","Text":"and our tension is going like so in this diagonal,"},{"Start":"04:46.685 ","End":"04:49.820","Text":"which means that its y component is going to be equal to"},{"Start":"04:49.820 ","End":"04:54.170","Text":"T multiplied by cosine of the angle,"},{"Start":"04:54.170 ","End":"04:57.368","Text":"which is 15 degrees."},{"Start":"04:57.368 ","End":"04:59.840","Text":"Then in the negative y-direction,"},{"Start":"04:59.840 ","End":"05:03.980","Text":"we have mg, so negative mg. As we said,"},{"Start":"05:03.980 ","End":"05:06.440","Text":"because our system is stationary,"},{"Start":"05:06.440 ","End":"05:09.485","Text":"that the sum of all of the forces or the net force"},{"Start":"05:09.485 ","End":"05:13.580","Text":"on the balls are going to be equal to 0."},{"Start":"05:13.580 ","End":"05:18.350","Text":"Now, let\u0027s do the same thing for our x components,"},{"Start":"05:18.350 ","End":"05:21.770","Text":"so the sum of all of the forces in the x-direction."},{"Start":"05:21.770 ","End":"05:23.870","Text":"Let\u0027s see, we have T,"},{"Start":"05:23.870 ","End":"05:27.685","Text":"which also has an x-component and we have our F over here."},{"Start":"05:27.685 ","End":"05:30.665","Text":"Our T, as we can see,"},{"Start":"05:30.665 ","End":"05:36.080","Text":"the x-component is going in the negative x-direction,"},{"Start":"05:36.080 ","End":"05:38.137","Text":"so let\u0027s put a negative over here."},{"Start":"05:38.137 ","End":"05:44.555","Text":"We can see that it\u0027s going to be equal to sine, the x-component"},{"Start":"05:44.555 ","End":"05:50.225","Text":"of our tension is going to be equal to T multiplied by sine of the angle,"},{"Start":"05:50.225 ","End":"05:53.830","Text":"which here is 15 degrees."},{"Start":"05:53.830 ","End":"05:58.490","Text":"Then in the positive x-direction, so plus,"},{"Start":"05:58.490 ","End":"06:04.280","Text":"we have our force from our electrostatic force,"},{"Start":"06:04.280 ","End":"06:06.755","Text":"which is given by our Coulomb\u0027s law,"},{"Start":"06:06.755 ","End":"06:08.075","Text":"which we worked out over here,"},{"Start":"06:08.075 ","End":"06:11.965","Text":"is kq^2 divided by r^2,"},{"Start":"06:11.965 ","End":"06:15.300","Text":"where r^2 is equal to this,"},{"Start":"06:15.300 ","End":"06:17.410","Text":"we already worked it out."},{"Start":"06:18.080 ","End":"06:23.710","Text":"Again, the net force is equal to 0."},{"Start":"06:24.410 ","End":"06:29.375","Text":"Now, all we want to do is we want to solve these 2 expressions."},{"Start":"06:29.375 ","End":"06:32.585","Text":"The first thing that we\u0027re going to do is we\u0027re going to isolate out"},{"Start":"06:32.585 ","End":"06:37.965","Text":"our T. We\u0027ll have the kq^2 divided by r^2,"},{"Start":"06:37.965 ","End":"06:40.635","Text":"we\u0027ll substitute in our r^2 later,"},{"Start":"06:40.635 ","End":"06:46.575","Text":"is equal to T multiplied by sine of 15 degrees."},{"Start":"06:46.575 ","End":"06:52.335","Text":"Therefore, we can say that our T=kq^2"},{"Start":"06:52.335 ","End":"06:59.290","Text":"divided by r^2 multiplied by sine of 15 degrees."},{"Start":"06:59.720 ","End":"07:05.960","Text":"Now we\u0027re going to substitute in this T into our first equation."},{"Start":"07:05.960 ","End":"07:08.735","Text":"We have T cosine of 15,"},{"Start":"07:08.735 ","End":"07:17.560","Text":"so therefore we have kq^2 divided by r^2 sine of 15 degrees,"},{"Start":"07:17.560 ","End":"07:27.050","Text":"multiplied by cosine of 15 degrees minus mg is equal to 0."