Gauss's Law
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[{"Name":"Gauss\u0027s Law","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Intro To Gauss_ Law","Duration":"15m 34s","ChapterTopicVideoID":21404,"CourseChapterTopicPlaylistID":246895,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/21404.jpeg","UploadDate":"2020-04-21T08:43:00.2200000","DurationForVideoObject":"PT15M34S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.985","Text":"Hello and welcome to our chapter about Gauss\u0027 laws."},{"Start":"00:05.985 ","End":"00:12.300","Text":"In this lesson, we\u0027re going to give a brief explanation on what Gauss\u0027 law is."},{"Start":"00:12.300 ","End":"00:15.720","Text":"First of all, Gauss\u0027 law is given like so."},{"Start":"00:15.720 ","End":"00:25.245","Text":"It\u0027s the integral on the electric field dot ds and that is,"},{"Start":"00:25.245 ","End":"00:30.405","Text":"so they say that the integral of the electric field dot ds is equal to 1"},{"Start":"00:30.405 ","End":"00:35.910","Text":"divided by Epsilon naught multiplied by Q_in,"},{"Start":"00:35.910 ","End":"00:47.125","Text":"where Q_in is generally defined to be the integral on Rho dv."},{"Start":"00:47.125 ","End":"00:50.555","Text":"Now, in this integral in Gauss\u0027 law,"},{"Start":"00:50.555 ","End":"00:57.900","Text":"the integral sign usually has a circle around it and soon we\u0027ll discuss what this means."},{"Start":"00:58.610 ","End":"01:01.770","Text":"How do I use Gauss\u0027 law?"},{"Start":"01:01.770 ","End":"01:04.920","Text":"Usually, I\u0027ll work out my left-hand side,"},{"Start":"01:04.920 ","End":"01:08.120","Text":"and then I\u0027ll work out my right-hand side."},{"Start":"01:08.120 ","End":"01:10.730","Text":"When one of the sides I have my unknown variable,"},{"Start":"01:10.730 ","End":"01:12.350","Text":"which I\u0027m trying to find out,"},{"Start":"01:12.350 ","End":"01:14.660","Text":"then I\u0027ll equate both the sides,"},{"Start":"01:14.660 ","End":"01:17.285","Text":"one to another and I\u0027ll find my unknown variable."},{"Start":"01:17.285 ","End":"01:19.415","Text":"Now, most of the time,"},{"Start":"01:19.415 ","End":"01:24.545","Text":"we use Gauss\u0027 law in order to find our electric field."},{"Start":"01:24.545 ","End":"01:27.710","Text":"Not always, but most of the questions that you will"},{"Start":"01:27.710 ","End":"01:31.890","Text":"encounter will be when our electric field is unknown."},{"Start":"01:32.930 ","End":"01:39.485","Text":"Now, let\u0027s explain how we work out or calculate our left-hand side of the equation,"},{"Start":"01:39.485 ","End":"01:43.020","Text":"and then afterwards, we\u0027ll speak about the right-hand side."},{"Start":"01:43.550 ","End":"01:46.940","Text":"The first thing is that on the left side,"},{"Start":"01:46.940 ","End":"01:53.135","Text":"whenever we do an integral along a close loop of E.ds,"},{"Start":"01:53.135 ","End":"01:57.540","Text":"so what we\u0027re working out is flux."},{"Start":"01:57.540 ","End":"02:02.585","Text":"Now specifically, because we\u0027re working out the flux within electric field,"},{"Start":"02:02.585 ","End":"02:06.570","Text":"so here we\u0027re working out electric flux."},{"Start":"02:07.610 ","End":"02:10.720","Text":"Now, whenever we\u0027re dealing with flux,"},{"Start":"02:10.720 ","End":"02:13.945","Text":"we\u0027re always going to need some thin,"},{"Start":"02:13.945 ","End":"02:18.650","Text":"flat surface, or rather a thin surface."},{"Start":"02:18.650 ","End":"02:20.325","Text":"It doesn\u0027t matter if it\u0027s flat or not."},{"Start":"02:20.325 ","End":"02:24.720","Text":"Here we can see that our ds is an integral along the area."},{"Start":"02:24.720 ","End":"02:33.710","Text":"Let\u0027s draw some surface that has an area and this is in 3-dimensions."},{"Start":"02:33.710 ","End":"02:36.415","Text":"It\u0027s a 3D shape,"},{"Start":"02:36.415 ","End":"02:39.750","Text":"but this is just area."},{"Start":"02:39.750 ","End":"02:43.870","Text":"It has no depth or width."},{"Start":"02:44.540 ","End":"02:52.280","Text":"What we can imagine is there is some sheet of paper blowing in the wind, for example."},{"Start":"02:53.120 ","End":"02:55.460","Text":"What is this integral?"},{"Start":"02:55.460 ","End":"03:00.395","Text":"We\u0027re taking lots of different points on this surface,"},{"Start":"03:00.395 ","End":"03:02.810","Text":"on this piece of paper flying on the wind,"},{"Start":"03:02.810 ","End":"03:06.725","Text":"and we\u0027re summing up the electric field."},{"Start":"03:06.725 ","End":"03:13.625","Text":"Now, we\u0027re not summing up the electric field parallel to the plane or on the plane,"},{"Start":"03:13.625 ","End":"03:19.020","Text":"we\u0027re summing the electric field that is perpendicular to the plane."},{"Start":"03:20.630 ","End":"03:27.990","Text":"For instance, if I have an electric field at this point pointing in this direction,"},{"Start":"03:27.990 ","End":"03:31.220","Text":"what I\u0027m going to do is I\u0027m going to find the component of"},{"Start":"03:31.220 ","End":"03:36.680","Text":"the electric field which is perpendicular to my plane at that exactly point,"},{"Start":"03:36.680 ","End":"03:39.170","Text":"it\u0027s E perpendicular,"},{"Start":"03:39.170 ","End":"03:42.910","Text":"and I\u0027m going to sum up on all of these E perpendiculars."},{"Start":"03:42.910 ","End":"03:46.355","Text":"We\u0027ll have this, which is E perpendicular,"},{"Start":"03:46.355 ","End":"03:52.170","Text":"and here, so perpendicular to that point on the plane."},{"Start":"03:53.240 ","End":"03:55.895","Text":"Where does all this come from?"},{"Start":"03:55.895 ","End":"04:01.745","Text":"How am I getting this perpendicular component?"},{"Start":"04:01.745 ","End":"04:06.600","Text":"It comes from the dot product going on over here."},{"Start":"04:07.300 ","End":"04:11.090","Text":"Our ds is a vector quantity,"},{"Start":"04:11.090 ","End":"04:13.430","Text":"which means it has size and direction."},{"Start":"04:13.430 ","End":"04:17.475","Text":"Its size is going to be some kind of area,"},{"Start":"04:17.475 ","End":"04:18.930","Text":"so dx, dy,"},{"Start":"04:18.930 ","End":"04:21.675","Text":"or id Theta, etc."},{"Start":"04:21.675 ","End":"04:26.900","Text":"Either the direction of this ds vector is going"},{"Start":"04:26.900 ","End":"04:32.345","Text":"to be in the perpendicular direction to that area."},{"Start":"04:32.345 ","End":"04:40.120","Text":"At that point, the direction of ds is going to be perpendicular to that point."},{"Start":"04:40.910 ","End":"04:47.150","Text":"Let\u0027s explain this letter about this perpendicular component and"},{"Start":"04:47.150 ","End":"04:53.600","Text":"let\u0027s say that we have our electric field going like so,"},{"Start":"04:53.600 ","End":"04:56.150","Text":"and it\u0027s also a vector quantity."},{"Start":"04:56.150 ","End":"04:59.930","Text":"Then I have my ds vector,"},{"Start":"04:59.930 ","End":"05:04.025","Text":"so it\u0027s going to be the size of this area and it\u0027s going to be"},{"Start":"05:04.025 ","End":"05:08.960","Text":"in this direction perpendicular to my area."},{"Start":"05:08.960 ","End":"05:12.020","Text":"This is my ds vector."},{"Start":"05:12.020 ","End":"05:16.714","Text":"Now what I want to do is I want to work out the dot product;"},{"Start":"05:16.714 ","End":"05:19.750","Text":"E, which is a vector,"},{"Start":"05:19.750 ","End":"05:21.680","Text":"dot ds, which is also a vector."},{"Start":"05:21.680 ","End":"05:23.945","Text":"That comes from this over here."},{"Start":"05:23.945 ","End":"05:29.090","Text":"As we know, this is equal to the size of"},{"Start":"05:29.090 ","End":"05:35.374","Text":"our electric field multiplied by the size of our ds vector,"},{"Start":"05:35.374 ","End":"05:38.360","Text":"so that means this area,"},{"Start":"05:38.360 ","End":"05:39.725","Text":"the size of the area,"},{"Start":"05:39.725 ","End":"05:45.665","Text":"multiplied by cosine of the angle between the two."},{"Start":"05:45.665 ","End":"05:53.070","Text":"Let\u0027s say that the angle between them is Alpha and it\u0027s this over here."},{"Start":"05:54.560 ","End":"05:58.540","Text":"Now it\u0027s important to know that if we wrote E ds"},{"Start":"05:58.540 ","End":"06:03.890","Text":"cosine Alpha or ds E cosine Alpha, it\u0027s the same thing."},{"Start":"06:03.890 ","End":"06:08.230","Text":"If we look at E multiplied by cosine Alpha,"},{"Start":"06:08.230 ","End":"06:12.280","Text":"we can see that that\u0027s going to be the,"},{"Start":"06:12.280 ","End":"06:14.200","Text":"here, y component,"},{"Start":"06:14.200 ","End":"06:18.475","Text":"for instance, or the component of our electric field,"},{"Start":"06:18.475 ","End":"06:22.750","Text":"which is in the same direction as our ds vector, or in other words,"},{"Start":"06:22.750 ","End":"06:24.910","Text":"the component of our electric field,"},{"Start":"06:24.910 ","End":"06:29.900","Text":"which is perpendicular to our area."},{"Start":"06:30.990 ","End":"06:36.418","Text":"This gray arrow over here is equal to,"},{"Start":"06:36.418 ","End":"06:38.120","Text":"see where I can write this."},{"Start":"06:38.120 ","End":"06:43.055","Text":"It\u0027s going to be equal to E cosine of Alpha."},{"Start":"06:43.055 ","End":"06:47.180","Text":"We can see that it\u0027s parallel to our ds vector,"},{"Start":"06:47.180 ","End":"06:53.520","Text":"which means that it\u0027s also perpendicular to our area or to our plane."},{"Start":"06:54.080 ","End":"07:00.935","Text":"In that case, we can see that E multiplied by cosine Alpha is simply equal to"},{"Start":"07:00.935 ","End":"07:08.060","Text":"our electric field perpendicular to the plane multiplied by the magnitude of ds,"},{"Start":"07:08.060 ","End":"07:10.100","Text":"which we can just write like so,"},{"Start":"07:10.100 ","End":"07:12.570","Text":"it\u0027s the size of our area."},{"Start":"07:13.250 ","End":"07:19.735","Text":"That means that for every small area on our surface,"},{"Start":"07:19.735 ","End":"07:28.035","Text":"so what I\u0027m doing is I\u0027m finding the electric field perpendicular to my plane."},{"Start":"07:28.035 ","End":"07:32.840","Text":"I\u0027m finding the electric field component which is perpendicular,"},{"Start":"07:32.840 ","End":"07:38.837","Text":"and I\u0027m multiplying it by this small area ds."},{"Start":"07:38.837 ","End":"07:43.135","Text":"That\u0027s how this integral works."},{"Start":"07:43.135 ","End":"07:46.700","Text":"Let\u0027s go back to our explanation."},{"Start":"07:46.890 ","End":"07:49.705","Text":"Now we\u0027ll notice that"},{"Start":"07:49.705 ","End":"07:55.195","Text":"our integration sign has this circle around it. What does that mean?"},{"Start":"07:55.195 ","End":"07:59.185","Text":"That means that I can\u0027t just take any surface in space."},{"Start":"07:59.185 ","End":"08:05.750","Text":"I have to calculate this integral on a closed surface."},{"Start":"08:05.850 ","End":"08:16.200","Text":"An example of a closed surface is going to be a spherical shell."},{"Start":"08:16.200 ","End":"08:21.000","Text":"Here we have our area of the spherical shell,"},{"Start":"08:21.000 ","End":"08:23.190","Text":"and then inside this sphere,"},{"Start":"08:23.190 ","End":"08:25.870","Text":"we have some volume."},{"Start":"08:27.000 ","End":"08:29.680","Text":"In this spherical shell,"},{"Start":"08:29.680 ","End":"08:34.945","Text":"I have some volume V. Now I can sum along this."},{"Start":"08:34.945 ","End":"08:39.520","Text":"Because obviously some open sheet of paper, let\u0027s say,"},{"Start":"08:39.520 ","End":"08:43.435","Text":"so I\u0027m not going to have any volume enclosed,"},{"Start":"08:43.435 ","End":"08:45.685","Text":"as we can see, whereas in the spherical shell,"},{"Start":"08:45.685 ","End":"08:49.100","Text":"I do have a volume enclosed."},{"Start":"08:49.170 ","End":"08:55.240","Text":"What I\u0027ll be doing is I\u0027ll be summing up all of"},{"Start":"08:55.240 ","End":"08:58.210","Text":"the perpendicular components of"},{"Start":"08:58.210 ","End":"09:03.295","Text":"the electric field in all the directions along my surface,"},{"Start":"09:03.295 ","End":"09:06.010","Text":"so long as the arrows where they\u0027re placed are"},{"Start":"09:06.010 ","End":"09:09.859","Text":"perpendicular to that point of the surface."},{"Start":"09:10.110 ","End":"09:14.740","Text":"That is how I will work out flux in general,"},{"Start":"09:14.740 ","End":"09:18.220","Text":"but here specifically, my electric flux."},{"Start":"09:18.220 ","End":"09:26.930","Text":"I just sum up all of the perpendicular electric field components along my closed surface."},{"Start":"09:27.960 ","End":"09:31.840","Text":"Now we\u0027ve finished speaking about the left side,"},{"Start":"09:31.840 ","End":"09:33.835","Text":"let\u0027s move on to our right side."},{"Start":"09:33.835 ","End":"09:36.850","Text":"Let\u0027s talk about what is our Q_in."},{"Start":"09:36.850 ","End":"09:41.410","Text":"Q_in is the total amount of charge which is"},{"Start":"09:41.410 ","End":"09:47.360","Text":"contained or enclosed within this closed surface."},{"Start":"09:48.360 ","End":"09:54.730","Text":"That\u0027s another reason why I need to be summing up on a closed surface so that I can"},{"Start":"09:54.730 ","End":"10:03.102","Text":"define what exactly or which charges exactly are contained within this space,"},{"Start":"10:03.102 ","End":"10:05.610","Text":"are contained within this volume."},{"Start":"10:05.610 ","End":"10:09.047","Text":"As we can see with this piece of paper,"},{"Start":"10:09.047 ","End":"10:11.460","Text":"I don\u0027t know which charges, if any,"},{"Start":"10:11.460 ","End":"10:17.325","Text":"are contained within this because there\u0027s no closed finite volume within this shape."},{"Start":"10:17.325 ","End":"10:19.020","Text":"Whereas within a spherical shell,"},{"Start":"10:19.020 ","End":"10:22.570","Text":"for instance, it\u0027s a finite space."},{"Start":"10:22.610 ","End":"10:25.244","Text":"As a brief little example,"},{"Start":"10:25.244 ","End":"10:27.780","Text":"if I have my closed surface,"},{"Start":"10:27.780 ","End":"10:29.340","Text":"which is a spherical shell,"},{"Start":"10:29.340 ","End":"10:32.445","Text":"and I have charged Q_1 inside."},{"Start":"10:32.445 ","End":"10:35.595","Text":"That means that my Q_in will be equal to Q_1."},{"Start":"10:35.595 ","End":"10:38.385","Text":"If I also have another charge Q_2 in,"},{"Start":"10:38.385 ","End":"10:41.970","Text":"so my Q_in will just be Q_1 plus Q_2,"},{"Start":"10:41.970 ","End":"10:43.440","Text":"and so on and so forth."},{"Start":"10:43.440 ","End":"10:46.425","Text":"If however, over here, I have a Q_3,"},{"Start":"10:46.425 ","End":"10:51.505","Text":"then my Q_in doesn\u0027t take in my Q_3 into consideration."},{"Start":"10:51.505 ","End":"10:59.960","Text":"My Q_in is still Q_1 and Q_2 because my Q_3 is not enclosed within this shell."},{"Start":"11:00.660 ","End":"11:07.480","Text":"What happens if I have my spherical shell and then I have some blob in here,"},{"Start":"11:07.480 ","End":"11:13.195","Text":"which has uniform charge density per volume of Rho."},{"Start":"11:13.195 ","End":"11:18.580","Text":"All I\u0027ll do is I\u0027ll do this integral of Rho multiplied by dv."},{"Start":"11:18.580 ","End":"11:22.675","Text":"The volume of this blob or this sphere,"},{"Start":"11:22.675 ","End":"11:24.970","Text":"and then I have my Q_in."},{"Start":"11:24.970 ","End":"11:28.945","Text":"Alternatively, if I have my sphere again,"},{"Start":"11:28.945 ","End":"11:31.165","Text":"but this time I have"},{"Start":"11:31.165 ","End":"11:38.005","Text":"a two-dimensional sheet that has a uniform charge density per unit area of Sigma."},{"Start":"11:38.005 ","End":"11:45.235","Text":"Then my Q_in over here will be the integral on Sigma ds."},{"Start":"11:45.235 ","End":"11:47.260","Text":"If my s is constant,"},{"Start":"11:47.260 ","End":"11:51.980","Text":"so it will just be Sigma multiplied by the dimensions."},{"Start":"11:52.440 ","End":"11:59.605","Text":"Similarly, if I have some rod with uniform charge density per unit length Lambda,"},{"Start":"11:59.605 ","End":"12:06.350","Text":"so then my Q_in will be the integral on Lambda dl."},{"Start":"12:07.130 ","End":"12:14.610","Text":"Now we\u0027ve explained where our Eds comes from and so on and so forth."},{"Start":"12:14.610 ","End":"12:20.270","Text":"Now let\u0027s give a little easy example of when we use Gauss\u0027 law."},{"Start":"12:20.270 ","End":"12:22.660","Text":"Now, most of the time,"},{"Start":"12:22.660 ","End":"12:29.360","Text":"I\u0027m going to be given a question where I\u0027m being asked to find the electric field."},{"Start":"12:29.550 ","End":"12:33.324","Text":"Now, the problem is if I\u0027m doing this integral,"},{"Start":"12:33.324 ","End":"12:38.860","Text":"I can\u0027t integrate along a function that I don\u0027t know what the function is."},{"Start":"12:38.860 ","End":"12:42.320","Text":"Then I have a problem."},{"Start":"12:42.570 ","End":"12:46.285","Text":"When are we going to use our Gauss\u0027 law?"},{"Start":"12:46.285 ","End":"12:54.385","Text":"It\u0027s when we\u0027re trying to find our value for E. We\u0027re being asked,"},{"Start":"12:54.385 ","End":"12:56.875","Text":"what is our E?"},{"Start":"12:56.875 ","End":"13:01.735","Text":"Also, we can only use it in very specific cases,"},{"Start":"13:01.735 ","End":"13:07.490","Text":"and that is when our E is a constant."},{"Start":"13:07.800 ","End":"13:14.060","Text":"That means that it\u0027s constant on our shell."},{"Start":"13:15.450 ","End":"13:21.505","Text":"If my electric field is constant about the shell,"},{"Start":"13:21.505 ","End":"13:24.610","Text":"so its magnitude is constant and its direction is"},{"Start":"13:24.610 ","End":"13:28.255","Text":"always perpendicular or we\u0027re using the perpendicular component,"},{"Start":"13:28.255 ","End":"13:33.115","Text":"then that means that I don\u0027t really have to integrate."},{"Start":"13:33.115 ","End":"13:37.420","Text":"Because my electric field is constant and then I just have to multiply it,"},{"Start":"13:37.420 ","End":"13:41.560","Text":"therefore, by the total area of my shape."},{"Start":"13:41.560 ","End":"13:44.335","Text":"Because the electric field is constant throughout."},{"Start":"13:44.335 ","End":"13:49.240","Text":"That means that my integral of"},{"Start":"13:49.240 ","End":"13:56.395","Text":"E.ds will become my electric field E,"},{"Start":"13:56.395 ","End":"13:59.155","Text":"which is constant on the shell,"},{"Start":"13:59.155 ","End":"14:07.780","Text":"multiplied by the total area of the shape or the total area of the shell."},{"Start":"14:07.780 ","End":"14:11.215","Text":"As we can see it, Gauss\u0027 law can only be used in"},{"Start":"14:11.215 ","End":"14:15.228","Text":"a specific case when the electric field is constant on the shell,"},{"Start":"14:15.228 ","End":"14:19.960","Text":"which means that there are 3 specific cases where we"},{"Start":"14:19.960 ","End":"14:24.850","Text":"can actually use Gauss\u0027 law and then all other questions or in all other cases,"},{"Start":"14:24.850 ","End":"14:26.830","Text":"we won\u0027t be able to use this."},{"Start":"14:26.830 ","End":"14:31.780","Text":"The first way that we can use Gauss\u0027 law is when we\u0027re dealing with"},{"Start":"14:31.780 ","End":"14:37.705","Text":"an infinite wire or an infinite cylinder or an infinite cylindrical shell."},{"Start":"14:37.705 ","End":"14:40.090","Text":"Notice that in all of these cases,"},{"Start":"14:40.090 ","End":"14:42.460","Text":"we have a case of symmetry."},{"Start":"14:42.460 ","End":"14:47.335","Text":"Because it\u0027s infinite so we only have the radial direction acting."},{"Start":"14:47.335 ","End":"14:52.220","Text":"We can see that in the radial direction, we have symmetry."},{"Start":"14:52.380 ","End":"14:54.805","Text":"Now, later in this chapter,"},{"Start":"14:54.805 ","End":"14:58.150","Text":"we\u0027ll see that how we solve questions for an infinite wire or"},{"Start":"14:58.150 ","End":"15:01.945","Text":"an infinite cylinder or cylindrical shell is very similar."},{"Start":"15:01.945 ","End":"15:09.230","Text":"The second time that we can use Gauss\u0027 law is when we\u0027re dealing with an infinite plane."},{"Start":"15:09.230 ","End":"15:14.140","Text":"The third case is when we\u0027re dealing with a sphere or a spherical shell."},{"Start":"15:14.140 ","End":"15:18.710","Text":"Again, both of these ideas are pretty much the same."},{"Start":"15:19.020 ","End":"15:21.170","Text":"In the next few lessons,"},{"Start":"15:21.170 ","End":"15:25.415","Text":"we\u0027ll be looking in detail at each one of these cases."},{"Start":"15:25.415 ","End":"15:28.205","Text":"We\u0027re going to also speak a little bit more"},{"Start":"15:28.205 ","End":"15:32.145","Text":"about the electric field or the electric flux."},{"Start":"15:32.145 ","End":"15:35.120","Text":"That\u0027s the end of our lesson."}],"ID":22358},{"Watched":false,"Name":"Flux","Duration":"5m 59s","ChapterTopicVideoID":21289,"CourseChapterTopicPlaylistID":246895,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/21289.jpeg","UploadDate":"2020-04-06T13:51:25.7570000","DurationForVideoObject":"PT5M59S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.790","Text":"Hello. In this lesson we\u0027re going to be"},{"Start":"00:02.790 ","End":"00:06.330","Text":"speaking a little bit more about the electric flux."},{"Start":"00:06.330 ","End":"00:09.600","Text":"Let\u0027s take an example of a spherical shell."},{"Start":"00:09.600 ","End":"00:12.750","Text":"So this is our spherical shell."},{"Start":"00:12.750 ","End":"00:14.490","Text":"It\u0027s given by area,"},{"Start":"00:14.490 ","End":"00:17.310","Text":"but it\u0027s a 3-dimensional shape."},{"Start":"00:17.310 ","End":"00:22.380","Text":"Now let\u0027s take a look at what is going to happen here with regards to the flux."},{"Start":"00:22.380 ","End":"00:28.005","Text":"Let\u0027s imagine that we have some charge over here, Q1."},{"Start":"00:28.005 ","End":"00:30.525","Text":"Due to Coulomb\u0027s law,"},{"Start":"00:30.525 ","End":"00:37.445","Text":"we know that our Q1 applies a force or an electric field in this direction."},{"Start":"00:37.445 ","End":"00:40.340","Text":"It hits the surface of"},{"Start":"00:40.340 ","End":"00:44.900","Text":"our spherical shell and there\u0027s an electric field going in this direction."},{"Start":"00:44.900 ","End":"00:48.425","Text":"Now, of course, it\u0027s not just at this point"},{"Start":"00:48.425 ","End":"00:51.660","Text":"that our charged Q1 is applying this electric field,"},{"Start":"00:51.660 ","End":"00:55.025","Text":"but also if we follow this arrow out,"},{"Start":"00:55.025 ","End":"01:01.325","Text":"there\u0027s also going to be an electric field at this point out of the sphere as well."},{"Start":"01:01.325 ","End":"01:03.980","Text":"We in fact have an electric field line."},{"Start":"01:03.980 ","End":"01:05.990","Text":"Let\u0027s say that our Q1 is positive."},{"Start":"01:05.990 ","End":"01:13.280","Text":"So we can see that our electric field line is going to follow this trajectory."},{"Start":"01:13.280 ","End":"01:17.285","Text":"Now, when I\u0027m trying to calculate my flux,"},{"Start":"01:17.285 ","End":"01:19.730","Text":"as we saw in the previous lesson,"},{"Start":"01:19.730 ","End":"01:23.570","Text":"I\u0027m summing up on the sum or on the component of"},{"Start":"01:23.570 ","End":"01:28.365","Text":"the electric field which is perpendicular to my surface."},{"Start":"01:28.365 ","End":"01:31.190","Text":"When I have an electric field going into my surface,"},{"Start":"01:31.190 ","End":"01:33.455","Text":"so my flux is going to be negative,"},{"Start":"01:33.455 ","End":"01:36.710","Text":"and when I have an electric field coming out of my surface,"},{"Start":"01:36.710 ","End":"01:40.050","Text":"my flux is going to be positive."},{"Start":"01:40.340 ","End":"01:42.575","Text":"As we can see,"},{"Start":"01:42.575 ","End":"01:48.770","Text":"I have flux smaller than 0 going in and flux larger than 0 going out."},{"Start":"01:48.770 ","End":"01:54.365","Text":"So we can see that the total flux that\u0027s affecting my sphere"},{"Start":"01:54.365 ","End":"02:00.990","Text":"is 0 because what is going in is exactly equal to what is going out."},{"Start":"02:01.280 ","End":"02:06.200","Text":"Now, if I carry on drawing my electric field lines"},{"Start":"02:06.200 ","End":"02:10.215","Text":"that hits other points on my spherical shell,"},{"Start":"02:10.215 ","End":"02:13.570","Text":"so we can have going like so."},{"Start":"02:13.570 ","End":"02:20.945","Text":"We can see again that whichever flux is going in is equal,"},{"Start":"02:20.945 ","End":"02:22.760","Text":"or whichever electric field, sorry,"},{"Start":"02:22.760 ","End":"02:25.935","Text":"is going in, is equal to the electric field going out."},{"Start":"02:25.935 ","End":"02:31.355","Text":"Therefore, we can see that for a charge which is"},{"Start":"02:31.355 ","End":"02:37.415","Text":"outside of my shell that I\u0027m trying to calculate or my shape or my surface,"},{"Start":"02:37.415 ","End":"02:43.970","Text":"so any charge outside is not going to affect the flux of the shape."},{"Start":"02:43.970 ","End":"02:50.405","Text":"Now alternatively, if I had a charge inside the shape, so Q2."},{"Start":"02:50.405 ","End":"02:53.750","Text":"If we draw its electric field lines,"},{"Start":"02:53.750 ","End":"02:57.190","Text":"so it\u0027s going to have lines going like so."},{"Start":"02:57.190 ","End":"03:01.130","Text":"Now let\u0027s draw its flux."},{"Start":"03:01.130 ","End":"03:03.980","Text":"So it\u0027s going to have an electric field out."},{"Start":"03:03.980 ","End":"03:07.100","Text":"That means that its flux here is bigger than 0."},{"Start":"03:07.100 ","End":"03:11.665","Text":"An arrow out means flux is bigger than 0,"},{"Start":"03:11.665 ","End":"03:14.415","Text":"and so on and so forth."},{"Start":"03:14.415 ","End":"03:18.140","Text":"We can see that there\u0027s no negative flux."},{"Start":"03:18.140 ","End":"03:22.055","Text":"There\u0027s no electric field coming in."},{"Start":"03:22.055 ","End":"03:27.710","Text":"Which means that a charge inside this shell,"},{"Start":"03:27.710 ","End":"03:33.630","Text":"this spherical shell, is going to affect the electric flux."},{"Start":"03:34.330 ","End":"03:40.520","Text":"So this is the idea about flux and why it\u0027s such a genius method."},{"Start":"03:40.520 ","End":"03:45.275","Text":"It means that only charges contained within our shell"},{"Start":"03:45.275 ","End":"03:51.150","Text":"are shaped by closed surface will affect the flux."},{"Start":"03:51.820 ","End":"03:58.940","Text":"Now an important note is that even though our external charge doesn\u0027t affect the flux,"},{"Start":"03:58.940 ","End":"04:05.135","Text":"because its net contribution to the flux is equal to 0."},{"Start":"04:05.135 ","End":"04:07.640","Text":"On our sphere over here."},{"Start":"04:07.640 ","End":"04:09.425","Text":"It will, however,"},{"Start":"04:09.425 ","End":"04:15.560","Text":"affect the electric field on the surface of our sphere, so for instance,"},{"Start":"04:15.560 ","End":"04:19.745","Text":"over here we can see that the electric field is pointing in this direction, however,"},{"Start":"04:19.745 ","End":"04:23.840","Text":"the electric field from our internal charge q2,"},{"Start":"04:23.840 ","End":"04:27.300","Text":"is pointing in the opposite direction."},{"Start":"04:27.300 ","End":"04:32.100","Text":"We can, we can see that the electric field on"},{"Start":"04:32.100 ","End":"04:38.285","Text":"these points over here is going to be less or might even be equal to 0."},{"Start":"04:38.285 ","End":"04:41.465","Text":"Might even cancel out completely."},{"Start":"04:41.465 ","End":"04:45.320","Text":"However, we can see that our red arrows over here on"},{"Start":"04:45.320 ","End":"04:49.620","Text":"the other side are in the same direction as our green arrows."},{"Start":"04:49.620 ","End":"04:54.020","Text":"That means that the electric field over"},{"Start":"04:54.020 ","End":"04:59.330","Text":"these points here on the surface of the sphere is going to grow."},{"Start":"04:59.330 ","End":"05:04.745","Text":"We can see that the electric field on the surface on this side is lessened or weakened,"},{"Start":"05:04.745 ","End":"05:09.330","Text":"and the electric field on this side is going to be strengthened."},{"Start":"05:09.950 ","End":"05:15.020","Text":"The external charges don\u0027t affect the flux,"},{"Start":"05:15.020 ","End":"05:17.375","Text":"but they do affect the E field,"},{"Start":"05:17.375 ","End":"05:20.340","Text":"the electric field on the surface."},{"Start":"05:21.110 ","End":"05:25.490","Text":"The flux is super useful because then we can just work"},{"Start":"05:25.490 ","End":"05:30.530","Text":"out what is going in on inside the shape,"},{"Start":"05:30.530 ","End":"05:33.050","Text":"and we don\u0027t have to take into consideration"},{"Start":"05:33.050 ","End":"05:36.080","Text":"external chargers and what\u0027s going on on the outside and how"},{"Start":"05:36.080 ","End":"05:42.055","Text":"our electric field everywhere is being changed due to these external charges."},{"Start":"05:42.055 ","End":"05:47.180","Text":"To conclude, our flux measures how many origins of"},{"Start":"05:47.180 ","End":"05:52.435","Text":"charge are contained within my shape, within my shell."},{"Start":"05:52.435 ","End":"05:56.495","Text":"This is a super useful technique in this course."},{"Start":"05:56.495 ","End":"05:59.460","Text":"Okay, that\u0027s the end of this lesson."}],"ID":21369},{"Watched":false,"Name":"Exercise 1","Duration":"14m 27s","ChapterTopicVideoID":21405,"CourseChapterTopicPlaylistID":246895,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/21405.jpeg","UploadDate":"2020-04-21T08:45:46.7400000","DurationForVideoObject":"PT14M27S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.785","Text":"Hello. In this lesson,"},{"Start":"00:01.785 ","End":"00:09.735","Text":"we\u0027re going to learn how to find the electric field due to charged infinite wire."},{"Start":"00:09.735 ","End":"00:12.660","Text":"Here we have a wire."},{"Start":"00:12.660 ","End":"00:15.015","Text":"It\u0027s infinitely long."},{"Start":"00:15.015 ","End":"00:18.225","Text":"Let\u0027s say that this is the z-axis."},{"Start":"00:18.225 ","End":"00:19.800","Text":"Now, because we\u0027re dealing with a wire,"},{"Start":"00:19.800 ","End":"00:20.850","Text":"it\u0027s in one dimension,"},{"Start":"00:20.850 ","End":"00:27.360","Text":"which means that it has a charge distribution per unit length of Lambda."},{"Start":"00:27.360 ","End":"00:31.500","Text":"Now we\u0027re going to choose some arbitrary point in space."},{"Start":"00:31.500 ","End":"00:33.210","Text":"Let\u0027s say over here."},{"Start":"00:33.210 ","End":"00:39.105","Text":"Let\u0027s say that it\u0027s a distance r away from our wire."},{"Start":"00:39.105 ","End":"00:45.270","Text":"Now, let\u0027s find out what the electric field is going to be at this point over here."},{"Start":"00:45.550 ","End":"00:48.725","Text":"Now the first thing that we can see is that"},{"Start":"00:48.725 ","End":"00:52.115","Text":"our electric field is going out in a radial direction,"},{"Start":"00:52.115 ","End":"00:56.670","Text":"and that as we move along our z-axis."},{"Start":"00:56.670 ","End":"00:59.460","Text":"Our electric field is going to be the same."