[{"Name":"Introduction to Calculating Charge Distribution","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Calculating Charge Distribution","Duration":"33m 22s","ChapterTopicVideoID":21299,"CourseChapterTopicPlaylistID":99470,"HasSubtitles":false,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/21299.jpeg","UploadDate":"2020-04-06T14:07:02.6130000","DurationForVideoObject":"PT33M22S","Description":null,"MetaTitle":"Calculating Charge Distribution: Video + Workbook | Proprep","MetaDescription":"Calculating Charge Distribution - Introduction to Calculating Charge Distribution. Watch the video made by an expert in the field. Download the workbook and maximize your learning.","Canonical":"https://www.proprep.uk/general-modules/all/physics-2-electricity-and-magnetism/calculating-charge-distribution/introduction-to-calculating-charge-distribution/vid21379","VideoComments":[],"Subtitles":[],"ID":21379},{"Watched":false,"Name":"A Jump In The Electric Field","Duration":"4m 56s","ChapterTopicVideoID":21300,"CourseChapterTopicPlaylistID":99470,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.040","Text":"Hello. In this lesson,"},{"Start":"00:02.040 ","End":"00:07.515","Text":"we\u0027ll see how we get to the equation of Sigma is equal Epsilon naught Delta E,"},{"Start":"00:07.515 ","End":"00:10.470","Text":"which is the jump in the electric field."},{"Start":"00:10.470 ","End":"00:16.695","Text":"Let\u0027s imagine that we have this plane with charge density per unit area of Sigma."},{"Start":"00:16.695 ","End":"00:23.325","Text":"Now, if we look at 2 points which are extremely close to the plane."},{"Start":"00:23.325 ","End":"00:27.269","Text":"So this distance is small."},{"Start":"00:27.269 ","End":"00:32.400","Text":"Then we can consider this plane to be infinite."},{"Start":"00:32.400 ","End":"00:35.175","Text":"Now that we can consider this plane as infinite,"},{"Start":"00:35.175 ","End":"00:40.490","Text":"we know that our electric field above an infinite plane is in"},{"Start":"00:40.490 ","End":"00:43.820","Text":"the upwards direction and then the electric field below"},{"Start":"00:43.820 ","End":"00:47.795","Text":"an infinite plane is in the downwards direction."},{"Start":"00:47.795 ","End":"00:54.590","Text":"If we\u0027re looking at this plane from the side, so this is the plane."},{"Start":"00:54.590 ","End":"00:56.315","Text":"Imagine this as a straight line."},{"Start":"00:56.315 ","End":"01:02.915","Text":"Then we\u0027ll see that we have our electric field lines going like so."},{"Start":"01:02.915 ","End":"01:09.170","Text":"Now let\u0027s imagine that we have point charges below the electric field."},{"Start":"01:09.170 ","End":"01:11.000","Text":"We have a q here,"},{"Start":"01:11.000 ","End":"01:12.950","Text":"a q here, and a q here."},{"Start":"01:12.950 ","End":"01:18.935","Text":"These are going to produce an electric field in space as well."},{"Start":"01:18.935 ","End":"01:20.930","Text":"If they\u0027re positive charges,"},{"Start":"01:20.930 ","End":"01:23.600","Text":"they are going to be producing an electric field"},{"Start":"01:23.600 ","End":"01:26.775","Text":"in the upwards direction below the plane."},{"Start":"01:26.775 ","End":"01:28.955","Text":"They\u0027re also going to be producing"},{"Start":"01:28.955 ","End":"01:32.270","Text":"an electric field in the upwards direction above the plane."},{"Start":"01:32.270 ","End":"01:39.950","Text":"So then the net electric field on both sides of the plane will be E_1 and E_2."},{"Start":"01:39.950 ","End":"01:41.750","Text":"Now obviously these charges could be"},{"Start":"01:41.750 ","End":"01:44.210","Text":"negative or they could be on the other side of the plane."},{"Start":"01:44.210 ","End":"01:45.943","Text":"It doesn\u0027t really matter."},{"Start":"01:45.943 ","End":"01:49.925","Text":"We just will switch over our signs or change the direction of the arrow."},{"Start":"01:49.925 ","End":"01:53.510","Text":"What matters is that now we\u0027re going to have 2 electric fields,"},{"Start":"01:53.510 ","End":"01:55.315","Text":"E_1 and E_2,"},{"Start":"01:55.315 ","End":"01:58.845","Text":"and now they\u0027re going to be uneven."},{"Start":"01:58.845 ","End":"02:01.010","Text":"Before they were equal,"},{"Start":"02:01.010 ","End":"02:03.260","Text":"just pointing in different directions."},{"Start":"02:03.260 ","End":"02:06.710","Text":"Now, they might still be pointing in different directions,"},{"Start":"02:06.710 ","End":"02:10.005","Text":"but their magnitude is not the same."},{"Start":"02:10.005 ","End":"02:14.715","Text":"Now what we\u0027re going to do is we\u0027re going to use a Gaussian surface."},{"Start":"02:14.715 ","End":"02:22.565","Text":"We\u0027re going to use a cube where the upper base has E_1 coming out of it,"},{"Start":"02:22.565 ","End":"02:27.960","Text":"and the lower base has E_2 coming through it."