Introduction to Dielectric Materials
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- Intro To Dielectric Materials
- Exercise 1
- Polarization Density
- Exercise 2
- Deriving Bound Charge Density Part 1
- Deriving Bound Charge Density Part 2
- Deriving Bound Charge Density Part 3
- Exercise 3
- Exercise 4
- Exercise 5
- Gauss_s Law and Displacement Vector
- Exercise 6
- Electric Field Vs Displacement Vector
- Linear Dielectric Materials
- Exercise 7

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[{"Name":"Introduction to Dielectric Materials","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Intro To Dielectric Materials","Duration":"17m 50s","ChapterTopicVideoID":21463,"CourseChapterTopicPlaylistID":99475,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/21463.jpeg","UploadDate":"2020-04-21T13:38:51.9330000","DurationForVideoObject":"PT17M50S","Description":null,"MetaTitle":"Intro To Dielectric Materials: Video + Workbook | Proprep","MetaDescription":"Dielectric Materials - Introduction to Dielectric Materials. 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/dielectric-materials/introduction-to-dielectric-materials/vid22311","VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.890","Text":"Hello. In this lesson,"},{"Start":"00:01.890 ","End":"00:05.130","Text":"we\u0027re going to be speaking about dielectric materials."},{"Start":"00:05.130 ","End":"00:10.560","Text":"Now in this course, we\u0027re going to be looking at the basis of dielectric materials,"},{"Start":"00:10.560 ","End":"00:13.125","Text":"and speaking about them in a very broad sense."},{"Start":"00:13.125 ","End":"00:17.295","Text":"In higher level courses,"},{"Start":"00:17.295 ","End":"00:21.551","Text":"you\u0027ll look at dielectric materials in a lot more detail."},{"Start":"00:21.551 ","End":"00:25.350","Text":"Here we have a dielectric material."},{"Start":"00:25.350 ","End":"00:30.330","Text":"The first thing to note is that dielectric materials are insulators."},{"Start":"00:30.330 ","End":"00:34.560","Text":"Now of course, if we place this in a large enough electric field,"},{"Start":"00:34.560 ","End":"00:42.800","Text":"so the dielectric will also conduct and there\u0027ll be some escape of charge or current."},{"Start":"00:42.800 ","End":"00:49.835","Text":"However, as it is in a weaker or relatively weak electric field,"},{"Start":"00:49.835 ","End":"00:52.445","Text":"a dielectric material is an insulator."},{"Start":"00:52.445 ","End":"00:54.800","Text":"Now, in dielectric materials,"},{"Start":"00:54.800 ","End":"00:59.185","Text":"there\u0027s lots and lots and lots of electric dipoles."},{"Start":"00:59.185 ","End":"01:05.795","Text":"Positive negative, and remember our arrow p points from the negative to the positive."},{"Start":"01:05.795 ","End":"01:12.860","Text":"They\u0027re just mixed up everywhere, like so randomly."},{"Start":"01:12.860 ","End":"01:16.205","Text":"As we know the dipole moment is"},{"Start":"01:16.205 ","End":"01:19.610","Text":"in the direction from the negative charge to the positive charge."},{"Start":"01:19.610 ","End":"01:26.784","Text":"But we know that the electric field is from the positive to the negative."},{"Start":"01:26.784 ","End":"01:30.755","Text":"That means that within the material,"},{"Start":"01:30.755 ","End":"01:35.824","Text":"we have lots and lots of little electric fields all over the material."},{"Start":"01:35.824 ","End":"01:41.600","Text":"However, because all of the dipoles are randomly oriented,"},{"Start":"01:41.600 ","End":"01:51.455","Text":"we know that the average electric field of the material is going to be equal to 0."},{"Start":"01:51.455 ","End":"01:56.120","Text":"Now, let\u0027s take a look what happens to our dielectric material and"},{"Start":"01:56.120 ","End":"02:01.390","Text":"the dipoles when we place this material in an electric field."},{"Start":"02:01.390 ","End":"02:05.420","Text":"Let\u0027s imagine that we\u0027re adding in here"},{"Start":"02:05.420 ","End":"02:11.810","Text":"an infinitely charged plane which has charge density Sigma."},{"Start":"02:11.810 ","End":"02:14.420","Text":"Now as we know, this infinite plane of"},{"Start":"02:14.420 ","End":"02:20.105","Text":"charge density Sigma is going to apply an electric field."},{"Start":"02:20.105 ","End":"02:23.765","Text":"We know it\u0027s going to be uniform from Gauss\u0027s law."},{"Start":"02:23.765 ","End":"02:29.720","Text":"We have our E field due to the plane pointing in this way."},{"Start":"02:29.720 ","End":"02:35.720","Text":"Now what\u0027s going to happen to the dipoles within the dielectric material?"},{"Start":"02:35.720 ","End":"02:39.220","Text":"Now from our lesson for our torque,"},{"Start":"02:39.220 ","End":"02:42.140","Text":"when we were speaking about electric dipoles and their torque,"},{"Start":"02:42.140 ","End":"02:47.705","Text":"we know that we\u0027re going to have a torque acting here due to this applied electric field,"},{"Start":"02:47.705 ","End":"02:51.070","Text":"that\u0027s going to look something like so."},{"Start":"02:51.070 ","End":"02:58.100","Text":"We can see that due to this torque our dipole moment or our dipole is going to"},{"Start":"02:58.100 ","End":"03:05.590","Text":"rotate so that the positive charges are in the direction of this external E field."},{"Start":"03:05.590 ","End":"03:07.170","Text":"As we can see,"},{"Start":"03:07.170 ","End":"03:11.480","Text":"our dipoles are going to line up such that the dipole moment"},{"Start":"03:11.480 ","End":"03:15.980","Text":"is in the direction of the external E field due to this charged plane."},{"Start":"03:15.980 ","End":"03:19.340","Text":"However, we can see that the many E field,"},{"Start":"03:19.340 ","End":"03:25.280","Text":"due to our dipoles having a negative and a positive charge is going to"},{"Start":"03:25.280 ","End":"03:31.500","Text":"be in the opposite direction to the external E field."