},{"Start":"00:59.460 ","End":"01:04.910","Text":"It\u0027s still going to be equal to e and in the radial direction."},{"Start":"01:04.910 ","End":"01:08.180","Text":"If we draw like so,"},{"Start":"01:08.180 ","End":"01:12.414","Text":"to represent that we\u0027re changing our angle Theta,"},{"Start":"01:12.414 ","End":"01:14.690","Text":"so we can see that the electric field is"},{"Start":"01:14.690 ","End":"01:20.009","Text":"still going to be continuing in a radial direction."},{"Start":"01:20.140 ","End":"01:28.355","Text":"Our electric field is only dependent on our radius and not on our angle Theta."},{"Start":"01:28.355 ","End":"01:32.820","Text":"I know this because we\u0027re dealing with a case of symmetry."},{"Start":"01:32.820 ","End":"01:34.010","Text":"At the end of this lesson,"},{"Start":"01:34.010 ","End":"01:37.830","Text":"we\u0027ll speak about it in some further detail."},{"Start":"01:37.840 ","End":"01:40.640","Text":"Because we can see that we have symmetry,"},{"Start":"01:40.640 ","End":"01:42.720","Text":"that means we can use Gauss\u0027 law."},{"Start":"01:42.720 ","End":"01:48.945","Text":"The first thing that I\u0027m going to do is I\u0027m going to choose my Gaussian surface."},{"Start":"01:48.945 ","End":"01:52.550","Text":"What I\u0027m going to do is I\u0027m going to choose a cylinder."},{"Start":"01:52.550 ","End":"01:56.465","Text":"This is the side of my cylinder."},{"Start":"01:56.465 ","End":"01:59.725","Text":"It goes around the wire."},{"Start":"01:59.725 ","End":"02:03.390","Text":"I\u0027m going to say that this length over here,"},{"Start":"02:03.390 ","End":"02:06.840","Text":"the length or the height of the cylinder, is l,"},{"Start":"02:06.840 ","End":"02:15.140","Text":"and the radius of the cylinder is going to be r. Just imagine,"},{"Start":"02:15.140 ","End":"02:16.610","Text":"I didn\u0027t draw this very well,"},{"Start":"02:16.610 ","End":"02:22.820","Text":"that the cylinder goes over the wire such that the wire is in the center of the cylinder."},{"Start":"02:22.820 ","End":"02:28.040","Text":"I chose this Gaussian surface because I can see that"},{"Start":"02:28.040 ","End":"02:35.545","Text":"the electric field is going to be constant at every point on this surface."},{"Start":"02:35.545 ","End":"02:38.445","Text":"Let\u0027s begin working this out."},{"Start":"02:38.445 ","End":"02:46.815","Text":"We know that our Gauss\u0027 law is given by the closed integral on E.ds."},{"Start":"02:46.815 ","End":"02:52.370","Text":"Because our electric field is uniform across our Gaussian surface,"},{"Start":"02:52.370 ","End":"03:01.170","Text":"that means that we can simply write it as E multiplied by S, the surface area."},{"Start":"03:01.170 ","End":"03:08.070","Text":"We know that that is going to be equal to Q_in divided by Epsilon naught."},{"Start":"03:08.360 ","End":"03:10.740","Text":"Let\u0027s begin by writing this."},{"Start":"03:10.740 ","End":"03:13.115","Text":"We have our E, which is what we\u0027re trying to find."},{"Start":"03:13.115 ","End":"03:17.120","Text":"We\u0027re multiplying it by the surface area of our Gaussian surface."},{"Start":"03:17.120 ","End":"03:22.085","Text":"What do we have to notice here is that our electric flux is"},{"Start":"03:22.085 ","End":"03:28.950","Text":"along the body of our Gaussian surface."},{"Start":"03:28.950 ","End":"03:32.020","Text":"It\u0027s not on the basis."},{"Start":"03:32.020 ","End":"03:36.710","Text":"In a few minutes, we\u0027ll speak about what\u0027s going on at the base of our cylinder."},{"Start":"03:36.710 ","End":"03:40.830","Text":"But right now we can see that it\u0027s the surface area of the body."},{"Start":"03:40.850 ","End":"03:44.160","Text":"That is where our electric flux is,"},{"Start":"03:44.160 ","End":"03:45.555","Text":"so that\u0027s what we have to multiply."},{"Start":"03:45.555 ","End":"03:48.515","Text":"We\u0027re going to have E multiplied by,"},{"Start":"03:48.515 ","End":"03:55.255","Text":"the surface area is going to be the perimeter of the circle."},{"Start":"03:55.255 ","End":"04:02.025","Text":"That is equal 2Pi multiplied the radius of our cylinder."},{"Start":"04:02.025 ","End":"04:06.810","Text":"Then we have to sum that up along the length of our cylinder,"},{"Start":"04:06.810 ","End":"04:08.280","Text":"so we multiply it by"},{"Start":"04:08.280 ","End":"04:16.100","Text":"l. Let\u0027s quickly discuss what\u0027s happening at the bases of our cylinder."},{"Start":"04:16.100 ","End":"04:19.160","Text":"Imagine that this is one of the bases,"},{"Start":"04:19.160 ","End":"04:23.030","Text":"and this is going to be the same for this base over here as well."},{"Start":"04:23.030 ","End":"04:28.100","Text":"We can see that our electric field coming from our wire that\u0027s going in the middle."},{"Start":"04:28.100 ","End":"04:31.400","Text":"Our electric field is going in the radial direction."},{"Start":"04:31.400 ","End":"04:35.270","Text":"It\u0027s going to look something like so."},{"Start":"04:35.270 ","End":"04:38.959","Text":"As we can see, our electric field at the base"},{"Start":"04:38.959 ","End":"04:44.765","Text":"is going to be parallel to our base\u0027s surface."},{"Start":"04:44.765 ","End":"04:49.130","Text":"Or in other words, it\u0027s going to be perpendicular to our ds."},{"Start":"04:49.130 ","End":"04:53.900","Text":"The problem with that is that we can only measure"},{"Start":"04:53.900 ","End":"04:59.355","Text":"our flux when the electric field is parallel to our ds."},{"Start":"04:59.355 ","End":"05:02.585","Text":"Here we can see it\u0027s perpendicular to our ds."},{"Start":"05:02.585 ","End":"05:07.805","Text":"That means that the electric flux through each base is going to be 0."},{"Start":"05:07.805 ","End":"05:10.320","Text":"We don\u0027t have to sum up on that."},{"Start":"05:11.360 ","End":"05:18.005","Text":"The only area where I have some contribution to the electric flux"},{"Start":"05:18.005 ","End":"05:24.335","Text":"is where my electric field is perpendicular to my surface or parallel to my ds."},{"Start":"05:24.335 ","End":"05:26.270","Text":"On the base, it\u0027s exactly the opposite."},{"Start":"05:26.270 ","End":"05:31.845","Text":"It\u0027s parallel to my surface and perpendicular to my ds. We don\u0027t add that."},{"Start":"05:31.845 ","End":"05:38.490","Text":"This is going to be equal to Q_in divided by Epsilon naught."},{"Start":"05:38.490 ","End":"05:42.940","Text":"Now let\u0027s see what our Q_in is equal to."},{"Start":"05:44.870 ","End":"05:51.830","Text":"Our Q_in is the amount of charge enclosed within our Gaussian surface."},{"Start":"05:51.830 ","End":"05:59.805","Text":"As we can see, we have a length of wire inside our Gaussian surface from here until here."},{"Start":"05:59.805 ","End":"06:02.810","Text":"We have this length of wire inside,"},{"Start":"06:02.810 ","End":"06:07.760","Text":"and we know that that\u0027s of length l. Our Q_in is simply going to be"},{"Start":"06:07.760 ","End":"06:12.830","Text":"equal to a uniform charge distribution per unit length,"},{"Start":"06:12.830 ","End":"06:15.140","Text":"which is this,"},{"Start":"06:15.140 ","End":"06:17.415","Text":"multiplied by the length that it takes up."},{"Start":"06:17.415 ","End":"06:19.545","Text":"That\u0027s going to be Lambda."},{"Start":"06:19.545 ","End":"06:21.125","Text":"Then the length is l,"},{"Start":"06:21.125 ","End":"06:27.600","Text":"because our Gaussian surface is of length l. Then divided by Epsilon naught."},{"Start":"06:27.600 ","End":"06:33.110","Text":"All the charge that\u0027s stored on the wire outside of our Gaussian surface doesn\u0027t"},{"Start":"06:33.110 ","End":"06:38.355","Text":"interest us because it isn\u0027t inside our Gaussian surface."},{"Start":"06:38.355 ","End":"06:40.335","Text":"That\u0027s why we have Q_in."},{"Start":"06:40.335 ","End":"06:44.020","Text":"What charge is inside our surface."},{"Start":"06:44.660 ","End":"06:49.100","Text":"Now we can see that the l\u0027s on both sides cancel out."},{"Start":"06:49.100 ","End":"06:54.605","Text":"We get that the electric field due to a uniformly charged wire"},{"Start":"06:54.605 ","End":"07:00.256","Text":"is equal to Lambda divided by 2 Pi Epsilon naught,"},{"Start":"07:00.256 ","End":"07:08.745","Text":"multiplied r, or we can switch out our 2Pi Epsilon naught and write it with"},{"Start":"07:08.745 ","End":"07:10.380","Text":"respect to k. That would be"},{"Start":"07:10.380 ","End":"07:18.007","Text":"2k Lambda divided by r. Both ways of writing this are correct."},{"Start":"07:18.007 ","End":"07:24.490","Text":"This is our electric field due to an infinite wire which is charged."},{"Start":"07:24.490 ","End":"07:25.810","Text":"Of course in Gauss\u0027s law,"},{"Start":"07:25.810 ","End":"07:28.330","Text":"we only get the magnitude of our electric field."},{"Start":"07:28.330 ","End":"07:30.955","Text":"If we want to find our electric field vector,"},{"Start":"07:30.955 ","End":"07:32.785","Text":"we denote our E with the vector,"},{"Start":"07:32.785 ","End":"07:35.230","Text":"and we just have to write down the direction."},{"Start":"07:35.230 ","End":"07:38.940","Text":"We already saw that our electric field is in the radial direction,"},{"Start":"07:38.940 ","End":"07:41.620","Text":"so we can just draw an I hat."},{"Start":"07:42.270 ","End":"07:45.625","Text":"This is the end of the lesson for this section."},{"Start":"07:45.625 ","End":"07:50.290","Text":"If however, you want to learn a little bit more about what we did here with"},{"Start":"07:50.290 ","End":"07:55.540","Text":"the idea of symmetry and how we got to know that we could solve this problem,"},{"Start":"07:55.540 ","End":"07:57.805","Text":"then carry on watching this video,"},{"Start":"07:57.805 ","End":"08:00.410","Text":"otherwise, you can just stop it now."},{"Start":"08:01.860 ","End":"08:08.665","Text":"How am I going to know that my electric field is only in this radial direction?"},{"Start":"08:08.665 ","End":"08:11.470","Text":"What I can do is I can use Coulomb\u0027s law,"},{"Start":"08:11.470 ","End":"08:13.525","Text":"which is what we learned in the previous chapter,"},{"Start":"08:13.525 ","End":"08:17.035","Text":"and I can split up my wire into lots of tiny little pieces."},{"Start":"08:17.035 ","End":"08:19.330","Text":"Let\u0027s choose 1 arbitrary piece."},{"Start":"08:19.330 ","End":"08:24.280","Text":"I have this arbitrary piece which is of length dl."},{"Start":"08:24.280 ","End":"08:26.253","Text":"It has some charge dq,"},{"Start":"08:26.253 ","End":"08:28.255","Text":"and because it\u0027s so small,"},{"Start":"08:28.255 ","End":"08:32.150","Text":"I can consider this as a point charge,"},{"Start":"08:32.150 ","End":"08:34.635","Text":"so I have a point charge in here."},{"Start":"08:34.635 ","End":"08:41.965","Text":"Now my point charge is going to exert an electric field in space,"},{"Start":"08:41.965 ","End":"08:46.270","Text":"and specifically, on this point over here that I\u0027m testing out."},{"Start":"08:46.270 ","End":"08:50.365","Text":"The electric field is going to go in a direction like this."},{"Start":"08:50.365 ","End":"08:58.405","Text":"I\u0027ve chosen that the distance of this dl along the z-axis is going to be some distance,"},{"Start":"08:58.405 ","End":"09:00.740","Text":"let\u0027s call it z."},{"Start":"09:02.130 ","End":"09:07.480","Text":"Then I\u0027m going to choose little piece of length on"},{"Start":"09:07.480 ","End":"09:13.510","Text":"the exact opposite side of this line over here that I drew,"},{"Start":"09:13.510 ","End":"09:17.364","Text":"and it\u0027s also going to be of length z just going in this direction."},{"Start":"09:17.364 ","End":"09:21.670","Text":"I\u0027m going to split my wire here into my little piece dl as well,"},{"Start":"09:21.670 ","End":"09:24.190","Text":"and so I can consider that a point charge,"},{"Start":"09:24.190 ","End":"09:33.830","Text":"and that\u0027s also going to exert an electric field like so at this point."},{"Start":"09:34.350 ","End":"09:37.885","Text":"Now due to the symmetry of the problem,"},{"Start":"09:37.885 ","End":"09:41.830","Text":"which means that both pieces are both of length dl."},{"Start":"09:41.830 ","End":"09:44.350","Text":"They both have the same charge on them,"},{"Start":"09:44.350 ","End":"09:45.970","Text":"which has some dq,"},{"Start":"09:45.970 ","End":"09:51.655","Text":"and they\u0027re both exactly the same distance away from this point over here in space."},{"Start":"09:51.655 ","End":"09:54.880","Text":"We can see that the electric fields are going to be the same,"},{"Start":"09:54.880 ","End":"09:57.080","Text":"just in the opposite diagonal."},{"Start":"09:57.080 ","End":"10:01.200","Text":"If we split up these diagonals into their components,"},{"Start":"10:01.200 ","End":"10:02.850","Text":"this is the z-component,"},{"Start":"10:02.850 ","End":"10:05.610","Text":"just because we said that this is the z-axis, it doesn\u0027t really matter."},{"Start":"10:05.610 ","End":"10:08.745","Text":"It could be the x-component or the y-component, doesn\u0027t really matter."},{"Start":"10:08.745 ","End":"10:14.350","Text":"We can see that the z-component of this diagonal arrow is going in"},{"Start":"10:14.350 ","End":"10:16.480","Text":"the left direction and the z-component of"},{"Start":"10:16.480 ","End":"10:20.800","Text":"this diagonal arrow is going in the positive z-direction."},{"Start":"10:20.800 ","End":"10:23.694","Text":"We can see that they\u0027re equal and opposite,"},{"Start":"10:23.694 ","End":"10:26.005","Text":"which means that they both cancel out."},{"Start":"10:26.005 ","End":"10:29.185","Text":"However, there are components going in the,"},{"Start":"10:29.185 ","End":"10:32.215","Text":"whatever you want to call this, x-direction, y-direction."},{"Start":"10:32.215 ","End":"10:34.090","Text":"We can see that they don\u0027t cancel out,"},{"Start":"10:34.090 ","End":"10:35.875","Text":"they\u0027re both going in the same direction,"},{"Start":"10:35.875 ","End":"10:41.440","Text":"so electric field is going to be in this direction,"},{"Start":"10:41.440 ","End":"10:44.240","Text":"let\u0027s say, the y direction."},{"Start":"10:44.820 ","End":"10:50.890","Text":"Now what we can see is that when we use Coulomb\u0027s law,"},{"Start":"10:50.890 ","End":"10:55.900","Text":"we can see that the electric field is only going to be in this one direction over here,"},{"Start":"10:55.900 ","End":"11:01.750","Text":"and we\u0027re not going to have any electric field components in the z-direction."},{"Start":"11:01.750 ","End":"11:08.390","Text":"This comes about from the fact that our wire is infinitely long."},{"Start":"11:09.210 ","End":"11:13.360","Text":"Now the next thing that we want to understand, that\u0027s fine,"},{"Start":"11:13.360 ","End":"11:15.460","Text":"we see that our electric field is only going to be"},{"Start":"11:15.460 ","End":"11:17.755","Text":"going in this direction because of symmetry."},{"Start":"11:17.755 ","End":"11:20.620","Text":"Perfect. But how come is the size of"},{"Start":"11:20.620 ","End":"11:28.670","Text":"the electric field going to be uniform along the length of this infinitely long wire?"},{"Start":"11:28.860 ","End":"11:34.630","Text":"This also comes about from the fact that our wire is infinitely long."},{"Start":"11:34.630 ","End":"11:39.385","Text":"Here specifically when I wanted to show the symmetry in the problem,"},{"Start":"11:39.385 ","End":"11:41.785","Text":"so I said that my origin was here."},{"Start":"11:41.785 ","End":"11:43.975","Text":"But because my wire is infinitely long,"},{"Start":"11:43.975 ","End":"11:47.785","Text":"I can shift it right or left, it doesn\u0027t really matter,"},{"Start":"11:47.785 ","End":"11:53.320","Text":"and I can choose any other arbitrary point and choose that to be my origin,"},{"Start":"11:53.320 ","End":"11:57.040","Text":"and it won\u0027t make any difference to the problem."},{"Start":"11:57.040 ","End":"12:02.305","Text":"Which means that if I could choose my electric field here to be equal to E,"},{"Start":"12:02.305 ","End":"12:07.435","Text":"that also here it\u0027s going to be equal to E and here and so on and so forth."},{"Start":"12:07.435 ","End":"12:12.235","Text":"As a rule, if our size or a problem"},{"Start":"12:12.235 ","End":"12:17.650","Text":"doesn\u0027t change along one of the axes in this infinitely long wire,"},{"Start":"12:17.650 ","End":"12:21.610","Text":"our problem doesn\u0027t change along the z-axis,"},{"Start":"12:21.610 ","End":"12:30.685","Text":"then that means that the derivative of every magnitude according to this axis,"},{"Start":"12:30.685 ","End":"12:32.980","Text":"here specifically it\u0027s z-axis,"},{"Start":"12:32.980 ","End":"12:36.040","Text":"is going to be equal to 0."},{"Start":"12:36.040 ","End":"12:41.800","Text":"Here specifically we\u0027re dealing with our electric field Er."},{"Start":"12:41.840 ","End":"12:46.275","Text":"That means that if we take the derivative of Er with respect to z,"},{"Start":"12:46.275 ","End":"12:50.820","Text":"it\u0027s going to be equal to 0 because the size or the magnitude of"},{"Start":"12:50.820 ","End":"12:58.060","Text":"our electric field that only has variable r isn\u0027t changing along the z-axis."},{"Start":"12:58.060 ","End":"13:06.220","Text":"We can see that if I take the derivative of my Er with respect to r,"},{"Start":"13:06.220 ","End":"13:09.655","Text":"then this derivative does not equal 0."},{"Start":"13:09.655 ","End":"13:11.635","Text":"Because as I move further away,"},{"Start":"13:11.635 ","End":"13:14.065","Text":"as my r increases,"},{"Start":"13:14.065 ","End":"13:18.505","Text":"my value for Er is going to change."},{"Start":"13:18.505 ","End":"13:22.015","Text":"Now another symmetry that we have in this problem,"},{"Start":"13:22.015 ","End":"13:23.575","Text":"or in this example,"},{"Start":"13:23.575 ","End":"13:26.860","Text":"is symmetry with respect to Theta."},{"Start":"13:26.860 ","End":"13:34.480","Text":"If we imagine that I rotate my wire at some angle Theta,"},{"Start":"13:34.480 ","End":"13:39.410","Text":"we can see that we\u0027re going to still be left with the exact same problem."},{"Start":"13:39.570 ","End":"13:46.465","Text":"We can see that if we take the derivative of our electric field,"},{"Start":"13:46.465 ","End":"13:48.115","Text":"which only has variable r,"},{"Start":"13:48.115 ","End":"13:50.395","Text":"and we take the derivative with respect to Theta,"},{"Start":"13:50.395 ","End":"13:52.420","Text":"we\u0027re still going to get 0."},{"Start":"13:52.420 ","End":"13:59.060","Text":"This means that we have symmetry in the z-direction and in the Theta direction."},{"Start":"13:59.280 ","End":"14:03.535","Text":"What we can see here is that our electric field is"},{"Start":"14:03.535 ","End":"14:07.300","Text":"only dependent on the distance that we have from the wire,"},{"Start":"14:07.300 ","End":"14:11.440","Text":"and it\u0027s unchanged if we move along z-axis or we"},{"Start":"14:11.440 ","End":"14:16.270","Text":"rotate our wire where we move around a wire."},{"Start":"14:16.270 ","End":"14:19.330","Text":"That is why our electric field is going to be"},{"Start":"14:19.330 ","End":"14:28.850","Text":"constant along our Gaussian surface of a cylinder."}],"ID":22359},{"Watched":false,"Name":"Exercise 2","Duration":"13m 52s","ChapterTopicVideoID":21406,"CourseChapterTopicPlaylistID":246895,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/21406.jpeg","UploadDate":"2020-04-21T08:48:13.8530000","DurationForVideoObject":"PT13M52S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"03:37.105","Text":"Hello."},{"Start":"03:37.105 ","End":"04:53.879","Text":"In"},{"Start":"04:53.879 ","End":"07:39.239","Text":"this"},{"Start":"07:39.239 ","End":"13:09.960","Text":"question"},{"Start":"13:09.960 ","End":"13:10.123","Text":", we\u0027re going to be using Gauss\u0027s law in order"},{"Start":"13:10.123 ","End":"13:10.300","Text":"to find the electric field of a spherical shell."},{"Start":"13:10.300 ","End":"13:10.464","Text":"Now what\u0027s important to note at this stage is"},{"Start":"13:10.464 ","End":"13:10.652","Text":"that when we\u0027re speaking about a spherical shell,"},{"Start":"13:10.652 ","End":"13:10.803","Text":"that means that we\u0027re just talking about"},{"Start":"13:10.803 ","End":"13:11.079","Text":"the surface area of the sphere and that inside the sphere there is nothing."},{"Start":"13:11.079 ","End":"13:11.204","Text":"This is a hollow spherical shell."},{"Start":"13:11.204 ","End":"13:11.477","Text":"We\u0027ve just learned Gauss\u0027s law and now we\u0027re going to learn how to apply it."},{"Start":"13:11.477 ","End":"13:11.681","Text":"Now, this sphere with the charge is a classic example of"},{"Start":"13:11.681 ","End":"13:11.895","Text":"how to use Gauss\u0027s law in order to find the electric field."},{"Start":"13:11.895 ","End":"13:12.128","Text":"Also throughout the entire course and also even in magnetism,"},{"Start":"13:12.128 ","End":"13:12.426","Text":"we\u0027re going to be using this idea, what we\u0027re going to be learning in this lesson."},{"Start":"13:12.426 ","End":"13:12.667","Text":"So it\u0027s very important that you understand what\u0027s going on here."},{"Start":"13:12.667 ","End":"13:12.960","Text":"This spherical shell has radius of R and the charge of the spherical shell is Q."},{"Start":"13:12.960 ","End":"13:13.135","Text":"That means that we have charge distribution or"},{"Start":"13:13.135 ","End":"13:13.304","Text":"a uniform charge distribution per unit area"},{"Start":"13:13.304 ","End":"13:13.537","Text":"because we\u0027re dealing with a spherical shell so that\u0027s area,"},{"Start":"13:13.537 ","End":"13:13.809","Text":"and it\u0027s distributed evenly or uniformly across this spherical shell,"},{"Start":"13:13.809 ","End":"13:13.946","Text":"and so the total charge is equal to Q."},{"Start":"13:13.946 ","End":"13:14.046","Text":"According to Gauss\u0027s law,"},{"Start":"13:14.046 ","End":"13:14.306","Text":"the first thing that I have to do right now is to find my electric flux,"},{"Start":"13:14.306 ","End":"13:14.525","Text":"which is given by the integral of my electric field dot DS."},{"Start":"13:14.525 ","End":"13:14.831","Text":"Now, I have a little bit of a problem here, because the electric field is unknown."},{"Start":"13:14.831 ","End":"13:15.128","Text":"This is exactly the variable that I\u0027m trying to calculate. How do I solve this?"},{"Start":"13:15.128 ","End":"13:15.246","Text":"I have to find some outer shell,"},{"Start":"13:15.246 ","End":"13:15.455","Text":"so if this is our shape, I have to find an external shell."},{"Start":"13:15.455 ","End":"13:15.671","Text":"Then I can sum up the electric field on my external shell."},{"Start":"13:15.671 ","End":"13:15.935","Text":"Remember it has to be a closed area or a closed shape or a closed shell."},{"Start":"13:15.935 ","End":"13:16.160","Text":"That\u0027s why we have this integral sign with a circle around."},{"Start":"13:16.160 ","End":"13:16.357","Text":"Then my electric field is going to be uniform across"},{"Start":"13:16.357 ","End":"13:16.586","Text":"that external casing or shallow however you want to call it."},{"Start":"13:16.586 ","End":"13:16.831","Text":"Then I\u0027ll be able to say that my electric flux will be equal to E.S,"},{"Start":"13:16.831 ","End":"13:16.973","Text":"where S is the total area of my casing."},{"Start":"13:16.973 ","End":"13:17.215","Text":"Here we\u0027re saying that E is uniform across our external shell."},{"Start":"13:17.215 ","End":"13:17.433","Text":"You have to differentiate between this spherical shell"},{"Start":"13:17.433 ","End":"13:17.674","Text":"and the external shell that we\u0027re going to define around this."},{"Start":"13:17.674 ","End":"13:17.971","Text":"Now, we\u0027re going to see that if our electric field is dependent on the radius,"},{"Start":"13:17.971 ","End":"13:18.083","Text":"we can still use Gauss\u0027s law."},{"Start":"13:18.083 ","End":"13:18.273","Text":"However, if our E field is dependent on our angle,"},{"Start":"13:18.273 ","End":"13:18.586","Text":"then we cannot use Gauss\u0027s law and that\u0027s because if it\u0027s dependent on the angle,"},{"Start":"13:18.586 ","End":"13:18.864","Text":"then there is no symmetry in the problem in which case we cannot use this."},{"Start":"13:18.864 ","End":"13:19.054","Text":"Whereas if it\u0027s dependent on radius then there is"},{"Start":"13:19.054 ","End":"13:19.252","Text":"symmetry in the problem and we can use Gauss\u0027s law."},{"Start":"13:19.252 ","End":"13:19.571","Text":"We\u0027re going to choose that our external shell is going to have to be some shape that"},{"Start":"13:19.571 ","End":"13:19.905","Text":"fits perfectly around this spherical shell and that the shape or the external shell,"},{"Start":"13:19.905 ","End":"13:20.028","Text":"external shape is also closed."},{"Start":"13:20.028 ","End":"13:20.230","Text":"We can obviously choose that or we can maybe see that"},{"Start":"13:20.230 ","End":"13:20.452","Text":"a natural shape to choose is some spherical outer shell."},{"Start":"13:20.452 ","End":"13:20.651","Text":"This is a spherical outer shell. We also have area."},{"Start":"13:20.651 ","End":"13:20.918","Text":"It\u0027s not just a circle over here, it\u0027s in 3-dimensions as well."},{"Start":"13:20.918 ","End":"13:21.071","Text":"We have our 3-dimensional shape."},{"Start":"13:21.071 ","End":"13:21.267","Text":"We can say that the radius of this external shell,"},{"Start":"13:21.267 ","End":"13:21.466","Text":"is going to be r. Now what we\u0027re going to do is we\u0027re"},{"Start":"13:21.466 ","End":"13:21.686","Text":"going to draw the electric field on our external shell."},{"Start":"13:21.686 ","End":"13:21.897","Text":"We can see that the electric field is still going to be"},{"Start":"13:21.897 ","End":"13:22.156","Text":"exiting this outer shell in the manner that it\u0027s perpendicular."},{"Start":"13:22.156 ","End":"13:22.280","Text":"Now we can see through this that"},{"Start":"13:22.280 ","End":"13:22.511","Text":"our electric field is going to be in the radial direction."},{"Start":"13:22.511 ","End":"13:22.799","Text":"Now this is super important. Every single time we\u0027re using Gauss\u0027s law,"},{"Start":"13:22.799 ","End":"13:22.982","Text":"when using a spherical shell or a full sphere,"},{"Start":"13:22.982 ","End":"13:23.323","Text":"it doesn\u0027t matter, our electric field is always going to be in the radial direction."},{"Start":"13:23.323 ","End":"13:23.478","Text":"This is very similar to a point charge."},{"Start":"13:23.478 ","End":"13:23.691","Text":"Now let\u0027s just do a little sanity check and check that"},{"Start":"13:23.691 ","End":"13:23.954","Text":"our electric field is in fact uniform across our external shell."},{"Start":"13:23.954 ","End":"13:24.277","Text":"Because we can see that our electric field is going out in the radial direction,"},{"Start":"13:24.277 ","End":"13:24.533","Text":"we can see that any which way we spin our inner spherical shell,"},{"Start":"13:24.533 ","End":"13:24.719","Text":"the problem is going to be exactly identical."},{"Start":"13:24.719 ","End":"13:24.919","Text":"There\u0027s no change if we rotate our system around."},{"Start":"13:24.919 ","End":"13:25.227","Text":"That means that we\u0027re working with the symmetry and that means that we have"},{"Start":"13:25.227 ","End":"13:25.554","Text":"a uniform electric field throughout and that means that we can use Gauss\u0027s law."},{"Start":"13:25.554 ","End":"13:25.873","Text":"Brilliant. Now we can move along and we can figure this out. Let\u0027s write this."},{"Start":"13:25.873 ","End":"13:25.953","Text":"We have that, our E,"},{"Start":"13:25.953 ","End":"13:26.238","Text":"we\u0027re still trying to find it, so E.S or multiplied by S. What is the S?"},{"Start":"13:26.238 ","End":"13:26.413","Text":"It\u0027s the total area of our external shell."},{"Start":"13:26.413 ","End":"13:26.753","Text":"Because we\u0027re dealing with the surface area of a sphere, it\u0027s going to be equal to,"},{"Start":"13:26.753 ","End":"13:26.973","Text":"the surface area of a sphere is equal to 4Pir^2,"},{"Start":"13:26.973 ","End":"13:27.158","Text":"where r is the radius of our external sphere."},{"Start":"13:27.158 ","End":"13:27.463","Text":"I wrote a little note that this S is the surface area of the external shell."},{"Start":"13:27.463 ","End":"13:27.723","Text":"Don\u0027t confuse it with the surface area of your internal shape."},{"Start":"13:27.723 ","End":"13:28.034","Text":"You\u0027re talking about your external shell that you designed for yourself."},{"Start":"13:28.034 ","End":"13:28.239","Text":"This is the left side of Gauss\u0027s law and as we know,"},{"Start":"13:28.239 ","End":"13:28.455","Text":"this has to be equal to the right side of Gauss\u0027s Law,"},{"Start":"13:28.455 ","End":"13:28.658","Text":"which is equal to Q_in divided by Epsilon_naught."},{"Start":"13:28.658 ","End":"13:28.723","Text":"What is our Q_in?"},{"Start":"13:28.723 ","End":"13:29.100","Text":"Specifically here we\u0027re told that the charge of this inner spherical shell is equal to Q."},{"Start":"13:29.100 ","End":"13:29.392","Text":"In our example over here, we have simply Q divided by Epsilon_naught."},{"Start":"13:29.392 ","End":"13:29.731","Text":"If we were given, for instance, that the uniform charge per surface area Sigma,"},{"Start":"13:29.731 ","End":"13:30.108","Text":"then we would have written Sigma multiplied by the surface area of this internal shell,"},{"Start":"13:30.108 ","End":"13:30.460","Text":"which would have been 4Pir^2 and then all of that divided by Epsilon_naught."},{"Start":"13:30.460 ","End":"13:30.689","Text":"Now because we\u0027re trying to find our electric field,"},{"Start":"13:30.689 ","End":"13:31.053","Text":"we can suddenly eyesight out our E. We\u0027re going to divide both sides by 4Pir^2."},{"Start":"13:31.053 ","End":"13:31.355","Text":"We\u0027ll have that E is equal to Q divided by 4Pir^2 Epsilon_naught."},{"Start":"13:31.355 ","End":"13:31.530","Text":"This is equal to kQ divided by r^2."},{"Start":"13:31.530 ","End":"13:31.647","Text":"Why kQ? Because as we know,"},{"Start":"13:31.647 ","End":"13:31.753","Text":"k is equal to 1 divided by"},{"Start":"13:31.753 ","End":"13:32.050","Text":"4Pi Epsilon_naught and here we have 1 divided by 4Pi Epsilon_naught."},{"Start":"13:32.050 ","End":"13:32.253","Text":"You can write it either way, both are correct."},{"Start":"13:32.253 ","End":"13:32.622","Text":"This or this, both are right, is the electric field of a spherical shell or a sphere."},{"Start":"13:32.622 ","End":"13:32.830","Text":"It\u0027s also the electric field of a point charge."},{"Start":"13:32.830 ","End":"13:33.039","Text":"When I\u0027m located outside of my external shell,"},{"Start":"13:33.039 ","End":"13:33.421","Text":"we\u0027ll see that the electric field for all those 3 are going to be exactly the same."},{"Start":"13:33.421 ","End":"13:33.680","Text":"Of course, the electric field is in the radial direction."},{"Start":"13:33.680 ","End":"13:33.891","Text":"If you study more advanced courses in physics,"},{"Start":"13:33.891 ","End":"13:34.153","Text":"you\u0027ll see that a point charge also doesn\u0027t really exist,"},{"Start":"13:34.153 ","End":"13:34.398","Text":"it\u0027s simply a super tiny spherical shell. There we go."},{"Start":"13:34.398 ","End":"13:34.518","Text":"It\u0027s the exact same thing."},{"Start":"13:34.518 ","End":"13:34.823","Text":"Now, Gauss\u0027s law always gives the magnitude of the electric field."