},{"Start":"02:27.960 ","End":"02:35.495","Text":"We can say that the surface area is going to be S. That\u0027s the surface area over here,"},{"Start":"02:35.495 ","End":"02:39.205","Text":"the surface area where it\u0027s on our surface."},{"Start":"02:39.205 ","End":"02:45.560","Text":"The cube is projected onto our surface and the surface area of the bottom side,"},{"Start":"02:45.560 ","End":"02:52.910","Text":"so that\u0027s S. Then we can say that the electric flux through our Gaussian surface is"},{"Start":"02:52.910 ","End":"03:00.480","Text":"going to be equal to E_1.S minus E_2.S."},{"Start":"03:00.480 ","End":"03:03.635","Text":"That\u0027s because our E_1 is a positive,"},{"Start":"03:03.635 ","End":"03:07.880","Text":"because our electric field is leaving our surface and"},{"Start":"03:07.880 ","End":"03:13.350","Text":"our E_2 is negative because our electric field is entering our Gaussian surface."},{"Start":"03:13.350 ","End":"03:15.740","Text":"That is going to be equal to,"},{"Start":"03:15.740 ","End":"03:19.610","Text":"if you remember what the electric field is when dealing with"},{"Start":"03:19.610 ","End":"03:24.245","Text":"a Gaussian surface and a cube along some kind of plane,"},{"Start":"03:24.245 ","End":"03:27.318","Text":"an infinite plane is going to be equal to 1 divided"},{"Start":"03:27.318 ","End":"03:30.770","Text":"by Epsilon naught multiplied by our Sigma,"},{"Start":"03:30.770 ","End":"03:35.105","Text":"or the charge density per unit area multiplied by"},{"Start":"03:35.105 ","End":"03:41.675","Text":"our surface area S. Now we can see that the S\u0027s on both sides cancel out."},{"Start":"03:41.675 ","End":"03:46.085","Text":"We, therefore, get that 1 divided by Epsilon naught multiplied by"},{"Start":"03:46.085 ","End":"03:51.590","Text":"Sigma is equal to E_1 minus E_2."},{"Start":"03:51.590 ","End":"03:53.630","Text":"Then if we isolate out our Sigma,"},{"Start":"03:53.630 ","End":"03:59.955","Text":"we can see that our Sigma is equal to Epsilon naught multiplied by E_1 minus E_2,"},{"Start":"03:59.955 ","End":"04:02.535","Text":"which is of course equal to Delta E,"},{"Start":"04:02.535 ","End":"04:04.340","Text":"and we got this equation over here."},{"Start":"04:04.340 ","End":"04:08.630","Text":"Now, what happens if we didn\u0027t have these charges over here,"},{"Start":"04:08.630 ","End":"04:11.675","Text":"and we were just left with an infinite plane?"},{"Start":"04:11.675 ","End":"04:17.645","Text":"This infinite plane is a specific case of this over here."},{"Start":"04:17.645 ","End":"04:22.940","Text":"In this case, our E_2 is simply going to be equal"},{"Start":"04:22.940 ","End":"04:28.775","Text":"to negative E_1 because the magnitude of both the fields are the same,"},{"Start":"04:28.775 ","End":"04:31.670","Text":"they\u0027re just pointing in different directions."},{"Start":"04:31.670 ","End":"04:37.625","Text":"In conclusion, our jump in the electric field is given by this equation over here."},{"Start":"04:37.625 ","End":"04:42.980","Text":"That means that our jump in the electric field comes only from our Sigma,"},{"Start":"04:42.980 ","End":"04:44.990","Text":"from our charge density."},{"Start":"04:44.990 ","End":"04:49.825","Text":"It doesn\u0027t matter if we have added charges in our system,"},{"Start":"04:49.825 ","End":"04:53.750","Text":"all that matters is the charge on our plane."},{"Start":"04:53.750 ","End":"04:56.940","Text":"That\u0027s the end of this lesson."}],"ID":21380},{"Watched":false,"Name":"Exercise 1","Duration":"22m 34s","ChapterTopicVideoID":21301,"CourseChapterTopicPlaylistID":99470,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.145","Text":"Hello. In this question,"},{"Start":"00:02.145 ","End":"00:05.790","Text":"we\u0027re being told that there\u0027s an electric field and space given"},{"Start":"00:05.790 ","End":"00:10.995","Text":"by c divided by r multiplied by,"},{"Start":"00:10.995 ","End":"00:14.340","Text":"say 1 in the r direction plus cosine Theta in"},{"Start":"00:14.340 ","End":"00:20.370","Text":"the Theta direction plus sine Theta cosine Phi in the Phi direction."},{"Start":"00:20.370 ","End":"00:21.900","Text":"Just to remind you,"},{"Start":"00:21.900 ","End":"00:31.360","Text":"if this is my z axis and if this is my r vector,"},{"Start":"00:31.550 ","End":"00:37.320","Text":"so my angle Phi is going to be the angle between my r vector and"},{"Start":"00:37.320 ","End":"00:43.475","Text":"the z-axis and my angle Theta is going to be,"},{"Start":"00:43.475 ","End":"00:49.850","Text":"if I draw the projection of my r vector on my x and y axis,"},{"Start":"00:49.850 ","End":"00:52.835","Text":"so let\u0027s say that this is y and this is x,"},{"Start":"00:52.835 ","End":"00:57.230","Text":"so my angle Theta is going to be the angle between"},{"Start":"00:57.230 ","End":"01:03.055","Text":"the x-axis and my projection of my r vector on the x, y plane."