},{"Start":"03:31.500 ","End":"03:34.610","Text":"Now what we want to do is we want to find out what"},{"Start":"03:34.610 ","End":"03:38.105","Text":"the total force is going to be equal to."},{"Start":"03:38.105 ","End":"03:43.610","Text":"Let\u0027s call it the electric fields due to the infinite plane E_0,"},{"Start":"03:43.610 ","End":"03:47.680","Text":"and the electric field due to each dipole as E tag."},{"Start":"03:47.680 ","End":"03:52.880","Text":"We can see that the total electric field that a particle will feel is going"},{"Start":"03:52.880 ","End":"03:58.255","Text":"to be a superposition between our E_naughts and our E tags."},{"Start":"03:58.255 ","End":"04:02.300","Text":"There is a mathematical connection which joins"},{"Start":"04:02.300 ","End":"04:07.010","Text":"the electric field that is due to this charged plane,"},{"Start":"04:07.010 ","End":"04:11.780","Text":"and the electric fields due to the dipoles."},{"Start":"04:11.780 ","End":"04:15.530","Text":"The total electric field, so this is E,"},{"Start":"04:15.530 ","End":"04:18.784","Text":"is going to be equal to E_naught,"},{"Start":"04:18.784 ","End":"04:24.260","Text":"which is the electric field due to the infinite plane, divided by Epsilon_r."},{"Start":"04:24.260 ","End":"04:25.680","Text":"Now what is Epsilon_r?"},{"Start":"04:25.680 ","End":"04:29.135","Text":"Epsilon_r will usually be given to us in our question,"},{"Start":"04:29.135 ","End":"04:36.064","Text":"and it is the dielectric constant of our dielectric material."},{"Start":"04:36.064 ","End":"04:38.575","Text":"This is an r of course."},{"Start":"04:38.575 ","End":"04:42.680","Text":"All I need to do in order to find my total E field due"},{"Start":"04:42.680 ","End":"04:47.240","Text":"to this infinitely charged plane and taking"},{"Start":"04:47.240 ","End":"04:50.600","Text":"into account the opposite E field due to"},{"Start":"04:50.600 ","End":"04:56.270","Text":"my dipoles is to find the electric field due to my infinitely charged plane,"},{"Start":"04:56.270 ","End":"04:58.310","Text":"and divide that by Epsilon_r."},{"Start":"04:58.310 ","End":"05:00.090","Text":"Just a little note,"},{"Start":"05:00.090 ","End":"05:03.870","Text":"in some textbooks they\u0027ll say that we use Epsilon_r."},{"Start":"05:03.870 ","End":"05:10.850","Text":"In other textbooks, Epsilon_r is simply denoted by the Greek letter Kappa."},{"Start":"05:10.850 ","End":"05:12.830","Text":"But these 2 mean the exact same thing,"},{"Start":"05:12.830 ","End":"05:14.900","Text":"and they\u0027re interchangeable over here."},{"Start":"05:14.900 ","End":"05:18.065","Text":"Sometimes we\u0027ll just be given Epsilon,"},{"Start":"05:18.065 ","End":"05:24.165","Text":"and Epsilon is equal to our Epsilon_naught multiplied by Epsilon_r or a Kappa."},{"Start":"05:24.165 ","End":"05:28.030","Text":"This equation just helps us find the total E field"},{"Start":"05:28.030 ","End":"05:31.870","Text":"inside this dielectric material due to the external E field,"},{"Start":"05:31.870 ","End":"05:34.645","Text":"and the internal dipole E field."},{"Start":"05:34.645 ","End":"05:35.990","Text":"Now a little side note,"},{"Start":"05:35.990 ","End":"05:42.970","Text":"our Epsilon_r or a Kappa is always going to be bigger or equal to 1."},{"Start":"05:42.970 ","End":"05:50.230","Text":"This makes sense because we can see that our total electric field is always going to"},{"Start":"05:50.230 ","End":"05:57.815","Text":"be smaller or equal to the electric field due to our externally charged plane."},{"Start":"05:57.815 ","End":"06:03.175","Text":"Because we\u0027re going to have our externally charged plane with our E_naught E field."},{"Start":"06:03.175 ","End":"06:10.955","Text":"Then we\u0027re going to always have to minus our E field due to the dipoles inside."},{"Start":"06:10.955 ","End":"06:15.605","Text":"Our E_naught divided by Epsilon_r,"},{"Start":"06:15.605 ","End":"06:22.805","Text":"this fraction always has to be smaller than the original E_naught."},{"Start":"06:22.805 ","End":"06:27.710","Text":"Now what we\u0027re going to do is we\u0027re going to be talking about something else,"},{"Start":"06:27.710 ","End":"06:30.455","Text":"which is called bound charges."},{"Start":"06:30.455 ","End":"06:35.105","Text":"It\u0027s signified by a Sigma_b, b for bound."},{"Start":"06:35.105 ","End":"06:38.870","Text":"We can also sometimes call it induced charge,"},{"Start":"06:38.870 ","End":"06:41.015","Text":"which will then be given by Sigma_i,"},{"Start":"06:41.015 ","End":"06:44.380","Text":"and both of these things mean the same thing."},{"Start":"06:44.380 ","End":"06:49.530","Text":"This bound charge is charge density,"},{"Start":"06:49.530 ","End":"06:52.520","Text":"and it\u0027s to do with the charge density,"},{"Start":"06:52.520 ","End":"06:58.175","Text":"which appears or is induced within our dielectric material."},{"Start":"06:58.175 ","End":"07:00.905","Text":"Now when we\u0027re dealing with a dielectric material,"},{"Start":"07:00.905 ","End":"07:05.240","Text":"we say that the total charge of the material is equal to 0."},{"Start":"07:05.240 ","End":"07:11.060","Text":"However, that doesn\u0027t mean that there aren\u0027t charges within the material."},{"Start":"07:11.060 ","End":"07:13.880","Text":"We can see that we have our electric dipole,"},{"Start":"07:13.880 ","End":"07:16.340","Text":"so we have a positive and a negative charge."},{"Start":"07:16.340 ","End":"07:19.490","Text":"But the total net charge of the material is"},{"Start":"07:19.490 ","End":"07:23.200","Text":"equal to 0 because all the charges cancel out."},{"Start":"07:23.200 ","End":"07:27.380","Text":"Our Q total for our dielectric material is equal to 0."},{"Start":"07:27.380 ","End":"07:31.505","Text":"However, when we apply an external electric field,"},{"Start":"07:31.505 ","End":"07:38.285","Text":"some of our charges will move up and some of our other charges will move down."},{"Start":"07:38.285 ","End":"07:43.205","Text":"Overall, the charge of our dielectric material is still going to be equal to 0."