},{"Start":"07:27.050 ","End":"07:31.205","Text":"Cosine over sine is equal to cot,"},{"Start":"07:31.205 ","End":"07:35.445","Text":"cotangent, where it\u0027s 1/tan."},{"Start":"07:35.445 ","End":"07:40.590","Text":"We have kq^2 divided by"},{"Start":"07:40.590 ","End":"07:47.400","Text":"r^2 multiplied by cot of 15 degrees,"},{"Start":"07:47.400 ","End":"07:50.220","Text":"which is equal to"},{"Start":"07:50.220 ","End":"07:59.270","Text":"mg. Now what we want to do is we want to go back to the question,"},{"Start":"07:59.270 ","End":"08:01.865","Text":"we\u0027re trying to find the charge on each ball."},{"Start":"08:01.865 ","End":"08:08.290","Text":"That means that we\u0027re going to want to isolate out our q^2 value over here."},{"Start":"08:08.290 ","End":"08:14.915","Text":"We\u0027re going to leave a q^2 over here and we\u0027re just going to divide by all of this."},{"Start":"08:14.915 ","End":"08:25.305","Text":"We\u0027ll have the q^2 is equal to mg divided by cot of 15 degrees,"},{"Start":"08:25.305 ","End":"08:27.975","Text":"multiplied by r^2,"},{"Start":"08:27.975 ","End":"08:32.820","Text":"and divided by k. Now,"},{"Start":"08:32.820 ","End":"08:36.560","Text":"1 over cot of 15 degrees is simply tan,"},{"Start":"08:36.560 ","End":"08:43.365","Text":"so we\u0027re going to have mgr^2 divided by k,"},{"Start":"08:43.365 ","End":"08:48.740","Text":"multiplied by tan of 15 degrees."},{"Start":"08:48.740 ","End":"08:52.820","Text":"Now, we can substitute in what our r^2 is."},{"Start":"08:52.820 ","End":"09:01.575","Text":"Get that this is mg divided by k multiplied by tan of 15 degrees,"},{"Start":"09:01.575 ","End":"09:04.275","Text":"multiplied by our r^2,"},{"Start":"09:04.275 ","End":"09:10.360","Text":"which is L^2 multiplied by 2 minus root 3."},{"Start":"09:10.700 ","End":"09:15.450","Text":"Now we know what our value for q is equal to,"},{"Start":"09:15.450 ","End":"09:20.450","Text":"it\u0027s simply going to be equal to the square root of all of this, so mg,"},{"Start":"09:20.450 ","End":"09:26.580","Text":"tan of 15 degrees divided by k,"},{"Start":"09:26.580 ","End":"09:32.310","Text":"L^2, 2 minus root 3."},{"Start":"09:32.310 ","End":"09:33.950","Text":"All of these variables,"},{"Start":"09:33.950 ","End":"09:37.070","Text":"we either know because they\u0027re"},{"Start":"09:37.070 ","End":"09:41.740","Text":"global variables or they\u0027ve been given to us in the question."},{"Start":"09:41.740 ","End":"09:47.555","Text":"Now we know what our charge is on each ball."},{"Start":"09:47.555 ","End":"09:50.220","Text":"That\u0027s the end of this question."}],"ID":22270},{"Watched":false,"Name":"Exercise 13","Duration":"16m 56s","ChapterTopicVideoID":21437,"CourseChapterTopicPlaylistID":99473,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.410 ","End":"00:03.080","Text":"Hello. In this question,"},{"Start":"00:03.080 ","End":"00:06.390","Text":"we\u0027re being told that electrons enter cathode ray tube at"},{"Start":"00:06.390 ","End":"00:10.110","Text":"a velocity of v. Inside the tube,"},{"Start":"00:10.110 ","End":"00:11.970","Text":"there is a constant electric field in"},{"Start":"00:11.970 ","End":"00:15.659","Text":"the 2 directions perpendicular to the velocity of the electrons."},{"Start":"00:15.659 ","End":"00:20.100","Text":"If we define the velocity of the electrons in the y direction,"},{"Start":"00:20.100 ","End":"00:26.325","Text":"that means that the 2 perpendicular electric fields will be in the x and z directions."},{"Start":"00:26.325 ","End":"00:29.460","Text":"The length of the tube is given and it\u0027s d."