},{"Start":"13:34.823 ","End":"13:35.048","Text":"However, if I want to find the vector quantities,"},{"Start":"13:35.048 ","End":"13:35.231","Text":"so that means magnitude and direction."},{"Start":"13:35.231 ","End":"13:35.292","Text":"First of all,"},{"Start":"13:35.292 ","End":"13:35.653","Text":"I denote my vector with this arrow on top and before we even began the question,"},{"Start":"13:35.653 ","End":"13:35.870","Text":"we already said that our E field is going to be in"},{"Start":"13:35.870 ","End":"13:36.165","Text":"the radial direction so I\u0027m just going to add my r-hat over here."},{"Start":"13:36.165 ","End":"13:36.299","Text":"Something super important,"},{"Start":"13:36.299 ","End":"13:36.670","Text":"this answer for my E field is going to be only correct so long as this outer shell,"},{"Start":"13:36.670 ","End":"13:37.072","Text":"this blue line, so this radius r is bigger than the radius of my inner spherical shell."},{"Start":"13:37.072 ","End":"13:37.431","Text":"That means that my blue line is truly outside of this inner spherical shell."},{"Start":"13:37.431 ","End":"13:37.802","Text":"Now what happens if my r is smaller than my R. Let\u0027s denote this internal shell."},{"Start":"13:37.802 ","End":"13:38.069","Text":"It\u0027s going to have radius of r. Our shell is just smaller"},{"Start":"13:38.069 ","End":"13:38.352","Text":"and it\u0027s within the spherical shell given in the question."},{"Start":"13:38.352 ","End":"13:38.732","Text":"Right now we\u0027re dealing with r is smaller than R. What is our E field going to be?"},{"Start":"13:38.732 ","End":"13:38.938","Text":"Our E field is still going to be, or rather,"},{"Start":"13:38.938 ","End":"13:39.158","Text":"let\u0027s write our flux is still going to be E.S,"},{"Start":"13:39.158 ","End":"13:39.475","Text":"which is still going to be equal to E multiplied by 4Pir^2."},{"Start":"13:39.475 ","End":"13:39.781","Text":"This is never changing because it\u0027s the same due to symmetry."},{"Start":"13:39.781 ","End":"13:40.051","Text":"We can see that if we would have an electric field here,"},{"Start":"13:40.051 ","End":"13:40.370","Text":"we could draw our arrows in the radial direction. That\u0027s great."},{"Start":"13:40.370 ","End":"13:40.740","Text":"Then we know that this is equal to Q_in divided by Epsilon_naught from here."},{"Start":"13:40.740 ","End":"13:41.027","Text":"Now here, because our charge Q is on the spherical shell,"},{"Start":"13:41.027 ","End":"13:41.435","Text":"that means that there\u0027s 0 charge inside because our spherical shell is hollow."},{"Start":"13:41.435 ","End":"13:41.559","Text":"This is a hollow sphere."},{"Start":"13:41.559 ","End":"13:41.664","Text":"Our Q_in is equal to 0,"},{"Start":"13:41.664 ","End":"13:42.111","Text":"therefore we\u0027ll get that our electric field within our spherical shell is equal to 0."},{"Start":"13:42.111 ","End":"13:42.358","Text":"Now, I can write that my E field is equal to 0 when r"},{"Start":"13:42.358 ","End":"13:42.634","Text":"is smaller than R and kQ divided by r^2"},{"Start":"13:42.634 ","End":"13:42.873","Text":"in the radial direction when r is bigger than"},{"Start":"13:42.873 ","End":"13:43.165","Text":"R. Now what I can do is I can draw this function."},{"Start":"13:43.165 ","End":"13:43.454","Text":"Now, we can see that this function isn\u0027t continuous."},{"Start":"13:43.454 ","End":"13:43.798","Text":"Let\u0027s say that this value is the radius of our spherical shell."},{"Start":"13:43.798 ","End":"13:43.958","Text":"We can see that our function,"},{"Start":"13:43.958 ","End":"13:44.289","Text":"our E field is equal to 0 so long as our R, this is the size of R,"},{"Start":"13:44.289 ","End":"13:44.813","Text":"so long as our R is smaller than R, is smaller than the radius of our spherical shell."},{"Start":"13:44.813 ","End":"13:45.177","Text":"But then as soon as our r is slightly bigger than this R,"},{"Start":"13:45.177 ","End":"13:45.396","Text":"so our function begins at some value."},{"Start":"13:45.396 ","End":"13:45.832","Text":"Then we can see that it approaches 0 as a function of 1 divided by r^2."},{"Start":"13:45.832 ","End":"13:46.346","Text":"It\u0027s going to look something like so. What is this point over here? What is its value?"},{"Start":"13:46.346 ","End":"13:46.751","Text":"If I simply substitute in R into this equation over here,"},{"Start":"13:46.751 ","End":"13:47.377","Text":"we\u0027ll see that my value in the y-direction over here is going to be kQ divided by R^2."},{"Start":"13:47.377 ","End":"13:47.687","Text":"What\u0027s important to know is that we can see that"},{"Start":"13:47.687 ","End":"13:48.133","Text":"our electric field equation is not a continuous equation and that"},{"Start":"13:48.133 ","End":"13:48.526","Text":"there\u0027s going to be a jump in the electric field when we go"},{"Start":"13:48.526 ","End":"13:48.944","Text":"from 0 and we\u0027re inside the spherical shell with charge Q."},{"Start":"13:48.944 ","End":"13:49.466","Text":"Then when we reach the surface of the shell and start moving away from"},{"Start":"13:49.466 ","End":"13:50.107","Text":"the spherical shell and our electric field is going to be reduced depending on"},{"Start":"13:50.107 ","End":"13:50.897","Text":"how far away we are from the spherical shell as a function of 1 divided by r^2."},{"Start":"13:50.897 ","End":"13:52.300","Text":"That\u0027s the end of this lesson."}],"ID":22360},{"Watched":false,"Name":"Exercise 3","Duration":"17m 42s","ChapterTopicVideoID":21407,"CourseChapterTopicPlaylistID":246895,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/21407.jpeg","UploadDate":"2020-04-21T08:51:37.4000000","DurationForVideoObject":"PT17M42S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.115","Text":"Hello, in this question,"},{"Start":"00:02.115 ","End":"00:09.645","Text":"we\u0027re being asked to find our electric field due to an infinitely long cylinder."},{"Start":"00:09.645 ","End":"00:15.560","Text":"We can say that its axis of symmetry is the z-axis and the cylinders infinitely long,"},{"Start":"00:15.560 ","End":"00:19.995","Text":"we\u0027re also being told that the cylinder is a solid cylinder and that"},{"Start":"00:19.995 ","End":"00:28.125","Text":"its charged distribution or its uniform charge density is given as Rho."},{"Start":"00:28.125 ","End":"00:33.135","Text":"Rho is the uniform charge distribution per unit volume."},{"Start":"00:33.135 ","End":"00:38.825","Text":"The radius of our cylinder"},{"Start":"00:38.825 ","End":"00:44.150","Text":"is given by R and we want to find the electric field."},{"Start":"00:44.150 ","End":"00:47.329","Text":"We\u0027re going to split this into 2 cases."},{"Start":"00:47.329 ","End":"00:52.055","Text":"1, when we\u0027re measuring the electric field"},{"Start":"00:52.055 ","End":"00:55.280","Text":"outside of our cylinder and another time when we\u0027re"},{"Start":"00:55.280 ","End":"00:59.490","Text":"measuring the electric field inside the cylinder."},{"Start":"01:00.530 ","End":"01:04.215","Text":"Now let\u0027s take a random point in space,"},{"Start":"01:04.215 ","End":"01:07.610","Text":"let\u0027s say it\u0027s over here and I\u0027m going to say that"},{"Start":"01:07.610 ","End":"01:12.500","Text":"its distance from the center of our solid cylinder is"},{"Start":"01:12.500 ","End":"01:16.850","Text":"r. Now what we\u0027re going to do is we\u0027re going to find"},{"Start":"01:16.850 ","End":"01:22.490","Text":"the electric field for when r is bigger than R,"},{"Start":"01:22.490 ","End":"01:27.030","Text":"so that means that we are right now located outside of the sphere."},{"Start":"01:27.860 ","End":"01:31.160","Text":"What I can see is just like when we were dealing"},{"Start":"01:31.160 ","End":"01:34.130","Text":"with the electric field due to an infinite wire."},{"Start":"01:34.130 ","End":"01:37.880","Text":"We can see that the electric field due to an infinite cylinder is"},{"Start":"01:37.880 ","End":"01:42.110","Text":"also going to have symmetry in the z-direction because it\u0027s"},{"Start":"01:42.110 ","End":"01:45.875","Text":"infinite and in the Theta direction because"},{"Start":"01:45.875 ","End":"01:50.915","Text":"our electric field is going out radially from a cylinder."},{"Start":"01:50.915 ","End":"01:55.490","Text":"We already know that our electric field is going to be in"},{"Start":"01:55.490 ","End":"01:59.840","Text":"the radial direction so we can draw it like that."},{"Start":"01:59.840 ","End":"02:03.109","Text":"Now, also, because we have symmetry,"},{"Start":"02:03.109 ","End":"02:05.660","Text":"we know that the electric field is going to be"},{"Start":"02:05.660 ","End":"02:11.206","Text":"uniform all the way along and around the cylinder."},{"Start":"02:11.206 ","End":"02:15.600","Text":"That means that I can choose my Gaussian surface like"},{"Start":"02:15.600 ","End":"02:22.840","Text":"so I can say that it is also a cylinder,"},{"Start":"02:23.350 ","End":"02:30.310","Text":"then we go like this and like so."},{"Start":"02:30.310 ","End":"02:35.330","Text":"This is a cylinder with radius r and this length,"},{"Start":"02:35.330 ","End":"02:43.695","Text":"the length of the cylinder is L. We\u0027re going to start by using Gauss\u0027s law."},{"Start":"02:43.695 ","End":"02:49.385","Text":"First, we\u0027re trying to find the electric flux and we know that that is given by"},{"Start":"02:49.385 ","End":"02:55.760","Text":"the closed integral E.ds."},{"Start":"02:55.760 ","End":"02:58.440","Text":"Now, of course, E, we don\u0027t know,"},{"Start":"02:58.440 ","End":"03:00.810","Text":"this is what we\u0027re trying to find."},{"Start":"03:00.810 ","End":"03:06.950","Text":"Now because our electric field is going to be uniform across our Gaussian surface,"},{"Start":"03:06.950 ","End":"03:11.985","Text":"so we know that we can write this rather as E.S,"},{"Start":"03:11.985 ","End":"03:18.410","Text":"so the electric field multiplied by the total surface area of our Gaussian surface."},{"Start":"03:18.410 ","End":"03:24.260","Text":"Here, our surface area is simply going to be the body of"},{"Start":"03:24.260 ","End":"03:31.070","Text":"our Gaussian surface which is just a large hollow cylinder and of course,"},{"Start":"03:31.070 ","End":"03:35.210","Text":"we don\u0027t have to take into consideration the flux coming out of"},{"Start":"03:35.210 ","End":"03:37.550","Text":"the basis for the same reason that we saw in"},{"Start":"03:37.550 ","End":"03:41.000","Text":"the video for finding the electric field due to an infinite wire."},{"Start":"03:41.000 ","End":"03:47.975","Text":"We saw that the electric field lines are parallel to our surfaces"},{"Start":"03:47.975 ","End":"03:50.240","Text":"or perpendicular to IDS so"},{"Start":"03:50.240 ","End":"03:55.550","Text":"their contribution to the electric flux is going through the basis,"},{"Start":"03:55.550 ","End":"03:57.710","Text":"is going to be equal to 0."},{"Start":"03:57.710 ","End":"04:04.175","Text":"Then we can write that this is equal to e and then the surface area of"},{"Start":"04:04.175 ","End":"04:12.515","Text":"the circular body section of a cylinder is equal to 2Pir which is the parameter,"},{"Start":"04:12.515 ","End":"04:15.859","Text":"and then multiplied by the length of the cylinder"},{"Start":"04:15.859 ","End":"04:21.965","Text":"L. Now we can write that this is equal to the right side of our Gauss\u0027s law,"},{"Start":"04:21.965 ","End":"04:27.120","Text":"which is equal to Q_in divided by Epsilon_naught."},{"Start":"04:27.120 ","End":"04:29.580","Text":"What is Q_in?"},{"Start":"04:29.580 ","End":"04:36.775","Text":"Q_in we can see that we have a section of cylinder this up until approximately"},{"Start":"04:36.775 ","End":"04:45.275","Text":"here which has some charge density per volume which is given by Rho and it\u0027s this length."},{"Start":"04:45.275 ","End":"04:47.855","Text":"Our Q_in, let\u0027s write it over here,"},{"Start":"04:47.855 ","End":"04:52.340","Text":"is going to be the integral on Rhodv,"},{"Start":"04:52.340 ","End":"04:56.584","Text":"so that is going to be equal to Rho and then multiplied"},{"Start":"04:56.584 ","End":"05:01.220","Text":"by the volume of this section of cylinder,"},{"Start":"05:01.260 ","End":"05:10.550","Text":"that is going to be equal to Pir^2 because the area of"},{"Start":"05:10.550 ","End":"05:15.410","Text":"this section is pi capital R^2 and then we have to"},{"Start":"05:15.410 ","End":"05:20.600","Text":"sum that up along the entire length that is enclosed in our Gaussian surface,"},{"Start":"05:20.600 ","End":"05:25.535","Text":"so multiplied by L. Now,"},{"Start":"05:25.535 ","End":"05:29.240","Text":"even though we still have uniform charge density on the sections of"},{"Start":"05:29.240 ","End":"05:32.870","Text":"this infinite cylinder which is outside our Gaussian surface,"},{"Start":"05:32.870 ","End":"05:38.285","Text":"we\u0027re specifically finding Q_in the charge enclosed within the Gaussian surface,"},{"Start":"05:38.285 ","End":"05:40.295","Text":"so that\u0027s why we do it like so."},{"Start":"05:40.295 ","End":"05:48.445","Text":"Now let\u0027s substitute in our Q_in over here so this is going to be RhoPir^2L."},{"Start":"05:48.445 ","End":"05:52.070","Text":"We can see that we can cross out our Pi in"},{"Start":"05:52.070 ","End":"05:56.315","Text":"both sides and our L on both sides and now we can isolate out our e,"},{"Start":"05:56.315 ","End":"05:58.105","Text":"which is what we\u0027re trying to find."},{"Start":"05:58.105 ","End":"06:01.280","Text":"We can see that our the magnitude of"},{"Start":"06:01.280 ","End":"06:05.105","Text":"our electric field because from Gauss\u0027s law we get the magnitude,"},{"Start":"06:05.105 ","End":"06:15.160","Text":"is going to be equal to Rho^2 divided by 2rEpsilon_naught."},{"Start":"06:15.160 ","End":"06:19.500","Text":"I forgot to substitute in the Epsilon_naught over here."},{"Start":"06:20.390 ","End":"06:25.130","Text":"Now we have the magnitude of our electric field when we\u0027re"},{"Start":"06:25.130 ","End":"06:29.637","Text":"located outside of our solid infinite cylinder."},{"Start":"06:29.637 ","End":"06:34.430","Text":"If we want to have this in vector quantities so that we have size and direction,"},{"Start":"06:34.430 ","End":"06:37.610","Text":"we can just add this n because we know that"},{"Start":"06:37.610 ","End":"06:41.790","Text":"the electric field is in the radial direction."},{"Start":"06:43.370 ","End":"06:46.910","Text":"Now, let\u0027s see what the electric field is equal"},{"Start":"06:46.910 ","End":"06:50.900","Text":"to when we\u0027re located inside the cylinder,"},{"Start":"06:50.900 ","End":"06:54.630","Text":"so that means that we\u0027re located somewhere here."},{"Start":"06:56.590 ","End":"06:59.780","Text":"Now what I\u0027m going to do is just like before when"},{"Start":"06:59.780 ","End":"07:02.330","Text":"I was located outside of my electric field."},{"Start":"07:02.330 ","End":"07:10.620","Text":"I\u0027m going to draw a Gaussian surface around this area."},{"Start":"07:10.620 ","End":"07:13.600","Text":"Here we go, here we have a Gaussian surface,"},{"Start":"07:13.600 ","End":"07:19.296","Text":"now imagine that it\u0027s located in the center."},{"Start":"07:19.296 ","End":"07:24.435","Text":"In blue, we see that we have our Gaussian surface of radius r,"},{"Start":"07:24.435 ","End":"07:28.020","Text":"and we\u0027re located inside of our cylinder."},{"Start":"07:28.020 ","End":"07:29.955","Text":"Now, just like before, what we saw,"},{"Start":"07:29.955 ","End":"07:37.125","Text":"is that our electric field is going to be pointing outwards in"},{"Start":"07:37.125 ","End":"07:41.565","Text":"the radial direction just like before and due to the symmetry"},{"Start":"07:41.565 ","End":"07:46.215","Text":"along the z-axis because it\u0027s an infinitely long cylinder,"},{"Start":"07:46.215 ","End":"07:49.560","Text":"and because when we rotate this around,"},{"Start":"07:49.560 ","End":"07:51.675","Text":"our problem remains exactly the same."},{"Start":"07:51.675 ","End":"07:53.430","Text":"We have symmetry, our electric field,"},{"Start":"07:53.430 ","End":"07:55.170","Text":"is going to be uniform,"},{"Start":"07:55.170 ","End":"07:57.179","Text":"and in the radial direction."},{"Start":"07:57.179 ","End":"08:00.780","Text":"Perfect, so we can use our Gauss\u0027s law."},{"Start":"08:00.780 ","End":"08:03.825","Text":"Let\u0027s work out our electric flux,"},{"Start":"08:03.825 ","End":"08:11.715","Text":"and this is equal to the closed integral on E.ds, just like before."},{"Start":"08:11.715 ","End":"08:17.295","Text":"As we just said, because of symmetry in the uniform charge across our Gaussian surface,"},{"Start":"08:17.295 ","End":"08:21.465","Text":"we can say that this is equal to E.S. What is this equal to?"},{"Start":"08:21.465 ","End":"08:23.670","Text":"we have E, which is what we\u0027re trying to find,"},{"Start":"08:23.670 ","End":"08:26.880","Text":"and then our surface area."},{"Start":"08:26.880 ","End":"08:30.570","Text":"Now again, we don\u0027t have to use the surface area of the bases,"},{"Start":"08:30.570 ","End":"08:33.825","Text":"because they contribute 0 to the flux,"},{"Start":"08:33.825 ","End":"08:37.710","Text":"so we\u0027re just working out of this curved body."},{"Start":"08:37.710 ","End":"08:44.400","Text":"The surface area of this is going to be equal to 2PirL."},{"Start":"08:44.400 ","End":"08:51.195","Text":"Just like before, just this time our r is smaller than R. Great."},{"Start":"08:51.195 ","End":"08:55.470","Text":"Now we know that this is going to be equal to Q_in divided by Epsilon_naught,"},{"Start":"08:55.470 ","End":"09:00.570","Text":"which is the right-side of our Gauss\u0027s law."},{"Start":"09:00.570 ","End":"09:05.175","Text":"Now let\u0027s see what our Q_in is going to be equal to."},{"Start":"09:05.175 ","End":"09:11.985","Text":"Here we are going to have a different Q_in to what we saw when our r is larger than iR."},{"Start":"09:11.985 ","End":"09:18.345","Text":"This is because our charge distribution is given in volume."},{"Start":"09:18.345 ","End":"09:20.699","Text":"As we sum up more volume,"},{"Start":"09:20.699 ","End":"09:23.040","Text":"we\u0027re going to have a larger Q_in."},{"Start":"09:23.040 ","End":"09:25.290","Text":"But right now, we can see that we\u0027re summing up"},{"Start":"09:25.290 ","End":"09:28.934","Text":"not the total volume of the entire cylinder,"},{"Start":"09:28.934 ","End":"09:36.085","Text":"but rather some section of it where our smallest cylinder has this r, as a radius."},{"Start":"09:36.085 ","End":"09:39.530","Text":"Let\u0027s write this out, so we know that our Q_in is equal to"},{"Start":"09:39.530 ","End":"09:42.895","Text":"Rhodv here because we\u0027re given Rho."},{"Start":"09:42.895 ","End":"09:44.820","Text":"That\u0027s equal to Rho,"},{"Start":"09:44.820 ","End":"09:53.280","Text":"and then the volume of this small sphere is going to be equal to Pir^2."},{"Start":"09:53.280 ","End":"10:00.350","Text":"Then again, let\u0027s say that the length of this is L as well."},{"Start":"10:00.350 ","End":"10:05.165","Text":"Then multiplied by L. Here we were multiplying it by"},{"Start":"10:05.165 ","End":"10:10.610","Text":"R^2 because we were summing up on the total charge in this entire cylinder,"},{"Start":"10:10.610 ","End":"10:15.840","Text":"but right now we have less charge because we\u0027re located inside the cylinder."},{"Start":"10:15.840 ","End":"10:18.626","Text":"Now let\u0027s substitute in our Q_in;"},{"Start":"10:18.626 ","End":"10:23.070","Text":"so that\u0027s RhoPir^2L,"},{"Start":"10:23.070 ","End":"10:27.645","Text":"and then divided by our Epsilon_naught."},{"Start":"10:27.645 ","End":"10:31.215","Text":"Now we can see that Pi crosses off,"},{"Start":"10:31.215 ","End":"10:33.255","Text":"one of our r crosses off,"},{"Start":"10:33.255 ","End":"10:35.340","Text":"and our L crosses off."},{"Start":"10:35.340 ","End":"10:39.435","Text":"Therefore, we can write that our magnitude of the electric field,"},{"Start":"10:39.435 ","End":"10:40.830","Text":"given by Gauss\u0027s law,"},{"Start":"10:40.830 ","End":"10:49.410","Text":"is going to be equal to Rhor divided by 2Epsilon_naught."},{"Start":"10:49.410 ","End":"10:52.245","Text":"Of course, if we want to write this as a vector quantity,"},{"Start":"10:52.245 ","End":"10:55.140","Text":"that means with magnitude and direction,"},{"Start":"10:55.140 ","End":"10:56.910","Text":"we denote it as a vector quantity,"},{"Start":"10:56.910 ","End":"11:00.100","Text":"and we know that it\u0027s in the radial direction."},{"Start":"11:03.350 ","End":"11:09.405","Text":"Now we worked out our electric field for when we\u0027re located outside of"},{"Start":"11:09.405 ","End":"11:12.915","Text":"our solid infinite cylinder and"},{"Start":"11:12.915 ","End":"11:18.555","Text":"the electric field for when we\u0027re located inside our solid infinite cylinder."},{"Start":"11:18.555 ","End":"11:26.290","Text":"We can see that if we\u0027re located right on the surface of our infinite cylinder, i.e."},{"Start":"11:26.290 ","End":"11:30.540","Text":"Our r=R, so we can substitute both of these in,"},{"Start":"11:30.540 ","End":"11:34.240","Text":"and we\u0027ll see that we\u0027ll get the same answer."},{"Start":"11:34.490 ","End":"11:42.249","Text":"We can see that our electric field function is going to be a constant function."},{"Start":"11:42.890 ","End":"11:45.600","Text":"Now what I\u0027m going to do is,"},{"Start":"11:45.600 ","End":"11:49.500","Text":"I\u0027m going to show how we find the electric field of an infinite cylinder,"},{"Start":"11:49.500 ","End":"11:53.520","Text":"where this time our cylinder is a cylindrical shell."},{"Start":"11:53.520 ","End":"11:57.720","Text":"We\u0027re going to have a uniform charge density per unit area."},{"Start":"11:57.720 ","End":"12:00.360","Text":"Sigma instead of Rho."},{"Start":"12:00.360 ","End":"12:04.140","Text":"That means that we\u0027re going to do the exact same thing, and what we\u0027ll see,"},{"Start":"12:04.140 ","End":"12:06.705","Text":"is that we\u0027ll have an electric field on"},{"Start":"12:06.705 ","End":"12:10.965","Text":"the outside when we\u0027re outside of our cylindrical shell."},{"Start":"12:10.965 ","End":"12:12.270","Text":"However, when we\u0027re inside,"},{"Start":"12:12.270 ","End":"12:15.090","Text":"we\u0027re going to have electric field which is equal to 0."},{"Start":"12:15.090 ","End":"12:18.690","Text":"Whoever sees this and thinks that this is obvious to them,"},{"Start":"12:18.690 ","End":"12:19.890","Text":"and they get the idea,"},{"Start":"12:19.890 ","End":"12:22.155","Text":"you can move on and this is the end of the lesson."},{"Start":"12:22.155 ","End":"12:24.090","Text":"However, if this isn\u0027t,"},{"Start":"12:24.090 ","End":"12:27.130","Text":"I\u0027ll show how we do that now."},{"Start":"12:28.790 ","End":"12:31.845","Text":"Now we have a cylindrical shell,"},{"Start":"12:31.845 ","End":"12:38.340","Text":"so we can say that we have a uniform charge density per unit area Sigma."},{"Start":"12:38.340 ","End":"12:41.085","Text":"Let\u0027s imagine that right now,"},{"Start":"12:41.085 ","End":"12:49.635","Text":"we\u0027re located at some point outside of our cylindrical shell at a distance of r away."},{"Start":"12:49.635 ","End":"12:51.945","Text":"We\u0027re located outside."},{"Start":"12:51.945 ","End":"12:54.450","Text":"Let\u0027s find our electric field."},{"Start":"12:54.450 ","End":"12:57.720","Text":"Now again, we have symmetry along the z-axis,"},{"Start":"12:57.720 ","End":"13:01.785","Text":"and width rotation with respect to some angle, let\u0027s say Theta."},{"Start":"13:01.785 ","End":"13:03.960","Text":"We can see that our electric field,"},{"Start":"13:03.960 ","End":"13:09.580","Text":"is again going to be in the radial direction."},{"Start":"13:10.460 ","End":"13:12.750","Text":"Now let\u0027s just write this,"},{"Start":"13:12.750 ","End":"13:15.210","Text":"we\u0027re going to do this super quickly because we\u0027ve already"},{"Start":"13:15.210 ","End":"13:18.465","Text":"spoken about a lot of the things in previous lessons."},{"Start":"13:18.465 ","End":"13:24.210","Text":"Our electric flux is equal to the integral on E.ds."},{"Start":"13:24.210 ","End":"13:25.650","Text":"Now again, because of symmetry,"},{"Start":"13:25.650 ","End":"13:29.850","Text":"we know that our electric field is going to be uniform throughout."},{"Start":"13:29.850 ","End":"13:33.105","Text":"We can just write it as E.S."},{"Start":"13:33.105 ","End":"13:34.785","Text":"Now what we\u0027re going to do,"},{"Start":"13:34.785 ","End":"13:37.725","Text":"is we\u0027re going to draw a Gaussian surface,"},{"Start":"13:37.725 ","End":"13:40.320","Text":"because we\u0027re dealing with an infinite cylinder,"},{"Start":"13:40.320 ","End":"13:48.070","Text":"so it\u0027s natural to draw a Gaussian surface which is itself a cylinder."},{"Start":"13:48.140 ","End":"13:51.370","Text":"Let\u0027s draw this like so."},{"Start":"13:51.740 ","End":"13:58.530","Text":"We can see that the radius of our cylinder is this r. As previously discussed,"},{"Start":"13:58.530 ","End":"14:02.100","Text":"we can see that the electric flux from the electric field"},{"Start":"14:02.100 ","End":"14:06.930","Text":"coming out of the bases of the cylinder is going to be equal to 0."},{"Start":"14:06.930 ","End":"14:11.280","Text":"We\u0027re just summing up the electric field and the radial direction,"},{"Start":"14:11.280 ","End":"14:14.880","Text":"coming out of this curved body of our cylinder."},{"Start":"14:14.880 ","End":"14:19.560","Text":"This is going to be equal to E multiplied by the surface area of that,"},{"Start":"14:19.560 ","End":"14:25.215","Text":"so we can see that that is going to be equal to 2Pir,"},{"Start":"14:25.215 ","End":"14:33.130","Text":"and let\u0027s say that the length of our cylinder is L, so 2PirL."},{"Start":"14:33.170 ","End":"14:39.495","Text":"Now we know that this is going to be equal to Q_in divided by Epsilon_naught."},{"Start":"14:39.495 ","End":"14:42.070","Text":"What is our Q_in?"},{"Start":"14:42.070 ","End":"14:47.975","Text":"It\u0027s our charge enclosed within our Gaussian surface."},{"Start":"14:47.975 ","End":"14:50.810","Text":"Because we\u0027re dealing with a cylindrical shell,"},{"Start":"14:50.810 ","End":"14:55.770","Text":"it\u0027s going to be the integral of Sigmads,"},{"Start":"14:55.770 ","End":"14:58.580","Text":"because we\u0027re working with units of area."},{"Start":"14:58.580 ","End":"15:02.060","Text":"That means that it\u0027s going to be equal to Sigma and then"},{"Start":"15:02.060 ","End":"15:08.040","Text":"the area of our cylindrical shell is going to be equal to 2Pir,"},{"Start":"15:08.150 ","End":"15:14.285","Text":"this time because that\u0027s the radius of our cylindrical shell multiplied by L."},{"Start":"15:14.285 ","End":"15:16.370","Text":"Because that is the length of"},{"Start":"15:16.370 ","End":"15:21.070","Text":"the cylindrical shell which is enclosed by our Gaussian surface."},{"Start":"15:21.070 ","End":"15:23.355","Text":"Now we can substitute in our Q_in,"},{"Start":"15:23.355 ","End":"15:29.320","Text":"so we\u0027re going to have that our Q_in is equal to 2SigmaPirL,"},{"Start":"15:30.110 ","End":"15:33.780","Text":"and then divided by our Epsilon_naught."},{"Start":"15:33.780 ","End":"15:36.515","Text":"Now we can see that our ls, are going to cancel out,"},{"Start":"15:36.515 ","End":"15:37.820","Text":"our Pis cancel out,"},{"Start":"15:37.820 ","End":"15:39.980","Text":"and our 2s cancel out."},{"Start":"15:39.980 ","End":"15:43.370","Text":"Therefore, we can isolate out our electric field,"},{"Start":"15:43.370 ","End":"15:51.955","Text":"and we can see that it\u0027s going to be equal to SigmaR divided by Epsilon_naughtr."},{"Start":"15:51.955 ","End":"15:56.615","Text":"This is of course the magnitude of the electric field given by Gauss\u0027s law."},{"Start":"15:56.615 ","End":"15:58.820","Text":"If we want it to be a vector quantity,"},{"Start":"15:58.820 ","End":"16:00.815","Text":"we have to add in our direction,"},{"Start":"16:00.815 ","End":"16:05.040","Text":"and as we know, our electric field is in the radial direction."},{"Start":"16:05.840 ","End":"16:11.300","Text":"This is the electric field when we\u0027re located out of our cylindrical shell."},{"Start":"16:11.300 ","End":"16:16.950","Text":"Now, what about when we\u0027re located within our cylindrical shell?"},{"Start":"16:17.750 ","End":"16:20.685","Text":"This time we\u0027re located inside,"},{"Start":"16:20.685 ","End":"16:22.430","Text":"so we\u0027re going to do the exact same thing."},{"Start":"16:22.430 ","End":"16:23.960","Text":"We\u0027re dealing again with a cylinder,"},{"Start":"16:23.960 ","End":"16:32.495","Text":"so we\u0027ll draw our Gaussian surface as a cylinder of length L,"},{"Start":"16:32.495 ","End":"16:39.290","Text":"and of radius r. Let\u0027s work out our electric flux E,"},{"Start":"16:39.290 ","End":"16:41.000","Text":"which I\u0027m just going to skip this,"},{"Start":"16:41.000 ","End":"16:44.130","Text":"it\u0027s going to be equal to E.S,"},{"Start":"16:45.310 ","End":"16:49.985","Text":"which is equal to E multiplied by just like here,"},{"Start":"16:49.985 ","End":"16:56.240","Text":"2PirL, the surface area of the curved section of our Gaussian surface."},{"Start":"16:56.240 ","End":"17:00.650","Text":"This is equal to Q_in divided by epsilon_naught."},{"Start":"17:00.650 ","End":"17:02.765","Text":"Now, what is our Q_in?"},{"Start":"17:02.765 ","End":"17:06.370","Text":"It\u0027s the integral of Sigmads."},{"Start":"17:06.370 ","End":"17:08.855","Text":"Because we\u0027re dealing with a cylindrical shell."},{"Start":"17:08.855 ","End":"17:11.690","Text":"However, we can see that the cylindrical shell is"},{"Start":"17:11.690 ","End":"17:14.975","Text":"located outside of our Gaussian surface,"},{"Start":"17:14.975 ","End":"17:18.760","Text":"but we\u0027re trying to work out what\u0027s inside,"},{"Start":"17:18.760 ","End":"17:21.710","Text":"Q_in what\u0027s inside our Gaussian surface,"},{"Start":"17:21.710 ","End":"17:25.865","Text":"which means that this integral is going to be equal to 0."},{"Start":"17:25.865 ","End":"17:28.295","Text":"Then we can write that this is equal to 0."},{"Start":"17:28.295 ","End":"17:32.090","Text":"Therefore, our electric field when we\u0027re"},{"Start":"17:32.090 ","End":"17:39.010","Text":"located inside our cylindrical shell is equal to 0."},{"Start":"17:39.110 ","End":"17:42.520","Text":"That\u0027s the end of this lesson."}],"ID":22361},{"Watched":false,"Name":"Exercise 4","Duration":"13m 39s","ChapterTopicVideoID":21408,"CourseChapterTopicPlaylistID":246895,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/21408.jpeg","UploadDate":"2020-04-21T08:55:32.8030000","DurationForVideoObject":"PT13M39S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.710","Text":"Hello. In this lesson,"},{"Start":"00:01.710 ","End":"00:04.770","Text":"we\u0027re going to be working out the electric field in"},{"Start":"00:04.770 ","End":"00:09.755","Text":"space due to an infinite charged plane."},{"Start":"00:09.755 ","End":"00:13.530","Text":"Here we have our plane and it\u0027s infinite."},{"Start":"00:13.530 ","End":"00:18.780","Text":"It\u0027s carrying on until infinity and all these directions."},{"Start":"00:18.780 ","End":"00:23.945","Text":"It has uniform charge density per unit area Sigma."},{"Start":"00:23.945 ","End":"00:29.100","Text":"Now let\u0027s work out the electric field at some point over here."},{"Start":"00:30.320 ","End":"00:34.550","Text":"Let\u0027s call this z at location z."