},{"Start":"01:03.055 ","End":"01:04.870","Text":"In question number 1,"},{"Start":"01:04.870 ","End":"01:08.540","Text":"we\u0027re being asked what is the charge density in space?"},{"Start":"01:08.540 ","End":"01:11.750","Text":"Whenever you\u0027re being asked about what is the charge density,"},{"Start":"01:11.750 ","End":"01:13.490","Text":"the first thing you\u0027re meant to say,"},{"Start":"01:13.490 ","End":"01:15.980","Text":"so question 1,"},{"Start":"01:15.980 ","End":"01:25.665","Text":"is that row I charged density per unit volume is equal to Epsilon Naught div E,"},{"Start":"01:25.665 ","End":"01:33.620","Text":"and that I charged density per unit area is equal to Epsilon Naught multiplied by"},{"Start":"01:33.620 ","End":"01:43.901","Text":"Delta E. This function over here for our electric field is a continuous function."},{"Start":"01:43.901 ","End":"01:47.900","Text":"That means that our jump in the electric field,"},{"Start":"01:47.900 ","End":"01:51.715","Text":"so our Delta E is going to be equal to 0."},{"Start":"01:51.715 ","End":"01:56.705","Text":"Because this function over here for our E field is continuous,"},{"Start":"01:56.705 ","End":"02:02.090","Text":"so that means that we\u0027re going to only have charged density per unit volume."},{"Start":"02:02.090 ","End":"02:05.640","Text":"This is the equation that interests us right now."},{"Start":"02:05.900 ","End":"02:11.645","Text":"At this point, once we\u0027ve written out these 2 equations and realized that"},{"Start":"02:11.645 ","End":"02:17.656","Text":"our electric field equation or function over here is continuous."},{"Start":"02:17.656 ","End":"02:20.645","Text":"Therefore, this will equal 0."},{"Start":"02:20.645 ","End":"02:23.225","Text":"Our physics in the question has ended,"},{"Start":"02:23.225 ","End":"02:28.745","Text":"and now we just have to use mathematics in order to actually solve these equations."},{"Start":"02:28.745 ","End":"02:32.060","Text":"Now, the mathematics that we\u0027re going to do here is going to be a"},{"Start":"02:32.060 ","End":"02:35.280","Text":"bit ugly and it\u0027s a bit confusing,"},{"Start":"02:35.280 ","End":"02:38.385","Text":"so I\u0027m confusing algebra but that\u0027s all that\u0027s left to do."},{"Start":"02:38.385 ","End":"02:42.180","Text":"The physics, the section of understanding has already be done."},{"Start":"02:43.820 ","End":"02:46.990","Text":"Let\u0027s see what I div E equation is."},{"Start":"02:46.990 ","End":"02:49.580","Text":"Now, because we\u0027re working in space and we"},{"Start":"02:49.580 ","End":"02:52.295","Text":"can see that we\u0027re working in spherical coordinates,"},{"Start":"02:52.295 ","End":"02:57.690","Text":"so I div E in spherical coordinates is given by this equation."},{"Start":"02:58.460 ","End":"03:03.117","Text":"As we can see, this equation is pretty long and complicated."},{"Start":"03:03.117 ","End":"03:06.020","Text":"You\u0027re not expected to remember this off by heart,"},{"Start":"03:06.020 ","End":"03:10.445","Text":"but I would suggest writing this on your equation sheets for"},{"Start":"03:10.445 ","End":"03:16.220","Text":"div E or div of any vector function when dealing with spherical coordinates."},{"Start":"03:16.220 ","End":"03:25.905","Text":"This is equal to 1 divided by r^2 multiplied by dr^2 the E component of dr plus 1 over"},{"Start":"03:25.905 ","End":"03:32.820","Text":"r sine Phi and then d by d Theta of the Theta component of e plus 1 divided"},{"Start":"03:32.820 ","End":"03:40.810","Text":"by r sine Phi d by d Phi the E component of Phi multiplied by sine of Phi."},{"Start":"03:41.810 ","End":"03:46.490","Text":"Remembering over here how we defined our Phi and our Theta,"},{"Start":"03:46.490 ","End":"03:51.260","Text":"so Phi is our angle with the z-axis and r Theta is between"},{"Start":"03:51.260 ","End":"03:56.545","Text":"the x-axis and the projection of the r vector on our x, y plane."},{"Start":"03:56.545 ","End":"04:02.910","Text":"In some books, the Theta and the Phi will be switched with 1 another,"},{"Start":"04:02.910 ","End":"04:05.810","Text":"so you just have to remember what this equation looks"},{"Start":"04:05.810 ","End":"04:09.600","Text":"like and then you can just switch it, however it\u0027s written."},{"Start":"04:14.180 ","End":"04:19.301","Text":"Now let\u0027s write out our equation with what we have."