},{"Start":"00:29.460 ","End":"00:32.790","Text":"We\u0027re being asked to calculate the electrons,"},{"Start":"00:32.790 ","End":"00:34.815","Text":"point of impact on the screen,"},{"Start":"00:34.815 ","End":"00:38.505","Text":"as in the x-z coordinates on the screen,"},{"Start":"00:38.505 ","End":"00:42.830","Text":"located a distance l from the tubes opening."},{"Start":"00:42.830 ","End":"00:46.250","Text":"We\u0027re being told to assume that the length of"},{"Start":"00:46.250 ","End":"00:49.970","Text":"the cathode ray tube is significantly smaller than"},{"Start":"00:49.970 ","End":"00:53.330","Text":"the distance between the opening of the tube to"},{"Start":"00:53.330 ","End":"00:57.560","Text":"the screen and that the electron mass is given,"},{"Start":"00:57.560 ","End":"01:07.350","Text":"as well as our lengths d and L. The charge of the electron is also given."},{"Start":"01:07.660 ","End":"01:10.790","Text":"We\u0027re going to start by just looking at"},{"Start":"01:10.790 ","End":"01:16.320","Text":"the x axis and later we\u0027ll see that the z axis is identical."},{"Start":"01:17.000 ","End":"01:21.785","Text":"If there\u0027s an electric field acting in the x direction,"},{"Start":"01:21.785 ","End":"01:25.295","Text":"the electrons are also going to move in the x direction."},{"Start":"01:25.295 ","End":"01:29.945","Text":"If the electric field in the x direction is constant"},{"Start":"01:29.945 ","End":"01:35.270","Text":"so the force acting on the electron is also going to be constant."},{"Start":"01:35.270 ","End":"01:42.090","Text":"Then we can work out the motion of the electrons inside the tube."},{"Start":"01:42.760 ","End":"01:47.930","Text":"Then when the electron reaches the edge of the tube,"},{"Start":"01:47.930 ","End":"01:51.395","Text":"it\u0027s going to have some kind of velocity in"},{"Start":"01:51.395 ","End":"01:55.535","Text":"the x direction or an x component of the velocity and also"},{"Start":"01:55.535 ","End":"02:00.620","Text":"some distance x that it moved from the center of"},{"Start":"02:00.620 ","End":"02:06.760","Text":"the tube to its position exiting the tube so some kind of displacement."},{"Start":"02:06.760 ","End":"02:09.890","Text":"Then we\u0027ll work out how far it travels with"},{"Start":"02:09.890 ","End":"02:14.330","Text":"this x component of velocity until it hits the screen."},{"Start":"02:14.330 ","End":"02:19.135","Text":"That\u0027s how we\u0027ll find our x-coordinate on the screen."},{"Start":"02:19.135 ","End":"02:23.030","Text":"Of course, throughout this distance L,"},{"Start":"02:23.030 ","End":"02:26.990","Text":"from the edge of the tube until we reach the screen,"},{"Start":"02:26.990 ","End":"02:31.790","Text":"there is no electric field acting on the electron,"},{"Start":"02:31.790 ","End":"02:33.575","Text":"so we don\u0027t have to take that into account,"},{"Start":"02:33.575 ","End":"02:36.200","Text":"and we can ignore the force due to"},{"Start":"02:36.200 ","End":"02:41.640","Text":"gravity because it\u0027s so small that we don\u0027t have to take it into account."},{"Start":"02:42.110 ","End":"02:46.500","Text":"Now let\u0027s begin our calculations."},{"Start":"02:46.500 ","End":"02:53.125","Text":"Let\u0027s write in the tube the calculations that we\u0027ll need."},{"Start":"02:53.125 ","End":"02:57.865","Text":"First of all, we know that the force due to"},{"Start":"02:57.865 ","End":"03:07.790","Text":"a charged particle is equal to particles charge multiplied by the electric field."},{"Start":"03:07.790 ","End":"03:12.790","Text":"Here\u0027s specifically, we\u0027re working in the x direction."