},{"Start":"00:34.550 ","End":"00:36.980","Text":"Now, when we do this calculation at the end,"},{"Start":"00:36.980 ","End":"00:39.320","Text":"we\u0027ll see that it doesn\u0027t matter what"},{"Start":"00:39.320 ","End":"00:42.635","Text":"our value for z is and how high above the plane we are."},{"Start":"00:42.635 ","End":"00:47.940","Text":"But in the meantime, let\u0027s put this in and then we\u0027ll see how it cancels out later on."},{"Start":"00:48.200 ","End":"00:53.570","Text":"The first thing that we can see is that our electric field is going to"},{"Start":"00:53.570 ","End":"00:59.225","Text":"be in the z-direction only because we\u0027re dealing with an infinite plane."},{"Start":"00:59.225 ","End":"01:04.865","Text":"If we shift our plane in the xyz-direction,"},{"Start":"01:04.865 ","End":"01:07.760","Text":"we can see that we\u0027re left with the exact same problem."},{"Start":"01:07.760 ","End":"01:10.790","Text":"We can see that our electric field will still remain"},{"Start":"01:10.790 ","End":"01:15.690","Text":"perpendicular to the plane in the z-direction."},{"Start":"01:15.940 ","End":"01:19.325","Text":"Now, whoever is having a bit of trouble"},{"Start":"01:19.325 ","End":"01:22.340","Text":"understanding this or picturing this in their mind,"},{"Start":"01:22.340 ","End":"01:23.555","Text":"at the end of the lesson,"},{"Start":"01:23.555 ","End":"01:26.360","Text":"I\u0027ll explain this via Coulomb\u0027s law more."},{"Start":"01:26.360 ","End":"01:29.000","Text":"But in the meantime, just know that the infinite plane,"},{"Start":"01:29.000 ","End":"01:31.145","Text":"if you shifted in the x or y-direction,"},{"Start":"01:31.145 ","End":"01:32.430","Text":"we\u0027re left with the same problem."},{"Start":"01:32.430 ","End":"01:36.455","Text":"Our electric field is going to be uniform in the z-direction."},{"Start":"01:36.455 ","End":"01:39.480","Text":"Later, I\u0027ll explain it in more detail."},{"Start":"01:39.770 ","End":"01:44.720","Text":"Of course, if we go to the underside of"},{"Start":"01:44.720 ","End":"01:52.745","Text":"our plane and we go z but in this downwards direction and the negatives z-direction."},{"Start":"01:52.745 ","End":"01:59.050","Text":"We can see that our electric field is going to be equal and opposite."},{"Start":"01:59.050 ","End":"02:04.200","Text":"We\u0027re also going to have an electric field in the negative z-direction."},{"Start":"02:05.630 ","End":"02:09.350","Text":"We can see that I have symmetry in my x and y directions,"},{"Start":"02:09.350 ","End":"02:13.760","Text":"that my electric field is in the z-direction and it\u0027s a uniform throughout,"},{"Start":"02:13.760 ","End":"02:16.010","Text":"which means that I can use Gauss\u0027s law."},{"Start":"02:16.010 ","End":"02:17.480","Text":"How am I going to do that?"},{"Start":"02:17.480 ","End":"02:20.169","Text":"First, I have to choose a Gaussian surface."},{"Start":"02:20.169 ","End":"02:23.870","Text":"Now, I\u0027m going to choose a cube."},{"Start":"02:23.870 ","End":"02:27.004","Text":"But the same thing would work if you choose a cylinder"},{"Start":"02:27.004 ","End":"02:30.080","Text":"that runs perpendicular to the plane."},{"Start":"02:30.080 ","End":"02:31.460","Text":"I\u0027m just going to choose a cube,"},{"Start":"02:31.460 ","End":"02:33.510","Text":"but it doesn\u0027t really matter."},{"Start":"02:33.530 ","End":"02:37.460","Text":"Let\u0027s draw this. Let\u0027s hope I draw this properly."},{"Start":"02:37.460 ","End":"02:40.795","Text":"As you know, my drawing skills aren\u0027t amazing."},{"Start":"02:40.795 ","End":"02:46.330","Text":"Then we go all the way down to here."},{"Start":"02:53.510 ","End":"02:56.950","Text":"Of course here,"},{"Start":"03:02.600 ","End":"03:07.635","Text":"this is our cube and it\u0027s of height 2z."},{"Start":"03:07.635 ","End":"03:11.300","Text":"The total distance from this base,"},{"Start":"03:11.300 ","End":"03:13.430","Text":"the upper base to the lower base is 2z."},{"Start":"03:13.430 ","End":"03:17.610","Text":"Because as we can see, we have a z here and z here."},{"Start":"03:19.220 ","End":"03:24.980","Text":"Now, I\u0027m going to say that the area of my base of"},{"Start":"03:24.980 ","End":"03:29.780","Text":"the cube is equal to A and the area of this space is also equal to A."},{"Start":"03:29.780 ","End":"03:31.730","Text":"My base is of course,"},{"Start":"03:31.730 ","End":"03:34.890","Text":"parallel to my plane."},{"Start":"03:35.450 ","End":"03:38.585","Text":"As we can see, if we would have drawn a cylinder,"},{"Start":"03:38.585 ","End":"03:42.170","Text":"we would have just works out that area differently, but it doesn\u0027t really matter."},{"Start":"03:42.170 ","End":"03:46.820","Text":"Now we want to find out our electric flux,"},{"Start":"03:46.820 ","End":"03:48.305","Text":"which as we know,"},{"Start":"03:48.305 ","End":"03:53.240","Text":"is the integral of E.ds."},{"Start":"03:53.240 ","End":"04:00.645","Text":"Now, because we can see that our electric field is perpendicular to our surface,"},{"Start":"04:00.645 ","End":"04:04.845","Text":"or it\u0027s parallel to our ds and it\u0027s a uniform throughout."},{"Start":"04:04.845 ","End":"04:08.130","Text":"We can simply write this as E.A,"},{"Start":"04:08.130 ","End":"04:15.570","Text":"where A is the area of each of our cube spaces."},{"Start":"04:16.250 ","End":"04:18.710","Text":"That\u0027s the electric flux through,"},{"Start":"04:18.710 ","End":"04:20.855","Text":"let\u0027s say, our upper base."},{"Start":"04:20.855 ","End":"04:25.145","Text":"Now let\u0027s write the electric flux through our lower base."},{"Start":"04:25.145 ","End":"04:29.310","Text":"We know that it\u0027s the exact same electric field."},{"Start":"04:29.310 ","End":"04:31.155","Text":"Let\u0027s write E,"},{"Start":"04:31.155 ","End":"04:38.035","Text":"but it\u0027s going in the opposite direction and the area that it\u0027s going through is also A."},{"Start":"04:38.035 ","End":"04:39.750","Text":"The area of the base."},{"Start":"04:39.750 ","End":"04:44.480","Text":"What do I do? Do I write a negative here or a positive here?"},{"Start":"04:44.480 ","End":"04:47.975","Text":"It\u0027s true our electric field here"},{"Start":"04:47.975 ","End":"04:51.850","Text":"underneath our infinite plane is going in the negative direction."},{"Start":"04:51.850 ","End":"04:54.620","Text":"It\u0027s easy to think that we should put a negative over here."},{"Start":"04:54.620 ","End":"04:58.805","Text":"However, wait, from our first lesson,"},{"Start":"04:58.805 ","End":"05:01.700","Text":"we learned that when we\u0027re dealing with flux,"},{"Start":"05:01.700 ","End":"05:09.019","Text":"flux exiting our surface or our Gaussian surface is going to be a positive."},{"Start":"05:09.019 ","End":"05:17.475","Text":"Whereas electric field rather entering our surface is going to be negative."},{"Start":"05:17.475 ","End":"05:21.560","Text":"Here we can see that all of our red arrows be them pointing upwards or"},{"Start":"05:21.560 ","End":"05:25.625","Text":"downwards they are all exiting our Gaussian surface,"},{"Start":"05:25.625 ","End":"05:29.395","Text":"which means that we put up positive here."},{"Start":"05:29.395 ","End":"05:34.420","Text":"That is going to be equal to 2 multiplied by EA."},{"Start":"05:36.200 ","End":"05:40.040","Text":"If for instance, 1 of our red arrows here at"},{"Start":"05:40.040 ","End":"05:44.405","Text":"the bottom was pointing upwards in the positive z-direction."},{"Start":"05:44.405 ","End":"05:48.695","Text":"Or if all the electric field lines are pointing in the opposite direction down here."},{"Start":"05:48.695 ","End":"05:54.815","Text":"We\u0027d have 0 flux because then we would have E.A minus E.A and we would have 0."},{"Start":"05:54.815 ","End":"05:57.380","Text":"But here everything is exiting our surface,"},{"Start":"05:57.380 ","End":"06:00.510","Text":"which means positive flux."},{"Start":"06:02.000 ","End":"06:08.330","Text":"Now what about the flux coming out of the other sides of our cube?"},{"Start":"06:08.330 ","End":"06:12.800","Text":"Out of this wall over here or this wall in the front over here."},{"Start":"06:12.800 ","End":"06:16.340","Text":"The flux over there is going to be equal to 0."},{"Start":"06:16.340 ","End":"06:20.420","Text":"That is because our electric field lines are going like"},{"Start":"06:20.420 ","End":"06:25.729","Text":"so upwards in this direction and down here there\u0027ll be going in this direction."},{"Start":"06:25.729 ","End":"06:27.500","Text":"As we can see,"},{"Start":"06:27.500 ","End":"06:33.220","Text":"that is going to be parallel to this wall of our cube."},{"Start":"06:33.220 ","End":"06:38.520","Text":"Or in other words, perpendicular to our ds vector."},{"Start":"06:38.710 ","End":"06:41.975","Text":"For this shape for our cube,"},{"Start":"06:41.975 ","End":"06:48.420","Text":"which means that our flux contribution from here is going to be equal to 0."},{"Start":"06:49.040 ","End":"06:52.910","Text":"That\u0027s great. We only had to work out the flux coming out"},{"Start":"06:52.910 ","End":"06:56.880","Text":"of our upper base and our lower base."},{"Start":"06:58.400 ","End":"07:03.590","Text":"Now let\u0027s move on to the right side of our Gauss\u0027s law,"},{"Start":"07:03.590 ","End":"07:07.250","Text":"which is equal to Q_in divided by Epsilon naught."},{"Start":"07:07.250 ","End":"07:09.960","Text":"What is our Q_in?"},{"Start":"07:10.270 ","End":"07:15.685","Text":"Our Q_in is our integral on Sigma ds,"},{"Start":"07:15.685 ","End":"07:18.905","Text":"because we\u0027re working with area over here."},{"Start":"07:18.905 ","End":"07:23.565","Text":"Let\u0027s just draw in green."},{"Start":"07:23.565 ","End":"07:27.170","Text":"We can see that enclosed within our Gaussian surface,"},{"Start":"07:27.170 ","End":"07:28.460","Text":"which here is a cube,"},{"Start":"07:28.460 ","End":"07:34.765","Text":"we have a certain section of our infinite plane enclosed within."},{"Start":"07:34.765 ","End":"07:44.220","Text":"Of course the area of this section enclosed is also equal to A as we can see."},{"Start":"07:44.220 ","End":"07:49.610","Text":"Because we know that our Sigma is uniform throughout our entire infinite plane and"},{"Start":"07:49.610 ","End":"07:55.710","Text":"also throughout this small section in my Gaussian surface."},{"Start":"07:55.710 ","End":"08:00.880","Text":"I can say that my Q_in is simply equal to Sigma A."},{"Start":"08:01.010 ","End":"08:03.405","Text":"Now let\u0027s plug this in."},{"Start":"08:03.405 ","End":"08:08.655","Text":"We\u0027re going to have that our Q_in is Sigma A and then divided by Epsilon naught."},{"Start":"08:08.655 ","End":"08:12.405","Text":"Now we can see that on both sides our A\u0027s cancel out."},{"Start":"08:12.405 ","End":"08:17.060","Text":"Therefore, we\u0027re left with the magnitude of electric field being"},{"Start":"08:17.060 ","End":"08:22.950","Text":"equal to Sigma divided by 2 Epsilon naught."},{"Start":"08:23.270 ","End":"08:27.810","Text":"Of course, if we want it as a vector quantity,"},{"Start":"08:27.880 ","End":"08:35.190","Text":"so then we can just add our directions and we can see that it\u0027s in the z-direction."},{"Start":"08:35.260 ","End":"08:42.195","Text":"This is of course in the positive direction when our z is bigger than 0."},{"Start":"08:42.195 ","End":"08:44.920","Text":"When we\u0027re above the plane, however,"},{"Start":"08:44.920 ","End":"08:48.370","Text":"when we\u0027re below the plane is just going to turn into"},{"Start":"08:48.370 ","End":"08:53.050","Text":"a minus sign because then we\u0027re going in the negative z-direction."},{"Start":"08:53.050 ","End":"08:56.480","Text":"This is when we\u0027re below the plane."},{"Start":"08:58.840 ","End":"09:05.350","Text":"This is something that you should write down in your equation papers."},{"Start":"09:05.350 ","End":"09:08.230","Text":"Also all of the other electric fields for"},{"Start":"09:08.230 ","End":"09:11.215","Text":"our basic shapes such as an infinite wire and infinite cylinder,"},{"Start":"09:11.215 ","End":"09:12.580","Text":"solid and shell,"},{"Start":"09:12.580 ","End":"09:14.710","Text":"and sphere, and everything else."},{"Start":"09:14.710 ","End":"09:18.755","Text":"We\u0027ll also speak about this we\u0027re going to use very often."},{"Start":"09:18.755 ","End":"09:21.305","Text":"Now a little note to remember at the beginning of the lesson,"},{"Start":"09:21.305 ","End":"09:24.875","Text":"I said that our electric field isn\u0027t going to be dependent on z."},{"Start":"09:24.875 ","End":"09:28.180","Text":"As we can see, our electric field is constant."},{"Start":"09:28.180 ","End":"09:33.970","Text":"No matter how high up above our infinite surface we are,"},{"Start":"09:33.970 ","End":"09:38.000","Text":"we\u0027re going to have the exact same value for our electric field."},{"Start":"09:38.000 ","End":"09:40.535","Text":"The same when we go below."},{"Start":"09:40.535 ","End":"09:43.340","Text":"The electric field is pointing in the opposite direction,"},{"Start":"09:43.340 ","End":"09:46.760","Text":"but the magnitude of the electric field is going to stay the same."},{"Start":"09:46.760 ","End":"09:48.050","Text":"It\u0027s independent of z."},{"Start":"09:48.050 ","End":"09:51.290","Text":"The only thing that does matter is just"},{"Start":"09:51.290 ","End":"09:55.715","Text":"to indicate if we\u0027re above the surface or below the surface."},{"Start":"09:55.715 ","End":"09:59.495","Text":"Because that will change the direction of our electric field,"},{"Start":"09:59.495 ","End":"10:02.370","Text":"but our magnitude remains the same."},{"Start":"10:02.870 ","End":"10:06.080","Text":"Now what I\u0027m going to do is I\u0027m going to speak"},{"Start":"10:06.080 ","End":"10:10.010","Text":"about why we can see that the electric field is"},{"Start":"10:10.010 ","End":"10:16.415","Text":"in fact in the z-direction and why there is symmetry in the x and y directions."},{"Start":"10:16.415 ","End":"10:20.060","Text":"If this is obvious to you or you don\u0027t want to watch the rest."},{"Start":"10:20.060 ","End":"10:23.105","Text":"This is the end of the lesson and you can move on to the next lesson."},{"Start":"10:23.105 ","End":"10:25.465","Text":"Otherwise, you can watch the explanation."},{"Start":"10:25.465 ","End":"10:29.645","Text":"How do we know that our electric field is in the z-direction?"},{"Start":"10:29.645 ","End":"10:35.555","Text":"Let\u0027s just choose this random point over here arbitrarily and space above our surface."},{"Start":"10:35.555 ","End":"10:41.060","Text":"Then let\u0027s choose this random piece of area."},{"Start":"10:41.060 ","End":"10:42.940","Text":"That\u0027s a small piece of area."},{"Start":"10:42.940 ","End":"10:47.420","Text":"We can say, or we can consider this as a point charge."},{"Start":"10:47.420 ","End":"10:50.720","Text":"We know that it is going to exert"},{"Start":"10:50.720 ","End":"10:56.030","Text":"an electric field on this point over here in this direction."},{"Start":"10:56.030 ","End":"11:00.140","Text":"Then let\u0027s work with symmetry and equal distance away,"},{"Start":"11:00.140 ","End":"11:04.475","Text":"we have another tiny area which we can consider a point charge."},{"Start":"11:04.475 ","End":"11:11.855","Text":"It is also going to exert an electric field on this area over here."},{"Start":"11:11.855 ","End":"11:16.820","Text":"From symmetry, because both of these have"},{"Start":"11:16.820 ","End":"11:18.890","Text":"the same charge on them and they\u0027re both"},{"Start":"11:18.890 ","End":"11:23.705","Text":"equally the same distance away from a point in space."},{"Start":"11:23.705 ","End":"11:30.815","Text":"We can say that the electric field here is a mirror image of the other arrow."},{"Start":"11:30.815 ","End":"11:33.950","Text":"Now we can split up the components so"},{"Start":"11:33.950 ","End":"11:36.680","Text":"we can see that the component in the x or y direction,"},{"Start":"11:36.680 ","End":"11:42.260","Text":"it doesn\u0027t really matter of both of these arrows is equal and opposite,"},{"Start":"11:42.260 ","End":"11:44.645","Text":"which means that they both cancel out."},{"Start":"11:44.645 ","End":"11:49.550","Text":"However, we can see that the z component of this arrow and"},{"Start":"11:49.550 ","End":"11:55.535","Text":"the z component of this arrow are in the same direction and are also equal."},{"Start":"11:55.535 ","End":"12:00.410","Text":"They add up and that is exactly in the z-direction."},{"Start":"12:00.410 ","End":"12:04.400","Text":"That means that we have our electric field in the z-direction and"},{"Start":"12:04.400 ","End":"12:10.080","Text":"our electric field in the other direction so x and y will cancel out."},{"Start":"12:11.270 ","End":"12:15.845","Text":"Now we\u0027ve discussed why our electric field is in the z-direction."},{"Start":"12:15.845 ","End":"12:19.445","Text":"But why does it have to be uniform throughout?"},{"Start":"12:19.445 ","End":"12:22.490","Text":"Because we can see that we have symmetry in"},{"Start":"12:22.490 ","End":"12:25.590","Text":"the x and y directions because our plane is infinite."},{"Start":"12:25.590 ","End":"12:29.840","Text":"If we shift our plane to the x-direction or the y-direction,"},{"Start":"12:29.840 ","End":"12:31.640","Text":"or we rotate it around,"},{"Start":"12:31.640 ","End":"12:36.260","Text":"we\u0027re going to be left with the exact same problem in front of us."},{"Start":"12:36.260 ","End":"12:41.465","Text":"We can see that the electric fields then has no reason to be different at"},{"Start":"12:41.465 ","End":"12:47.690","Text":"this arbitrary point or at this arbitrary point a bit further along."},{"Start":"12:47.690 ","End":"12:50.300","Text":"That\u0027s because of the symmetry in the problem."},{"Start":"12:50.300 ","End":"12:56.150","Text":"A slightly more mathematical way of discussing this is if we say that"},{"Start":"12:56.150 ","End":"12:58.745","Text":"our new x position is equal to"},{"Start":"12:58.745 ","End":"13:03.095","Text":"our original x position plus some distance A that we moved."},{"Start":"13:03.095 ","End":"13:08.495","Text":"If we see that we have the exact same situation going on."},{"Start":"13:08.495 ","End":"13:12.200","Text":"Then we can say that the derivative of our electric field in"},{"Start":"13:12.200 ","End":"13:18.265","Text":"the z-direction with respect to x is going to be equal to 0."},{"Start":"13:18.265 ","End":"13:20.570","Text":"That\u0027s the same with our y-direction."},{"Start":"13:20.570 ","End":"13:22.970","Text":"The derivative of our electric field in"},{"Start":"13:22.970 ","End":"13:27.200","Text":"the z-direction with respect to y is also equal to 0,"},{"Start":"13:27.200 ","End":"13:31.225","Text":"which means that we have symmetry in the x and y directions."},{"Start":"13:31.225 ","End":"13:37.115","Text":"We can see that our electric field is going to be uniform throughout."},{"Start":"13:37.115 ","End":"13:40.530","Text":"That\u0027s the end of this lesson."}],"ID":22362},{"Watched":false,"Name":"Exercise 5","Duration":"14m 45s","ChapterTopicVideoID":21409,"CourseChapterTopicPlaylistID":246895,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/21409.jpeg","UploadDate":"2020-04-21T08:59:17.9600000","DurationForVideoObject":"PT14M45S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.265","Text":"Hello. In this lesson,"},{"Start":"00:02.265 ","End":"00:06.210","Text":"we\u0027re going to see how to work out the electric field of"},{"Start":"00:06.210 ","End":"00:11.475","Text":"a solid sphere that has non-uniform charge density."},{"Start":"00:11.475 ","End":"00:18.435","Text":"Here we can see that our charge density is changing as a function of r,"},{"Start":"00:18.435 ","End":"00:23.520","Text":"where r is the distance that we are from the center of the sphere,"},{"Start":"00:23.520 ","End":"00:25.565","Text":"and it\u0027s divided by R,"},{"Start":"00:25.565 ","End":"00:29.345","Text":"which is the radius of our solid sphere."},{"Start":"00:29.345 ","End":"00:32.705","Text":"Let\u0027s see how we\u0027re going to do this."},{"Start":"00:32.705 ","End":"00:37.125","Text":"The first thing that we\u0027re going to do is we\u0027re going to divide this into 2 subsections."},{"Start":"00:37.125 ","End":"00:41.120","Text":"We have the subsection where we\u0027re located somewhere within"},{"Start":"00:41.120 ","End":"00:43.175","Text":"the sphere and then we have the subsection"},{"Start":"00:43.175 ","End":"00:46.385","Text":"where we\u0027re located somewhere outside of the sphere."},{"Start":"00:46.385 ","End":"00:50.600","Text":"Let\u0027s start with when we\u0027re located outside of the sphere."},{"Start":"00:50.600 ","End":"00:57.440","Text":"We\u0027re at some random point in space of radius r and we\u0027re outside the sphere."},{"Start":"00:57.440 ","End":"01:01.740","Text":"That means that r is bigger than R."},{"Start":"01:01.940 ","End":"01:05.600","Text":"The first thing that I\u0027m going to do is I\u0027m going to check if"},{"Start":"01:05.600 ","End":"01:08.810","Text":"I can use Gauss\u0027s law in order to solve this question."},{"Start":"01:08.810 ","End":"01:11.675","Text":"Now, we said that with Gauss\u0027s law,"},{"Start":"01:11.675 ","End":"01:15.680","Text":"our electric field has to always be perpendicular to our surface."},{"Start":"01:15.680 ","End":"01:17.375","Text":"As we can see here,"},{"Start":"01:17.375 ","End":"01:22.340","Text":"our electric field is going to be in the radial direction."},{"Start":"01:22.340 ","End":"01:25.010","Text":"We\u0027re only going to have a radial component for"},{"Start":"01:25.010 ","End":"01:26.540","Text":"our electric field because we\u0027re dealing with"},{"Start":"01:26.540 ","End":"01:29.240","Text":"a sphere and we\u0027ve seen this in previous questions."},{"Start":"01:29.240 ","End":"01:33.155","Text":"Now, the next thing that we have to do is we have to make sure that"},{"Start":"01:33.155 ","End":"01:40.170","Text":"our electric field is uniform as we travel over the sphere."},{"Start":"01:40.610 ","End":"01:46.600","Text":"Because we have symmetry in the Theta and Phi direction,"},{"Start":"01:46.600 ","End":"01:49.310","Text":"then we can see that our electric field,"},{"Start":"01:49.310 ","End":"01:51.620","Text":"if we move along the sphere,"},{"Start":"01:51.620 ","End":"01:54.320","Text":"is going to be uniform throughout."},{"Start":"01:54.320 ","End":"02:00.140","Text":"Why is that? We can see that our charged density is only dependent on our radius,"},{"Start":"02:00.140 ","End":"02:03.665","Text":"on our distance away from the center of the sphere."},{"Start":"02:03.665 ","End":"02:07.055","Text":"We can see that our charge,"},{"Start":"02:07.055 ","End":"02:11.915","Text":"if we spin our sphere around and either the Theta direction,"},{"Start":"02:11.915 ","End":"02:14.767","Text":"so something like so,"},{"Start":"02:14.767 ","End":"02:17.315","Text":"or in the Phi direction,"},{"Start":"02:17.315 ","End":"02:18.654","Text":"which is something like so,"},{"Start":"02:18.654 ","End":"02:23.315","Text":"we\u0027re going to have the exact same question in front of us."},{"Start":"02:23.315 ","End":"02:28.415","Text":"We haven\u0027t changed anything and our values for the electric field aren\u0027t going to change."},{"Start":"02:28.415 ","End":"02:32.000","Text":"That means that we have symmetry and that we\u0027re only going"},{"Start":"02:32.000 ","End":"02:35.724","Text":"to have an electric field component in the radial direction."},{"Start":"02:35.724 ","End":"02:39.740","Text":"That also means that the electric field along the surface of"},{"Start":"02:39.740 ","End":"02:43.855","Text":"the sphere is going to be uniform so we can use Gauss\u0027s law."},{"Start":"02:43.855 ","End":"02:45.710","Text":"Because we\u0027re dealing with a sphere,"},{"Start":"02:45.710 ","End":"02:49.145","Text":"it\u0027s natural to use a Gaussian surface,"},{"Start":"02:49.145 ","End":"02:51.755","Text":"which is also a sphere,"},{"Start":"02:51.755 ","End":"02:54.425","Text":"but it\u0027s going to be a surface or a spherical shell,"},{"Start":"02:54.425 ","End":"02:56.630","Text":"not a solid sphere."},{"Start":"02:56.630 ","End":"03:00.560","Text":"Then we can see that the electric field is going to"},{"Start":"03:00.560 ","End":"03:05.255","Text":"be the same in the radial direction at any point we look at."},{"Start":"03:05.255 ","End":"03:07.755","Text":"Now, of course,"},{"Start":"03:07.755 ","End":"03:09.660","Text":"this is an area,"},{"Start":"03:09.660 ","End":"03:15.100","Text":"we\u0027re dealing with this external shell over here."},{"Start":"03:16.370 ","End":"03:21.110","Text":"We\u0027ve seen that I can use Gauss\u0027s law so, let\u0027s do that."},{"Start":"03:21.110 ","End":"03:25.220","Text":"The electric flux is, as we know,"},{"Start":"03:25.220 ","End":"03:29.540","Text":"the integral on E.ds."},{"Start":"03:29.540 ","End":"03:37.080","Text":"Now because we saw that our electric field is uniform all along this Gaussian surface,"},{"Start":"03:37.080 ","End":"03:40.265","Text":"so we can simply write this as E,"},{"Start":"03:40.265 ","End":"03:41.705","Text":"which is where we\u0027re trying to find,"},{"Start":"03:41.705 ","End":"03:45.925","Text":"multiplied by S, which is the surface area of our Gaussian shape."},{"Start":"03:45.925 ","End":"03:50.915","Text":"That is going to be equal to E multiplied by the surface area"},{"Start":"03:50.915 ","End":"03:55.835","Text":"of a sphere is equal to 4 Pi r^2."},{"Start":"03:55.835 ","End":"04:00.730","Text":"We\u0027re using this r because that\u0027s the radius of our Gaussian surface."},{"Start":"04:00.730 ","End":"04:04.250","Text":"That was the left side of Gauss\u0027s law."},{"Start":"04:04.250 ","End":"04:06.260","Text":"Now, let\u0027s go into the right side."},{"Start":"04:06.260 ","End":"04:12.010","Text":"We know that the right side is equal to Q_in divided by Epsilon naught."},{"Start":"04:12.010 ","End":"04:15.305","Text":"Up until now, we\u0027ve been dealing with questions where"},{"Start":"04:15.305 ","End":"04:20.450","Text":"our Q_in was given by some charge density,"},{"Start":"04:20.450 ","End":"04:22.025","Text":"which is a uniform throughout."},{"Start":"04:22.025 ","End":"04:24.514","Text":"Now here\u0027s specifically in our question,"},{"Start":"04:24.514 ","End":"04:30.435","Text":"we have a charge density which changes according to our radius. Let\u0027s write this."},{"Start":"04:30.435 ","End":"04:35.700","Text":"We have that our Q_in is going to be the integral of Rho dV."},{"Start":"04:36.920 ","End":"04:39.110","Text":"In previous questions,"},{"Start":"04:39.110 ","End":"04:41.794","Text":"we saw that our Rho was constant,"},{"Start":"04:41.794 ","End":"04:44.045","Text":"didn\u0027t have any variables."},{"Start":"04:44.045 ","End":"04:49.220","Text":"We could just multiply our Rho value by the total volume of the sphere."},{"Start":"04:49.220 ","End":"04:52.305","Text":"However here, our Rho has variable,"},{"Start":"04:52.305 ","End":"04:56.740","Text":"r. Now we\u0027re going to have to actually integrate along this."},{"Start":"04:56.740 ","End":"04:58.900","Text":"Because we\u0027re integrating along volumes,"},{"Start":"04:58.900 ","End":"05:05.605","Text":"we\u0027re going to have a triple integral and then our Rho value is equal to"},{"Start":"05:05.605 ","End":"05:13.365","Text":"Rho 0 or Rho naught multiplied by r divided by R,"},{"Start":"05:13.365 ","End":"05:15.720","Text":"and then our dv."},{"Start":"05:15.720 ","End":"05:20.030","Text":"Our dv, we\u0027re dealing right now with spherical coordinates,"},{"Start":"05:20.030 ","End":"05:27.950","Text":"where this angle, if we rotate our sphere in this direction,"},{"Start":"05:27.950 ","End":"05:29.720","Text":"this is our Phi angle."},{"Start":"05:29.720 ","End":"05:33.595","Text":"If we rotate our sphere in this direction,"},{"Start":"05:33.595 ","End":"05:35.850","Text":"that\u0027s our Theta angle."},{"Start":"05:35.850 ","End":"05:38.480","Text":"Then we have to add in our Jacobian,"},{"Start":"05:38.480 ","End":"05:45.675","Text":"which is equal to r^2 sine of Phi and then we have di,"},{"Start":"05:45.675 ","End":"05:48.945","Text":"d Theta, d Phi."},{"Start":"05:48.945 ","End":"05:53.070","Text":"Now let\u0027s put in our borders or our bounds."},{"Start":"05:53.070 ","End":"05:56.870","Text":"We know that we have to sum along this entire sphere."},{"Start":"05:56.870 ","End":"05:58.445","Text":"For our dr,"},{"Start":"05:58.445 ","End":"06:01.225","Text":"we\u0027re summing from 0 to r,"},{"Start":"06:01.225 ","End":"06:03.485","Text":"the maximum radius of our sphere."},{"Start":"06:03.485 ","End":"06:05.120","Text":"Then for our Theta,"},{"Start":"06:05.120 ","End":"06:07.550","Text":"we\u0027re summing from until 2 Pi,"},{"Start":"06:07.550 ","End":"06:10.535","Text":"so a full rotation in our Theta direction."},{"Start":"06:10.535 ","End":"06:13.289","Text":"Then for our Phi direction,"},{"Start":"06:13.289 ","End":"06:15.870","Text":"we\u0027re going from zero until Pi,"},{"Start":"06:15.870 ","End":"06:18.760","Text":"so half a circle."},{"Start":"06:19.760 ","End":"06:22.135","Text":"Now let\u0027s integrate."},{"Start":"06:22.135 ","End":"06:25.100","Text":"We can see that we have no variable Theta,"},{"Start":"06:25.100 ","End":"06:28.025","Text":"so we can automatically do the integration on Theta."},{"Start":"06:28.025 ","End":"06:30.410","Text":"We\u0027re just going to multiply everything by 2Pi."},{"Start":"06:30.410 ","End":"06:36.900","Text":"Then our sine Phi is our only variable here that has Phi inside."},{"Start":"06:36.900 ","End":"06:39.825","Text":"We can also substitute that in."},{"Start":"06:39.825 ","End":"06:44.355","Text":"We\u0027ll get negative cosine Phi,"},{"Start":"06:44.355 ","End":"06:48.460","Text":"and then we substitute in our zero and Pi."},{"Start":"06:48.620 ","End":"06:55.590","Text":"Then we\u0027ll be left with our integral on r. We\u0027ll have zero until R,"},{"Start":"06:55.590 ","End":"07:06.025","Text":"and then we\u0027ll have Rho 0 r^3 because we have this r multiplied by r^2 divided by R."},{"Start":"07:06.025 ","End":"07:10.095","Text":"Then from our integral along Theta and Phi,"},{"Start":"07:10.095 ","End":"07:15.645","Text":"we\u0027ll have that this is multiplied by 4 Pi and then dr. Now,"},{"Start":"07:15.645 ","End":"07:18.495","Text":"when we integrate along our r,"},{"Start":"07:18.495 ","End":"07:23.385","Text":"we\u0027ll get that we have 4 Pi Rho 0 divided by"},{"Start":"07:23.385 ","End":"07:30.150","Text":"R multiplied by R^4 divided by 4."},{"Start":"07:30.150 ","End":"07:35.460","Text":"Now, we can see that our 4\u0027s cancel out,"},{"Start":"07:35.460 ","End":"07:37.680","Text":"and here we\u0027ll have 3 and this cancels out."},{"Start":"07:37.680 ","End":"07:39.930","Text":"We\u0027ll get that our Q_in,"},{"Start":"07:39.930 ","End":"07:48.285","Text":"therefore is equal to Pi Rho 0 R^3."},{"Start":"07:48.285 ","End":"07:51.990","Text":"Perfect.Now, let\u0027s plug that back into our Gauss\u0027s law."},{"Start":"07:51.990 ","End":"08:02.055","Text":"We got that the electric flux was equal to E multiplied by 4 Pi r^2,"},{"Start":"08:02.055 ","End":"08:04.755","Text":"and that\u0027s equal to our Q_in,"},{"Start":"08:04.755 ","End":"08:10.995","Text":"which we just got now is equal to Pi Rho 0 R^3,"},{"Start":"08:10.995 ","End":"08:14.790","Text":"and then divided by Epsilon naught."