},{"Start":"04:19.301 ","End":"04:26.500","Text":"Our div E soon we\u0027ll plug in our Epsilon Naught spin means I\u0027m div E,"},{"Start":"04:26.500 ","End":"04:32.895","Text":"so that\u0027s going to be equal to 1 divided by r^2 and then d"},{"Start":"04:32.895 ","End":"04:40.355","Text":"by dr of r^2 multiplied by our r component of our electric fields."},{"Start":"04:40.355 ","End":"04:44.195","Text":"c divided by r is multiplying everything over here,"},{"Start":"04:44.195 ","End":"04:48.105","Text":"so it\u0027s c divided by r in the r direction,"},{"Start":"04:48.105 ","End":"04:53.690","Text":"so we\u0027re just going to write in here c divided by r. Now we can see that"},{"Start":"04:53.690 ","End":"04:59.665","Text":"this r and 1 of the r\u0027s over here cancels out and we\u0027re going to differentiate soon,"},{"Start":"04:59.665 ","End":"05:03.480","Text":"let\u0027s just carry on onto our Theta components,"},{"Start":"05:03.480 ","End":"05:08.850","Text":"so now we have 1 divided by r sine of Phi,"},{"Start":"05:08.850 ","End":"05:16.655","Text":"and then we have d by d Theta of Theta component of our electric fields."},{"Start":"05:16.655 ","End":"05:21.225","Text":"Which is going to be c divided by r cosine Theta,"},{"Start":"05:21.225 ","End":"05:28.065","Text":"so we have c divided by r cosine of Theta."},{"Start":"05:28.065 ","End":"05:29.970","Text":"Now our Phi components,"},{"Start":"05:29.970 ","End":"05:35.640","Text":"so it\u0027s again 1 divided by r sine of Phi,"},{"Start":"05:35.640 ","End":"05:42.000","Text":"d by d Phi and then we have our Phi components,"},{"Start":"05:42.000 ","End":"05:52.267","Text":"so that\u0027s c divided by r multiplied by sine Theta cosine Phi."},{"Start":"05:52.267 ","End":"05:59.460","Text":"Then multiplied by sine Phi over here from the equation."},{"Start":"06:01.880 ","End":"06:10.260","Text":"This is our equation for div E. Let\u0027s differentiate everything."},{"Start":"06:10.260 ","End":"06:15.595","Text":"We have 1 divided by r^2 and then we have to differentiate,"},{"Start":"06:15.595 ","End":"06:18.870","Text":"r multiplied by c with respect to r."},{"Start":"06:18.870 ","End":"06:22.550","Text":"Then our r will cancel out and we\u0027ll just be left with our c,"},{"Start":"06:22.550 ","End":"06:27.715","Text":"so we\u0027ll just have that this is equal to c divided by r squared."},{"Start":"06:27.715 ","End":"06:30.120","Text":"Now let\u0027s move on to this."},{"Start":"06:30.120 ","End":"06:34.295","Text":"Here, we\u0027re differentiating with respect to Theta only,"},{"Start":"06:34.295 ","End":"06:39.430","Text":"so our c divided by r are constants in this term."},{"Start":"06:39.430 ","End":"06:41.475","Text":"We can move them out of the differential,"},{"Start":"06:41.475 ","End":"06:50.880","Text":"so that\u0027s going to be c divided by our r^2 sine Phi."},{"Start":"06:50.880 ","End":"06:55.040","Text":"Then when we differentiate cosine of Theta,"},{"Start":"06:55.040 ","End":"07:00.507","Text":"we\u0027re going to get negative sine of Theta."},{"Start":"07:00.507 ","End":"07:03.465","Text":"Now, the next term,"},{"Start":"07:03.465 ","End":"07:04.680","Text":"again, we\u0027ll do the same."},{"Start":"07:04.680 ","End":"07:06.555","Text":"We\u0027ll take out all of our constants,"},{"Start":"07:06.555 ","End":"07:10.420","Text":"so here we\u0027re differentiating with respect to Phi."},{"Start":"07:11.870 ","End":"07:17.355","Text":"We can take out a c divided by r and a sine Theta,"},{"Start":"07:17.355 ","End":"07:19.230","Text":"because those are constants."},{"Start":"07:19.230 ","End":"07:25.559","Text":"We\u0027ll have a plus c sine Theta"},{"Start":"07:25.559 ","End":"07:32.355","Text":"divided by r^2 sine Phi."},{"Start":"07:32.355 ","End":"07:37.140","Text":"Now, we just have to differentiate cosine Phi sine Phi."},{"Start":"07:37.140 ","End":"07:45.255","Text":"Now, we can use a trig identity which says that cosine Phi sine Phi,"},{"Start":"07:45.255 ","End":"07:52.605","Text":"so what we have over here is equal to half sine of 2 Phi."},{"Start":"07:52.605 ","End":"07:56.070","Text":"This is super useful and it\u0027s going to make"},{"Start":"07:56.070 ","End":"08:00.930","Text":"our differentiation here a lot better, or our derivative."},{"Start":"08:00.930 ","End":"08:04.830","Text":"What we\u0027ll get is we\u0027ll multiply this by a 1/2."},{"Start":"08:04.830 ","End":"08:07.206","Text":"Now we\u0027re differentiating."},{"Start":"08:07.206 ","End":"08:10.725","Text":"Then sine of 2 Phi,"},{"Start":"08:10.725 ","End":"08:15.705","Text":"the derivative of that is cosine of 2 Phi."},{"Start":"08:15.705 ","End":"08:20.325","Text":"Then we have to multiply it by the anti-derivative, which is 2."},{"Start":"08:20.325 ","End":"08:25.360","Text":"Then our half and 2 cancel out and we\u0027re left with this."