},{"Start":"03:12.790 ","End":"03:16.465","Text":"Therefore, in order to find the size,"},{"Start":"03:16.465 ","End":"03:21.135","Text":"we just want the magnitude of our force in the x direction,"},{"Start":"03:21.135 ","End":"03:25.010","Text":"this is going to be equal to the charge of"},{"Start":"03:25.010 ","End":"03:31.500","Text":"the electron multiplied by the electric field in the x-direction."},{"Start":"03:32.930 ","End":"03:40.160","Text":"Now, we also know that our force in the x direction has to be equal to,"},{"Start":"03:40.160 ","End":"03:45.245","Text":"from Newton, mass times the acceleration in the x direction."},{"Start":"03:45.245 ","End":"03:51.170","Text":"Therefore, we can get that the acceleration in the x direction is equal"},{"Start":"03:51.170 ","End":"03:57.110","Text":"to our force in the x direction divided by the mass of the electron,"},{"Start":"03:57.110 ","End":"04:01.820","Text":"which is equal to our force in the x direction we saw as the absolute value of"},{"Start":"04:01.820 ","End":"04:07.340","Text":"the electric charge multiplied by our electric field in the x direction,"},{"Start":"04:07.340 ","End":"04:11.250","Text":"and then divided by the mass of the electron."},{"Start":"04:12.260 ","End":"04:18.170","Text":"Now we can see that our acceleration in the x direction is a constant."},{"Start":"04:18.170 ","End":"04:20.015","Text":"Because we have constant acceleration,"},{"Start":"04:20.015 ","End":"04:23.390","Text":"we can use this equation from kinematics,"},{"Start":"04:23.390 ","End":"04:27.680","Text":"which is that our displacement in the x direction is equal"},{"Start":"04:27.680 ","End":"04:33.080","Text":"to our original position in the x direction plus"},{"Start":"04:33.080 ","End":"04:38.610","Text":"our original velocity in the x direction multiplied"},{"Start":"04:38.610 ","End":"04:44.920","Text":"by t plus 1/2 at^2."},{"Start":"04:44.920 ","End":"04:48.760","Text":"Now, because of how we can define our axis,"},{"Start":"04:48.760 ","End":"04:52.705","Text":"we can say that our original x position was at the origin."},{"Start":"04:52.705 ","End":"04:57.205","Text":"In that case, our x_0 is equal to 0."},{"Start":"04:57.205 ","End":"05:00.055","Text":"Also, we\u0027re being told that"},{"Start":"05:00.055 ","End":"05:04.795","Text":"the electrons enter the cathode ray tube and at a velocity of v,"},{"Start":"05:04.795 ","End":"05:08.800","Text":"which we\u0027re being told is only in the y direction because we"},{"Start":"05:08.800 ","End":"05:12.689","Text":"know that our electric fields are in the perpendicular direction,"},{"Start":"05:12.689 ","End":"05:15.790","Text":"and we said that there in the x and z direction so that"},{"Start":"05:15.790 ","End":"05:19.535","Text":"means that our velocity can only be in the y direction."},{"Start":"05:19.535 ","End":"05:25.355","Text":"That means that our x component of our initial velocity is also equal to 0."},{"Start":"05:25.355 ","End":"05:27.500","Text":"This whole term is equal to 0."},{"Start":"05:27.500 ","End":"05:32.630","Text":"Then we\u0027re left with our x position is going to be equal to"},{"Start":"05:32.630 ","End":"05:40.050","Text":"1/2 multiplied by the acceleration in the x direction multiplied by t^2."},{"Start":"05:41.720 ","End":"05:47.555","Text":"Now what we\u0027re trying to find is our time value."},{"Start":"05:47.555 ","End":"05:52.345","Text":"How long is our electron is going to be in this tube?"},{"Start":"05:52.345 ","End":"05:58.590","Text":"As we know, velocity is equal"},{"Start":"05:58.590 ","End":"06:07.630","Text":"to our distance"},{"Start":"06:07.630 ","End":"06:09.450","Text":"divided by our time."