},{"Start":"08:14.790 ","End":"08:19.815","Text":"First of all, we can see that our Pi\u0027s are going to cancel out,"},{"Start":"08:19.815 ","End":"08:22.725","Text":"and that is all that cancels out right now."},{"Start":"08:22.725 ","End":"08:26.205","Text":"Therefore, we can get that our E field,"},{"Start":"08:26.205 ","End":"08:28.830","Text":"when we\u0027re located outside of our sphere,"},{"Start":"08:28.830 ","End":"08:34.739","Text":"is going to be equal to Rho 0 R^3 divided"},{"Start":"08:34.739 ","End":"08:42.730","Text":"by 4 Epsilon naught r^2."},{"Start":"08:42.920 ","End":"08:45.960","Text":"This is our accurate electric field,"},{"Start":"08:45.960 ","End":"08:48.375","Text":"taking into account our Rho."},{"Start":"08:48.375 ","End":"08:51.390","Text":"However, I wanted to just show you something."},{"Start":"08:51.390 ","End":"08:55.260","Text":"If we go back to our E.S,"},{"Start":"08:55.260 ","End":"08:59.010","Text":"so E.4 Pi r^2,"},{"Start":"08:59.010 ","End":"09:05.055","Text":"and we leave this as Q_in divided by Epsilon naught."},{"Start":"09:05.055 ","End":"09:08.775","Text":"Then we don\u0027t substitute in what we found our Q_in to be,"},{"Start":"09:08.775 ","End":"09:10.830","Text":"let\u0027s just leave it as Q_in."},{"Start":"09:10.830 ","End":"09:14.265","Text":"Now, if we want to isolate out our E,"},{"Start":"09:14.265 ","End":"09:20.325","Text":"we\u0027ll get that our E is equal to Q_in divided by"},{"Start":"09:20.325 ","End":"09:28.070","Text":"4 Pi Epsilon naught multiplied by r^2,"},{"Start":"09:28.070 ","End":"09:29.570","Text":"or in other words,"},{"Start":"09:29.570 ","End":"09:35.520","Text":"that\u0027s equal to kQ_in divided by r^2."},{"Start":"09:35.520 ","End":"09:42.190","Text":"As we can see, this is the electric field due to a point charge."},{"Start":"09:42.290 ","End":"09:44.970","Text":"If we just play around with the numbers,"},{"Start":"09:44.970 ","End":"09:47.445","Text":"we\u0027ll get that these 2 are equal."},{"Start":"09:47.445 ","End":"09:50.895","Text":"As I said, this is the electric field due to a point charge."},{"Start":"09:50.895 ","End":"09:54.915","Text":"We can see that when we\u0027re located out some charged sphere,"},{"Start":"09:54.915 ","End":"09:57.318","Text":"it doesn\u0027t matter what\u0027s going on inside the sphere,"},{"Start":"09:57.318 ","End":"09:59.370","Text":"it will have some Q_in."},{"Start":"09:59.370 ","End":"10:02.940","Text":"But what\u0027s important to know is that when we\u0027re located outside of it,"},{"Start":"10:02.940 ","End":"10:06.660","Text":"our electric field is going to look like that of a point charge,"},{"Start":"10:06.660 ","End":"10:10.500","Text":"and that\u0027s very useful to remember,"},{"Start":"10:10.500 ","End":"10:14.470","Text":"and that\u0027s always true when we\u0027re dealing with a sphere."},{"Start":"10:16.280 ","End":"10:21.225","Text":"That\u0027s the electric field when we are located outside of the sphere."},{"Start":"10:21.225 ","End":"10:25.080","Text":"Now, let\u0027s look at what the electric field is when we\u0027re inside the sphere."},{"Start":"10:25.080 ","End":"10:27.840","Text":"Notice that once we\u0027re inside the sphere,"},{"Start":"10:27.840 ","End":"10:33.520","Text":"our electric field is not going to have anything to do with a point charge."},{"Start":"10:34.490 ","End":"10:37.395","Text":"Let\u0027s do this really quickly."},{"Start":"10:37.395 ","End":"10:45.748","Text":"When our r is smaller than R. We\u0027re located somewhere inside our sphere."},{"Start":"10:45.748 ","End":"10:47.130","Text":"Again, just like before,"},{"Start":"10:47.130 ","End":"10:51.235","Text":"our electric field is going to be uniform and in the radial direction."},{"Start":"10:51.235 ","End":"10:57.357","Text":"We can draw a small Gaussian surface within our charge sphere."},{"Start":"10:57.357 ","End":"11:02.080","Text":"Just remember that this Gaussian surface is 3-dimensional."},{"Start":"11:02.510 ","End":"11:04.770","Text":"Now, let\u0027s write."},{"Start":"11:04.770 ","End":"11:08.850","Text":"Our electric flux is going to be equal to, again,"},{"Start":"11:08.850 ","End":"11:14.850","Text":"because we have uniform electric field in the radial direction,"},{"Start":"11:14.850 ","End":"11:19.755","Text":"we can just simply say without writing the integral that this is equal to E.S,"},{"Start":"11:19.755 ","End":"11:28.410","Text":"which is going to be E multiplied by 4 Pi r^2 up until this r over here."},{"Start":"11:28.410 ","End":"11:33.720","Text":"This is of course equal to Q_in divided by Epsilon naught."},{"Start":"11:33.720 ","End":"11:36.165","Text":"What is our Q_in?"},{"Start":"11:36.165 ","End":"11:40.185","Text":"Again, it\u0027s going to be the integral of Rho, dV,"},{"Start":"11:40.185 ","End":"11:44.670","Text":"which is going to be the triple integral of our Rho,"},{"Start":"11:44.670 ","End":"11:53.040","Text":"so that\u0027s Rho 0 r divided by R. Then with our Jacobian,"},{"Start":"11:53.040 ","End":"11:54.945","Text":"because we\u0027re in spherical coordinates,"},{"Start":"11:54.945 ","End":"12:03.990","Text":"so this is going to be r^2 sin Phi dr d Theta d Phi."},{"Start":"12:03.990 ","End":"12:09.915","Text":"Our bounds are going to be pretty much exactly the same, so let\u0027s see."},{"Start":"12:09.915 ","End":"12:14.340","Text":"On Theta, we\u0027re again between 0 and 2 Pi and for Phi,"},{"Start":"12:14.340 ","End":"12:17.205","Text":"we\u0027re again between 0 and Pi."},{"Start":"12:17.205 ","End":"12:20.355","Text":"Our only difference is when we\u0027re integrating"},{"Start":"12:20.355 ","End":"12:23.685","Text":"for our dr. We\u0027re going to be going from 0,"},{"Start":"12:23.685 ","End":"12:25.845","Text":"and this time not until R,"},{"Start":"12:25.845 ","End":"12:31.065","Text":"we\u0027re going to stop at whichever radius our Gaussian surface is at."},{"Start":"12:31.065 ","End":"12:35.430","Text":"Here specifically, we\u0027re located at a radius of r."},{"Start":"12:35.430 ","End":"12:40.410","Text":"We\u0027re summing up the field up until this arbitrary r,"},{"Start":"12:40.410 ","End":"12:44.850","Text":"which is located somewhere within our charged sphere."},{"Start":"12:44.850 ","End":"12:47.925","Text":"Then what we\u0027ll see is,"},{"Start":"12:47.925 ","End":"12:53.715","Text":"our answer is going to be very similar to what we got last time but this time"},{"Start":"12:53.715 ","End":"12:55.995","Text":"it\u0027s going to be equal to"},{"Start":"12:55.995 ","End":"13:01.780","Text":"4 Pi Rho 0"},{"Start":"13:01.970 ","End":"13:07.830","Text":"multiplied by r^4,"},{"Start":"13:07.830 ","End":"13:13.950","Text":"then divided by 4 multiplied by R. We can see that we get"},{"Start":"13:13.950 ","End":"13:21.090","Text":"pretty much the same answer just because our final integration bound is this r,"},{"Start":"13:21.090 ","End":"13:24.360","Text":"so the things cancel out a little bit differently."},{"Start":"13:24.360 ","End":"13:27.840","Text":"What we can cancel out here are those 4\u0027s."},{"Start":"13:27.840 ","End":"13:32.190","Text":"It\u0027s a little bit different but really the only difference is our bound"},{"Start":"13:32.190 ","End":"13:36.060","Text":"when we\u0027re integrating along r. Therefore,"},{"Start":"13:36.060 ","End":"13:43.650","Text":"our electric field.4 Pi r^2 is equal to Q_in,"},{"Start":"13:43.650 ","End":"13:51.450","Text":"which we said is equal to Pi Rho 0 r^4 divided by R,"},{"Start":"13:51.450 ","End":"13:54.690","Text":"and then divided by Epsilon naught."},{"Start":"13:54.690 ","End":"13:57.270","Text":"We can add in our Epsilon naught over here."},{"Start":"13:57.270 ","End":"14:03.345","Text":"Now, we can divide both sides by r^2 and both sides by Pi,"},{"Start":"14:03.345 ","End":"14:06.030","Text":"and then we will get therefore that"},{"Start":"14:06.030 ","End":"14:09.960","Text":"our electric field when we\u0027re located within the sphere,"},{"Start":"14:09.960 ","End":"14:16.200","Text":"is going to be equal to Rho 0 r^2 divided by"},{"Start":"14:16.200 ","End":"14:25.055","Text":"4 Epsilon naught R. That\u0027s the end of the lesson."},{"Start":"14:25.055 ","End":"14:28.730","Text":"This is the electric field due to a sphere of"},{"Start":"14:28.730 ","End":"14:33.465","Text":"non-uniform charged density when we\u0027re located outside of the sphere,"},{"Start":"14:33.465 ","End":"14:36.080","Text":"and this is the electric field due to"},{"Start":"14:36.080 ","End":"14:43.280","Text":"a non-uniformly charged sphere when we\u0027re located within the sphere itself."},{"Start":"14:43.280 ","End":"14:45.870","Text":"That\u0027s the end of this lesson."}],"ID":22363},{"Watched":false,"Name":"Exercise 6","Duration":"25m 44s","ChapterTopicVideoID":21410,"CourseChapterTopicPlaylistID":246895,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/21410.jpeg","UploadDate":"2020-04-21T09:08:51.8930000","DurationForVideoObject":"PT25M44S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.300","Text":"Hello, in this question we have some surface of area A."},{"Start":"00:06.300 ","End":"00:08.535","Text":"Let\u0027s write this here, of area A,"},{"Start":"00:08.535 ","End":"00:10.965","Text":"and it has a width over here,"},{"Start":"00:10.965 ","End":"00:14.595","Text":"d. This surface isn\u0027t infinite."},{"Start":"00:14.595 ","End":"00:17.790","Text":"In the whole volume enclosed,"},{"Start":"00:17.790 ","End":"00:23.130","Text":"we have charged density per unit volume, Rho."},{"Start":"00:23.130 ","End":"00:25.695","Text":"Question number 1 is,"},{"Start":"00:25.695 ","End":"00:30.945","Text":"what is the electric field at a point far from the plane?"},{"Start":"00:30.945 ","End":"00:34.960","Text":"Let\u0027s say we choose this point over here."},{"Start":"00:35.450 ","End":"00:40.810","Text":"If we imagine that this point is very far away from this,"},{"Start":"00:40.810 ","End":"00:46.370","Text":"we can also consider it as if we\u0027re standing at this point we"},{"Start":"00:46.370 ","End":"00:54.060","Text":"see this plane or this cube as some point charge."},{"Start":"00:54.200 ","End":"00:58.175","Text":"Then, if we can see this as some point charge,"},{"Start":"00:58.175 ","End":"01:01.340","Text":"then we can say using Coulomb\u0027s law that"},{"Start":"01:01.340 ","End":"01:05.870","Text":"the electric field of a point charge is going to be equal"},{"Start":"01:05.870 ","End":"01:14.680","Text":"to kq divided by r squared in the r direction."},{"Start":"01:15.230 ","End":"01:21.355","Text":"Now what we need to do is we need to calculate what this q is equal to."},{"Start":"01:21.355 ","End":"01:25.655","Text":"We know that q is going to be equal to,"},{"Start":"01:25.655 ","End":"01:27.514","Text":"because we\u0027re dealing with volume,"},{"Start":"01:27.514 ","End":"01:34.150","Text":"our integral on our charge density multiplied by the volume, so dV."},{"Start":"01:34.150 ","End":"01:38.900","Text":"Now here, because our Rho is constant,"},{"Start":"01:38.900 ","End":"01:41.855","Text":"that means that we don\u0027t have to integrate and we can simply"},{"Start":"01:41.855 ","End":"01:47.375","Text":"multiply our Rho by our total volume enclosed."},{"Start":"01:47.375 ","End":"01:56.755","Text":"Which over here is simply going to be row multiplied by d multiplied by A."},{"Start":"01:56.755 ","End":"02:02.600","Text":"Then we can say that our electric field is equal to k multiplied by q,"},{"Start":"02:02.600 ","End":"02:11.460","Text":"which is Rho dA divided by r squared in the r direction."},{"Start":"02:11.720 ","End":"02:15.155","Text":"Now, question number 2 is asking us,"},{"Start":"02:15.155 ","End":"02:19.085","Text":"what is the electric field very close to the plane,"},{"Start":"02:19.085 ","End":"02:24.190","Text":"so very close to its surface or inside it?"},{"Start":"02:24.190 ","End":"02:30.785","Text":"Now, if I\u0027m located right above the area of the plane."},{"Start":"02:30.785 ","End":"02:33.230","Text":"If we draw it like this,"},{"Start":"02:33.230 ","End":"02:37.170","Text":"and if I\u0027m located very close."},{"Start":"02:38.200 ","End":"02:43.010","Text":"Well, we can imagine that our plane is infinite because"},{"Start":"02:43.010 ","End":"02:46.985","Text":"we\u0027re so close if we\u0027re something like here."},{"Start":"02:46.985 ","End":"02:50.075","Text":"This plane relative to"},{"Start":"02:50.075 ","End":"02:55.915","Text":"my distance away from the plane of my point is going to look infinite."},{"Start":"02:55.915 ","End":"03:01.355","Text":"That means that if my plane is infinite relative to my point,"},{"Start":"03:01.355 ","End":"03:08.995","Text":"that means that my surface area A doesn\u0027t interest me because A is equal to infinity."},{"Start":"03:08.995 ","End":"03:12.725","Text":"Now what we\u0027re going to do is we\u0027re going to try and find"},{"Start":"03:12.725 ","End":"03:17.285","Text":"our E field from an infinite plane that has some width,"},{"Start":"03:17.285 ","End":"03:26.000","Text":"which we know is d. We know from Coulomb\u0027s law or from symmetry that if we choose,"},{"Start":"03:26.000 ","End":"03:28.340","Text":"let\u0027s say this point over here,"},{"Start":"03:28.340 ","End":"03:35.167","Text":"we\u0027re going to have an electric field going in this direction from a point over here,"},{"Start":"03:35.167 ","End":"03:38.915","Text":"then we\u0027ll have an equal and opposite electric fields"},{"Start":"03:38.915 ","End":"03:43.115","Text":"going in this direction from this point over here."},{"Start":"03:43.115 ","End":"03:51.450","Text":"In total, our electric field is simply going to be in this direction."},{"Start":"03:52.040 ","End":"03:59.139","Text":"Then similarly our electric field at the bottom is going to be pointing downwards."},{"Start":"03:59.139 ","End":"04:02.700","Text":"Because our Rho is constant throughout the width,"},{"Start":"04:02.700 ","End":"04:06.830","Text":"our electric field is going to be coming out of our surface."},{"Start":"04:06.830 ","End":"04:11.680","Text":"In that case, I can work out the electric field by using Gauss\u0027s law."},{"Start":"04:11.680 ","End":"04:13.390","Text":"What I\u0027m going to do is,"},{"Start":"04:13.390 ","End":"04:20.505","Text":"I\u0027m going to draw my Gauss shape as some cube."},{"Start":"04:20.505 ","End":"04:27.160","Text":"Like so, where I have the surface area over here is S. That"},{"Start":"04:27.160 ","End":"04:33.945","Text":"means that the surface area over here is also S. Now,"},{"Start":"04:33.945 ","End":"04:41.190","Text":"we\u0027re going to find out our electric flux through this surface."},{"Start":"04:41.890 ","End":"04:46.685","Text":"Our electric flux is going to be equal to."},{"Start":"04:46.685 ","End":"04:48.935","Text":"As we saw before,"},{"Start":"04:48.935 ","End":"04:53.630","Text":"we have our electric field going in this direction,"},{"Start":"04:53.630 ","End":"04:56.785","Text":"and in this direction."},{"Start":"04:56.785 ","End":"05:00.635","Text":"That means that our flux is going to be equal to"},{"Start":"05:00.635 ","End":"05:09.375","Text":"2 times E times the surface area of a Gaussian envelope."},{"Start":"05:09.375 ","End":"05:11.240","Text":"Now as we know,"},{"Start":"05:11.240 ","End":"05:15.395","Text":"the electric field lines which are parallel to"},{"Start":"05:15.395 ","End":"05:22.475","Text":"our Gaussian envelope are going to cancel out or not add any electric flux,"},{"Start":"05:22.475 ","End":"05:24.680","Text":"exactly because they\u0027re parallel."},{"Start":"05:24.680 ","End":"05:27.770","Text":"It doesn\u0027t matter in which direction they\u0027re pointing,"},{"Start":"05:27.770 ","End":"05:31.265","Text":"the point is that they\u0027re parallel and we can only"},{"Start":"05:31.265 ","End":"05:36.125","Text":"measure the electric field which is perpendicular to our Gaussian surface."},{"Start":"05:36.125 ","End":"05:43.620","Text":"That\u0027s why we only can feel the electric fields over here and over here."},{"Start":"05:44.330 ","End":"05:52.745","Text":"As we know, this is going to be equal to Q_in divided by Epsilon naught."},{"Start":"05:52.745 ","End":"05:58.230","Text":"Now let\u0027s work out what our Q_in is equal to."},{"Start":"05:58.230 ","End":"06:03.770","Text":"We, first of all, know that it\u0027s going to be the integral on Rho dV."},{"Start":"06:03.770 ","End":"06:06.590","Text":"Because just like in the previous example we saw that"},{"Start":"06:06.590 ","End":"06:09.590","Text":"our Rho is uniform throughout the volume,"},{"Start":"06:09.590 ","End":"06:15.715","Text":"we can simply write that this is equal to Rho multiplied by our volume."},{"Start":"06:15.715 ","End":"06:18.075","Text":"What exactly is our volume?"},{"Start":"06:18.075 ","End":"06:25.255","Text":"Our volume is the volume of this shape of the plane,"},{"Start":"06:25.255 ","End":"06:30.470","Text":"which is enclosed within our Gaussian surface or Gaussian envelope."},{"Start":"06:30.470 ","End":"06:36.590","Text":"As we can see that we\u0027ll"},{"Start":"06:36.590 ","End":"06:44.040","Text":"have this volume over here enclosed."},{"Start":"06:44.270 ","End":"06:50.950","Text":"The dimensions of the volume is going to be equal to our surface area S"},{"Start":"06:50.950 ","End":"06:56.980","Text":"because it\u0027s projected onto our plane multiplied by this width over here,"},{"Start":"06:56.980 ","End":"07:03.120","Text":"which as we know is equal to d. Now let\u0027s plug everything in."},{"Start":"07:03.120 ","End":"07:08.580","Text":"We\u0027ll get that our 2ES is equal to Q_in,"},{"Start":"07:08.580 ","End":"07:13.150","Text":"which is Rho times our width,"},{"Start":"07:13.150 ","End":"07:18.310","Text":"times our surface area for our Gaussian surface divided"},{"Start":"07:18.310 ","End":"07:23.825","Text":"by Epsilon naught or 1 over Epsilon naught multiplied."},{"Start":"07:23.825 ","End":"07:27.095","Text":"Now we can see that we can divide both sides by S,"},{"Start":"07:27.095 ","End":"07:32.250","Text":"which is great because our S was some arbitrary surface area,"},{"Start":"07:32.250 ","End":"07:34.145","Text":"so it\u0027s good that there\u0027s cancels out."},{"Start":"07:34.145 ","End":"07:39.290","Text":"Then we can see that our electric field is simply going to be equal"},{"Start":"07:39.290 ","End":"07:46.000","Text":"to Rho d divided by 2 Epsilon naught."},{"Start":"07:46.000 ","End":"07:49.685","Text":"This is the magnitude of the electric field."},{"Start":"07:49.685 ","End":"07:51.395","Text":"However, as we saw,"},{"Start":"07:51.395 ","End":"07:53.105","Text":"if we want to find the direction,"},{"Start":"07:53.105 ","End":"07:58.550","Text":"we have our electric field going in the positive z direction when we\u0027re above the plane,"},{"Start":"07:58.550 ","End":"08:02.410","Text":"and in the negative z direction when we\u0027re below the plane."},{"Start":"08:02.410 ","End":"08:06.440","Text":"What does this mean above and below the plane?"},{"Start":"08:06.440 ","End":"08:14.475","Text":"What do we have to do is we have to define some axes over here at the origin."},{"Start":"08:14.475 ","End":"08:21.410","Text":"We can say that our y-axis is right in the middle of our plane."},{"Start":"08:21.410 ","End":"08:24.455","Text":"Then we can say if so,"},{"Start":"08:24.455 ","End":"08:28.040","Text":"that halfway through the y-axis,"},{"Start":"08:28.040 ","End":"08:33.670","Text":"so the top half of this plane is going to be d divided by 2,"},{"Start":"08:33.670 ","End":"08:40.120","Text":"and in the bottom half of the plane is also going to be d divided by 2."},{"Start":"08:40.120 ","End":"08:44.435","Text":"We\u0027re going right through the center of the width of the plane."},{"Start":"08:44.435 ","End":"08:50.780","Text":"Once we\u0027re located somewhere above this d divided by 2 over here,"},{"Start":"08:50.780 ","End":"08:52.775","Text":"then we\u0027re above the plane."},{"Start":"08:52.775 ","End":"08:58.175","Text":"When we\u0027re located somewhere below negative d divided by 2,"},{"Start":"08:58.175 ","End":"09:01.040","Text":"because this is negative d divided by 2,"},{"Start":"09:01.040 ","End":"09:03.350","Text":"then we\u0027re located below the plane."},{"Start":"09:03.350 ","End":"09:06.260","Text":"I\u0027m just going to draw this and big over here."},{"Start":"09:06.260 ","End":"09:10.580","Text":"Here we have the center of our plane."},{"Start":"09:10.580 ","End":"09:13.010","Text":"Then going up until here,"},{"Start":"09:13.010 ","End":"09:14.915","Text":"up until this point over here,"},{"Start":"09:14.915 ","End":"09:16.415","Text":"this is equal to z,"},{"Start":"09:16.415 ","End":"09:18.875","Text":"is equal to d divided by 2."},{"Start":"09:18.875 ","End":"09:22.505","Text":"Then this point over here is equal to z,"},{"Start":"09:22.505 ","End":"09:26.870","Text":"is equal to negative d divided by 2."},{"Start":"09:26.870 ","End":"09:31.885","Text":"If this is where z is equal to 0 in this red dotted line."},{"Start":"09:31.885 ","End":"09:37.400","Text":"Then, we can say that what is our electric field equal to?"},{"Start":"09:37.400 ","End":"09:40.970","Text":"We can say that it\u0027s equal to Rho d divided by"},{"Start":"09:40.970 ","End":"09:45.380","Text":"2 Epsilon naughts in the positive z direction,"},{"Start":"09:45.380 ","End":"09:48.785","Text":"when z is somewhere above the plane,"},{"Start":"09:48.785 ","End":"09:52.770","Text":"which means somewhere above d divided by 2."},{"Start":"09:53.430 ","End":"10:00.550","Text":"Then, we can say that our electric field is Rho d"},{"Start":"10:00.550 ","End":"10:08.365","Text":"divided by 2 Epsilon naught in the negative z direction."},{"Start":"10:08.365 ","End":"10:11.320","Text":"That is when we\u0027re below the plane."},{"Start":"10:11.320 ","End":"10:14.440","Text":"That is when z is below the plane,"},{"Start":"10:14.440 ","End":"10:19.855","Text":"so it\u0027s smaller than negative d divided by 2."},{"Start":"10:19.855 ","End":"10:27.070","Text":"This is our electric field when we\u0027re outside of our plane with the width."},{"Start":"10:27.070 ","End":"10:30.790","Text":"When we\u0027re either above the plane or below the plane."},{"Start":"10:30.790 ","End":"10:37.390","Text":"But what happens if we\u0027re trying to measure the electric field within the plane."},{"Start":"10:37.390 ","End":"10:43.420","Text":"I\u0027ve redrawn this picture of our plane with width from the side view."},{"Start":"10:43.420 ","End":"10:47.065","Text":"We\u0027re seeing this side of the plane over here,"},{"Start":"10:47.065 ","End":"10:51.070","Text":"and I\u0027ll remind you that going in this direction,"},{"Start":"10:51.070 ","End":"10:55.600","Text":"we have still our y-axis."},{"Start":"10:55.600 ","End":"11:00.880","Text":"We can see that it\u0027s going right through the middle of the plane."},{"Start":"11:00.880 ","End":"11:06.820","Text":"Now, just like what I did in order to find the electric field outside of the plane."},{"Start":"11:06.820 ","End":"11:11.005","Text":"I\u0027m going to again draw a Gaussian surface over here."},{"Start":"11:11.005 ","End":"11:15.560","Text":"Again, it\u0027s going to be this cube,"},{"Start":"11:15.690 ","End":"11:19.840","Text":"just like we had before."},{"Start":"11:19.840 ","End":"11:25.180","Text":"But this time, our cube is located inside the plane."},{"Start":"11:25.180 ","End":"11:27.580","Text":"Now, just like before,"},{"Start":"11:27.580 ","End":"11:31.480","Text":"we know that we\u0027re going to have some electric field in"},{"Start":"11:31.480 ","End":"11:35.905","Text":"the z-direction going out from this side of our cube,"},{"Start":"11:35.905 ","End":"11:42.955","Text":"and an electric field in this direction coming out from this side of the cube."},{"Start":"11:42.955 ","End":"11:49.495","Text":"Again, of course, no electric field is going to be coming out perpendicularly over here."},{"Start":"11:49.495 ","End":"11:51.670","Text":"Because we know that it\u0027s perpendicular to"},{"Start":"11:51.670 ","End":"11:56.005","Text":"this plane or this side of the Gaussian surface."},{"Start":"11:56.005 ","End":"12:01.870","Text":"We\u0027re going to not be able to measure the electric field coming through."},{"Start":"12:01.870 ","End":"12:06.910","Text":"We\u0027re just dealing like before with the top and bottom sides of the cube."},{"Start":"12:06.910 ","End":"12:11.290","Text":"Now we can say that the height of this cube,"},{"Start":"12:11.290 ","End":"12:13.285","Text":"so this distance over here,"},{"Start":"12:13.285 ","End":"12:15.130","Text":"is some height z."},{"Start":"12:15.130 ","End":"12:22.140","Text":"Similarly, because we\u0027ve drawn this cube in a symmetrical way relative to our y-axis."},{"Start":"12:22.140 ","End":"12:26.420","Text":"This height over here is also going to be z."},{"Start":"12:26.420 ","End":"12:29.005","Text":"Now what I\u0027m going to do is,"},{"Start":"12:29.005 ","End":"12:31.525","Text":"I\u0027m going to use Gauss\u0027s law."},{"Start":"12:31.525 ","End":"12:38.980","Text":"Here we can see that just like before when we had our cube pointing outside of our plane."},{"Start":"12:38.980 ","End":"12:45.835","Text":"We see that we have E multiplied by the surface area of the cube,"},{"Start":"12:45.835 ","End":"12:49.810","Text":"which is S. But then we have 2 of those."},{"Start":"12:49.810 ","End":"12:52.285","Text":"We\u0027ll have 2ES,"},{"Start":"12:52.285 ","End":"12:59.650","Text":"which is equal to 1 divided by Epsilon naught multiplied by Q_ in."},{"Start":"12:59.650 ","End":"13:02.185","Text":"What is our Q_ in?"},{"Start":"13:02.185 ","End":"13:11.860","Text":"Let\u0027s see, our Q_ in is equal to our Rho multiplied by our volume,"},{"Start":"13:11.860 ","End":"13:17.530","Text":"which is equal to Rho multiplied by the surface area of our cube,"},{"Start":"13:17.530 ","End":"13:22.630","Text":"so that\u0027s S. Now we have to multiply by this length over here."},{"Start":"13:22.630 ","End":"13:25.375","Text":"Now, before when we use Gauss\u0027s law,"},{"Start":"13:25.375 ","End":"13:29.050","Text":"we multiply it by d. Because we took"},{"Start":"13:29.050 ","End":"13:38.125","Text":"some Gaussian surface that went above our plane."},{"Start":"13:38.125 ","End":"13:43.089","Text":"Then, when we were summing up how much charge was included within the surface,"},{"Start":"13:43.089 ","End":"13:47.215","Text":"we had to multiply by this total width, d,"},{"Start":"13:47.215 ","End":"13:50.230","Text":"because that would be the volume multiplied"},{"Start":"13:50.230 ","End":"13:53.440","Text":"by our surface area S which is some arbitrary value."},{"Start":"13:53.440 ","End":"13:58.345","Text":"However, here our Gaussian surface is of length 2z,"},{"Start":"13:58.345 ","End":"14:01.210","Text":"as we can see over here, z and z,"},{"Start":"14:01.210 ","End":"14:05.770","Text":"and it\u0027s always going to be smaller than d because our Gaussian surface"},{"Start":"14:05.770 ","End":"14:12.475","Text":"is aiding us and finding out the electric field within the surface itself."},{"Start":"14:12.475 ","End":"14:18.085","Text":"We\u0027re going to be multiplying this by 2z, z plus z."},{"Start":"14:18.085 ","End":"14:20.890","Text":"Now let\u0027s rewrite this."},{"Start":"14:20.890 ","End":"14:22.510","Text":"We\u0027ll have that this is, therefore,"},{"Start":"14:22.510 ","End":"14:27.595","Text":"equal to 1 divided by Epsilon naught multiplied by Q_ in."},{"Start":"14:27.595 ","End":"14:31.240","Text":"Our Q_ in is Rho times"},{"Start":"14:31.240 ","End":"14:39.420","Text":"2z times S. Now we can see that our 2 and our S can cross it off."},{"Start":"14:39.420 ","End":"14:45.310","Text":"Then we\u0027ll get therefore that our electric field is going to be equal"},{"Start":"14:45.310 ","End":"14:51.645","Text":"to Rho z divided by Epsilon naught."},{"Start":"14:51.645 ","End":"14:55.470","Text":"Of course, that is going to be in the z direction"},{"Start":"14:55.470 ","End":"15:00.250","Text":"when we\u0027re located within our plane with width."},{"Start":"15:00.250 ","End":"15:10.540","Text":"That\u0027s going to be when z is between negative d divided by 2 and positive d divided by 2."},{"Start":"15:10.540 ","End":"15:14.560","Text":"Then what we can do is for our final answer,"},{"Start":"15:14.560 ","End":"15:18.935","Text":"we can just take this and plug it in over here."},{"Start":"15:18.935 ","End":"15:24.210","Text":"Now, here we have to deal with our signs because the sign of the electric field above"},{"Start":"15:24.210 ","End":"15:29.760","Text":"the plane was positive and the sign of the electric field below the plane was negative."},{"Start":"15:29.760 ","End":"15:34.670","Text":"Here, however, we don\u0027t have to deal with whether we\u0027re above this center line."},{"Start":"15:34.670 ","End":"15:36.310","Text":"Why don\u0027t we need to do that?"},{"Start":"15:36.310 ","End":"15:39.580","Text":"It\u0027s because our variable here is z."},{"Start":"15:39.580 ","End":"15:42.760","Text":"We can see that if we\u0027re at a positive z,"},{"Start":"15:42.760 ","End":"15:44.965","Text":"so above this dotted line."},{"Start":"15:44.965 ","End":"15:47.740","Text":"We\u0027ll get a positive value for our electric field."},{"Start":"15:47.740 ","End":"15:50.499","Text":"If however, our z value over here is negative,"},{"Start":"15:50.499 ","End":"15:53.140","Text":"that means that will be under the dotted line."},{"Start":"15:53.140 ","End":"15:57.805","Text":"Then our value for our electric field will be a negative."},{"Start":"15:57.805 ","End":"16:00.685","Text":"This is our answer to question number 2."},{"Start":"16:00.685 ","End":"16:04.150","Text":"Now let\u0027s take a look at question number 3."},{"Start":"16:04.150 ","End":"16:08.380","Text":"Question number 3 is saying that an electron is"},{"Start":"16:08.380 ","End":"16:13.405","Text":"placed at z is smaller than d divided by 2."},{"Start":"16:13.405 ","End":"16:18.010","Text":"That means that we\u0027re located somewhere around here,"},{"Start":"16:18.010 ","End":"16:23.140","Text":"is our point z_ 0 inside our plane."},{"Start":"16:23.140 ","End":"16:28.720","Text":"We\u0027re being told that if the charge density Rho over here is positive,"},{"Start":"16:28.720 ","End":"16:32.935","Text":"what is the position of the electron as a function of time?"},{"Start":"16:32.935 ","End":"16:39.025","Text":"What we\u0027re trying to do is we\u0027re trying to find our position r(t)."},{"Start":"16:39.025 ","End":"16:40.855","Text":"In order to do that,"},{"Start":"16:40.855 ","End":"16:47.270","Text":"I\u0027m going to need to find my positions along the x, y, and z(t)."},{"Start":"16:47.940 ","End":"16:54.580","Text":"The way that I\u0027m going to do that is I\u0027m going to say that my force,"},{"Start":"16:54.580 ","End":"16:56.710","Text":"I\u0027m going to use the ideal force,"},{"Start":"16:56.710 ","End":"16:57.925","Text":"which is equal to,"},{"Start":"16:57.925 ","End":"17:04.130","Text":"as we know, q multiplied by the electric field."},{"Start":"17:04.590 ","End":"17:07.180","Text":"Let\u0027s substitute in."},{"Start":"17:07.180 ","End":"17:09.580","Text":"Our charge is q,"},{"Start":"17:09.580 ","End":"17:12.370","Text":"and because we\u0027re dealing with an electron,"},{"Start":"17:12.370 ","End":"17:16.720","Text":"so we\u0027re going to take the absolute value of the charge of an electron,"},{"Start":"17:16.720 ","End":"17:19.750","Text":"and an electron has a minus sign,"},{"Start":"17:19.750 ","End":"17:25.735","Text":"so we\u0027ll have a minus out over here multiplied by the electric field."