},{"Start":"07:43.205 ","End":"07:49.075","Text":"However, there\u0027s some splits due to our electric dipoles."},{"Start":"07:49.075 ","End":"07:54.080","Text":"What we can see with our electric dipoles is that we have a positive charge at the top,"},{"Start":"07:54.080 ","End":"07:55.970","Text":"and a negative charge at the bottom."},{"Start":"07:55.970 ","End":"08:00.470","Text":"What we can see is that our dipoles are going to align where we\u0027ll have negative charge,"},{"Start":"08:00.470 ","End":"08:03.085","Text":"positive, negative, positive, negative, positive."},{"Start":"08:03.085 ","End":"08:07.430","Text":"We\u0027ll just have a bunch of dipoles stacked 1 on top of the other."},{"Start":"08:07.430 ","End":"08:11.870","Text":"We can see the next to each positive charge we\u0027ll have a negative charge just above,"},{"Start":"08:11.870 ","End":"08:14.750","Text":"and then above that negative charge we\u0027ll have another positive charge,"},{"Start":"08:14.750 ","End":"08:16.280","Text":"and so on and so forth."},{"Start":"08:16.280 ","End":"08:19.865","Text":"Now the only area where this positive,"},{"Start":"08:19.865 ","End":"08:20.990","Text":"the negative, then positive,"},{"Start":"08:20.990 ","End":"08:28.492","Text":"then negative aspect stops working is when we get to the boundary of our material."},{"Start":"08:28.492 ","End":"08:31.140","Text":"What we\u0027ll see is that we\u0027ll have at"},{"Start":"08:31.140 ","End":"08:34.590","Text":"the bottom a bunch of negative charges along the bottom."},{"Start":"08:34.590 ","End":"08:36.570","Text":"Then we\u0027ll have positive, negative, positive,"},{"Start":"08:36.570 ","End":"08:44.955","Text":"negative until we get to the top upper boundary of the dielectric material."},{"Start":"08:44.955 ","End":"08:50.420","Text":"Then we\u0027ll see a bunch of positive charges lining the top."},{"Start":"08:50.420 ","End":"08:53.150","Text":"These stacks of dipoles,"},{"Start":"08:53.150 ","End":"08:59.885","Text":"1 on top of the other is what we call a bound charges or our induced charge."},{"Start":"08:59.885 ","End":"09:04.440","Text":"A bound charge is the charge density due to"},{"Start":"09:04.440 ","End":"09:10.785","Text":"the arrangement of the electric dipoles within the dielectric material."},{"Start":"09:10.785 ","End":"09:14.460","Text":"Why do we sometimes call it the induced charge?"},{"Start":"09:14.460 ","End":"09:20.850","Text":"That\u0027s because we have to induce this type of arrangement within the dielectric material."},{"Start":"09:20.850 ","End":"09:22.235","Text":"How is this induced?"},{"Start":"09:22.235 ","End":"09:25.535","Text":"It\u0027s induced by an external electric field."},{"Start":"09:25.535 ","End":"09:29.315","Text":"Then why do we sometimes call it the bound charge, Sigma_b?"},{"Start":"09:29.315 ","End":"09:33.050","Text":"That\u0027s because our dielectric material is still an insulator,"},{"Start":"09:33.050 ","End":"09:36.240","Text":"which means that these charges can move around too much,"},{"Start":"09:36.240 ","End":"09:40.470","Text":"they can just rearrange within the dielectric material."},{"Start":"09:40.470 ","End":"09:45.684","Text":"Our charges are bound inside the dielectric material."},{"Start":"09:45.684 ","End":"09:48.120","Text":"Now of course, with everything in physics,"},{"Start":"09:48.120 ","End":"09:50.975","Text":"there\u0027s always going to be a mathematical equation in"},{"Start":"09:50.975 ","End":"09:54.520","Text":"order to calculate what our bound charge is."},{"Start":"09:54.520 ","End":"09:58.635","Text":"Now let\u0027s take a look at what our equations are and how to use them."},{"Start":"09:58.635 ","End":"10:00.585","Text":"Before we look at the equation,"},{"Start":"10:00.585 ","End":"10:05.872","Text":"our bound charge is always denoted by Sigma_b or a Sigma_I,"},{"Start":"10:05.872 ","End":"10:07.550","Text":"it\u0027s the same thing."},{"Start":"10:07.550 ","End":"10:10.650","Text":"In order to not get confused with this Sigma,"},{"Start":"10:10.650 ","End":"10:15.290","Text":"when we\u0027re dealing with a dielectric material and an induced electric field,"},{"Start":"10:15.290 ","End":"10:21.085","Text":"this Sigma sometimes becomes Sigma free or Sigma_f,"},{"Start":"10:21.085 ","End":"10:23.250","Text":"so let\u0027s say it\u0027s called Sigma free,"},{"Start":"10:23.250 ","End":"10:25.445","Text":"but it can also be Sigma_f."},{"Start":"10:25.445 ","End":"10:30.930","Text":"This is because the charges on here are not bound."},{"Start":"10:30.930 ","End":"10:36.270","Text":"They\u0027re free to move because usually this infinite plane or whatever might be"},{"Start":"10:36.270 ","End":"10:42.550","Text":"inducing this electric field is going to be made out of a conductor."},{"Start":"10:42.990 ","End":"10:45.525","Text":"As we know in a conductor,"},{"Start":"10:45.525 ","End":"10:48.705","Text":"the charges within the conductor can move around."},{"Start":"10:48.705 ","End":"10:50.885","Text":"They\u0027re free to move around."},{"Start":"10:50.885 ","End":"10:55.460","Text":"This Sigma over here will be called free so that we don\u0027t get confused."},{"Start":"10:55.460 ","End":"10:58.680","Text":"The Sigma when we\u0027re talking about the charged density"},{"Start":"10:58.680 ","End":"11:02.220","Text":"within the dielectric is going to be the bound charge because"},{"Start":"11:02.220 ","End":"11:05.350","Text":"the dielectric is an insulator or"},{"Start":"11:05.350 ","End":"11:10.365","Text":"the induced charge because it\u0027s induced by this external electric field."},{"Start":"11:10.365 ","End":"11:14.450","Text":"Now what I\u0027m going to do is I\u0027m going to draw this picture in"},{"Start":"11:14.450 ","End":"11:19.500","Text":"2-dimensions in order for the diagram to be clearer."