},{"Start":"06:09.450 ","End":"06:15.120","Text":"Our distance inside the tube is actually d. It\u0027s denoted by the letter d,"},{"Start":"06:15.120 ","End":"06:17.185","Text":"so I\u0027ll just make this a little bit more clear."},{"Start":"06:17.185 ","End":"06:20.720","Text":"Distance divided by time."},{"Start":"06:21.800 ","End":"06:26.125","Text":"Now we know that our velocity inside the tube."},{"Start":"06:26.125 ","End":"06:29.230","Text":"We know that our electrons are moving through the tube in"},{"Start":"06:29.230 ","End":"06:33.075","Text":"the y direction with a velocity of v."},{"Start":"06:33.075 ","End":"06:41.125","Text":"The only component of velocity that we have is in the y direction and it\u0027s v. Therefore,"},{"Start":"06:41.125 ","End":"06:49.580","Text":"we can say that our t is equal to our distance traveled through the tube,"},{"Start":"06:49.580 ","End":"06:51.095","Text":"which is of length d,"},{"Start":"06:51.095 ","End":"06:55.160","Text":"divided by our velocity in the y direction,"},{"Start":"06:55.160 ","End":"06:59.180","Text":"which is given to us as v. This is our value for"},{"Start":"06:59.180 ","End":"07:04.770","Text":"t. Then we can plug this in to our equation over here."},{"Start":"07:06.100 ","End":"07:11.520","Text":"Then we\u0027ll get that our displacement in the x direction is equal"},{"Start":"07:11.520 ","End":"07:16.670","Text":"to 1/2 multiplied by our acceleration in our x direction,"},{"Start":"07:16.670 ","End":"07:20.285","Text":"so that was equal to this over here,"},{"Start":"07:20.285 ","End":"07:24.525","Text":"which is the absolute value of our electron charge,"},{"Start":"07:24.525 ","End":"07:28.910","Text":"multiplied by our E field in the x direction divided by"},{"Start":"07:28.910 ","End":"07:33.810","Text":"the mass of our electron multiplied by t^2,"},{"Start":"07:33.810 ","End":"07:36.470","Text":"where t we calculated over here,"},{"Start":"07:36.470 ","End":"07:42.210","Text":"multiplied by d divided by v^2."},{"Start":"07:43.050 ","End":"07:47.080","Text":"Now what we\u0027ve found is this distance"},{"Start":"07:47.080 ","End":"07:52.510","Text":"x over here that our electron has moved in the x direction."},{"Start":"07:52.510 ","End":"07:56.860","Text":"The x displacement of the electron from when it"},{"Start":"07:56.860 ","End":"08:01.660","Text":"entered the tube until when it exits the tube."},{"Start":"08:01.660 ","End":"08:03.715","Text":"I\u0027ll write it in blue."},{"Start":"08:03.715 ","End":"08:09.355","Text":"This is the x we see over here in this diagram."},{"Start":"08:09.355 ","End":"08:12.505","Text":"The shift of the electron in the x direction"},{"Start":"08:12.505 ","End":"08:17.180","Text":"due to the electric field in the x direction."},{"Start":"08:18.120 ","End":"08:24.205","Text":"Now what we want to do is we\u0027re moving to the second stage of our calculations,"},{"Start":"08:24.205 ","End":"08:27.220","Text":"which is working out the velocity in"},{"Start":"08:27.220 ","End":"08:32.155","Text":"the x direction upon exiting the tube until we hit the screen."},{"Start":"08:32.155 ","End":"08:36.880","Text":"Now we\u0027re calculating the motion between the tube and the screen."},{"Start":"08:36.880 ","End":"08:40.345","Text":"What we\u0027re trying to find is this v_x over here,"},{"Start":"08:40.345 ","End":"08:48.910","Text":"a velocity in the x direction due to this electric field when we\u0027re exiting the tubes."},{"Start":"08:48.910 ","End":"08:57.800","Text":"Our v_x is equal to our acceleration in the x direction multiplied by time."