},{"Start":"08:26.870 ","End":"08:34.298","Text":"Now we can say that our row from this over here is equal to"},{"Start":"08:34.298 ","End":"08:41.250","Text":"Epsilon Naught multiplied by our div E. Let\u0027s take out all of our like terms,"},{"Start":"08:41.250 ","End":"08:45.480","Text":"so Epsilon Naught multiplies everything here so we can put it on the outside."},{"Start":"08:45.480 ","End":"08:50.445","Text":"We also have a common term of c divided by r^2 squared."},{"Start":"08:50.445 ","End":"08:55.200","Text":"We can also take out our c divided by r^2 squared and now we can"},{"Start":"08:55.200 ","End":"09:00.780","Text":"write this out of 1 minus sine of"},{"Start":"09:00.780 ","End":"09:06.855","Text":"Theta divided by sine Phi plus"},{"Start":"09:06.855 ","End":"09:11.205","Text":"sine Theta cosine of"},{"Start":"09:11.205 ","End":"09:18.966","Text":"2 Phi divided by sine Phi."},{"Start":"09:18.966 ","End":"09:22.815","Text":"This has our answer to question Number 1."},{"Start":"09:22.815 ","End":"09:25.845","Text":"Now let\u0027s move on to question Number 2."},{"Start":"09:25.845 ","End":"09:27.330","Text":"Now in question Number 2,"},{"Start":"09:27.330 ","End":"09:30.210","Text":"we\u0027re asked to find through integration and charge"},{"Start":"09:30.210 ","End":"09:34.410","Text":"density the amount of charge located inside a sphere of"},{"Start":"09:34.410 ","End":"09:39.495","Text":"radius r. We can imagine that we have a sphere"},{"Start":"09:39.495 ","End":"09:46.590","Text":"and it is of radius R. This is a 3-dimensional shape,"},{"Start":"09:46.590 ","End":"09:50.445","Text":"you can see from amazing artwork."},{"Start":"09:50.445 ","End":"09:55.440","Text":"What we\u0027re meant to do is we\u0027re meant to find the amount of charge located inside of"},{"Start":"09:55.440 ","End":"10:01.860","Text":"this sphere through integration and we\u0027re going to integrate over our charge density."},{"Start":"10:01.860 ","End":"10:04.710","Text":"In actual fact, what they\u0027re asking us is,"},{"Start":"10:04.710 ","End":"10:06.135","Text":"what is our Q_in."},{"Start":"10:06.135 ","End":"10:09.225","Text":"Q_in or, q total how much charges in here?"},{"Start":"10:09.225 ","End":"10:13.560","Text":"As we know our Q_in is equal to the integral and then when we\u0027re"},{"Start":"10:13.560 ","End":"10:18.015","Text":"dealing with volume is equal to Rho dv."},{"Start":"10:18.015 ","End":"10:19.965","Text":"What does that equal to here?"},{"Start":"10:19.965 ","End":"10:22.515","Text":"First of all, because we have volume."},{"Start":"10:22.515 ","End":"10:29.145","Text":"This is going to turn into a triple integral and then our row is what we have over here,"},{"Start":"10:29.145 ","End":"10:31.305","Text":"multiplied by dv,"},{"Start":"10:31.305 ","End":"10:34.035","Text":"so because we\u0027re dealing with spherical coordinates,"},{"Start":"10:34.035 ","End":"10:41.010","Text":"our dv is equal to r^2 sine of Phi,"},{"Start":"10:41.010 ","End":"10:44.505","Text":"where Phi is the angle with the z axis,"},{"Start":"10:44.505 ","End":"10:48.870","Text":"dr, d Theta, d Phi."},{"Start":"10:48.870 ","End":"10:54.090","Text":"Now what we have to do is we have to fill in our integrating bounds."},{"Start":"10:54.090 ","End":"10:57.915","Text":"For r, when we\u0027re integrating along the radius,"},{"Start":"10:57.915 ","End":"11:01.200","Text":"because we\u0027re going through the entire volume of the sphere,"},{"Start":"11:01.200 ","End":"11:07.200","Text":"our radius is going from 0 until capital R, the radius of the sphere."},{"Start":"11:07.200 ","End":"11:11.670","Text":"Our Theta is going full circle from 0"},{"Start":"11:11.670 ","End":"11:17.055","Text":"until 2 Pi and our Phi so that we don\u0027t sum up twice,"},{"Start":"11:17.055 ","End":"11:21.555","Text":"we\u0027re going for a semicircle, 0 to Pi."},{"Start":"11:21.555 ","End":"11:24.480","Text":"Now let\u0027s begin our integration."},{"Start":"11:24.480 ","End":"11:26.490","Text":"Now the first thing we can see is that"},{"Start":"11:26.490 ","End":"11:31.170","Text":"this r^2 cancels out with this r^2 in the denominator."},{"Start":"11:31.170 ","End":"11:34.395","Text":"Now if we integrate along our r,"},{"Start":"11:34.395 ","End":"11:36.975","Text":"we\u0027re simply going to get i."},{"Start":"11:36.975 ","End":"11:39.795","Text":"Let\u0027s begin by writing this."},{"Start":"11:39.795 ","End":"11:46.785","Text":"Our Q_in is therefore going to be equal to Epsilon Naught multiplied by c,"},{"Start":"11:46.785 ","End":"11:51.670","Text":"multiplied by this R and we\u0027ve integrated along rd."},{"Start":"11:51.710 ","End":"11:55.755","Text":"Now, let\u0027s leave our Phi."},{"Start":"11:55.755 ","End":"11:57.540","Text":"Let\u0027s get to that in seconds,"},{"Start":"11:57.540 ","End":"11:59.985","Text":"so the bounds for our Phi from 0 to Pi."