},{"Start":"11:19.500 ","End":"11:21.980","Text":"Let\u0027s say that our Sigma free,"},{"Start":"11:21.980 ","End":"11:26.235","Text":"which is our charge distribution per unit area,"},{"Start":"11:26.235 ","End":"11:32.673","Text":"for our infinitely charged plane over here at the bottom in the blue line,"},{"Start":"11:32.673 ","End":"11:36.395","Text":"let\u0027s say that it\u0027s filled with positive charges."},{"Start":"11:36.395 ","End":"11:43.660","Text":"We have lots and lots and lots of positive charges due to this Sigma free."},{"Start":"11:43.660 ","End":"11:46.395","Text":"These positive charges, as we can see,"},{"Start":"11:46.395 ","End":"11:52.985","Text":"are going to attract the negative charges of our dipole within the dielectric material."},{"Start":"11:52.985 ","End":"11:57.035","Text":"Here we can see all of the negative dipole charges."},{"Start":"11:57.035 ","End":"12:01.865","Text":"Then we\u0027re going to have stacks of positive, negative, positive, negative."},{"Start":"12:01.865 ","End":"12:03.365","Text":"Then here at the top,"},{"Start":"12:03.365 ","End":"12:05.760","Text":"we\u0027re going to have the top of the stack,"},{"Start":"12:05.760 ","End":"12:11.555","Text":"which is going to be lots of little positive charges,"},{"Start":"12:11.555 ","End":"12:15.965","Text":"which are the positive side of the electric dipole."},{"Start":"12:15.965 ","End":"12:21.540","Text":"Now what I want to do is I want to calculate my Sigmas."},{"Start":"12:21.540 ","End":"12:27.735","Text":"We know that we can get the value from Sigma due to the jump in the electric field."},{"Start":"12:27.735 ","End":"12:32.685","Text":"You know that the equation for Sigma is equal to Epsilon_naught multiplied by"},{"Start":"12:32.685 ","End":"12:37.860","Text":"the jump in the electric field perpendicular to our shapes."},{"Start":"12:37.860 ","End":"12:39.615","Text":"If we have an infinite plane,"},{"Start":"12:39.615 ","End":"12:43.310","Text":"electric field is going to be perpendicular to the plane."},{"Start":"12:43.310 ","End":"12:47.190","Text":"What we\u0027re going to do is we\u0027re going to measure on a point"},{"Start":"12:47.190 ","End":"12:51.040","Text":"very close to our infinite plane,"},{"Start":"12:51.040 ","End":"12:54.005","Text":"our electric field over here."},{"Start":"12:54.005 ","End":"12:58.205","Text":"Then slightly within the dielectric material,"},{"Start":"12:58.205 ","End":"13:04.012","Text":"very close to the lower boundary of dielectric material shape,"},{"Start":"13:04.012 ","End":"13:08.925","Text":"we\u0027re going to measure the electric field over here,"},{"Start":"13:08.925 ","End":"13:11.535","Text":"and this is E plus."},{"Start":"13:11.535 ","End":"13:19.260","Text":"Then our equation for Sigma is going to become Epsilon_naught multiplied by"},{"Start":"13:19.260 ","End":"13:24.405","Text":"the electric field slightly within the dielectric material minus"},{"Start":"13:24.405 ","End":"13:30.185","Text":"the electric field below are infinitely charged plane."},{"Start":"13:30.185 ","End":"13:33.150","Text":"Great. Now we have this equation for Sigma,"},{"Start":"13:33.150 ","End":"13:34.785","Text":"but which Sigma are we talking about?"},{"Start":"13:34.785 ","End":"13:38.880","Text":"Are we talking about our bound charge or are we talking about our Sigma free?"},{"Start":"13:38.880 ","End":"13:40.875","Text":"When we use this equation,"},{"Start":"13:40.875 ","End":"13:44.015","Text":"because we\u0027re checking slightly below"},{"Start":"13:44.015 ","End":"13:48.260","Text":"our infinite plane and slightly within our dielectric material,"},{"Start":"13:48.260 ","End":"13:50.890","Text":"we\u0027re going to get both of the Sigmas."},{"Start":"13:50.890 ","End":"13:54.780","Text":"We\u0027re going to get a combination of Sigma free and Sigma_b."},{"Start":"13:54.780 ","End":"13:56.850","Text":"This Sigma is called,"},{"Start":"13:56.850 ","End":"13:59.325","Text":"let\u0027s call it Sigma total."},{"Start":"13:59.325 ","End":"14:07.990","Text":"Sigma total is going to be equal to Sigma free plus Sigma_b."},{"Start":"14:08.120 ","End":"14:15.995","Text":"This is the equation or these 2 equations you should be writing your equation notes."},{"Start":"14:15.995 ","End":"14:20.790","Text":"Here we find that the total charge distribution,"},{"Start":"14:20.790 ","End":"14:27.760","Text":"which is made out of the charge distribution of our infinitely charged plane,"},{"Start":"14:27.760 ","End":"14:31.543","Text":"which is inducing an electric field and"},{"Start":"14:31.543 ","End":"14:37.290","Text":"the charge distribution of the bound charges within the dielectric material."},{"Start":"14:37.290 ","End":"14:42.090","Text":"That means that what we have here over here,"},{"Start":"14:42.090 ","End":"14:45.635","Text":"so this is going to be the real electric field"},{"Start":"14:45.635 ","End":"14:49.640","Text":"that we feel because this is the total electric field,"},{"Start":"14:49.640 ","End":"14:53.280","Text":"taking into account the external electric field"},{"Start":"14:53.280 ","End":"14:56.440","Text":"and the induced electric field within the dielectric,"},{"Start":"14:56.440 ","End":"14:59.840","Text":"which will act to reduce the external field."},{"Start":"14:59.840 ","End":"15:03.260","Text":"If we\u0027re standing here and we\u0027re measuring the electric field,"},{"Start":"15:03.260 ","End":"15:10.290","Text":"this is the field that we will measure and not the fields due to our Sigma free."},{"Start":"15:10.290 ","End":"15:17.780","Text":"That\u0027s great. But now what happens if I want to calculate my Sigma free, only that?"},{"Start":"15:17.780 ","End":"15:22.035","Text":"Or if I want to just calculate my Sigma_b,"},{"Start":"15:22.035 ","End":"15:26.115","Text":"and not the total Sigma? How would we do that?"},{"Start":"15:26.115 ","End":"15:30.390","Text":"My Sigma free is easily calculated as"},{"Start":"15:30.390 ","End":"15:35.460","Text":"Epsilon_naught multiplied by the jump in the electric field,"},{"Start":"15:35.460 ","End":"15:37.970","Text":"but the original electric fields."