},{"Start":"08:58.440 ","End":"09:03.595","Text":"We already know what our acceleration in the x direction is."},{"Start":"09:03.595 ","End":"09:05.500","Text":"We worked it out over here."},{"Start":"09:05.500 ","End":"09:09.880","Text":"So, a is the absolute value of the electron charge multiplied by"},{"Start":"09:09.880 ","End":"09:14.425","Text":"the electric field in the x direction divided by the mass of the electron,"},{"Start":"09:14.425 ","End":"09:16.210","Text":"and then multiplied by t,"},{"Start":"09:16.210 ","End":"09:18.505","Text":"which we also calculated over here,"},{"Start":"09:18.505 ","End":"09:25.540","Text":"is this distance d divided by our velocity in the y-direction"},{"Start":"09:25.540 ","End":"09:31.660","Text":"v. This is the velocity of"},{"Start":"09:31.660 ","End":"09:39.230","Text":"the electrons in the x-direction at the moment that they exit the cathode ray tube."},{"Start":"09:41.760 ","End":"09:45.970","Text":"We saw that our exiting velocity is this v_x,"},{"Start":"09:45.970 ","End":"09:51.595","Text":"and because we don\u0027t have any forces acting on the electron once it exits the tube,"},{"Start":"09:51.595 ","End":"09:54.325","Text":"this velocity is going to remain constant."},{"Start":"09:54.325 ","End":"09:56.515","Text":"It\u0027s not going to change or accelerate."},{"Start":"09:56.515 ","End":"10:00.370","Text":"That means that we have this same v_x as"},{"Start":"10:00.370 ","End":"10:05.725","Text":"the electron travels in this space up until the screen."},{"Start":"10:05.725 ","End":"10:08.755","Text":"Now what we want to know is,"},{"Start":"10:08.755 ","End":"10:13.610","Text":"we saw that there\u0027s an initial displacement of where our electron use to be,"},{"Start":"10:13.620 ","End":"10:16.240","Text":"its x position, where it used to be,"},{"Start":"10:16.240 ","End":"10:21.535","Text":"which was at 0 until it got to this point over here due to our electric field."},{"Start":"10:21.535 ","End":"10:24.880","Text":"Now what we want to see is from this point over here,"},{"Start":"10:24.880 ","End":"10:28.300","Text":"how much the electron is going to be displaced,"},{"Start":"10:28.300 ","End":"10:33.835","Text":"due to this x component of velocity until it hits the screen."},{"Start":"10:33.835 ","End":"10:40.030","Text":"Now we\u0027re going to add on another x component, x_2."},{"Start":"10:40.030 ","End":"10:46.690","Text":"This is the second x displacement in this area between the tube and the screen,"},{"Start":"10:46.690 ","End":"10:48.370","Text":"which we\u0027re going to add on to"},{"Start":"10:48.370 ","End":"10:53.815","Text":"our original x displacement in the tube due to our electric field."},{"Start":"10:53.815 ","End":"10:57.700","Text":"Our x_2, we\u0027re just going to rearrange this equation."},{"Start":"10:57.700 ","End":"11:04.720","Text":"It\u0027s simply going to be our velocity in the x-direction multiplied"},{"Start":"11:04.720 ","End":"11:12.460","Text":"by the time that it\u0027s spent in this space between the exit of the tube and the screen."},{"Start":"11:12.460 ","End":"11:18.610","Text":"This is simply speed is equal to distance divided by time rearranged,"},{"Start":"11:18.610 ","End":"11:21.960","Text":"and of course this is t_2."},{"Start":"11:21.960 ","End":"11:24.775","Text":"It\u0027s going to be a different t to this."},{"Start":"11:24.775 ","End":"11:29.275","Text":"Now let\u0027s calculate what our t_2 is equal to."