},{"Start":"17:25.735 ","End":"17:30.265","Text":"We can see that our force on our electron is going to be"},{"Start":"17:30.265 ","End":"17:35.620","Text":"acting in the opposite direction to that of our electric field."},{"Start":"17:35.620 ","End":"17:40.135","Text":"We know that our electric field is pointing"},{"Start":"17:40.135 ","End":"17:44.320","Text":"in this direction and from this equation over here,"},{"Start":"17:44.320 ","End":"17:49.840","Text":"we know that our force is pointing in this direction,"},{"Start":"17:49.840 ","End":"17:53.320","Text":"in the opposite direction to our electric fields."},{"Start":"17:53.320 ","End":"17:54.670","Text":"Now, in the question,"},{"Start":"17:54.670 ","End":"18:01.105","Text":"we\u0027re being told that the electron is placed at this position z_ 0 over here."},{"Start":"18:01.105 ","End":"18:06.920","Text":"That means that our initial velocity is also going to be equal to 0."},{"Start":"18:07.650 ","End":"18:12.760","Text":"As we can see, our electron is going to go and move in"},{"Start":"18:12.760 ","End":"18:18.535","Text":"this downwards direction towards the center of our plane."},{"Start":"18:18.535 ","End":"18:23.485","Text":"Once it gets to below the center of our plane over here."},{"Start":"18:23.485 ","End":"18:30.640","Text":"We know that our electric field over here is pointing in the opposite direction,"},{"Start":"18:30.640 ","End":"18:38.050","Text":"so here our E field is in this direction because we\u0027re below the center line."},{"Start":"18:38.050 ","End":"18:45.115","Text":"Then, once our electron gets to this point if the electric field is pointing downwards,"},{"Start":"18:45.115 ","End":"18:47.424","Text":"as we can see from this equation,"},{"Start":"18:47.424 ","End":"18:51.505","Text":"our force is going to be pointing upwards."},{"Start":"18:51.505 ","End":"18:54.925","Text":"Back towards the center of the plane."},{"Start":"18:54.925 ","End":"19:02.750","Text":"We can see that we\u0027re going to have some harmonic motion about the center of the plane."},{"Start":"19:03.660 ","End":"19:10.505","Text":"Let\u0027s begin substituting in our values to this equation and solving,"},{"Start":"19:10.505 ","End":"19:14.335","Text":"and we\u0027ll see that we really do get harmonic motion."},{"Start":"19:14.335 ","End":"19:17.485","Text":"We have our negative,"},{"Start":"19:17.485 ","End":"19:20.050","Text":"the size of our charge,"},{"Start":"19:20.050 ","End":"19:23.920","Text":"so that\u0027s E multiplied by our electric field."},{"Start":"19:23.920 ","End":"19:28.615","Text":"We\u0027re assuming that we\u0027re always going to be somewhere in the middle of our plane."},{"Start":"19:28.615 ","End":"19:33.580","Text":"We\u0027re going to have to use the electric field that is relevant."},{"Start":"19:33.580 ","End":"19:36.400","Text":"That\u0027s going to be this section over here."},{"Start":"19:36.400 ","End":"19:40.210","Text":"Will have that this is multiplied by Rho z,"},{"Start":"19:40.210 ","End":"19:43.165","Text":"divided by Epsilon naught."},{"Start":"19:43.165 ","End":"19:49.060","Text":"Now, of course, because we only have an electric field component in the z-direction."},{"Start":"19:49.060 ","End":"19:55.000","Text":"Of course, we\u0027re only going to have our force acting in the z-direction."},{"Start":"19:55.000 ","End":"19:57.340","Text":"Then according to Newton,"},{"Start":"19:57.340 ","End":"20:02.950","Text":"this all has to be equal to our mass multiplied by acceleration."},{"Start":"20:02.950 ","End":"20:10.070","Text":"Now our acceleration is also only going to be in the z-direction."},{"Start":"20:10.530 ","End":"20:18.265","Text":"We have no components of acceleration or electric field in either the x or y direction,"},{"Start":"20:18.265 ","End":"20:20.660","Text":"only in the z-direction."},{"Start":"20:21.180 ","End":"20:27.490","Text":"What is another way of rising out my acceleration in the z-direction?"},{"Start":"20:27.490 ","End":"20:31.690","Text":"I can write that this is equal to z double dot."},{"Start":"20:31.690 ","End":"20:36.940","Text":"Now we can see that we have a differential equation because here I have a variable z,"},{"Start":"20:36.940 ","End":"20:41.360","Text":"and here I have the second derivative of that variable."},{"Start":"20:41.360 ","End":"20:47.840","Text":"Let\u0027s move over here and continue with our equation."},{"Start":"20:47.840 ","End":"20:51.355","Text":"All of this, as we can see,"},{"Start":"20:51.355 ","End":"20:58.195","Text":"my electron charge, Epsilon naught and Rho are constants in the question."},{"Start":"20:58.195 ","End":"21:03.565","Text":"I can call this constant K effective."},{"Start":"21:03.565 ","End":"21:05.965","Text":"Now let\u0027s write this out."},{"Start":"21:05.965 ","End":"21:08.665","Text":"I have negative K effective."},{"Start":"21:08.665 ","End":"21:18.835","Text":"Negative K_eff multiplied by z is equal to mz double dot."},{"Start":"21:18.835 ","End":"21:26.140","Text":"Now we can already recognize this equation as the equation for some spring."},{"Start":"21:26.140 ","End":"21:29.320","Text":"That means that we can already see that this is going to"},{"Start":"21:29.320 ","End":"21:31.930","Text":"be harmonic motion when dealing with a spring,"},{"Start":"21:31.930 ","End":"21:39.790","Text":"and we know that the frequency is going to be equal to,"},{"Start":"21:39.790 ","End":"21:50.240","Text":"so Omega is equal to the square root of K divided by m. Where here our K is K effective,"},{"Start":"21:51.030 ","End":"21:55.570","Text":"which now we can substitute in what our K effective is."},{"Start":"21:55.570 ","End":"21:58.420","Text":"We can say that our frequency of oscillations,"},{"Start":"21:58.420 ","End":"22:01.960","Text":"it\u0027s going to be equal to the size of"},{"Start":"22:01.960 ","End":"22:06.880","Text":"our electron charge multiplied by our charge density per"},{"Start":"22:06.880 ","End":"22:11.455","Text":"unit volume divided by Epsilon naught times"},{"Start":"22:11.455 ","End":"22:17.545","Text":"m. Now that we have our Omega,"},{"Start":"22:17.545 ","End":"22:23.800","Text":"so we can write our equation for a position along z as a function of time, which,"},{"Start":"22:23.800 ","End":"22:26.305","Text":"as we know, because we\u0027re dealing with harmonic motion,"},{"Start":"22:26.305 ","End":"22:30.280","Text":"and also we have to solve this differential equation."},{"Start":"22:30.280 ","End":"22:36.490","Text":"We know that our solution is going to be equal to A cosine of Omega,"},{"Start":"22:36.490 ","End":"22:40.315","Text":"which we have over here, t plus Phi."},{"Start":"22:40.315 ","End":"22:43.360","Text":"Now what we have to do is we have to find out what A is"},{"Start":"22:43.360 ","End":"22:47.060","Text":"equal to and what our Phi is equal to."},{"Start":"22:47.820 ","End":"22:51.775","Text":"In order to find out what our A and Phi is equal to,"},{"Start":"22:51.775 ","End":"22:55.210","Text":"we have to use our initial conditions."},{"Start":"22:55.210 ","End":"22:57.475","Text":"What we have to do, is we have to find"},{"Start":"22:57.475 ","End":"23:01.870","Text":"our position along the z-axis at t is equal to 0,"},{"Start":"23:01.870 ","End":"23:08.630","Text":"and our velocity in the z direction at t is equal to 0."},{"Start":"23:09.060 ","End":"23:12.970","Text":"My position at t is equal to 0,"},{"Start":"23:12.970 ","End":"23:16.400","Text":"is here at z_0."},{"Start":"23:16.500 ","End":"23:19.045","Text":"That\u0027s my initial position,"},{"Start":"23:19.045 ","End":"23:25.630","Text":"and my initial velocity we\u0027re being told that the electron is placed at this position,"},{"Start":"23:25.630 ","End":"23:29.390","Text":"which means that our initial velocity is equal to 0."},{"Start":"23:30.110 ","End":"23:33.705","Text":"First of all, before we begin with this,"},{"Start":"23:33.705 ","End":"23:39.175","Text":"let\u0027s see what our z dot as a function of t is equal to."},{"Start":"23:39.175 ","End":"23:43.030","Text":"We\u0027re simply taking the derivative of this equation over here,"},{"Start":"23:43.030 ","End":"23:54.230","Text":"and we\u0027ll get that that is equal to negative Omega A sine of Omega t plus Phi."},{"Start":"23:54.570 ","End":"23:57.340","Text":"Now let\u0027s plug that in."},{"Start":"23:57.340 ","End":"24:00.205","Text":"We know that 0 has to be equal to"},{"Start":"24:00.205 ","End":"24:09.205","Text":"negative Omega A sine of Omega t plus Phi when t is equal to 0."},{"Start":"24:09.205 ","End":"24:11.995","Text":"That\u0027s going to be sine of Phi."},{"Start":"24:11.995 ","End":"24:14.050","Text":"If this has to be equal to 0,"},{"Start":"24:14.050 ","End":"24:18.955","Text":"we know that our Omega is n equal to 0 and our A is n equal to 0."},{"Start":"24:18.955 ","End":"24:22.225","Text":"That means r sine Phi must be equal to 0."},{"Start":"24:22.225 ","End":"24:25.750","Text":"When is sine Phi equal to 0?"},{"Start":"24:25.750 ","End":"24:29.905","Text":"That happens when our Phi is equal to 0."},{"Start":"24:29.905 ","End":"24:33.250","Text":"There we have that our Phi is equal to 0."},{"Start":"24:33.250 ","End":"24:35.575","Text":"Now we can plug that in here."},{"Start":"24:35.575 ","End":"24:41.065","Text":"Now when our z at t is equal to 0, is equal to z_0."},{"Start":"24:41.065 ","End":"24:45.730","Text":"Then we get that that is equal to A cosine of"},{"Start":"24:45.730 ","End":"24:51.190","Text":"Omega t plus Phi when our t is equal to 0 and our Phi is equal to 0,"},{"Start":"24:51.190 ","End":"24:52.923","Text":"so cosine of 0."},{"Start":"24:52.923 ","End":"24:55.900","Text":"Cosine of 0 is equal to 1."},{"Start":"24:55.900 ","End":"25:02.995","Text":"That means that our z_0 is A or A is equal to 0."},{"Start":"25:02.995 ","End":"25:07.300","Text":"Then, therefore, let\u0027s scroll down."},{"Start":"25:07.300 ","End":"25:16.090","Text":"We can say that our position as a function of time is equal to z_0 multiplied"},{"Start":"25:16.090 ","End":"25:25.450","Text":"by cosine of Omega t. This is the answer to question number 3."},{"Start":"25:25.450 ","End":"25:27.160","Text":"If we scroll back,"},{"Start":"25:27.160 ","End":"25:30.355","Text":"we saw that we wanted to find our position as a function of time,"},{"Start":"25:30.355 ","End":"25:35.785","Text":"and we can see that our electron is only moving in the z-direction,"},{"Start":"25:35.785 ","End":"25:41.335","Text":"which means that this will be its position as a function of time."},{"Start":"25:41.335 ","End":"25:44.210","Text":"That\u0027s the end of this lesson."}],"ID":22364},{"Watched":false,"Name":"Differential Form Of Gauss_s Law","Duration":"6m 41s","ChapterTopicVideoID":21411,"CourseChapterTopicPlaylistID":246895,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/21411.jpeg","UploadDate":"2020-04-21T09:11:27.6470000","DurationForVideoObject":"PT6M41S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.695","Text":"Hello. In this lesson,"},{"Start":"00:01.695 ","End":"00:05.880","Text":"we\u0027re going to be speaking about the differential form of Gauss\u0027s law."},{"Start":"00:05.880 ","End":"00:10.305","Text":"This equation is the differential form of Gauss\u0027s law and it\u0027s equal to"},{"Start":"00:10.305 ","End":"00:15.480","Text":"the divergence of E is equal to Rho divided by Epsilon naught,"},{"Start":"00:15.480 ","End":"00:21.900","Text":"where E is our electric field and Rho is the charge density per unit volume."},{"Start":"00:21.900 ","End":"00:25.830","Text":"Of course, Epsilon naught is our constant."},{"Start":"00:25.830 ","End":"00:31.155","Text":"The first thing that we\u0027re going to do is we\u0027re going to see how"},{"Start":"00:31.155 ","End":"00:34.050","Text":"this equation over here is connected"},{"Start":"00:34.050 ","End":"00:38.170","Text":"to the equation that we already know for Gauss\u0027s law."},{"Start":"00:38.170 ","End":"00:44.630","Text":"What we\u0027ve been talking about up until now is the integral form of Gauss\u0027s law."},{"Start":"00:44.630 ","End":"00:51.209","Text":"We know that that\u0027s this closed integral of E.ds,"},{"Start":"00:51.209 ","End":"00:57.065","Text":"and we said that that is equal to Qn divided by Epsilon naught."},{"Start":"00:57.065 ","End":"00:59.810","Text":"Then as a little side note,"},{"Start":"00:59.810 ","End":"01:10.190","Text":"our Qn can also be written as the integral of Rho dv."},{"Start":"01:10.190 ","End":"01:14.260","Text":"We can already see this Rho divided by Epsilon naught over here,"},{"Start":"01:14.260 ","End":"01:16.910","Text":"so we can see what\u0027s coming."},{"Start":"01:17.780 ","End":"01:24.625","Text":"Let\u0027s integrate both sides of our differential form of Gauss\u0027s law."},{"Start":"01:24.625 ","End":"01:29.160","Text":"We\u0027re integrating dv E"},{"Start":"01:29.160 ","End":"01:35.905","Text":"and that is of course equal to the integral of Rho divided by Epsilon naught."},{"Start":"01:35.905 ","End":"01:37.480","Text":"Because both sides are equal."},{"Start":"01:37.480 ","End":"01:39.070","Text":"Now because we have a Rho here,"},{"Start":"01:39.070 ","End":"01:44.450","Text":"we\u0027re integrating along the volume and so here as well."},{"Start":"01:45.330 ","End":"01:50.560","Text":"We can already see that this integral is going to simply"},{"Start":"01:50.560 ","End":"01:52.690","Text":"equal what this is equal to so let\u0027s just write"},{"Start":"01:52.690 ","End":"01:54.970","Text":"it out so that it\u0027s a little bit more clear."},{"Start":"01:54.970 ","End":"01:58.210","Text":"We\u0027re going to have that our Qn is equal to Rho dv,"},{"Start":"01:58.210 ","End":"02:01.055","Text":"the integral of that divided by Epsilon naught."},{"Start":"02:01.055 ","End":"02:05.570","Text":"We can already see that this integral is going to be equal to that."},{"Start":"02:05.570 ","End":"02:08.810","Text":"But what about our left-hand side over here?"},{"Start":"02:08.810 ","End":"02:13.220","Text":"Now something that\u0027s important to know is that anytime we have an"},{"Start":"02:13.220 ","End":"02:18.604","Text":"integral along a volume of the divergence of any function,"},{"Start":"02:18.604 ","End":"02:21.020","Text":"not just a function for the electric field,"},{"Start":"02:21.020 ","End":"02:23.000","Text":"but absolutely any function."},{"Start":"02:23.000 ","End":"02:28.580","Text":"If we have a volumetric integral along a divergence of some function,"},{"Start":"02:28.580 ","End":"02:32.000","Text":"it\u0027s always going to be equal to the integral along"},{"Start":"02:32.000 ","End":"02:40.085","Text":"the closed loop of E.ds."},{"Start":"02:40.085 ","End":"02:41.945","Text":"This is correct always."},{"Start":"02:41.945 ","End":"02:43.595","Text":"Now what is this s over here?"},{"Start":"02:43.595 ","End":"02:46.984","Text":"It\u0027s the total area of the surface."},{"Start":"02:46.984 ","End":"02:52.920","Text":"This integral is to show that our surface is a closed surface."},{"Start":"02:53.590 ","End":"03:00.800","Text":"Here we\u0027re integrating along the volume enclosed within some shape and here we\u0027re"},{"Start":"03:00.800 ","End":"03:08.245","Text":"integrating along the surface area which encloses this volume."},{"Start":"03:08.245 ","End":"03:13.220","Text":"This equality comes from vector calculus and it\u0027s"},{"Start":"03:13.220 ","End":"03:18.665","Text":"called the divergence theorem and also the Gauss theorem."},{"Start":"03:18.665 ","End":"03:21.620","Text":"It relates the electric flux,"},{"Start":"03:21.620 ","End":"03:23.255","Text":"which is this over here."},{"Start":"03:23.255 ","End":"03:28.520","Text":"That\u0027s the flow of a vector field through some closed surface,"},{"Start":"03:28.520 ","End":"03:32.270","Text":"to the behavior of this vector field,"},{"Start":"03:32.270 ","End":"03:35.405","Text":"whichever it might be, here specifically it\u0027s the electric field."},{"Start":"03:35.405 ","End":"03:38.120","Text":"It relates the behavior of this electric field"},{"Start":"03:38.120 ","End":"03:41.885","Text":"within that enclosed shape and the volume of that enclosed shape."},{"Start":"03:41.885 ","End":"03:47.180","Text":"Now it\u0027s pretty clear to see that if our integral"},{"Start":"03:47.180 ","End":"03:52.639","Text":"on dv E is equal to what we know over here is our electric flux,"},{"Start":"03:52.639 ","End":"03:56.355","Text":"which comes from our integral form of Gauss\u0027s law,"},{"Start":"03:56.355 ","End":"03:59.180","Text":"and that that is therefore equal to this,"},{"Start":"03:59.180 ","End":"04:00.530","Text":"which is equal to this,"},{"Start":"04:00.530 ","End":"04:05.365","Text":"which is equal to what we saw in previous lessons so we can see how it all relates."},{"Start":"04:05.365 ","End":"04:09.746","Text":"What we\u0027re seeing here is the differential form of Gauss\u0027s law,"},{"Start":"04:09.746 ","End":"04:12.949","Text":"whereas when we\u0027re speaking about this version,"},{"Start":"04:12.949 ","End":"04:16.415","Text":"this is either just Gauss\u0027s law or the integral version."},{"Start":"04:16.415 ","End":"04:23.405","Text":"Now a lot of the time people will speak about this equation as 1 of Maxwell\u0027s laws."},{"Start":"04:23.405 ","End":"04:25.835","Text":"What Maxwell did is he took"},{"Start":"04:25.835 ","End":"04:34.400","Text":"4 very useful mathematical equations and they\u0027re very useful for a broad range of topics,"},{"Start":"04:34.400 ","End":"04:37.085","Text":"both in this course and many other courses that you\u0027ll do."},{"Start":"04:37.085 ","End":"04:39.470","Text":"Those are Maxwell\u0027s equations,"},{"Start":"04:39.470 ","End":"04:43.265","Text":"and this is Maxwell\u0027s first equation."},{"Start":"04:43.265 ","End":"04:47.945","Text":"Here we have Maxwell\u0027s first equation in the differential form,"},{"Start":"04:47.945 ","End":"04:53.615","Text":"and here we have Maxwell\u0027s first equation in the integral form."},{"Start":"04:53.615 ","End":"04:56.750","Text":"Now the real question is,"},{"Start":"04:56.750 ","End":"05:00.950","Text":"why do we have this differential form?"},{"Start":"05:00.950 ","End":"05:04.940","Text":"We\u0027ve been using the integral form and it\u0027s worked great."},{"Start":"05:04.940 ","End":"05:10.790","Text":"The main use is to find our charge density."},{"Start":"05:10.790 ","End":"05:16.360","Text":"The integral form is super useful when we\u0027re trying to find our electric field,"},{"Start":"05:16.360 ","End":"05:21.964","Text":"and the differential form is super useful when we\u0027re trying to find our charge density."},{"Start":"05:21.964 ","End":"05:26.295","Text":"Let\u0027s give a brief little explanation."},{"Start":"05:26.295 ","End":"05:27.900","Text":"In future questions,"},{"Start":"05:27.900 ","End":"05:31.985","Text":"we\u0027ll really use this and work it out but let\u0027s just see what this means."},{"Start":"05:31.985 ","End":"05:37.925","Text":"We can take the divergence of some electric field."},{"Start":"05:37.925 ","End":"05:39.260","Text":"What does that mean?"},{"Start":"05:39.260 ","End":"05:40.835","Text":"In Cartesian coordinates,"},{"Start":"05:40.835 ","End":"05:46.550","Text":"we\u0027re going to have d our x component of our electric field by dx"},{"Start":"05:46.550 ","End":"05:53.330","Text":"plus the differential of our electric fields y component dy,"},{"Start":"05:53.330 ","End":"06:00.335","Text":"plus the differential of our electric fields z component dz."},{"Start":"06:00.335 ","End":"06:05.765","Text":"Then that is going to be equal to our charge density divided by Epsilon naught,"},{"Start":"06:05.765 ","End":"06:09.097","Text":"where Epsilon naught is just some constant value,"},{"Start":"06:09.097 ","End":"06:10.510","Text":"it doesn\u0027t really matter."},{"Start":"06:10.510 ","End":"06:16.325","Text":"Now what we can see from here is that we\u0027re going to get a scalar number."},{"Start":"06:16.325 ","End":"06:18.020","Text":"This isn\u0027t a vector quantity,"},{"Start":"06:18.020 ","End":"06:19.685","Text":"this is a scalar quantity."},{"Start":"06:19.685 ","End":"06:25.000","Text":"Now of course, this is the divergence of an electric field in Cartesian coordinates,"},{"Start":"06:25.000 ","End":"06:29.310","Text":"if we have a spherical or a cylindrical coordinate."},{"Start":"06:29.310 ","End":"06:33.050","Text":"Our divergence is going to look slightly differently and"},{"Start":"06:33.050 ","End":"06:38.900","Text":"that you can look up and you can write that in your equation notes."},{"Start":"06:38.900 ","End":"06:42.330","Text":"That\u0027s the end of this lesson."}],"ID":22365},{"Watched":false,"Name":"Exercise 7","Duration":"20m 30s","ChapterTopicVideoID":21412,"CourseChapterTopicPlaylistID":246895,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/21412.jpeg","UploadDate":"2020-04-21T09:17:34.4630000","DurationForVideoObject":"PT20M30S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.770","Text":"Hello, in this question,"},{"Start":"00:01.770 ","End":"00:06.735","Text":"we\u0027re being told that an infinite plane of width 2d is charged with"},{"Start":"00:06.735 ","End":"00:12.075","Text":"a charge density of 7 times Az^6,"},{"Start":"00:12.075 ","End":"00:14.580","Text":"where A is some constant."},{"Start":"00:14.580 ","End":"00:18.780","Text":"We\u0027re being told that the z-axis is perpendicular to the plane,"},{"Start":"00:18.780 ","End":"00:21.930","Text":"and located at its center."},{"Start":"00:21.930 ","End":"00:26.955","Text":"Now question 1 is what is the electric field?"},{"Start":"00:26.955 ","End":"00:29.750","Text":"In order to find the electric field,"},{"Start":"00:29.750 ","End":"00:33.095","Text":"we\u0027re going to be using Gauss\u0027s law."},{"Start":"00:33.095 ","End":"00:35.825","Text":"Gauss\u0027s law goes like so."},{"Start":"00:35.825 ","End":"00:42.780","Text":"We have the integral along E.ds is equal"},{"Start":"00:42.780 ","End":"00:50.585","Text":"to 1 divided by Epsilon naught multiplied by the integral of Rho dv,"},{"Start":"00:50.585 ","End":"00:53.110","Text":"where v is the volume."},{"Start":"00:53.110 ","End":"00:56.625","Text":"Let\u0027s see what this is."},{"Start":"00:56.625 ","End":"01:01.265","Text":"We know from previous videos that when we\u0027re dealing with a plane,"},{"Start":"01:01.265 ","End":"01:04.250","Text":"we can simply use a Gaussian surface,"},{"Start":"01:04.250 ","End":"01:08.140","Text":"which is a cube."},{"Start":"01:08.300 ","End":"01:11.535","Text":"Here, we have our cube over here."},{"Start":"01:11.535 ","End":"01:14.750","Text":"We know from symmetry that we\u0027re going to have"},{"Start":"01:14.750 ","End":"01:19.775","Text":"an electric field above the plane going in this direction,"},{"Start":"01:19.775 ","End":"01:24.830","Text":"and an electric fields of equal magnitude,"},{"Start":"01:24.830 ","End":"01:26.600","Text":"but in the opposite direction,"},{"Start":"01:26.600 ","End":"01:30.990","Text":"so pointing down when we\u0027re below the plane."},{"Start":"01:31.790 ","End":"01:40.625","Text":"Now we can see that our integral along E.ds is simply going to be equal to E"},{"Start":"01:40.625 ","End":"01:49.555","Text":"multiplied by the surface area of this surface plus the surface area of this surface."},{"Start":"01:49.555 ","End":"01:54.550","Text":"If we say that the surface area of both of these are S,"},{"Start":"01:54.550 ","End":"01:59.150","Text":"so it\u0027s going to be e multiplied by 2S."},{"Start":"02:00.470 ","End":"02:06.400","Text":"Now, let\u0027s see what our integral of Rho dv is equal to."},{"Start":"02:06.400 ","End":"02:12.835","Text":"We have E multiplied by 2S is equal to 1 divided by Epsilon naught,"},{"Start":"02:12.835 ","End":"02:16.630","Text":"multiplied by, so this is a volume integral,"},{"Start":"02:16.630 ","End":"02:19.755","Text":"so it\u0027s going to be a triple integral over here,"},{"Start":"02:19.755 ","End":"02:29.459","Text":"and then our Rho is given by 7AZ^6, and then dv."},{"Start":"02:29.459 ","End":"02:35.080","Text":"So we can have dx, dy, dz."},{"Start":"02:36.350 ","End":"02:43.550","Text":"We know that this integral of Rho dv is going to equal to our Q in."},{"Start":"02:43.550 ","End":"02:47.420","Text":"Our Q in is our charge inside,"},{"Start":"02:47.420 ","End":"02:51.020","Text":"so our charge enclosed by a Gaussian surface."},{"Start":"02:51.020 ","End":"02:58.745","Text":"Now, obviously, our charge is only distributed along our plane with width."},{"Start":"02:58.745 ","End":"03:04.370","Text":"That means that only the section that is blue over here has charge,"},{"Start":"03:04.370 ","End":"03:07.845","Text":"the rest doesn\u0027t have charge."},{"Start":"03:07.845 ","End":"03:09.950","Text":"When we\u0027re integrating along z,"},{"Start":"03:09.950 ","End":"03:18.410","Text":"we\u0027re integrating up until d over here and until minus d over here."},{"Start":"03:18.410 ","End":"03:22.740","Text":"Or in total a length of 2d."},{"Start":"03:22.870 ","End":"03:27.290","Text":"Now, when we\u0027re integrating along our dx, dy,"},{"Start":"03:27.290 ","End":"03:36.020","Text":"we can see that that is simply going to be this area over here, which as we know,"},{"Start":"03:36.020 ","End":"03:45.800","Text":"is simply our S. We know that this surface area is S in order to integrate."},{"Start":"03:45.800 ","End":"03:49.040","Text":"For our volume, we have to do dx, dy, dz."},{"Start":"03:49.040 ","End":"03:54.780","Text":"Our dz is going to go from negative d to positive d, and our dx,"},{"Start":"03:54.780 ","End":"03:58.610","Text":"dy is simply the surface area over here,"},{"Start":"03:58.610 ","End":"04:04.890","Text":"S. We can just say S over here."},{"Start":"04:05.570 ","End":"04:08.460","Text":"This is important to note,"},{"Start":"04:08.460 ","End":"04:12.620","Text":"and this is also because we can see that our function for"},{"Start":"04:12.620 ","End":"04:17.330","Text":"our charge distribution is independent of x and y."},{"Start":"04:17.330 ","End":"04:20.530","Text":"It\u0027s only dependent on z."},{"Start":"04:20.600 ","End":"04:26.015","Text":"Of course, as discussed our bounds for our z is from"},{"Start":"04:26.015 ","End":"04:32.600","Text":"minus d over here until positive d over here."},{"Start":"04:32.600 ","End":"04:36.320","Text":"Now let\u0027s integrate. First of all,"},{"Start":"04:36.320 ","End":"04:37.775","Text":"we have our dx, dy,"},{"Start":"04:37.775 ","End":"04:40.030","Text":"which we\u0027re just integrating,"},{"Start":"04:40.030 ","End":"04:42.495","Text":"and our total area is going to be S,"},{"Start":"04:42.495 ","End":"04:47.210","Text":"so we will have S divided by our Epsilon naught over here."},{"Start":"04:47.210 ","End":"04:50.510","Text":"Now we\u0027re integrating between negative d and"},{"Start":"04:50.510 ","End":"05:00.080","Text":"d for 7AZ^6 dz."},{"Start":"05:00.080 ","End":"05:01.705","Text":"Let\u0027s do this integral."},{"Start":"05:01.705 ","End":"05:04.965","Text":"We\u0027ll have S divided by Epsilon naught,"},{"Start":"05:04.965 ","End":"05:08.780","Text":"and then we\u0027ll have that this is multiplied by 7A,"},{"Start":"05:08.780 ","End":"05:15.225","Text":"multiplied by Z^7, divided by 7."},{"Start":"05:15.225 ","End":"05:16.830","Text":"These 7s cancel out,"},{"Start":"05:16.830 ","End":"05:20.960","Text":"and then of course we have to substitute in our bounds of negative d"},{"Start":"05:20.960 ","End":"05:25.865","Text":"and d. Once we substitute these in,"},{"Start":"05:25.865 ","End":"05:31.070","Text":"we\u0027ll have that E multiplied by 2S is going to"},{"Start":"05:31.070 ","End":"05:36.770","Text":"be equal to S divided by Epsilon naught multiplied by,"},{"Start":"05:36.770 ","End":"05:43.085","Text":"so here we\u0027ll have d^7 minus negative d^7,"},{"Start":"05:43.085 ","End":"05:52.160","Text":"which is simply going to be a total of 2d^7."},{"Start":"05:52.160 ","End":"05:57.525","Text":"Now we can see that 2S cancels out on both sides,"},{"Start":"05:57.525 ","End":"06:02.030","Text":"and therefore I will get that our electric field is equal"},{"Start":"06:02.030 ","End":"06:08.365","Text":"to d^7 divided by Epsilon naught."},{"Start":"06:08.365 ","End":"06:11.580","Text":"Of course, there\u0027s an A here as well,"},{"Start":"06:11.580 ","End":"06:14.640","Text":"so multiplied by A."},{"Start":"06:14.640 ","End":"06:24.080","Text":"Now, we can say that our electric field in vector form is going to be equal to,"},{"Start":"06:24.080 ","End":"06:31.250","Text":"so d^7 multiplied by A divided by Epsilon naught,"},{"Start":"06:31.250 ","End":"06:39.920","Text":"and it\u0027s going to be in the positive z-direction when we\u0027re located at some value Z,"},{"Start":"06:39.920 ","End":"06:45.900","Text":"which is bigger than d. That means that we\u0027re above the plane."},{"Start":"06:46.270 ","End":"06:49.970","Text":"Or our electric field is going to be equal"},{"Start":"06:49.970 ","End":"06:55.730","Text":"to d^7 multiplied by A divided by Epsilon naught,"},{"Start":"06:55.730 ","End":"07:00.070","Text":"but this time in the negative z-direction,"},{"Start":"07:00.070 ","End":"07:06.035","Text":"and that is going to happen when we\u0027re located below the plane or when we\u0027re located at Z"},{"Start":"07:06.035 ","End":"07:12.540","Text":"is smaller than negative d. That\u0027s great."},{"Start":"07:12.540 ","End":"07:16.445","Text":"That is our electric field when we\u0027re located outside of the plane."},{"Start":"07:16.445 ","End":"07:19.910","Text":"But what happens when we\u0027re located within the plane?"},{"Start":"07:19.910 ","End":"07:22.640","Text":"When z is between d and"},{"Start":"07:22.640 ","End":"07:29.840","Text":"negative d. If we wanted to work out what our electric field is at the center."},{"Start":"07:29.840 ","End":"07:33.455","Text":"Just like before, and also just like in the previous video,"},{"Start":"07:33.455 ","End":"07:43.095","Text":"we can draw a Gaussian surface that simply is located within our plane."},{"Start":"07:43.095 ","End":"07:46.670","Text":"Just like before, we\u0027re going to have that we have"},{"Start":"07:46.670 ","End":"07:52.670","Text":"an electric field at the top side over here pointing in the positive z-direction,"},{"Start":"07:52.670 ","End":"07:58.720","Text":"and our electric field at the bottom pointing in the negative z-direction."},{"Start":"07:58.720 ","End":"08:01.640","Text":"Now, just like what we did before,"},{"Start":"08:01.640 ","End":"08:03.695","Text":"we\u0027re going to use Gauss\u0027s law."},{"Start":"08:03.695 ","End":"08:07.715","Text":"Again, we\u0027re going to have that our E.ds"},{"Start":"08:07.715 ","End":"08:13.585","Text":"integral will simply be equal to E multiplied by 2S,"},{"Start":"08:13.585 ","End":"08:17.420","Text":"and our integral over here,"},{"Start":"08:17.420 ","End":"08:20.960","Text":"instead of being this over here,"},{"Start":"08:20.960 ","End":"08:25.576","Text":"where we\u0027re integrating along our z from minus d until d."},{"Start":"08:25.576 ","End":"08:31.760","Text":"We\u0027re taking all the charge included in our large Gaussian surface,"},{"Start":"08:31.760 ","End":"08:36.435","Text":"which was larger than this width."},{"Start":"08:36.435 ","End":"08:41.300","Text":"Now we\u0027re simply integrating along this section over here,"},{"Start":"08:41.300 ","End":"08:45.395","Text":"which is enclosed by this Gaussian surface which is smaller"},{"Start":"08:45.395 ","End":"08:50.570","Text":"than d. Instead of integrating between negative d and d,"},{"Start":"08:50.570 ","End":"08:57.150","Text":"we\u0027ll be integrating between negative z and z."