},{"Start":"15:37.970 ","End":"15:41.760","Text":"That we can simply do via Gauss\u0027s law just like we saw."},{"Start":"15:41.760 ","End":"15:45.270","Text":"We imagine that this dielectric material doesn\u0027t exist,"},{"Start":"15:45.270 ","End":"15:50.855","Text":"it isn\u0027t here and we have an infinitely charged plane and then we use Gauss\u0027s law and we"},{"Start":"15:50.855 ","End":"15:53.640","Text":"find the jump in the electric field just above"},{"Start":"15:53.640 ","End":"15:57.165","Text":"the infinitely charged plane and just below that infinitely charged plane."},{"Start":"15:57.165 ","End":"15:59.625","Text":"Then we\u0027ll get our Sigma_f."},{"Start":"15:59.625 ","End":"16:02.435","Text":"It\u0027s easy just via Gauss\u0027s law."},{"Start":"16:02.435 ","End":"16:08.040","Text":"This E naught is the E field when the dielectric material is not present."},{"Start":"16:08.040 ","End":"16:10.540","Text":"We\u0027re imagining that it\u0027s non-present and we\u0027re just measuring"},{"Start":"16:10.540 ","End":"16:13.365","Text":"the E field dye to Sigma free."},{"Start":"16:13.365 ","End":"16:17.010","Text":"Now we have to be really careful with this terminology"},{"Start":"16:17.010 ","End":"16:21.450","Text":"because we know that this infinitely charged plane has"},{"Start":"16:21.450 ","End":"16:26.490","Text":"some charge density Sigma free and this charge density can"},{"Start":"16:26.490 ","End":"16:32.715","Text":"be altered or changed slightly when we placed the dielectric material on top of it."},{"Start":"16:32.715 ","End":"16:37.250","Text":"When we\u0027re talking about the E field without the dielectric material,"},{"Start":"16:37.250 ","End":"16:43.530","Text":"we\u0027re imagining as if we pause time and we took away the dielectric material,"},{"Start":"16:43.530 ","End":"16:45.555","Text":"but that our Sigma free,"},{"Start":"16:45.555 ","End":"16:51.165","Text":"whatever it looks like when the dielectric material is still there, that doesn\u0027t change."},{"Start":"16:51.165 ","End":"16:54.010","Text":"Just we\u0027re imagining that the dielectric material isn\u0027t there,"},{"Start":"16:54.010 ","End":"17:00.835","Text":"but the Sigma free is acting exactly the same as if the dielectric material was there."},{"Start":"17:00.835 ","End":"17:05.174","Text":"Because in reality, if we take away the dielectric material,"},{"Start":"17:05.174 ","End":"17:10.175","Text":"the charge distribution over here on the infinitely charged plane will change."},{"Start":"17:10.175 ","End":"17:13.350","Text":"We\u0027re imagining that we\u0027re just measuring"},{"Start":"17:13.350 ","End":"17:19.775","Text":"the charge density Sigma free when the dielectric material is on top."},{"Start":"17:19.775 ","End":"17:22.585","Text":"Just we\u0027re taking our E field measurements,"},{"Start":"17:22.585 ","End":"17:25.210","Text":"not including the dielectric material"},{"Start":"17:25.210 ","End":"17:29.760","Text":"itself and the charge densities within the dielectric material."},{"Start":"17:29.760 ","End":"17:33.260","Text":"In short, it\u0027s to take the dielectric material off"},{"Start":"17:33.260 ","End":"17:37.750","Text":"the infinitely charged plane without changing our Sigma free."},{"Start":"17:37.750 ","End":"17:42.355","Text":"Then of course, in order to find our Sigma_b,"},{"Start":"17:42.355 ","End":"17:49.210","Text":"it\u0027s simply going to be equal to sigma t minus our sigma free."},{"Start":"17:49.210 ","End":"17:51.840","Text":"That\u0027s the end of this lesson."}],"ID":22311},{"Watched":false,"Name":"Exercise 1","Duration":"20m 46s","ChapterTopicVideoID":21302,"CourseChapterTopicPlaylistID":99475,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.205","Text":"Hello. In this question,"},{"Start":"00:02.205 ","End":"00:08.730","Text":"we have a conducting spherical shell with a charge on it of Q."},{"Start":"00:08.730 ","End":"00:11.730","Text":"We\u0027re told that the radius of the spherical shell is"},{"Start":"00:11.730 ","End":"00:14.955","Text":"this capital R. Around the spherical shell,"},{"Start":"00:14.955 ","End":"00:21.077","Text":"we have a dielectric material with the dielectric constant of Er,"},{"Start":"00:21.077 ","End":"00:24.285","Text":"so all of this where there\u0027s the yellow highlighter."},{"Start":"00:24.285 ","End":"00:29.280","Text":"Its thickness and the radius from the center of the sphere"},{"Start":"00:29.280 ","End":"00:35.105","Text":"until the outer edge of our dielectric material is a radius b,"},{"Start":"00:35.105 ","End":"00:39.755","Text":"which means that the thickness of the dielectric material is going to be b minus"},{"Start":"00:39.755 ","End":"00:46.655","Text":"R. The yellow highlighter is symbolizing that the dielectric material has width."},{"Start":"00:46.655 ","End":"00:50.795","Text":"We\u0027re being asked to find the electric field."},{"Start":"00:50.795 ","End":"00:54.235","Text":"Obviously, it\u0027s going to be in the radial direction."},{"Start":"00:54.235 ","End":"00:55.910","Text":"How are we going to do this?"},{"Start":"00:55.910 ","End":"01:01.535","Text":"First of all, we have to split our system over here into different regions."},{"Start":"01:01.535 ","End":"01:10.175","Text":"As we know, the electric field when we\u0027re located inside our spherical shell,"},{"Start":"01:10.175 ","End":"01:13.970","Text":"from what we know from Gauss\u0027s law our E field there is going to be equal to"},{"Start":"01:13.970 ","End":"01:18.710","Text":"0 because obviously we have no charge inside there."},{"Start":"01:18.710 ","End":"01:23.525","Text":"So that\u0027s done. Now, let\u0027s look when we\u0027re in the region between"},{"Start":"01:23.525 ","End":"01:29.630","Text":"our charged spherical shell and we\u0027re in the middle of our dielectric."},{"Start":"01:29.630 ","End":"01:34.430","Text":"Now we\u0027re located here,"},{"Start":"01:34.430 ","End":"01:37.595","Text":"somewhere in the region with a yellow highlighter."