},{"Start":"11:59.985 ","End":"12:03.105","Text":"Now let\u0027s integrate along Theta."},{"Start":"12:03.105 ","End":"12:06.585","Text":"Let\u0027s go term by term and integrate."},{"Start":"12:06.585 ","End":"12:09.090","Text":"In this term over here where we have r1,"},{"Start":"12:09.090 ","End":"12:10.590","Text":"we don\u0027t have a Theta value,"},{"Start":"12:10.590 ","End":"12:14.610","Text":"so we can just simply plug in 2 Pi."},{"Start":"12:14.900 ","End":"12:19.870","Text":"Great. 1d Theta between 0 and 2 Pi is equal to 2 Pi."},{"Start":"12:19.870 ","End":"12:21.615","Text":"Now, let\u0027s look at this."},{"Start":"12:21.615 ","End":"12:26.010","Text":"We have negative sine Theta divided by this constant."},{"Start":"12:26.010 ","End":"12:30.864","Text":"Right now, this is a constant because we\u0027re integrating via Theta."},{"Start":"12:30.864 ","End":"12:35.655","Text":"Sine of Theta when integrated between 0 and 2 Pi,"},{"Start":"12:35.655 ","End":"12:37.575","Text":"we learned this a few lessons ago,"},{"Start":"12:37.575 ","End":"12:41.940","Text":"is always going to be equal to 0 and also with cosine Theta,"},{"Start":"12:41.940 ","End":"12:44.010","Text":"when integrating between 0 and 2 Pi,"},{"Start":"12:44.010 ","End":"12:45.315","Text":"that will be equal to 0."},{"Start":"12:45.315 ","End":"12:49.200","Text":"Here we have a 0 and then here again we have sine Theta"},{"Start":"12:49.200 ","End":"12:54.250","Text":"integrated from 0 to 2 Pi multiplied by some constant."},{"Start":"12:54.800 ","End":"12:59.145","Text":"We can see that this is also going to be equal to 0."},{"Start":"12:59.145 ","End":"13:02.715","Text":"Great. Now we\u0027ve dealt with rd Theta."},{"Start":"13:02.715 ","End":"13:06.465","Text":"Great. Now let\u0027s get back to integrating along Phi."},{"Start":"13:06.465 ","End":"13:10.110","Text":"We have our 2 Pi from integrating along Theta."},{"Start":"13:10.110 ","End":"13:16.830","Text":"Now we have to multiply this by sine of Phi d Phi."},{"Start":"13:16.830 ","End":"13:21.135","Text":"Now of course our 2 Pi is also a constant so we can move it out."},{"Start":"13:21.135 ","End":"13:27.060","Text":"Then we have to integrate sine Phi according to Phi between 0 and Pi,"},{"Start":"13:27.060 ","End":"13:31.380","Text":"so that\u0027s going to be equal to 2. Now we can rewrite it."},{"Start":"13:31.380 ","End":"13:33.450","Text":"We have 2 times 2 Pi,"},{"Start":"13:33.450 ","End":"13:38.985","Text":"so that\u0027s 4 Pi multiplied by Epsilon Naught CR."},{"Start":"13:38.985 ","End":"13:43.935","Text":"This is our answer to Question 2."},{"Start":"13:43.935 ","End":"13:48.765","Text":"This is the total charge located inside the sphere."},{"Start":"13:48.765 ","End":"13:52.260","Text":"Now let\u0027s go on to Question 3."},{"Start":"13:52.260 ","End":"13:58.947","Text":"Question 3 is asking to again find the amount of charge inside the sphere."},{"Start":"13:58.947 ","End":"14:02.760","Text":"Again, we\u0027re finding our Q_in in the same sphere of radius r,"},{"Start":"14:02.760 ","End":"14:06.075","Text":"but this time via Gauss\u0027s law."},{"Start":"14:06.075 ","End":"14:12.955","Text":"Using the electric flux and Gauss\u0027s law that we learned earlier in our course."},{"Start":"14:12.955 ","End":"14:14.630","Text":"Let\u0027s see how we do that."},{"Start":"14:14.630 ","End":"14:18.050","Text":"Now I\u0027ve kept the answer that we got for Question 2."},{"Start":"14:18.050 ","End":"14:22.885","Text":"Let\u0027s see if we get the same answer when we use this other method."},{"Start":"14:22.885 ","End":"14:26.270","Text":"Gauss\u0027s law is as follows."},{"Start":"14:26.270 ","End":"14:33.560","Text":"We have the closed loop integral of E.ds and"},{"Start":"14:33.560 ","End":"14:41.650","Text":"we say that this is equal to the charge inside of a shape divided by Epsilon Naught."},{"Start":"14:41.650 ","End":"14:49.839","Text":"If we have our sphere and it\u0027s in 3-dimensions."},{"Start":"14:49.839 ","End":"14:54.084","Text":"This means that the electric field,"},{"Start":"14:54.084 ","End":"14:58.420","Text":"which is perpendicular to our surface."},{"Start":"14:58.420 ","End":"15:03.400","Text":"Imagine lots of little arrows which are all perpendicular to the surface."},{"Start":"15:03.400 ","End":"15:08.470","Text":"That multiplied by the total surface area is going to be equal"},{"Start":"15:08.470 ","End":"15:15.670","Text":"to Q_in divided by Epsilon Naught and Q_in is our variable which we\u0027re trying to find."},{"Start":"15:15.670 ","End":"15:19.000","Text":"Now let\u0027s write out our integral."