},{"Start":"11:29.275 ","End":"11:35.185","Text":"We know that it\u0027s this distance between the exit of the tube until the screen,"},{"Start":"11:35.185 ","End":"11:38.965","Text":"which we were told is a distance l,"},{"Start":"11:38.965 ","End":"11:43.660","Text":"divided by the velocity that it\u0027s traveling in,"},{"Start":"11:43.660 ","End":"11:46.495","Text":"in the y direction,"},{"Start":"11:46.495 ","End":"11:52.450","Text":"which we were told in our question is equal to v. This is"},{"Start":"11:52.450 ","End":"11:54.550","Text":"the amount of time that our electron will"},{"Start":"11:54.550 ","End":"11:58.330","Text":"take from exiting the tube until reaching the screen."},{"Start":"11:58.330 ","End":"12:02.845","Text":"It\u0027s the total distance that it needs to cover divided by"},{"Start":"12:02.845 ","End":"12:07.630","Text":"the velocity that it\u0027s traveling in that direction of travel,"},{"Start":"12:07.630 ","End":"12:09.920","Text":"which is the y-direction."},{"Start":"12:11.490 ","End":"12:16.760","Text":"Now we can plug in our values for x_2."},{"Start":"12:16.830 ","End":"12:22.555","Text":"X_2 is equal to our velocity in the x-direction,"},{"Start":"12:22.555 ","End":"12:25.244","Text":"which we saw was equal to this."},{"Start":"12:25.244 ","End":"12:29.320","Text":"So, this is the absolute value of our electron charge multiplied"},{"Start":"12:29.320 ","End":"12:34.510","Text":"by x divided by the mass of the electron,"},{"Start":"12:34.510 ","End":"12:38.310","Text":"multiplied by our time,"},{"Start":"12:38.310 ","End":"12:41.385","Text":"which is d divided by v,"},{"Start":"12:41.385 ","End":"12:45.210","Text":"and then multiplied by our t_2,"},{"Start":"12:45.210 ","End":"12:47.520","Text":"which we calculated over here,"},{"Start":"12:47.520 ","End":"12:51.310","Text":"is equal to l/v."},{"Start":"12:54.210 ","End":"13:01.030","Text":"Now we can see that we\u0027ve got in our displacement over here, which is our x_2,"},{"Start":"13:01.030 ","End":"13:03.670","Text":"which is the amount that our electron has"},{"Start":"13:03.670 ","End":"13:07.225","Text":"shifted in the x-direction once it exited the tube."},{"Start":"13:07.225 ","End":"13:12.085","Text":"Then we also have to add on this value x,"},{"Start":"13:12.085 ","End":"13:17.560","Text":"which has the amount that the electron shifted inside the tube from once it entered"},{"Start":"13:17.560 ","End":"13:24.805","Text":"the tube and traveled under the influence of the electric field in the x-direction."},{"Start":"13:24.805 ","End":"13:30.114","Text":"That means that our total x displacement from"},{"Start":"13:30.114 ","End":"13:36.025","Text":"the time that our electron entered the cathode ray tube until it hits the screen,"},{"Start":"13:36.025 ","End":"13:43.629","Text":"is equal to our first x plus our x2,"},{"Start":"13:43.629 ","End":"13:51.895","Text":"which is equal to our, from here,"},{"Start":"13:51.895 ","End":"13:59.544","Text":"half of absolute value of our electron charge multiplied by our E field,"},{"Start":"13:59.544 ","End":"14:01.780","Text":"divided by the mass of the electron,"},{"Start":"14:01.780 ","End":"14:08.455","Text":"multiplied by d divided by v^2 plus our x_2,"},{"Start":"14:08.455 ","End":"14:09.970","Text":"which is what we have here."},{"Start":"14:09.970 ","End":"14:13.450","Text":"So, this is the absolute value of"},{"Start":"14:13.450 ","End":"14:17.740","Text":"our electron charge multiplied by the E field in the x-direction"},{"Start":"14:17.740 ","End":"14:26.900","Text":"divided by m multiplied by d/v, multiplied by l/v."