},{"Start":"01:37.595 ","End":"01:41.090","Text":"As we know, our E field due to"},{"Start":"01:41.090 ","End":"01:47.765","Text":"a charged spherical shell is going to be equal from Gauss\u0027s law,"},{"Start":"01:47.765 ","End":"01:56.615","Text":"kQ divided by r^2 but now what we want to do is we want to find our real electric fields."},{"Start":"01:56.615 ","End":"01:58.535","Text":"As we saw in previous lesson,"},{"Start":"01:58.535 ","End":"02:03.760","Text":"our real electric field comes from the electric field due to our spherical shell,"},{"Start":"02:03.760 ","End":"02:06.755","Text":"but it also comes from the electric field"},{"Start":"02:06.755 ","End":"02:10.395","Text":"due to the dipoles within the dielectric material."},{"Start":"02:10.395 ","End":"02:11.630","Text":"In our last lesson,"},{"Start":"02:11.630 ","End":"02:15.590","Text":"we saw that in order to find our real electric field,"},{"Start":"02:15.590 ","End":"02:19.310","Text":"that\u0027s going to be equal to our electric field due to"},{"Start":"02:19.310 ","End":"02:25.865","Text":"the charged spherical shell divided by the dielectric constant Epsilon Naught."},{"Start":"02:25.865 ","End":"02:34.595","Text":"We\u0027ll get that our real electric field is equal to kQ divided by Epsilon r, r^2."},{"Start":"02:34.595 ","End":"02:40.455","Text":"Of course, this is in the radial direction."},{"Start":"02:40.455 ","End":"02:46.175","Text":"Now, what happens when we\u0027re outside or so of the dielectric material?"},{"Start":"02:46.175 ","End":"02:49.285","Text":"We\u0027re located at r is bigger than b."},{"Start":"02:49.285 ","End":"02:57.695","Text":"Now we can see that we\u0027re only going to feel the electric field due to a Sigma free."},{"Start":"02:57.695 ","End":"03:04.170","Text":"That\u0027s the charge density on our conductor."},{"Start":"03:04.730 ","End":"03:08.600","Text":"This is because we know that overall,"},{"Start":"03:08.600 ","End":"03:11.959","Text":"even though when we\u0027re within our dielectric material,"},{"Start":"03:11.959 ","End":"03:15.080","Text":"there\u0027s going to be some kind of electric field due to the dipoles"},{"Start":"03:15.080 ","End":"03:18.530","Text":"lining up the way that they do but once we\u0027re outside,"},{"Start":"03:18.530 ","End":"03:21.455","Text":"we still know that the net charge of"},{"Start":"03:21.455 ","End":"03:26.335","Text":"all of this dielectric material is always going to be equal to 0."},{"Start":"03:26.335 ","End":"03:29.295","Text":"Let\u0027s even write that."},{"Start":"03:29.295 ","End":"03:35.495","Text":"Net Q is always equal to 0 when we\u0027re outside, when we\u0027re within."},{"Start":"03:35.495 ","End":"03:37.490","Text":"So we\u0027re going to have some kind of charge that will"},{"Start":"03:37.490 ","End":"03:40.250","Text":"feel or some electric field but outside,"},{"Start":"03:40.250 ","End":"03:44.810","Text":"our charge from the dielectric material is equal to 0."},{"Start":"03:44.810 ","End":"03:50.975","Text":"Because we have spherical symmetry once we\u0027re located outside of our spherical shell,"},{"Start":"03:50.975 ","End":"03:52.775","Text":"out of the dielectric material,"},{"Start":"03:52.775 ","End":"03:58.105","Text":"we\u0027re going to feel the electric field due to a spherical shell inside."},{"Start":"03:58.105 ","End":"04:01.055","Text":"Our E field is going to be equal to,"},{"Start":"04:01.055 ","End":"04:04.430","Text":"again kQ divided by r^2,"},{"Start":"04:04.430 ","End":"04:10.500","Text":"where this time r^2 is just located outside of our radius b and of course,"},{"Start":"04:10.500 ","End":"04:12.375","Text":"it\u0027s in the radial direction."},{"Start":"04:12.375 ","End":"04:16.205","Text":"When I\u0027m within my charged spherical shell,"},{"Start":"04:16.205 ","End":"04:20.345","Text":"the electric field I experience is equal to 0 because there\u0027s no QN."},{"Start":"04:20.345 ","End":"04:22.580","Text":"Just like we know from Gauss\u0027s law."},{"Start":"04:22.580 ","End":"04:25.460","Text":"Then when I\u0027m outside of my charged spherical shell,"},{"Start":"04:25.460 ","End":"04:28.760","Text":"but within my dielectric material due to the lining up of"},{"Start":"04:28.760 ","End":"04:32.545","Text":"the dipoles within the dielectric material,"},{"Start":"04:32.545 ","End":"04:37.040","Text":"I\u0027m going to get net field and net E field,"},{"Start":"04:37.040 ","End":"04:41.825","Text":"which is slightly smaller than the E field due to my charge spherical shell,"},{"Start":"04:41.825 ","End":"04:45.470","Text":"and that\u0027s given by this equation over here,"},{"Start":"04:45.470 ","End":"04:47.660","Text":"and so this will be my E field."},{"Start":"04:47.660 ","End":"04:48.830","Text":"Then when I\u0027m outside,"},{"Start":"04:48.830 ","End":"04:52.715","Text":"everything by E field is going to be again due to Gauss\u0027s law,"},{"Start":"04:52.715 ","End":"04:56.270","Text":"because the net charge of my dielectric material,"},{"Start":"04:56.270 ","End":"05:00.410","Text":"taking the whole material into account is always equal to 0."},{"Start":"05:00.410 ","End":"05:02.840","Text":"This will be my E field."},{"Start":"05:02.840 ","End":"05:05.795","Text":"Let\u0027s say that this was question Number 1,"},{"Start":"05:05.795 ","End":"05:08.030","Text":"and here\u0027s the answer to question Number 1."},{"Start":"05:08.030 ","End":"05:13.415","Text":"Now, our question Number 2 is what is our Sigma i equal to?"},{"Start":"05:13.415 ","End":"05:17.825","Text":"That\u0027s our induced charge density inside"},{"Start":"05:17.825 ","End":"05:22.970","Text":"our dielectric material due to this external or here internal,"},{"Start":"05:22.970 ","End":"05:27.230","Text":"but our external electric field due to this charge spherical shell."},{"Start":"05:27.230 ","End":"05:29.195","Text":"Now, in our previous lesson,"},{"Start":"05:29.195 ","End":"05:37.080","Text":"we saw that our Sigma i only appears on the surfaces of our dielectric."