},{"Start":"09:00.230 ","End":"09:03.990","Text":"Then, we can make this a tag to"},{"Start":"09:03.990 ","End":"09:08.265","Text":"differentiate between this z and what we\u0027re integrating against."},{"Start":"09:08.265 ","End":"09:11.490","Text":"Here again, we have between z and negative z."},{"Start":"09:11.490 ","End":"09:15.338","Text":"Add in the tags, and again,"},{"Start":"09:15.338 ","End":"09:20.955","Text":"and then what we\u0027ll get is that here,"},{"Start":"09:20.955 ","End":"09:27.490","Text":"instead of having d^7, we\u0027ll have z^7."},{"Start":"09:27.530 ","End":"09:30.023","Text":"Also over here,"},{"Start":"09:30.023 ","End":"09:36.435","Text":"and so that means that our electric field when we\u0027re located within"},{"Start":"09:36.435 ","End":"09:44.100","Text":"is simply going to be equal to A divided by Epsilon naught,"},{"Start":"09:44.100 ","End":"09:46.830","Text":"and then instead of d^7,"},{"Start":"09:46.830 ","End":"09:51.640","Text":"we\u0027re going to have that this is z^7."},{"Start":"09:51.800 ","End":"09:57.090","Text":"This is of course going to be in the z-direction."},{"Start":"09:57.090 ","End":"10:01.545","Text":"This is when z is smaller than d but"},{"Start":"10:01.545 ","End":"10:06.795","Text":"bigger than negative d. Now here I don\u0027t have to differentiate"},{"Start":"10:06.795 ","End":"10:11.430","Text":"between whether I\u0027m above my center or"},{"Start":"10:11.430 ","End":"10:17.153","Text":"below with the signs of my arrows like I did over here,"},{"Start":"10:17.153 ","End":"10:20.460","Text":"because as we can see here, my electric field is in the positive direction,"},{"Start":"10:20.460 ","End":"10:22.335","Text":"and here in the negative direction,"},{"Start":"10:22.335 ","End":"10:24.855","Text":"and that\u0027s because it all depends on my z."},{"Start":"10:24.855 ","End":"10:27.012","Text":"If my z is a positive number,"},{"Start":"10:27.012 ","End":"10:29.655","Text":"then z^7 will be positive,"},{"Start":"10:29.655 ","End":"10:31.725","Text":"and then as we expected,"},{"Start":"10:31.725 ","End":"10:34.140","Text":"we\u0027re pointing in the positive z-direction."},{"Start":"10:34.140 ","End":"10:41.880","Text":"If z is negative number to the power of an odd number,"},{"Start":"10:41.880 ","End":"10:47.265","Text":"so 7, we\u0027re also going to have a negative value."},{"Start":"10:47.265 ","End":"10:53.220","Text":"Then, that will show that if we\u0027re located below this center line,"},{"Start":"10:53.220 ","End":"10:59.009","Text":"that our electric field will be pointing in the negative z-direction as expected,"},{"Start":"10:59.009 ","End":"11:01.930","Text":"and then it all works out."},{"Start":"11:01.940 ","End":"11:05.745","Text":"This is our answer to question number 1."},{"Start":"11:05.745 ","End":"11:09.010","Text":"Let\u0027s go on to question number 2."},{"Start":"11:10.910 ","End":"11:16.905","Text":"Now we\u0027re going to be dealing with question number 2."},{"Start":"11:16.905 ","End":"11:25.180","Text":"Question number 2 is to show that Gauss\u0027s law in its differential form is applied here."},{"Start":"11:25.970 ","End":"11:31.455","Text":"Gauss\u0027s law in differential form is given as"},{"Start":"11:31.455 ","End":"11:40.895","Text":"the divergence of E is equal to Rho divided by Epsilon naught."},{"Start":"11:40.895 ","End":"11:46.305","Text":"Now let\u0027s write out what our divergence of E is equal to."},{"Start":"11:46.305 ","End":"11:49.395","Text":"The divergence of E is equal to"},{"Start":"11:49.395 ","End":"11:57.255","Text":"the partial differential of the x component of our electric field by dx,"},{"Start":"11:57.255 ","End":"12:01.485","Text":"partial differential of the y component of our E field dy,"},{"Start":"12:01.485 ","End":"12:08.380","Text":"plus the partial differential of our z component of our E field dz."},{"Start":"12:10.190 ","End":"12:13.335","Text":"Let\u0027s see what this means."},{"Start":"12:13.335 ","End":"12:19.095","Text":"When we\u0027re located along our Gaussian surface,"},{"Start":"12:19.095 ","End":"12:21.615","Text":"so let\u0027s draw it again,"},{"Start":"12:21.615 ","End":"12:27.030","Text":"this cube going like so."},{"Start":"12:27.030 ","End":"12:30.480","Text":"When we\u0027re located above our surface,"},{"Start":"12:30.480 ","End":"12:32.430","Text":"we can see that our electric field,"},{"Start":"12:32.430 ","End":"12:34.335","Text":"either above or below,"},{"Start":"12:34.335 ","End":"12:39.480","Text":"is going to be equal to d^7"},{"Start":"12:39.480 ","End":"12:45.315","Text":"multiplied by A divided by Epsilon naught in either the z or the minus z-direction."},{"Start":"12:45.315 ","End":"12:54.760","Text":"We can see therefore that when z is bigger than d or when z is smaller than negative d,"},{"Start":"12:55.310 ","End":"12:58.740","Text":"our electric field is constant."},{"Start":"12:58.740 ","End":"13:01.320","Text":"If our electric field is constant,"},{"Start":"13:01.320 ","End":"13:06.000","Text":"the divergence of it is going to be equal to 0."},{"Start":"13:06.000 ","End":"13:09.960","Text":"Over here, in these areas,"},{"Start":"13:09.960 ","End":"13:12.405","Text":"our divergence is equal to 0."},{"Start":"13:12.405 ","End":"13:17.580","Text":"Which therefore means since our divergence is equal to Rho divided by Epsilon naught,"},{"Start":"13:17.580 ","End":"13:19.530","Text":"that means that our Rho,"},{"Start":"13:19.530 ","End":"13:26.475","Text":"which is our charge per unit volume or our charge distribution is equal to 0."},{"Start":"13:26.475 ","End":"13:32.670","Text":"Which is what we would expect because when we\u0027re located outside of our plane width,"},{"Start":"13:32.670 ","End":"13:38.320","Text":"width, we can see that there is no charge in that area."},{"Start":"13:38.360 ","End":"13:46.155","Text":"Now let\u0027s take a look what happens when we\u0027re located inside our plane width, width."},{"Start":"13:46.155 ","End":"13:52.604","Text":"Now we\u0027re located in the region of z is between"},{"Start":"13:52.604 ","End":"14:00.659","Text":"negative d and d. Now let\u0027s work out the divergence of our electric field."},{"Start":"14:00.659 ","End":"14:04.110","Text":"We can see that our electric field over here is"},{"Start":"14:04.110 ","End":"14:08.070","Text":"dependent on z and it\u0027s only in the z-direction."},{"Start":"14:08.070 ","End":"14:11.700","Text":"Our electric field here only has a z component."},{"Start":"14:11.700 ","End":"14:15.780","Text":"This component over here of our divergence is equal to"},{"Start":"14:15.780 ","End":"14:20.278","Text":"0 and also with respect to y is also equal to 0,"},{"Start":"14:20.278 ","End":"14:23.030","Text":"and now we just have to take the derivative of"},{"Start":"14:23.030 ","End":"14:26.750","Text":"our z component of the electric field with respect to z."},{"Start":"14:26.750 ","End":"14:33.075","Text":"We\u0027ll see that this is going to be equal to 7 divided by"},{"Start":"14:33.075 ","End":"14:40.230","Text":"Epsilon naught multiplied by Az^6."},{"Start":"14:40.230 ","End":"14:43.230","Text":"Then as given over here,"},{"Start":"14:43.230 ","End":"14:47.625","Text":"this is equal to Rho divided by Epsilon naught."},{"Start":"14:47.625 ","End":"14:52.185","Text":"Then we can see that if we multiply both sides by Epsilon naught,"},{"Start":"14:52.185 ","End":"14:57.660","Text":"we rarely get that Rho is equal to 7Az^6,"},{"Start":"14:57.660 ","End":"15:02.169","Text":"which is exactly what we were given in our question."},{"Start":"15:02.900 ","End":"15:10.275","Text":"We can see from both of our answers to our Gauss\u0027s equation in the differential form,"},{"Start":"15:10.275 ","End":"15:16.650","Text":"we can see that they really do describe the physical quality that we have going on over"},{"Start":"15:16.650 ","End":"15:23.300","Text":"here where above or below our plane with width we have 0 charge."},{"Start":"15:23.300 ","End":"15:26.340","Text":"Our charged density there is equal to 0 and"},{"Start":"15:26.340 ","End":"15:29.745","Text":"within our plane where we do actually have a charge density,"},{"Start":"15:29.745 ","End":"15:35.145","Text":"we get the exact same charge density that we were given in our question."},{"Start":"15:35.145 ","End":"15:38.085","Text":"Now let\u0027s go on to question number 3."},{"Start":"15:38.085 ","End":"15:42.015","Text":"Question number 3 is asking us to find the curl,"},{"Start":"15:42.015 ","End":"15:46.575","Text":"which is defined as Nabla cross E of the electric field,"},{"Start":"15:46.575 ","End":"15:49.035","Text":"and to then explain the result."},{"Start":"15:49.035 ","End":"15:53.070","Text":"The first thing that we\u0027re going to do is we\u0027re"},{"Start":"15:53.070 ","End":"15:57.760","Text":"going to write out our equation for the curl."},{"Start":"15:58.070 ","End":"16:04.455","Text":"Here, we have our equation for the curl of the electric field."},{"Start":"16:04.455 ","End":"16:09.105","Text":"Now, again, our equation is given in Cartesian coordinates,"},{"Start":"16:09.105 ","End":"16:10.860","Text":"and that\u0027s because we can see that that\u0027s"},{"Start":"16:10.860 ","End":"16:14.160","Text":"the coordinate system that we used throughout the question."},{"Start":"16:14.160 ","End":"16:18.240","Text":"What we can see is that we\u0027re taking"},{"Start":"16:18.240 ","End":"16:21.690","Text":"the derivative of each component of the electric field"},{"Start":"16:21.690 ","End":"16:29.250","Text":"with respect to some other coordinates and then we put that in vector format."},{"Start":"16:29.250 ","End":"16:32.805","Text":"The curl of the electric field is itself a vector,"},{"Start":"16:32.805 ","End":"16:37.780","Text":"whereas the divergence of the electric field is a scalar number."},{"Start":"16:37.970 ","End":"16:45.795","Text":"As we can see, our curl is going to be equal to 0. Why is that?"},{"Start":"16:45.795 ","End":"16:52.350","Text":"We can see that our electric field is always in the z-direction no matter where we are."},{"Start":"16:52.350 ","End":"16:56.738","Text":"We only have a z component for the electric field,"},{"Start":"16:56.738 ","End":"16:58.560","Text":"and we can see that each time that we take"},{"Start":"16:58.560 ","End":"17:01.650","Text":"the derivative of the z component of our electric field,"},{"Start":"17:01.650 ","End":"17:07.710","Text":"we\u0027re taking it with respect to either y or x,"},{"Start":"17:07.710 ","End":"17:10.140","Text":"and we can see that there\u0027s no x or y"},{"Start":"17:10.140 ","End":"17:15.060","Text":"variable in our z components for the electric field."},{"Start":"17:15.060 ","End":"17:21.340","Text":"That means that all of this is simply going to be equal to 0."},{"Start":"17:22.370 ","End":"17:26.670","Text":"We can see that the curl of our electric field is going to be equal to"},{"Start":"17:26.670 ","End":"17:31.860","Text":"0 when we\u0027re outside of our plane,"},{"Start":"17:31.860 ","End":"17:36.240","Text":"and also when we\u0027re located within the plane itself."},{"Start":"17:36.240 ","End":"17:38.910","Text":"Why is that? Now,"},{"Start":"17:38.910 ","End":"17:42.495","Text":"outside it\u0027s obvious that it\u0027s going to be equal to 0"},{"Start":"17:42.495 ","End":"17:46.510","Text":"because our electric field there is constant,"},{"Start":"17:46.510 ","End":"17:49.755","Text":"so when we take the derivative it\u0027s just going to be equal to 0."},{"Start":"17:49.755 ","End":"17:56.955","Text":"However, why does the curl of the electric field equal 0 also inside the plane itself?"},{"Start":"17:56.955 ","End":"18:02.375","Text":"As we know, the electric field is formed by lots of little charges,"},{"Start":"18:02.375 ","End":"18:07.070","Text":"because we have our charged distribution given like so."},{"Start":"18:07.070 ","End":"18:10.310","Text":"We know that there\u0027s lots of little different charges."},{"Start":"18:10.310 ","End":"18:14.420","Text":"Now, a field which is produced by lots of"},{"Start":"18:14.420 ","End":"18:19.133","Text":"charges is always going to be a conserved field,"},{"Start":"18:19.133 ","End":"18:26.200","Text":"and a conserved field is defined by a field where its rotor is equal to 0."},{"Start":"18:26.200 ","End":"18:31.955","Text":"From now on, we can know automatically that whenever we have some field,"},{"Start":"18:31.955 ","End":"18:35.105","Text":"which is the product of charges,"},{"Start":"18:35.105 ","End":"18:39.905","Text":"the rotor of that field is always going to be equal to 0."},{"Start":"18:39.905 ","End":"18:41.930","Text":"Now another way of thinking about this,"},{"Start":"18:41.930 ","End":"18:44.990","Text":"which is a way that we\u0027re going to look at in more detail"},{"Start":"18:44.990 ","End":"18:49.345","Text":"slightly later on is by using Maxwell\u0027s third equation."},{"Start":"18:49.345 ","End":"18:52.745","Text":"Maxwell\u0027s third equation is like so."},{"Start":"18:52.745 ","End":"18:58.490","Text":"It says that the rotor of our E field is equal"},{"Start":"18:58.490 ","End":"19:05.705","Text":"to the negative change in our B field with respect to time."},{"Start":"19:05.705 ","End":"19:10.820","Text":"Now, the B field is the magnetic field and we haven\u0027t yet spoken about"},{"Start":"19:10.820 ","End":"19:12.860","Text":"the magnetic field and its relation to"},{"Start":"19:12.860 ","End":"19:16.640","Text":"the electric field but just know that there is this equation,"},{"Start":"19:16.640 ","End":"19:19.495","Text":"it\u0027s Maxwell\u0027s third equation."},{"Start":"19:19.495 ","End":"19:25.580","Text":"Just to take this back why this equation is useful or so for our question over here."},{"Start":"19:25.580 ","End":"19:30.155","Text":"We can see that in our question we don\u0027t have a magnetic field."},{"Start":"19:30.155 ","End":"19:35.150","Text":"In which case, if we take the time derivative of our non-existent magnetic field,"},{"Start":"19:35.150 ","End":"19:38.480","Text":"we\u0027ll get that this is equal to 0 and then we get that"},{"Start":"19:38.480 ","End":"19:43.555","Text":"the rotor of our E field is again equal to 0."},{"Start":"19:43.555 ","End":"19:49.045","Text":"There are two ways to know what the curl of the electric field is equal to."},{"Start":"19:49.045 ","End":"19:53.050","Text":"The one way is to know automatically that it\u0027s equal to 0 because"},{"Start":"19:53.050 ","End":"19:57.490","Text":"our E field is created by charge distribution,"},{"Start":"19:57.490 ","End":"19:59.440","Text":"which means it\u0027s created by charges,"},{"Start":"19:59.440 ","End":"20:02.545","Text":"which means that it\u0027s going to be a conservative field,"},{"Start":"20:02.545 ","End":"20:06.310","Text":"which means that the curl of the electric field will be equal to 0."},{"Start":"20:06.310 ","End":"20:11.110","Text":"The other way is remembering Maxwell\u0027s third equation and knowing that the curl of"},{"Start":"20:11.110 ","End":"20:16.128","Text":"the electric field is equal to the negative time derivative of the magnetic field."},{"Start":"20:16.128 ","End":"20:18.235","Text":"Of course, in our question over here,"},{"Start":"20:18.235 ","End":"20:19.900","Text":"we have no magnetic field,"},{"Start":"20:19.900 ","End":"20:23.845","Text":"which means that the time derivative is equal to 0 and there again,"},{"Start":"20:23.845 ","End":"20:25.670","Text":"we get that our curl,"},{"Start":"20:25.670 ","End":"20:28.770","Text":"the E field is equal to 0."},{"Start":"20:28.770 ","End":"20:31.839","Text":"That\u0027s the end of this lesson."}],"ID":22366},{"Watched":false,"Name":"Exercise 8","Duration":"35m 50s","ChapterTopicVideoID":21413,"CourseChapterTopicPlaylistID":246895,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/21413.jpeg","UploadDate":"2020-04-21T09:29:59.7500000","DurationForVideoObject":"PT35M50S","Description":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.980","Text":"Hello. In this question,"},{"Start":"00:01.980 ","End":"00:04.740","Text":"we\u0027re given an infinite plane of width d,"},{"Start":"00:04.740 ","End":"00:07.890","Text":"which has a charge density of Rho as a function of z,"},{"Start":"00:07.890 ","End":"00:10.935","Text":"which is equal to Az."},{"Start":"00:10.935 ","End":"00:14.550","Text":"This is dependent on the distance away from the center of the plane."},{"Start":"00:14.550 ","End":"00:16.080","Text":"A is a constant,"},{"Start":"00:16.080 ","End":"00:21.700","Text":"and what is the electric field formed because of this charge density?"},{"Start":"00:21.700 ","End":"00:30.555","Text":"As we can see, we have this plane which is infinite in the xy-directions,"},{"Start":"00:30.555 ","End":"00:39.875","Text":"and then we have our z-axis going up perpendicular to the plane where the origin is,"},{"Start":"00:39.875 ","End":"00:41.645","Text":"as we\u0027re being told in the question,"},{"Start":"00:41.645 ","End":"00:44.195","Text":"at the center of the plane."},{"Start":"00:44.195 ","End":"00:47.510","Text":"That means that each distance over here,"},{"Start":"00:47.510 ","End":"00:54.360","Text":"so this is going to be d over 2 and this distance is also going to be d over 2,"},{"Start":"00:54.360 ","End":"00:55.850","Text":"but this value over here,"},{"Start":"00:55.850 ","End":"01:01.950","Text":"because it\u0027s below the xy-plane will be negative d over 2."},{"Start":"01:02.350 ","End":"01:04.790","Text":"Now, at first glance,"},{"Start":"01:04.790 ","End":"01:09.394","Text":"this question might look exactly identical to the previous questions"},{"Start":"01:09.394 ","End":"01:14.785","Text":"that we\u0027ve solved that deal with an infinite plane that has some width."},{"Start":"01:14.785 ","End":"01:17.944","Text":"However, this question is a little bit different,"},{"Start":"01:17.944 ","End":"01:19.685","Text":"still we\u0027ll go into y,"},{"Start":"01:19.685 ","End":"01:24.700","Text":"and therefore we can\u0027t solve it in the way that we\u0027ve previously seen."},{"Start":"01:24.700 ","End":"01:29.780","Text":"The way that we previously saw is that we\u0027d have some infinite plane with width,"},{"Start":"01:29.780 ","End":"01:36.464","Text":"and then we\u0027d put a Gaussian surface over this,"},{"Start":"01:36.464 ","End":"01:38.279","Text":"that looks something like this,"},{"Start":"01:38.279 ","End":"01:46.934","Text":"some cube, where it was a height z from the center,"},{"Start":"01:46.934 ","End":"01:50.970","Text":"above and an height z below the center, going down."},{"Start":"01:50.970 ","End":"01:55.950","Text":"Then we would say that there\u0027s some E field coming out from the top,"},{"Start":"01:55.950 ","End":"01:59.224","Text":"and from symmetry we\u0027d have an equal but"},{"Start":"01:59.224 ","End":"02:03.824","Text":"oppositely directed E field coming from the bottom."},{"Start":"02:03.824 ","End":"02:12.830","Text":"Then we would use Gauss\u0027s law and we would say that E multiplied by 2s,"},{"Start":"02:12.830 ","End":"02:18.139","Text":"S being the surface area of the upper and lower sides,"},{"Start":"02:18.139 ","End":"02:27.240","Text":"is equal to the total charge inside our Gaussian surface divided by Epsilon_naught."},{"Start":"02:27.340 ","End":"02:31.129","Text":"Why could we use Gauss\u0027s law?"},{"Start":"02:31.129 ","End":"02:34.070","Text":"It\u0027s because usually we were being told that"},{"Start":"02:34.070 ","End":"02:40.415","Text":"the charge distribution inside our plane was either uniform,"},{"Start":"02:40.415 ","End":"02:47.249","Text":"or symmetrical throughout, then we could use that,"},{"Start":"02:47.249 ","End":"02:51.575","Text":"the fact that it\u0027s either uniform or symmetrically distributed"},{"Start":"02:51.575 ","End":"02:57.090","Text":"along the xy-plane above and below,"},{"Start":"02:57.090 ","End":"02:58.745","Text":"then we could use Gauss\u0027s law."},{"Start":"02:58.745 ","End":"03:04.355","Text":"However, we can see here that we don\u0027t have symmetry. Why is that?"},{"Start":"03:04.355 ","End":"03:06.740","Text":"When we\u0027re above the xy-plane,"},{"Start":"03:06.740 ","End":"03:10.460","Text":"so in the positive section of our z-axis."},{"Start":"03:10.460 ","End":"03:15.035","Text":"Charge distribution will be A times some positive number."},{"Start":"03:15.035 ","End":"03:18.670","Text":"We\u0027ll have some positive charge distribution."},{"Start":"03:18.670 ","End":"03:22.290","Text":"If we\u0027re located below the xy-plane,"},{"Start":"03:22.290 ","End":"03:26.539","Text":"in the region where z-axis is minus,"},{"Start":"03:26.539 ","End":"03:30.150","Text":"so our z-value will also be minus."},{"Start":"03:30.150 ","End":"03:36.499","Text":"Then our charge distribution will be some constant multiplied by a negative number."},{"Start":"03:36.499 ","End":"03:39.790","Text":"Our charge distribution will be negative."},{"Start":"03:39.790 ","End":"03:43.174","Text":"Now we can see that we don\u0027t have the symmetry and therefore,"},{"Start":"03:43.174 ","End":"03:47.110","Text":"we can\u0027t exactly use these laws."},{"Start":"03:47.110 ","End":"03:53.460","Text":"What we have here is charge density, which is anti-symmetric."},{"Start":"03:53.460 ","End":"03:59.605","Text":"That means that above this center line,"},{"Start":"03:59.605 ","End":"04:02.539","Text":"the xy-plane, we have positive charges,"},{"Start":"04:02.539 ","End":"04:07.525","Text":"and below the xy-plane we have negative charges."},{"Start":"04:07.525 ","End":"04:10.579","Text":"Because we have this anti-symmetry,"},{"Start":"04:10.579 ","End":"04:13.189","Text":"that means if we flip our problem around,"},{"Start":"04:13.189 ","End":"04:15.094","Text":"because there\u0027s no symmetry,"},{"Start":"04:15.094 ","End":"04:18.430","Text":"we\u0027ll have a different problem at hand,"},{"Start":"04:18.430 ","End":"04:21.559","Text":"because then the minuses will be on top and the positives"},{"Start":"04:21.559 ","End":"04:24.830","Text":"at the bottom and the direction of the E field will be different."},{"Start":"04:24.830 ","End":"04:27.019","Text":"Because we can\u0027t do that,"},{"Start":"04:27.019 ","End":"04:32.380","Text":"that means that we can use Gauss\u0027s law like so."},{"Start":"04:32.380 ","End":"04:34.279","Text":"Now just as a side note,"},{"Start":"04:34.279 ","End":"04:37.925","Text":"a way of writing out the E field above and below,"},{"Start":"04:37.925 ","End":"04:40.024","Text":"because we have this anti-symmetry,"},{"Start":"04:40.024 ","End":"04:43.954","Text":"we can say that E field at z plus,"},{"Start":"04:43.954 ","End":"04:52.399","Text":"so at the positive side of the z-axis is equal to minus the E field at z minus,"},{"Start":"04:52.399 ","End":"04:56.070","Text":"so the negative side of the z-axis."},{"Start":"04:57.020 ","End":"05:01.909","Text":"As we can see, these are 2 different Es."},{"Start":"05:01.909 ","End":"05:05.329","Text":"Then in order for us to draw this shape,"},{"Start":"05:05.329 ","End":"05:09.920","Text":"this E up over here is not equal to this E over here."},{"Start":"05:09.920 ","End":"05:15.314","Text":"This could be equal to E_1 and this can be equal to E_2."},{"Start":"05:15.314 ","End":"05:24.125","Text":"Then we can\u0027t use this equation because this E is a different E. We can cross that out,"},{"Start":"05:24.125 ","End":"05:28.940","Text":"and then what we\u0027d have to write is that E_1 multiplied by S,"},{"Start":"05:28.940 ","End":"05:30.740","Text":"the surface area over here,"},{"Start":"05:30.740 ","End":"05:33.990","Text":"plus E_2 multiplied by S"},{"Start":"05:33.990 ","End":"05:42.424","Text":"is equal to our Q_in divided by Epsilon_naught."},{"Start":"05:42.424 ","End":"05:48.485","Text":"However, now, it\u0027s going to be very difficult to work out these electric fields."},{"Start":"05:48.485 ","End":"05:51.279","Text":"How are we going to do this?"},{"Start":"05:51.279 ","End":"05:54.079","Text":"Due to these issues that we\u0027ve seen,"},{"Start":"05:54.079 ","End":"05:58.370","Text":"we\u0027re not going to use this method of solving this question."},{"Start":"05:58.370 ","End":"06:00.230","Text":"What I\u0027m going to do now is I\u0027m going to rub"},{"Start":"06:00.230 ","End":"06:03.430","Text":"everything out and we\u0027re going to start again."},{"Start":"06:03.430 ","End":"06:08.705","Text":"The method that we\u0027re going to look at now works also with"},{"Start":"06:08.705 ","End":"06:16.320","Text":"uniform charge density or when we have symmetry and also when we have anti-symmetry."},{"Start":"06:16.370 ","End":"06:22.520","Text":"What do we do in these types of questions is that we split our infinite plane with"},{"Start":"06:22.520 ","End":"06:29.449","Text":"width into lots of tiny infinite planes,"},{"Start":"06:29.449 ","End":"06:32.460","Text":"which are very thin."},{"Start":"06:32.460 ","End":"06:40.125","Text":"We can say that the width of this infinite plane is of width dz,"},{"Start":"06:40.125 ","End":"06:45.630","Text":"and it\u0027s located at a position z."},{"Start":"06:46.880 ","End":"06:53.760","Text":"This is very similar to Coulomb\u0027s law and we\u0027re using the idea of superposition."},{"Start":"06:54.130 ","End":"06:59.690","Text":"Then what I do is I just split up my plane"},{"Start":"06:59.690 ","End":"07:04.834","Text":"with width into tiny infinite planes with a very small width,"},{"Start":"07:04.834 ","End":"07:07.504","Text":"I work out their electric field,"},{"Start":"07:07.504 ","End":"07:11.060","Text":"the electric field from each thin plane,"},{"Start":"07:11.060 ","End":"07:14.060","Text":"and I sum them all up."},{"Start":"07:14.060 ","End":"07:19.669","Text":"We know that the electric field of a thin plane"},{"Start":"07:19.669 ","End":"07:25.775","Text":"is equal to Sigma divided by 2Epsilon_naughts,"},{"Start":"07:25.775 ","End":"07:29.149","Text":"where this is a positive when we\u0027re"},{"Start":"07:29.149 ","End":"07:34.664","Text":"above the infinite plane and it\u0027s negative when we\u0027re below."},{"Start":"07:34.664 ","End":"07:38.880","Text":"Guess we\u0027ve seen this earlier on, in this chapter."},{"Start":"07:38.880 ","End":"07:43.639","Text":"Now what we have to do is we simply have to find out what Sigma is."},{"Start":"07:43.639 ","End":"07:49.319","Text":"Where Sigma is the uniform charge density or the charge density,"},{"Start":"07:49.319 ","End":"07:51.066","Text":"it\u0027s not necessarily uniform,"},{"Start":"07:51.066 ","End":"07:52.324","Text":"well, here it will be,"},{"Start":"07:52.324 ","End":"07:56.699","Text":"uniform charge density per unit area."},{"Start":"07:56.970 ","End":"07:59.154","Text":"Now in this question,"},{"Start":"07:59.154 ","End":"08:03.160","Text":"we don\u0027t necessarily have to work out what our Sigma is."},{"Start":"08:03.160 ","End":"08:06.490","Text":"Let\u0027s see how we do this."},{"Start":"08:06.490 ","End":"08:13.299","Text":"We know that our infinite plane over"},{"Start":"08:13.299 ","End":"08:20.170","Text":"here has a positive charge because it\u0027s located at some plus z,"},{"Start":"08:20.170 ","End":"08:24.354","Text":"which as we know from our charge density per unit volume,"},{"Start":"08:24.354 ","End":"08:30.340","Text":"is a positive when we\u0027re above our xy-plane."},{"Start":"08:30.340 ","End":"08:33.309","Text":"Now as we know from our question,"},{"Start":"08:33.309 ","End":"08:38.320","Text":"our charge density is independent of our x and y."},{"Start":"08:38.320 ","End":"08:41.975","Text":"That means that on this infinite plane,"},{"Start":"08:41.975 ","End":"08:46.429","Text":"which is then our charge density is uniform because it\u0027s"},{"Start":"08:46.429 ","End":"08:51.720","Text":"only dependent on z and we\u0027re located at 1 value of z."},{"Start":"08:51.720 ","End":"08:53.420","Text":"Along the entire plane,"},{"Start":"08:53.420 ","End":"08:57.709","Text":"it\u0027s going to be uniform in the x and y-directions."},{"Start":"08:57.709 ","End":"09:05.045","Text":"Now let\u0027s say that I want to work out the electric field at some points above my plane."},{"Start":"09:05.045 ","End":"09:07.525","Text":"Let\u0027s say at this point over here."},{"Start":"09:07.525 ","End":"09:11.675","Text":"We already know that my electric field is going to be in this direction,"},{"Start":"09:11.675 ","End":"09:17.700","Text":"and it\u0027s going to be equal to positive because we\u0027re above the infinite plane,"},{"Start":"09:17.700 ","End":"09:23.765","Text":"so we\u0027ll just have positive Sigma divided by 2Epsilon_naughts."},{"Start":"09:23.765 ","End":"09:30.109","Text":"Now what we can actually write is that our Sigma is as a function of z,"},{"Start":"09:30.109 ","End":"09:32.060","Text":"because as we know,"},{"Start":"09:32.060 ","End":"09:39.575","Text":"this is going to be the charge density per unit area when we\u0027re located at this z."},{"Start":"09:39.575 ","End":"09:47.735","Text":"But as we know, as we move up and down the z-axis our charge density is going to change."},{"Start":"09:47.735 ","End":"09:51.140","Text":"We know that our Sigma is dependent on z,"},{"Start":"09:51.140 ","End":"09:55.522","Text":"on where we are along the z-axis."},{"Start":"09:55.522 ","End":"10:02.875","Text":"Now what we\u0027re going to do is we\u0027re going to choose another infinite thin plane,"},{"Start":"10:02.875 ","End":"10:05.950","Text":"but on the opposite side of the z-axis."},{"Start":"10:05.950 ","End":"10:10.690","Text":"We\u0027re going to go down a distance z,"},{"Start":"10:10.690 ","End":"10:12.700","Text":"but in the opposite direction."},{"Start":"10:12.700 ","End":"10:14.875","Text":"This is going to be minus z."},{"Start":"10:14.875 ","End":"10:22.030","Text":"Then we\u0027re going to draw our thin infinite plane."},{"Start":"10:22.030 ","End":"10:24.730","Text":"We\u0027re using symmetry here for the picture,"},{"Start":"10:24.730 ","End":"10:29.140","Text":"so the width of this infinite plane is also dz."},{"Start":"10:29.140 ","End":"10:33.040","Text":"It\u0027s also located a magnitude of z away from the x,"},{"Start":"10:33.040 ","End":"10:36.715","Text":"y plane, but it\u0027s just in the opposite direction."},{"Start":"10:36.715 ","End":"10:42.024","Text":"That means, as we know from what we are given for our charge density,"},{"Start":"10:42.024 ","End":"10:47.770","Text":"that here we\u0027re going to have minus charges over here."},{"Start":"10:47.770 ","End":"10:51.229","Text":"We\u0027re going to have a negative charge density."},{"Start":"10:51.540 ","End":"10:55.270","Text":"Now, what we can say from what we know,"},{"Start":"10:55.270 ","End":"11:02.815","Text":"is that our charge density at a point over here minus z is"},{"Start":"11:02.815 ","End":"11:11.300","Text":"equal to the negative charge density of our plane at z."},{"Start":"11:12.060 ","End":"11:15.669","Text":"Just to give an example of what this means,"},{"Start":"11:15.669 ","End":"11:23.330","Text":"let\u0027s say we\u0027re located at d divided by 2 and then minus d divided by 2."},{"Start":"11:23.940 ","End":"11:31.587","Text":"Our charge density at 1/2d or d divided by 2,"},{"Start":"11:31.587 ","End":"11:36.700","Text":"so over here at the top is going to be equal to what we\u0027re given here,"},{"Start":"11:36.700 ","End":"11:40.045","Text":"a multiplied by our height,"},{"Start":"11:40.045 ","End":"11:42.625","Text":"which is d divided by 2."},{"Start":"11:42.625 ","End":"11:46.990","Text":"Now what about the other side of our plane with width?"},{"Start":"11:46.990 ","End":"11:50.230","Text":"This is at negative d divided by 2."