},{"Start":"15:19.000 ","End":"15:26.530","Text":"We have that the closed loop integral an E.ds is"},{"Start":"15:26.530 ","End":"15:33.760","Text":"going to be equal to our integral on E. We already have what our E field is equal to,"},{"Start":"15:33.760 ","End":"15:35.830","Text":"so right now I\u0027m not going to write out that whole thing,"},{"Start":"15:35.830 ","End":"15:40.945","Text":"so let\u0027s just write E for this dot our ds."},{"Start":"15:40.945 ","End":"15:43.330","Text":"What exactly is our ds?"},{"Start":"15:43.330 ","End":"15:51.276","Text":"Our ds is a small piece of area which is located on the surface of our sphere."},{"Start":"15:51.276 ","End":"15:58.300","Text":"It looks like some curved square where 1 of these curves comes"},{"Start":"15:58.300 ","End":"16:03.210","Text":"from r sine Phi d Theta and the other curve comes"},{"Start":"16:03.210 ","End":"16:09.135","Text":"from r d Phi and this we spoke about in the lesson about coordinates."},{"Start":"16:09.135 ","End":"16:14.510","Text":"Our ds has those dimensions,"},{"Start":"16:15.480 ","End":"16:18.370","Text":"so when we put them together,"},{"Start":"16:18.370 ","End":"16:20.200","Text":"it\u0027s equal to the Jacobian,"},{"Start":"16:20.200 ","End":"16:24.710","Text":"which is r^2 sine Phi d Theta d Phi"},{"Start":"16:25.890 ","End":"16:30.760","Text":"and the direction of our ds vector is,"},{"Start":"16:30.760 ","End":"16:35.545","Text":"as we said, always perpendicular to the surface area."},{"Start":"16:35.545 ","End":"16:39.475","Text":"That means that because we\u0027re dealing with a sphere,"},{"Start":"16:39.475 ","End":"16:42.505","Text":"so in order to be always perpendicular,"},{"Start":"16:42.505 ","End":"16:47.365","Text":"it\u0027s going to be in the radial direction."},{"Start":"16:47.365 ","End":"16:55.900","Text":"Now we can see that we have our E field dot product with our ds,"},{"Start":"16:55.900 ","End":"16:59.305","Text":"where our ds vector is in the r hat direction."},{"Start":"16:59.305 ","End":"17:00.880","Text":"When we\u0027re doing the dot product,"},{"Start":"17:00.880 ","End":"17:05.950","Text":"we can see that only dot r hat is"},{"Start":"17:05.950 ","End":"17:11.110","Text":"going to give us some answer or some addition or subtraction,"},{"Start":"17:11.110 ","End":"17:16.510","Text":"but here specifically addition to our integral because we know that Theta"},{"Start":"17:16.510 ","End":"17:23.050","Text":"hat.r hat is equal to 0 and phi hat dot.r hat is also equal to 0."},{"Start":"17:23.050 ","End":"17:27.970","Text":"We\u0027re only going to be left with termini E field,"},{"Start":"17:27.970 ","End":"17:31.760","Text":"which is also in the radial direction."},{"Start":"17:33.510 ","End":"17:37.165","Text":"That\u0027s just coming from the rules of the dot product."},{"Start":"17:37.165 ","End":"17:42.429","Text":"Now another way of thinking of this is that we know that we\u0027re summing up the flux"},{"Start":"17:42.429 ","End":"17:48.370","Text":"which is coming out of our surface or out of our shape our body that has charge in it."},{"Start":"17:48.370 ","End":"17:53.440","Text":"It has to be coming out perpendicular to our Gaussian surface,"},{"Start":"17:53.440 ","End":"17:56.215","Text":"or at least to the body where the charge in it."},{"Start":"17:56.215 ","End":"17:59.695","Text":"If it\u0027s coming out perpendicularly when we\u0027re dealing with a sphere,"},{"Start":"17:59.695 ","End":"18:02.110","Text":"then we know that our electric field or"},{"Start":"18:02.110 ","End":"18:07.340","Text":"electric flux is also going to be in the radial direction."},{"Start":"18:07.620 ","End":"18:14.155","Text":"From both ways, either intuition or by using the rules of the dot product,"},{"Start":"18:14.155 ","End":"18:18.400","Text":"we can see that this integral is therefore going to"},{"Start":"18:18.400 ","End":"18:22.690","Text":"be the integral on the radial component of"},{"Start":"18:22.690 ","End":"18:31.870","Text":"our E field multiplied by r^2 sine Phi d Theta d Phi."},{"Start":"18:31.870 ","End":"18:39.290","Text":"Our r hat crossed out because r hat dot product with the r hat is simply equal to 1."},{"Start":"18:39.290 ","End":"18:41.275","Text":"Of course, we have to add in"},{"Start":"18:41.275 ","End":"18:47.035","Text":"another integration sign because we\u0027re integrating for area d Theta d Phi,"},{"Start":"18:47.035 ","End":"18:48.835","Text":"so it\u0027s a double integral."},{"Start":"18:48.835 ","End":"18:51.115","Text":"Now let\u0027s substitute this in,"},{"Start":"18:51.115 ","End":"18:53.215","Text":"we have our double integral,"},{"Start":"18:53.215 ","End":"18:59.110","Text":"r radial component of the electric field is simply c divided by r,"},{"Start":"18:59.110 ","End":"19:08.305","Text":"and then multiply by r^2 sine Phi d Theta d Phi."