},{"Start":"05:37.080 ","End":"05:39.740","Text":"Here it\u0027s only going to appear on"},{"Start":"05:39.740 ","End":"05:44.945","Text":"this boundary between our charged spherical shell in our dielectric material,"},{"Start":"05:44.945 ","End":"05:53.685","Text":"and on the outer boundary between our dielectric material and just general space outside."},{"Start":"05:53.685 ","End":"05:59.210","Text":"Because inside all of our dipoles just cancel each other out with plus, minus, plus,"},{"Start":"05:59.210 ","End":"06:02.150","Text":"minus alternating and the only areas where there\u0027s"},{"Start":"06:02.150 ","End":"06:08.555","Text":"no negative charge to cancel out the positive charge is on here,"},{"Start":"06:08.555 ","End":"06:14.900","Text":"let\u0027s say the outer edge of our dielectric material and the only place where there is"},{"Start":"06:14.900 ","End":"06:18.410","Text":"no positive charge to cancel out our negative charge"},{"Start":"06:18.410 ","End":"06:23.800","Text":"inside the dielectric material is on our inner edge over here."},{"Start":"06:23.800 ","End":"06:27.740","Text":"Let\u0027s imagine that our Q here is a positive charge,"},{"Start":"06:27.740 ","End":"06:34.325","Text":"so that means that I\u0027m going to have lots of little pluses on the surface"},{"Start":"06:34.325 ","End":"06:42.015","Text":"of my charged conducting sphere and that means that in that case,"},{"Start":"06:42.015 ","End":"06:48.025","Text":"around this boundary, I\u0027m going to have in my dielectric or my minuses."},{"Start":"06:48.025 ","End":"06:54.650","Text":"Here I have all my little minuses which I attracted to this positive charge over here,"},{"Start":"06:54.650 ","End":"06:57.755","Text":"and then that means that on this side,"},{"Start":"06:57.755 ","End":"07:02.480","Text":"I\u0027ll have all of my pluses within the dielectric material."},{"Start":"07:02.480 ","End":"07:05.420","Text":"Due to the dipole, stacking up one on top of each"},{"Start":"07:05.420 ","End":"07:09.185","Text":"other so I\u0027ll have all my positive charges at the edge."},{"Start":"07:09.185 ","End":"07:16.492","Text":"Let\u0027s say that my induced charge on this edge is Sigma iR,"},{"Start":"07:16.492 ","End":"07:21.935","Text":"because it\u0027s located at a radius r around the brown circle."},{"Start":"07:21.935 ","End":"07:28.535","Text":"Then the Sigma i at the end of my dielectric material over here,"},{"Start":"07:28.535 ","End":"07:30.185","Text":"we\u0027ll call sigma ib."},{"Start":"07:30.185 ","End":"07:35.360","Text":"Now notice that our Sigma i at R is not going to be equal to our sigma"},{"Start":"07:35.360 ","End":"07:40.610","Text":"i at b because the total area is going to be different."},{"Start":"07:40.610 ","End":"07:44.600","Text":"Obviously, here we have a smaller surface area because"},{"Start":"07:44.600 ","End":"07:50.045","Text":"our radius is smaller and here we have a larger surface area,"},{"Start":"07:50.045 ","End":"07:54.545","Text":"so our Sigma I is going to be slightly less dense."},{"Start":"07:54.545 ","End":"07:58.850","Text":"However, the total is going to cancel out because"},{"Start":"07:58.850 ","End":"08:03.395","Text":"our net charge for everything has to be equal to 0."},{"Start":"08:03.395 ","End":"08:05.945","Text":"Soon we\u0027ll see how to deal with this."},{"Start":"08:05.945 ","End":"08:08.780","Text":"There are few ways to work out our Sigma i\u0027s,"},{"Start":"08:08.780 ","End":"08:10.550","Text":"but let\u0027s work with this way."},{"Start":"08:10.550 ","End":"08:13.130","Text":"As we know in our previous lesson,"},{"Start":"08:13.130 ","End":"08:17.975","Text":"we saw that our total Sigma due to our Sigma free and Sigma i"},{"Start":"08:17.975 ","End":"08:24.035","Text":"is going to be equal to Epsilon Naught multiplied by the jump in the electric field,"},{"Start":"08:24.035 ","End":"08:32.165","Text":"which is going to be equal to our Sigma i at radius R plus"},{"Start":"08:32.165 ","End":"08:40.925","Text":"our Sigma free due to the electron or the charges on our charged spherical shell."},{"Start":"08:40.925 ","End":"08:43.910","Text":"In order to find the jump in my electric field,"},{"Start":"08:43.910 ","End":"08:51.500","Text":"I\u0027m going to measure slightly above the radius of this charged spherical shell,"},{"Start":"08:51.500 ","End":"08:52.970","Text":"and slightly below it,"},{"Start":"08:52.970 ","End":"08:55.630","Text":"the radius of the child spherical shell."},{"Start":"08:55.630 ","End":"08:58.395","Text":"Let\u0027s write that out."},{"Start":"08:58.395 ","End":"09:04.715","Text":"My Sigma total is going to be equal to Epsilon Naught multiplied by"},{"Start":"09:04.715 ","End":"09:08.015","Text":"the electric field just above my R radius"},{"Start":"09:08.015 ","End":"09:13.302","Text":"minus the electric field just below my R radius."},{"Start":"09:13.302 ","End":"09:16.140","Text":"Now let\u0027s see what that is equal to."},{"Start":"09:16.140 ","End":"09:18.730","Text":"Just above my R radius,"},{"Start":"09:18.730 ","End":"09:22.545","Text":"the electric field is going to be in this region over here."},{"Start":"09:22.545 ","End":"09:28.915","Text":"All I have to do is I have to sub in my R radius into here."},{"Start":"09:28.915 ","End":"09:38.585","Text":"I\u0027ll have Epsilon Naught multiplied by KQ divided by Epsilon R capital R^2."},{"Start":"09:38.585 ","End":"09:41.200","Text":"Here I\u0027m in the region just above my R radius,"},{"Start":"09:41.200 ","End":"09:46.830","Text":"but my radius is still capital R. I substitute that in here."},{"Start":"09:46.830 ","End":"09:50.680","Text":"Then I\u0027m minusing the electric field when I\u0027m located"},{"Start":"09:50.680 ","End":"09:54.350","Text":"just below the R. I substitute in my capital R,"},{"Start":"09:54.350 ","End":"09:59.345","Text":"my R radius into the equation for the region here just below the R,"},{"Start":"09:59.345 ","End":"10:01.715","Text":"which as we can see as always equal to 0,"},{"Start":"10:01.715 ","End":"10:06.605","Text":"so minus 0, and this is my Sigma total."