},{"Start":"11:50.230 ","End":"12:00.609","Text":"Our Rho at negative 1/2d is going to be equal to a multiplied by our z value,"},{"Start":"12:00.609 ","End":"12:05.330","Text":"which is negative d divided by 2."},{"Start":"12:05.880 ","End":"12:14.600","Text":"Then we can see that we have a minus over here and that works out."},{"Start":"12:15.870 ","End":"12:19.795","Text":"Now that we can see that are charged densities"},{"Start":"12:19.795 ","End":"12:25.210","Text":"are of equal magnitude but pointing in the opposite direction."},{"Start":"12:25.210 ","End":"12:28.225","Text":"Back to our original example over here."},{"Start":"12:28.225 ","End":"12:35.049","Text":"We saw that our electric field due to this positively charged infinite plane is going in"},{"Start":"12:35.049 ","End":"12:39.159","Text":"the positive z-direction and is equal to Sigma as"},{"Start":"12:39.159 ","End":"12:44.140","Text":"a function of z divided by 2Epsilon_naught."},{"Start":"12:44.140 ","End":"12:48.880","Text":"Now we\u0027re going to add in to that point because we can see that"},{"Start":"12:48.880 ","End":"12:57.040","Text":"our electric field is independent of our distance away from the plane itself."},{"Start":"12:57.040 ","End":"13:03.970","Text":"Our charge density is dependent on which strip we\u0027re taking from our plane with width."},{"Start":"13:03.970 ","End":"13:05.935","Text":"However, the electric field,"},{"Start":"13:05.935 ","End":"13:09.340","Text":"if we\u0027re using this strip to work out the electric field,"},{"Start":"13:09.340 ","End":"13:11.724","Text":"or electric field at this point up here,"},{"Start":"13:11.724 ","End":"13:13.330","Text":"or at this point over here,"},{"Start":"13:13.330 ","End":"13:16.070","Text":"is going to be the same."},{"Start":"13:17.220 ","End":"13:22.480","Text":"Now we\u0027re going to add in the electric field from this plane with width."},{"Start":"13:22.480 ","End":"13:26.455","Text":"Again, it\u0027s going to be equal to Sigma divided by 2Epsilon_naught,"},{"Start":"13:26.455 ","End":"13:31.660","Text":"however, with a minus sign because our Sigma is negatively charged."},{"Start":"13:31.660 ","End":"13:35.409","Text":"What does that mean? It just means that our electric field is"},{"Start":"13:35.409 ","End":"13:40.120","Text":"pointing in the opposite direction but has equal magnitude."},{"Start":"13:40.120 ","End":"13:42.775","Text":"Again, it\u0027s going to be Sigma,"},{"Start":"13:42.775 ","End":"13:49.460","Text":"which is a function of z divided by 2Epsilon_naught."},{"Start":"13:49.620 ","End":"13:52.224","Text":"As we can see,"},{"Start":"13:52.224 ","End":"13:56.080","Text":"our electric fields up here are going to cross"},{"Start":"13:56.080 ","End":"14:01.165","Text":"out because they are of equal magnitude but in the opposite direction."},{"Start":"14:01.165 ","End":"14:03.250","Text":"Therefore, we\u0027ll get"},{"Start":"14:03.250 ","End":"14:12.159","Text":"that our total electric field outside of our infinite plane is going to be equal to 0."},{"Start":"14:12.159 ","End":"14:16.075","Text":"Again, this is because our electric fields,"},{"Start":"14:16.075 ","End":"14:20.769","Text":"when we\u0027re located outside of our plane with width,"},{"Start":"14:20.769 ","End":"14:24.550","Text":"is independent of the distance away from the plane."},{"Start":"14:24.550 ","End":"14:28.825","Text":"If we\u0027re calculating the electric field here or here,"},{"Start":"14:28.825 ","End":"14:32.229","Text":"we\u0027re going to have the exact same value for the electric field,"},{"Start":"14:32.229 ","End":"14:38.930","Text":"even though our Sigma is dependent on z within the plane."},{"Start":"14:39.030 ","End":"14:45.474","Text":"Then we can see that for every positively charged infinite plane,"},{"Start":"14:45.474 ","End":"14:49.634","Text":"we have a negatively charged infinite plane,"},{"Start":"14:49.634 ","End":"14:53.975","Text":"which has the exact same charge density Sigma,"},{"Start":"14:53.975 ","End":"14:57.850","Text":"however, just with a minus sign in front of it."},{"Start":"14:57.850 ","End":"15:01.915","Text":"That means that at every point we choose outside of the plane,"},{"Start":"15:01.915 ","End":"15:03.969","Text":"be it above or below,"},{"Start":"15:03.969 ","End":"15:10.960","Text":"our 2 electric fields due to the 2 planes are going to cancel out."},{"Start":"15:10.960 ","End":"15:15.445","Text":"What do we can say therefore,"},{"Start":"15:15.445 ","End":"15:23.755","Text":"is that our electric field is equal to 0 when we\u0027re above our infinite plane with width,"},{"Start":"15:23.755 ","End":"15:28.359","Text":"which means that z is bigger than d divided by"},{"Start":"15:28.359 ","End":"15:32.979","Text":"2 or when we\u0027re below our infinite plane with width because"},{"Start":"15:32.979 ","End":"15:42.685","Text":"we\u0027ll have the same thing happen or when z is smaller than negative d divided by 2."},{"Start":"15:42.685 ","End":"15:48.700","Text":"Now another way of writing out these 2 inequalities is to just write that"},{"Start":"15:48.700 ","End":"15:56.540","Text":"our absolute value of z is bigger than d divided by 2."},{"Start":"15:56.550 ","End":"16:00.710","Text":"This means this."},{"Start":"16:02.040 ","End":"16:05.770","Text":"Now we\u0027ve found what the electric field is"},{"Start":"16:05.770 ","End":"16:10.225","Text":"everywhere outside of this infinite plane with width."},{"Start":"16:10.225 ","End":"16:16.839","Text":"But now, what is the electric field when we\u0027re located within the infinite plane,"},{"Start":"16:16.839 ","End":"16:20.440","Text":"when we\u0027re located in the region where the absolute value of"},{"Start":"16:20.440 ","End":"16:25.150","Text":"z is smaller than d divided by 2?"},{"Start":"16:25.150 ","End":"16:29.030","Text":"That means we\u0027re somewhere inside over here."},{"Start":"16:29.790 ","End":"16:32.455","Text":"Now what we\u0027re trying to do,"},{"Start":"16:32.455 ","End":"16:36.699","Text":"is we\u0027re trying to figure out what the electric field is going to be when"},{"Start":"16:36.699 ","End":"16:41.455","Text":"we\u0027re located at some point within our plane."},{"Start":"16:41.455 ","End":"16:44.425","Text":"Let\u0027s say we\u0027re located at that point over there,"},{"Start":"16:44.425 ","End":"16:46.975","Text":"which has a value of z."},{"Start":"16:46.975 ","End":"16:52.569","Text":"This height z is independent of these small infinite strips."},{"Start":"16:52.569 ","End":"16:55.749","Text":"We\u0027re just located at some z within"},{"Start":"16:55.749 ","End":"16:58.060","Text":"our plane and we\u0027re trying to work out"},{"Start":"16:58.060 ","End":"17:01.825","Text":"the electric field at that point in green over here."},{"Start":"17:01.825 ","End":"17:04.840","Text":"The first thing I\u0027m going to tell you is that we\u0027ll see"},{"Start":"17:04.840 ","End":"17:10.089","Text":"that our electric field within the plane is not going to be equal to 0."},{"Start":"17:10.089 ","End":"17:13.030","Text":"There\u0027s a few ways that we can work this out."},{"Start":"17:13.030 ","End":"17:16.119","Text":"One of the ways is really by working out the electric field at"},{"Start":"17:16.119 ","End":"17:20.814","Text":"that point and summing up along all of these infinite planes."},{"Start":"17:20.814 ","End":"17:25.645","Text":"We can do that, it\u0027s a very long and tedious way to do it,"},{"Start":"17:25.645 ","End":"17:32.170","Text":"or we can use our knowledge that the electric field outside of the plane is"},{"Start":"17:32.170 ","End":"17:40.010","Text":"equal to 0 and that\u0027s going to significantly shorten the way we work this out."},{"Start":"17:40.020 ","End":"17:42.519","Text":"Now, just before we begin,"},{"Start":"17:42.519 ","End":"17:45.670","Text":"if you would have thought that we could use inside"},{"Start":"17:45.670 ","End":"17:51.435","Text":"a Gaussian surface in order to work out the electric field,"},{"Start":"17:51.435 ","End":"17:54.279","Text":"again, we can\u0027t use it in this question just like we"},{"Start":"17:54.279 ","End":"17:57.790","Text":"couldn\u0027t use it when we were working out the electric field outside."},{"Start":"17:57.790 ","End":"18:01.989","Text":"That is because the electric field coming out of the top of"},{"Start":"18:01.989 ","End":"18:04.810","Text":"our Gaussian surface will not be the same as"},{"Start":"18:04.810 ","End":"18:08.035","Text":"the electric field coming from the bottom of our Gaussian surface,"},{"Start":"18:08.035 ","End":"18:09.939","Text":"in this question, specifically,"},{"Start":"18:09.939 ","End":"18:15.265","Text":"because we have anti-symmetry due to our charge distribution."},{"Start":"18:15.265 ","End":"18:19.880","Text":"Again, we can\u0027t use Gauss\u0027s law here."},{"Start":"18:20.850 ","End":"18:23.380","Text":"What can I do?"},{"Start":"18:23.380 ","End":"18:28.270","Text":"What I\u0027m going to do is I\u0027m going to use some Gauss\u0027s surface,"},{"Start":"18:28.270 ","End":"18:30.264","Text":"but not the one that we\u0027re used to."},{"Start":"18:30.264 ","End":"18:32.980","Text":"I have the top end."},{"Start":"18:32.980 ","End":"18:37.600","Text":"I have an end of my Gaussian cube over here,"},{"Start":"18:37.600 ","End":"18:40.720","Text":"but then instead of drawing it downwards,"},{"Start":"18:40.720 ","End":"18:44.360","Text":"I\u0027m going to draw it upwards."},{"Start":"18:44.880 ","End":"18:54.410","Text":"It\u0027s going to be like so, excuse my drawing."},{"Start":"18:54.810 ","End":"19:00.250","Text":"Now the surface area of the top is going to be as per usual,"},{"Start":"19:00.250 ","End":"19:03.250","Text":"some arbitrary surface area S,"},{"Start":"19:03.250 ","End":"19:05.619","Text":"and we\u0027re going to have"},{"Start":"19:05.619 ","End":"19:12.235","Text":"our electric field and here we\u0027re going to draw it in the positive direction."},{"Start":"19:12.235 ","End":"19:13.810","Text":"Maybe it\u0027s a negative direction,"},{"Start":"19:13.810 ","End":"19:18.100","Text":"but then we\u0027ll just get a minus when we work out our answer."},{"Start":"19:18.100 ","End":"19:27.367","Text":"Here we have our electric field going up in this positive z-direction."},{"Start":"19:27.367 ","End":"19:30.999","Text":"Now what I want to work out is I want to work out"},{"Start":"19:30.999 ","End":"19:35.545","Text":"my electric flux through this Gaussian surface."},{"Start":"19:35.545 ","End":"19:44.120","Text":"Let\u0027s write the electric flux from this lower side of our cube."},{"Start":"19:44.220 ","End":"19:47.349","Text":"We\u0027re going to have that it\u0027s equal to"},{"Start":"19:47.349 ","End":"19:53.244","Text":"the electric field multiplied by the surface area, which as we know as S,"},{"Start":"19:53.244 ","End":"19:55.779","Text":"it\u0027s S up here and S at the bottom over here,"},{"Start":"19:55.779 ","End":"19:59.274","Text":"so E times S. Now we can see"},{"Start":"19:59.274 ","End":"20:04.550","Text":"that our electric field lines are going in to our Gaussian surface."},{"Start":"20:04.800 ","End":"20:10.659","Text":"That means that our flux is going in to our Gaussian surface,"},{"Start":"20:10.659 ","End":"20:12.910","Text":"which means that there\u0027s a minus sign."},{"Start":"20:12.910 ","End":"20:17.935","Text":"Remember, when our electric field lines are coming out of the Gaussian surface,"},{"Start":"20:17.935 ","End":"20:20.170","Text":"our flux is positive."},{"Start":"20:20.170 ","End":"20:24.249","Text":"When our electric field lines are going into our Gaussian surface,"},{"Start":"20:24.249 ","End":"20:27.384","Text":"then our flux is negative."},{"Start":"20:27.384 ","End":"20:29.740","Text":"Now of course, as usual,"},{"Start":"20:29.740 ","End":"20:35.079","Text":"the electric flux on the side panels of the cube is going to be equal to"},{"Start":"20:35.079 ","End":"20:41.094","Text":"0 because the panels are parallel to our electric field lines,"},{"Start":"20:41.094 ","End":"20:49.700","Text":"so there\u0027s going to be no additional flux through these sides of the cube."},{"Start":"20:50.250 ","End":"20:55.944","Text":"This is the electric flux for the bottom side of our cube."},{"Start":"20:55.944 ","End":"21:01.539","Text":"But now, what about the electric flux due to the top side of the cube?"},{"Start":"21:01.539 ","End":"21:08.515","Text":"Now, here is where we use the electric field outside of the plane is equal to 0."},{"Start":"21:08.515 ","End":"21:15.280","Text":"That means that our electric field over here is equal to 0."},{"Start":"21:15.280 ","End":"21:19.089","Text":"Then we can know that from the top side of the cube,"},{"Start":"21:19.089 ","End":"21:27.025","Text":"we\u0027re not receiving and we\u0027re not giving out any electric flux."},{"Start":"21:27.025 ","End":"21:30.527","Text":"Here the electric flux is going to be equal to 0,"},{"Start":"21:30.527 ","End":"21:38.125","Text":"because we\u0027ll have 0 times S which is equal to 0."},{"Start":"21:38.125 ","End":"21:42.474","Text":"This is our electric flux from the bottom side,"},{"Start":"21:42.474 ","End":"21:46.839","Text":"and then we\u0027re adding our electric flux from,"},{"Start":"21:46.839 ","End":"21:55.389","Text":"let\u0027s call this B bottom and this will be T top multiplied by"},{"Start":"21:55.389 ","End":"22:05.514","Text":"S. Then we can see that our E top is equal to 0 so this whole thing is equal to 0."},{"Start":"22:05.514 ","End":"22:09.354","Text":"Then as we know from Gauss\u0027s law, that this,"},{"Start":"22:09.354 ","End":"22:13.224","Text":"our electric flux is equal to our charge"},{"Start":"22:13.224 ","End":"22:18.980","Text":"inside our Gaussian surface divided by Epsilon_naught."},{"Start":"22:19.830 ","End":"22:25.900","Text":"Now what we want to do is we want to work out what our Q_in is equal to."},{"Start":"22:25.900 ","End":"22:28.075","Text":"What is our charge enclosed?"},{"Start":"22:28.075 ","End":"22:32.530","Text":"Now what I\u0027m going to do is I\u0027m going to draw this Gaussian surface but in"},{"Start":"22:32.530 ","End":"22:37.930","Text":"2-dimensions so that our picture is easier to understand."},{"Start":"22:37.930 ","End":"22:43.329","Text":"We\u0027re located at this z and our Gaussian surface is something like this,"},{"Start":"22:43.329 ","End":"22:50.170","Text":"then all of the charge enclosed is going to be all of this space over here."},{"Start":"22:50.170 ","End":"22:54.234","Text":"From z, from this green dot,"},{"Start":"22:54.234 ","End":"22:59.680","Text":"until the upper edge of our infinite plane."},{"Start":"22:59.680 ","End":"23:01.960","Text":"Now of course we also have depth,"},{"Start":"23:01.960 ","End":"23:04.209","Text":"because we\u0027re dealing with volume over here."},{"Start":"23:04.209 ","End":"23:06.294","Text":"But in 2-dimensions,"},{"Start":"23:06.294 ","End":"23:09.800","Text":"this is the charge enclosed."},{"Start":"23:10.080 ","End":"23:15.580","Text":"What\u0027s important to note is that when we\u0027re dealing with a question with anti-symmetry,"},{"Start":"23:15.580 ","End":"23:18.700","Text":"we\u0027re drawing our Gaussian surface like this."},{"Start":"23:18.700 ","End":"23:25.150","Text":"That means that we\u0027re summing up on all of the charge enclosed in this direction."},{"Start":"23:25.150 ","End":"23:29.725","Text":"From the point where we want to work out our electric field, we\u0027re summing upwards."},{"Start":"23:29.725 ","End":"23:31.854","Text":"Whereas in previous questions,"},{"Start":"23:31.854 ","End":"23:35.319","Text":"we\u0027ve been drawing some kind of Gaussian surface"},{"Start":"23:35.319 ","End":"23:40.550","Text":"inside and we\u0027ve been summing along that, like so."},{"Start":"23:41.850 ","End":"23:47.210","Text":"Now we\u0027re doing a different type of integral."},{"Start":"23:47.820 ","End":"23:53.480","Text":"Let us see now what our Q_in is therefore equal to."},{"Start":"23:53.760 ","End":"23:58.195","Text":"Now let\u0027s work out what our Q_in is equal to."},{"Start":"23:58.195 ","End":"24:03.205","Text":"We have that Q_in is equal to our usual equation which is Rho,"},{"Start":"24:03.205 ","End":"24:05.900","Text":"which here is as a function of zdv,"},{"Start":"24:06.020 ","End":"24:08.594","Text":"where v is volume."},{"Start":"24:08.594 ","End":"24:11.594","Text":"Now because we\u0027re dealing with Cartesian coordinates,"},{"Start":"24:11.594 ","End":"24:14.159","Text":"this is going to be a triple"},{"Start":"24:14.159 ","End":"24:18.880","Text":"integral and our Rho is a function of z which is equal to Az."},{"Start":"24:19.590 ","End":"24:29.870","Text":"Then our dv will be equal to in Cartesian coordinates dxdyydz."},{"Start":"24:29.960 ","End":"24:35.765","Text":"Now let\u0027s talk about what our integration bounds are going to be."},{"Start":"24:35.765 ","End":"24:39.910","Text":"Let\u0027s first deal with our bounds on the z-axis."},{"Start":"24:39.910 ","End":"24:45.310","Text":"As we can see, we\u0027re integrating from our point z where we\u0027re trying to work out"},{"Start":"24:45.310 ","End":"24:53.140","Text":"our electric field until we reach the upper edge or the edge of our infinite plane."},{"Start":"24:53.140 ","End":"24:57.520","Text":"We know that the edge of our infinite plane is d divided by 2."},{"Start":"24:57.520 ","End":"25:05.470","Text":"That means that we\u0027re integrating from our height z until d divided by 2."},{"Start":"25:05.470 ","End":"25:07.749","Text":"Let\u0027s put that in,"},{"Start":"25:07.749 ","End":"25:11.060","Text":"z until d divided by 2."},{"Start":"25:11.400 ","End":"25:15.489","Text":"Now because we have this z in our bounds,"},{"Start":"25:15.489 ","End":"25:22.060","Text":"we have to differentiate it from our integration variables."},{"Start":"25:22.060 ","End":"25:29.740","Text":"Here our z in the integral is our location that we\u0027re summing up"},{"Start":"25:29.740 ","End":"25:37.270","Text":"within this space over here of the red colored in rectangle over here."},{"Start":"25:37.270 ","End":"25:41.665","Text":"We\u0027re going to differentiate it by just putting tags over here."},{"Start":"25:41.665 ","End":"25:47.155","Text":"Of course, this is z is the location of our green dot."},{"Start":"25:47.155 ","End":"25:49.255","Text":"Where we\u0027re trying to work out"},{"Start":"25:49.255 ","End":"25:53.414","Text":"the electric field or what the electric field is equal to,"},{"Start":"25:53.414 ","End":"25:56.395","Text":"so that point is this z over here."},{"Start":"25:56.395 ","End":"26:04.550","Text":"These z\u0027s represent what we\u0027re summing up in this space over here."},{"Start":"26:05.160 ","End":"26:08.530","Text":"Now what about our x and y?"},{"Start":"26:08.530 ","End":"26:13.764","Text":"We\u0027re simply summing up for a total surface area of S,"},{"Start":"26:13.764 ","End":"26:16.720","Text":"which was some arbitrary value."},{"Start":"26:16.720 ","End":"26:21.970","Text":"Because our charge density is independent on x and y,"},{"Start":"26:21.970 ","End":"26:26.380","Text":"that means that our integral can just be summing up on S,"},{"Start":"26:26.380 ","End":"26:30.880","Text":"this arbitrary surface area of our Gaussian surface."},{"Start":"26:30.880 ","End":"26:33.670","Text":"Then when we integrate along x and y,"},{"Start":"26:33.670 ","End":"26:37.910","Text":"we\u0027ll just get this S-value over here."},{"Start":"26:37.920 ","End":"26:41.140","Text":"Now let\u0027s integrate."},{"Start":"26:41.140 ","End":"26:46.224","Text":"We\u0027ll get that by integrating on dxdy, we\u0027ll get S,"},{"Start":"26:46.224 ","End":"26:52.487","Text":"and then we\u0027ll have multiplied by Az tag dz,"},{"Start":"26:52.487 ","End":"27:01.015","Text":"and we\u0027re integrating between z and up until d divided by 2."},{"Start":"27:01.015 ","End":"27:05.469","Text":"Then this is simply going to be equal to"},{"Start":"27:05.469 ","End":"27:10.645","Text":"SA z tag^2 divided by"},{"Start":"27:10.645 ","End":"27:18.740","Text":"2 between the bounds of z and d divided by 2."},{"Start":"27:19.950 ","End":"27:24.408","Text":"Once we substitute in our bounds,"},{"Start":"27:24.408 ","End":"27:32.545","Text":"our Q_in is going to be equal to 1/2 S times A multiplied by d"},{"Start":"27:32.545 ","End":"27:41.824","Text":"divided by 2^2 minus z^2."},{"Start":"27:41.824 ","End":"27:51.030","Text":"Now we can substitute in our Q_in into our equation."},{"Start":"27:51.030 ","End":"27:58.350","Text":"Now we have that negative E multiplied by S,"},{"Start":"27:58.350 ","End":"28:03.195","Text":"now we don\u0027t have to use that it\u0027s at the bottom because we\u0027ve understood,"},{"Start":"28:03.195 ","End":"28:11.986","Text":"is equal to our Q_in which is 1/2 multiplied by SA (d divided by"},{"Start":"28:11.986 ","End":"28:18.749","Text":"2)^2 minus"},{"Start":"28:18.749 ","End":"28:22.425","Text":"z^2 divided by Epsilon_naught."},{"Start":"28:22.425 ","End":"28:29.310","Text":"We can plop in our Epsilon_naught over here and now we can see that we can divide our S"},{"Start":"28:29.310 ","End":"28:32.880","Text":"from both sides to cancel them out which"},{"Start":"28:32.880 ","End":"28:36.540","Text":"is great because our S was just some arbitrary surface area,"},{"Start":"28:36.540 ","End":"28:39.060","Text":"so that\u0027s great that it cancels out."},{"Start":"28:39.060 ","End":"28:41.625","Text":"Now we can rewrite our electric field,"},{"Start":"28:41.625 ","End":"28:45.000","Text":"so I\u0027ll move the minus to the other side and it\u0027s going to be"},{"Start":"28:45.000 ","End":"28:48.809","Text":"equal to negative A divided by"},{"Start":"28:48.809 ","End":"28:52.916","Text":"2Epsilon_naught multiplied by (d divided"},{"Start":"28:52.916 ","End":"29:00.990","Text":"by 2)^2 minus z^2."},{"Start":"29:00.990 ","End":"29:04.850","Text":"Now of course, if we want to write our electric field in vector form,"},{"Start":"29:04.850 ","End":"29:07.340","Text":"so we know that it\u0027s going to be, of course,"},{"Start":"29:07.340 ","End":"29:10.504","Text":"in the z-direction and we can see that we"},{"Start":"29:10.504 ","End":"29:14.425","Text":"have it going in the minus or in the negative z-direction."},{"Start":"29:14.425 ","End":"29:17.534","Text":"These green arrows denoting the electric field"},{"Start":"29:17.534 ","End":"29:21.000","Text":"are meant to actually be pointing downwards."},{"Start":"29:21.000 ","End":"29:25.900","Text":"Soon we\u0027re going to speak about why that makes complete sense."},{"Start":"29:26.150 ","End":"29:29.850","Text":"I\u0027m just reminding that this answer for"},{"Start":"29:29.850 ","End":"29:35.159","Text":"the electric field is only correct when well located within the plane itself,"},{"Start":"29:35.159 ","End":"29:42.570","Text":"so when the absolute value of z is smaller than d divided by 2 and everywhere else,"},{"Start":"29:42.570 ","End":"29:44.429","Text":"so that means above or below the plane"},{"Start":"29:44.429 ","End":"29:48.405","Text":"our electric field is equal to 0 as we worked out before."},{"Start":"29:48.405 ","End":"29:54.090","Text":"Now, this value for the electric field is also correct for when our z is over here,"},{"Start":"29:54.090 ","End":"29:56.865","Text":"but it\u0027s also correct for every single value of z,"},{"Start":"29:56.865 ","End":"30:01.409","Text":"so that includes our negatives zs and that\u0027s just because if we\u0027re"},{"Start":"30:01.409 ","End":"30:06.060","Text":"trying to work out a z located over here at the negative z,"},{"Start":"30:06.060 ","End":"30:09.959","Text":"so we\u0027re going to draw the exact same Gaussian surface,"},{"Start":"30:09.959 ","End":"30:14.411","Text":"which is going to be some cube going like so and"},{"Start":"30:14.411 ","End":"30:18.810","Text":"then we\u0027re just going to sum along all of this surface area."},{"Start":"30:18.810 ","End":"30:26.775","Text":"So again, we\u0027re going from some value of z and summing up until d divided by 2."},{"Start":"30:26.775 ","End":"30:29.745","Text":"We\u0027re going to get the exact same integral,"},{"Start":"30:29.745 ","End":"30:33.359","Text":"which means that this is the electric field when"},{"Start":"30:33.359 ","End":"30:38.800","Text":"we\u0027re measuring the electric field from within the plane."},{"Start":"30:40.190 ","End":"30:45.090","Text":"Now we have our answer and let\u0027s see why it"},{"Start":"30:45.090 ","End":"30:49.470","Text":"makes sense that our electric field is pointing downwards,"},{"Start":"30:49.470 ","End":"30:52.440","Text":"so if we\u0027re measuring the electric field at this point over here,"},{"Start":"30:52.440 ","End":"30:58.695","Text":"that our electric field lines are pointing in the negative z-direction."},{"Start":"30:58.695 ","End":"31:04.830","Text":"Imagine that we\u0027re looking at this point over here and then we\u0027re going to have lots of"},{"Start":"31:04.830 ","End":"31:10.605","Text":"little infinite planes over here that have all a positive charge,"},{"Start":"31:10.605 ","End":"31:15.270","Text":"so they\u0027re going to be producing an electric field pointing upwards."},{"Start":"31:15.270 ","End":"31:17.565","Text":"Now in the equal and opposite direction,"},{"Start":"31:17.565 ","End":"31:19.890","Text":"we\u0027re going to have lots and lots of infinite planes over"},{"Start":"31:19.890 ","End":"31:23.969","Text":"here that all have negative charges and they\u0027re"},{"Start":"31:23.969 ","End":"31:27.300","Text":"going to be producing an electric field in"},{"Start":"31:27.300 ","End":"31:31.810","Text":"an equal and opposite direction, so downwards."},{"Start":"31:32.270 ","End":"31:39.135","Text":"That means that all of this section of our plane with width,"},{"Start":"31:39.135 ","End":"31:42.834","Text":"so all of this over here is going to cancel out,"},{"Start":"31:42.834 ","End":"31:47.460","Text":"because here we\u0027re going to have positive charges forming an electric field upwards and"},{"Start":"31:47.460 ","End":"31:52.260","Text":"here negative charges forming electric field downwards and they will cancel out."},{"Start":"31:52.260 ","End":"31:55.260","Text":"That\u0027s exactly what we saw over here when working out"},{"Start":"31:55.260 ","End":"32:00.585","Text":"the electric field outside of the plane itself."},{"Start":"32:00.585 ","End":"32:04.980","Text":"We can look at this point over here as outside of"},{"Start":"32:04.980 ","End":"32:10.030","Text":"this infinite plane over here enclosed in gray."},{"Start":"32:10.250 ","End":"32:15.405","Text":"Now what happens right above this point over here?"},{"Start":"32:15.405 ","End":"32:21.030","Text":"Let\u0027s look over here at this infinite plane in blue with the positive charges."},{"Start":"32:21.030 ","End":"32:26.430","Text":"This plane has lots of positive charges which are"},{"Start":"32:26.430 ","End":"32:32.395","Text":"forming an electric field or producing electric field in the downwards direction."},{"Start":"32:32.395 ","End":"32:38.420","Text":"Now we have the infinite planes on the equal and opposite side,"},{"Start":"32:38.420 ","End":"32:44.659","Text":"so located here at the negative Zs and this is negatively charged,"},{"Start":"32:44.659 ","End":"32:48.679","Text":"which means that this electric field produced from"},{"Start":"32:48.679 ","End":"32:54.010","Text":"this infinite plane is also pointing in the downwards direction."},{"Start":"32:54.010 ","End":"32:57.915","Text":"Here we have 2 arrows pointing in the downwards direction,"},{"Start":"32:57.915 ","End":"33:01.859","Text":"meaning that the total electric field is always going to"},{"Start":"33:01.859 ","End":"33:06.910","Text":"be in the downwards direction from superposition."},{"Start":"33:08.240 ","End":"33:14.309","Text":"As I take my point for this positively charged infinite plane to be"},{"Start":"33:14.309 ","End":"33:20.640","Text":"closer and closer to the center of my infinitely charged plane with width,"},{"Start":"33:20.640 ","End":"33:25.510","Text":"so closer and closer to the xy-plane over here,"},{"Start":"33:25.510 ","End":"33:29.249","Text":"we can see that my values for z are going to become"},{"Start":"33:29.249 ","End":"33:33.360","Text":"smaller or the absolute value of z is going to become smaller and if"},{"Start":"33:33.360 ","End":"33:39.825","Text":"the absolute value of z is going to become smaller than my value here for z^2 is also"},{"Start":"33:39.825 ","End":"33:46.980","Text":"going to become smaller and that means that as we move closer to z=0,"},{"Start":"33:46.980 ","End":"33:52.725","Text":"we\u0027re going to get that our electric field increases in size."},{"Start":"33:52.725 ","End":"33:55.799","Text":"That\u0027s an explanation for why it makes"},{"Start":"33:55.799 ","End":"33:59.325","Text":"sense that our electric field is always pointing downwards."},{"Start":"33:59.325 ","End":"34:01.754","Text":"Now also as we can see,"},{"Start":"34:01.754 ","End":"34:07.395","Text":"when our z value is equal to d divided by 2,"},{"Start":"34:07.395 ","End":"34:12.449","Text":"so we can see that our electric field becomes 0, which makes sense."},{"Start":"34:12.449 ","End":"34:14.729","Text":"We can see that our electric field above"},{"Start":"34:14.729 ","End":"34:18.885","Text":"our infinite plane is also equal to 0, up until infinity."},{"Start":"34:18.885 ","End":"34:24.510","Text":"This makes sense and this shows us that our electric field is continuous,"},{"Start":"34:24.510 ","End":"34:28.665","Text":"so it\u0027s reducing as we hit"},{"Start":"34:28.665 ","End":"34:30.539","Text":"this upper edge of"},{"Start":"34:30.539 ","End":"34:32.939","Text":"our infinite plane so we\u0027ll get"},{"Start":"34:32.939 ","End":"34:35.730","Text":"that our electric field at that point is exactly equal to 0,"},{"Start":"34:35.730 ","End":"34:39.179","Text":"and then it carries on being equal to 0 up until infinity."},{"Start":"34:39.179 ","End":"34:46.605","Text":"Now this happens because we don\u0027t have any charge density per unit area,"},{"Start":"34:46.605 ","End":"34:52.034","Text":"and when we don\u0027t have any charge density per unit area or Sigma,"},{"Start":"34:52.034 ","End":"34:56.715","Text":"so that means that there\u0027s going to be no jump in the electric field,"},{"Start":"34:56.715 ","End":"35:00.340","Text":"meaning that our electric field is continuous."},{"Start":"35:00.350 ","End":"35:03.885","Text":"Every time we have Rho,"},{"Start":"35:03.885 ","End":"35:07.034","Text":"so charge density per unit volume,"},{"Start":"35:07.034 ","End":"35:10.499","Text":"then we know that our electric field is always going to be"},{"Start":"35:10.499 ","End":"35:15.360","Text":"continuous and whenever we have charged density per unit area,"},{"Start":"35:15.360 ","End":"35:18.540","Text":"so Sigma, we know that our electric field is not"},{"Start":"35:18.540 ","End":"35:23.080","Text":"going to be continuous and that there will be a jump in the electric field."},{"Start":"35:23.840 ","End":"35:26.570","Text":"We can see that everything works out,"},{"Start":"35:26.570 ","End":"35:30.179","Text":"that we were indeed given charge density for a unit volume and"},{"Start":"35:30.179 ","End":"35:34.425","Text":"we really do get a continuous electric field."},{"Start":"35:34.425 ","End":"35:40.200","Text":"This is the answer for the value of the electric field outside of the infinite plane,"},{"Start":"35:40.200 ","End":"35:47.190","Text":"and this is the answer for the value of the electric field inside the infinite plane."},{"Start":"35:47.190 ","End":"35:50.259","Text":"That\u0027s the end of this lesson."}],"ID":22367}],"Thumbnail":null,"ID":246895}]

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