},{"Start":"14:27.630 ","End":"14:30.310","Text":"Then this is equal to,"},{"Start":"14:30.310 ","End":"14:33.010","Text":"if we simplify our like terms,"},{"Start":"14:33.010 ","End":"14:37.510","Text":"the absolute value of e multiplied by the E field in the x direction,"},{"Start":"14:37.510 ","End":"14:41.785","Text":"multiplied by d divided by"},{"Start":"14:41.785 ","End":"14:49.525","Text":"mv^2 multiplied by d/2+l."},{"Start":"14:49.525 ","End":"14:51.490","Text":"Now, in our question,"},{"Start":"14:51.490 ","End":"14:55.345","Text":"we\u0027re being told that we can assume that our distance d,"},{"Start":"14:55.345 ","End":"15:00.775","Text":"so the length of the tube is significantly smaller than our length l,"},{"Start":"15:00.775 ","End":"15:04.465","Text":"which is the length between the edge of the tube to the screen."},{"Start":"15:04.465 ","End":"15:07.615","Text":"If our d is significantly smaller than l,"},{"Start":"15:07.615 ","End":"15:10.150","Text":"let\u0027s say l is a 1000,000 and d is 1,"},{"Start":"15:10.150 ","End":"15:13.585","Text":"so 1000,000+1, doesn\u0027t really make much of a difference."},{"Start":"15:13.585 ","End":"15:15.295","Text":"We could just leave it at 1000,000."},{"Start":"15:15.295 ","End":"15:18.550","Text":"That means that we can cross out this term,"},{"Start":"15:18.550 ","End":"15:21.250","Text":"and say that it\u0027s almost equal to 0,"},{"Start":"15:21.250 ","End":"15:23.755","Text":"that we don\u0027t have to take it into account,"},{"Start":"15:23.755 ","End":"15:25.630","Text":"because the difference between a 1000,000,"},{"Start":"15:25.630 ","End":"15:28.985","Text":"and 1000,001 is something that we don\u0027t need to focus on."},{"Start":"15:28.985 ","End":"15:32.735","Text":"Then we can say that our total displacement in the x-direction,"},{"Start":"15:32.735 ","End":"15:36.920","Text":"once we\u0027ve taken into account that d is significantly smaller than l,"},{"Start":"15:36.920 ","End":"15:39.980","Text":"is approximately equal to the absolute value of"},{"Start":"15:39.980 ","End":"15:43.910","Text":"our electron charge multiplied by our E field in the x direction,"},{"Start":"15:43.910 ","End":"15:52.230","Text":"multiplied by d multiplied by l divided by mv^2."},{"Start":"15:53.910 ","End":"16:02.245","Text":"Now if you want to find out the displacement of the electron in the z direction,"},{"Start":"16:02.245 ","End":"16:06.370","Text":"all you have to do is follow this exact same train of thoughts,"},{"Start":"16:06.370 ","End":"16:11.395","Text":"but simply substituting in instead of our electric field in the x-direction,"},{"Start":"16:11.395 ","End":"16:15.040","Text":"you will substitute in the electric field in the z-direction."},{"Start":"16:15.040 ","End":"16:21.040","Text":"The total z displacement is approximately,"},{"Start":"16:21.040 ","End":"16:26.830","Text":"once you use this approximation that d is significantly smaller than l,"},{"Start":"16:26.830 ","End":"16:29.980","Text":"you\u0027ll get that it is equal to the absolute value of"},{"Start":"16:29.980 ","End":"16:34.015","Text":"the electron charge multiplied by the E field in the z-direction,"},{"Start":"16:34.015 ","End":"16:35.185","Text":"multiplied by d,"},{"Start":"16:35.185 ","End":"16:40.315","Text":"multiplied by l divided by mv^2."},{"Start":"16:40.315 ","End":"16:44.990","Text":"Feel free to stop this video and work out"},{"Start":"16:44.990 ","End":"16:47.360","Text":"the displacement in the z-direction of"},{"Start":"16:47.360 ","End":"16:51.870","Text":"the electron on your own and see if you get to this same answer."},{"Start":"16:52.940 ","End":"16:56.380","Text":"That\u0027s the end of this lesson."}],"ID":22271}],"Thumbnail":null,"ID":99473}]

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