},{"Start":"19:08.305 ","End":"19:11.530","Text":"The bounds for our integration as usual,"},{"Start":"19:11.530 ","End":"19:13.270","Text":"when we\u0027re integrating along Theta,"},{"Start":"19:13.270 ","End":"19:14.890","Text":"we do a full circle,"},{"Start":"19:14.890 ","End":"19:22.810","Text":"so 0 till 2Pi and that\u0027s of course here when how we defined our Theta angle,"},{"Start":"19:22.810 ","End":"19:29.155","Text":"it\u0027s between x and the projection of our r vector on the x, y plane."},{"Start":"19:29.155 ","End":"19:31.240","Text":"Sometimes Theta and Phi are switch,"},{"Start":"19:31.240 ","End":"19:36.000","Text":"so you have to notice in the question which angle they\u0027re speaking about."},{"Start":"19:36.000 ","End":"19:39.810","Text":"Then when we\u0027re integrating along Phi as we defined it,"},{"Start":"19:39.810 ","End":"19:42.975","Text":"where Phi is the angle between the z axis and our r vector,"},{"Start":"19:42.975 ","End":"19:46.625","Text":"we integrate from 0 until Pi."},{"Start":"19:46.625 ","End":"19:50.725","Text":"Now, we can cross out this,"},{"Start":"19:50.725 ","End":"19:52.690","Text":"r^2 will cross out with this r,"},{"Start":"19:52.690 ","End":"19:56.130","Text":"so I will just be left with 1 r over here."},{"Start":"19:56.130 ","End":"20:00.600","Text":"Now, because we\u0027re looking at how much charge"},{"Start":"20:00.600 ","End":"20:05.325","Text":"is inside and because we\u0027re only integrating along area,"},{"Start":"20:05.325 ","End":"20:09.915","Text":"so we\u0027re located right now on the surface of our sphere."},{"Start":"20:09.915 ","End":"20:14.560","Text":"The total surface area of the sphere and what we\u0027re trying to do"},{"Start":"20:14.560 ","End":"20:19.420","Text":"is the charge enclosed within the total surface of our sphere."},{"Start":"20:19.420 ","End":"20:24.370","Text":"That means that the r that we\u0027re going to be looking at, instead of small r,"},{"Start":"20:24.370 ","End":"20:26.800","Text":"we just substitute in capital R,"},{"Start":"20:26.800 ","End":"20:30.850","Text":"which was the radius of our sphere."},{"Start":"20:30.850 ","End":"20:34.930","Text":"Because we\u0027re not looking at integrating according to volume,"},{"Start":"20:34.930 ","End":"20:38.155","Text":"we\u0027re simply integrating according to area."},{"Start":"20:38.155 ","End":"20:40.135","Text":"We\u0027re on the surface area,"},{"Start":"20:40.135 ","End":"20:46.525","Text":"which means we\u0027re at a radius of r is equal to capital R. Now,"},{"Start":"20:46.525 ","End":"20:48.580","Text":"we can rewrite this,"},{"Start":"20:48.580 ","End":"20:53.560","Text":"so we\u0027re integrating from 0 to Pi and then from 0"},{"Start":"20:53.560 ","End":"21:02.410","Text":"until Pi of C capital R sine Phi d Theta d Phi."},{"Start":"21:02.410 ","End":"21:06.670","Text":"We can already see that we have no variable Theta here,"},{"Start":"21:06.670 ","End":"21:13.030","Text":"so when we integrate along 1dTheta from 0-2Pi we\u0027ll get 2Pi."},{"Start":"21:13.030 ","End":"21:21.715","Text":"Then when we integrate sine Phi d Phi from 0 to Pi, we get 2."},{"Start":"21:21.715 ","End":"21:28.165","Text":"This can become 4Pi multiplied by CR,"},{"Start":"21:28.165 ","End":"21:31.195","Text":"which is a constant."},{"Start":"21:31.195 ","End":"21:34.735","Text":"Now we can see, let\u0027s scroll down a little bit."},{"Start":"21:34.735 ","End":"21:39.880","Text":"We can see that the left side of the equation is equal to 4Pi CR,"},{"Start":"21:39.880 ","End":"21:44.905","Text":"and then that\u0027s equal to Q_in divided by Epsilon Naught."},{"Start":"21:44.905 ","End":"21:48.070","Text":"What we want to isolate out is our Q_in,"},{"Start":"21:48.070 ","End":"21:51.699","Text":"that\u0027s going to be equal to Epsilon Naught multiplied"},{"Start":"21:51.699 ","End":"21:55.480","Text":"by the closed-loop integral of E.ds,"},{"Start":"21:55.480 ","End":"21:57.130","Text":"which is what we just worked out."},{"Start":"21:57.130 ","End":"22:03.145","Text":"That\u0027s going to be Epsilon Naught multiplied by 4Pi CR."},{"Start":"22:03.145 ","End":"22:07.795","Text":"Then we\u0027ll simply get 4Pi Epsilon Naught CR,"},{"Start":"22:07.795 ","End":"22:14.335","Text":"which how exciting is exactly what we got for our answer in question 2."},{"Start":"22:14.335 ","End":"22:21.595","Text":"Now we can see that whenever we\u0027re being asked to find the charge inside some shape,"},{"Start":"22:21.595 ","End":"22:26.860","Text":"we can use the idea of calculating the charge density and we"},{"Start":"22:26.860 ","End":"22:32.590","Text":"can also use Gauss\u0027s law if our electric field is given to us."},{"Start":"22:32.590 ","End":"22:35.360","Text":"That\u0027s the end of this lesson."}],"ID":21381}],"Thumbnail":null,"ID":99470}]

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