},{"Start":"10:06.605 ","End":"10:10.695","Text":"Now what I want to work out is what is my Sigma free?"},{"Start":"10:10.695 ","End":"10:13.940","Text":"My Sigma free is equal to,"},{"Start":"10:13.940 ","End":"10:16.565","Text":"again, the jump in the electric field."},{"Start":"10:16.565 ","End":"10:20.600","Text":"Epsilon Naught multiplied by the jump in my electric field,"},{"Start":"10:20.600 ","End":"10:22.275","Text":"but E Naught,"},{"Start":"10:22.275 ","End":"10:27.140","Text":"where this E Naught represents the jump in the electric field due to"},{"Start":"10:27.140 ","End":"10:31.340","Text":"my charged spherical shell and not taking into account"},{"Start":"10:31.340 ","End":"10:36.680","Text":"the jump in the electric field also to my dielectric material over here."},{"Start":"10:36.680 ","End":"10:40.485","Text":"Again, that\u0027s going to be equal to Epsilon Naught"},{"Start":"10:40.485 ","End":"10:45.150","Text":"multiplied by our original field at R,"},{"Start":"10:45.150 ","End":"10:49.055","Text":"plus so just above R radius minus the E,"},{"Start":"10:49.055 ","End":"10:54.270","Text":"the original E field just below our R radius."},{"Start":"10:54.270 ","End":"10:58.385","Text":"That is going to be equal to Epsilon Naught"},{"Start":"10:58.385 ","End":"11:02.490","Text":"multiplied by our original E-field at R plus,"},{"Start":"11:02.490 ","End":"11:10.390","Text":"so that\u0027s in this region and our E Naught we already wrote down is KQ divided by R^2,"},{"Start":"11:10.390 ","End":"11:15.570","Text":"where we just substitute in our capital R. KQ divided"},{"Start":"11:15.570 ","End":"11:20.850","Text":"by capital R^2 and now minus our original E-field at R minus,"},{"Start":"11:20.850 ","End":"11:23.495","Text":"which as we can see is this region here and it\u0027s always going"},{"Start":"11:23.495 ","End":"11:26.625","Text":"to be equal to 0 because there\u0027s no Q_in."},{"Start":"11:26.625 ","End":"11:35.505","Text":"Now our Sigma i is simply going to be equal to our Sigma total minus our Sigma free,"},{"Start":"11:35.505 ","End":"11:41.620","Text":"which is simply going to be equal to Epsilon Naught KQ divided"},{"Start":"11:41.620 ","End":"11:49.170","Text":"by R^2 multiplied by 1 divided by Epsilon r minus 1."},{"Start":"11:49.170 ","End":"11:54.705","Text":"Of course, this is the Sigma i at R. This here in blue."},{"Start":"11:54.705 ","End":"11:58.485","Text":"Now, what about Sigma i at B?"},{"Start":"11:58.485 ","End":"12:03.005","Text":"Our Sigma i at B is simply going to be equal to"},{"Start":"12:03.005 ","End":"12:07.835","Text":"our Sigma total but this is a different Sigma total to this."},{"Start":"12:07.835 ","End":"12:10.855","Text":"This was the Sigma total for this side of the question."},{"Start":"12:10.855 ","End":"12:14.190","Text":"Now, how come it\u0027s going to be equal to Sigma total?"},{"Start":"12:14.190 ","End":"12:20.905","Text":"Because our dielectric over here doesn\u0027t have any free electrons moving."},{"Start":"12:20.905 ","End":"12:24.095","Text":"There\u0027s not going to be a Sigma free."},{"Start":"12:24.095 ","End":"12:28.935","Text":"This is going to be equal to 0 because there\u0027s no free charges moving around."},{"Start":"12:28.935 ","End":"12:32.820","Text":"Because it\u0027s a dielectric material which has an insulator."},{"Start":"12:32.820 ","End":"12:35.630","Text":"Here we had Sigma free as well,"},{"Start":"12:35.630 ","End":"12:40.020","Text":"because we were on the boundary between an insulator,"},{"Start":"12:40.020 ","End":"12:43.660","Text":"which was this, and a conductor,"},{"Start":"12:43.660 ","End":"12:47.985","Text":"which was this over here but here we only have an insulator,"},{"Start":"12:47.985 ","End":"12:50.160","Text":"which means that we have no Sigma free."},{"Start":"12:50.160 ","End":"12:55.085","Text":"Let\u0027s work out our Sigma total and so that is going to be"},{"Start":"12:55.085 ","End":"13:00.440","Text":"simply equal to Epsilon Naught multiplied by the jump in the electric field."},{"Start":"13:00.440 ","End":"13:05.970","Text":"Our jump in the electric field is going to be equal to the electric field just"},{"Start":"13:05.970 ","End":"13:11.750","Text":"above our b radius minus the electric fields just below our b radius."},{"Start":"13:11.750 ","End":"13:13.810","Text":"Because now we\u0027re located here and we want to find"},{"Start":"13:13.810 ","End":"13:17.145","Text":"the electric field here and here at this point."},{"Start":"13:17.145 ","End":"13:19.350","Text":"Then we\u0027ll find it."},{"Start":"13:19.350 ","End":"13:23.850","Text":"This is simply going to be equal to"},{"Start":"13:23.850 ","End":"13:28.560","Text":"Epsilon Naught multiplied by in the region just above our b."},{"Start":"13:28.560 ","End":"13:32.045","Text":"That\u0027s the electric field over here, this region."},{"Start":"13:32.045 ","End":"13:37.550","Text":"We\u0027ll just substitute in b in place of R. We have KQ"},{"Start":"13:37.550 ","End":"13:43.610","Text":"divided by b^2 minus the electric field just below our b radius,"},{"Start":"13:43.610 ","End":"13:46.575","Text":"which is this region over here."},{"Start":"13:46.575 ","End":"13:50.470","Text":"We just substitute into this equation over here,"},{"Start":"13:50.470 ","End":"13:59.710","Text":"b instead of R. Minus KQ divided by Epsilon rb^2."},{"Start":"13:59.710 ","End":"14:01.835","Text":"This is our Sigma i."},{"Start":"14:01.835 ","End":"14:06.770","Text":"Used Sigma i bound charge at our radius b over here."},{"Start":"14:06.770 ","End":"14:10.005","Text":"Now a little point over here, our Sigma free."},{"Start":"14:10.005 ","End":"14:16.440","Text":"Here we got that it\u0027s equal to Epsilon Naught multiplied by KQ divided by R^2."},{"Start":"14:16.440 ","End":"14:22.535","Text":"Because we know that our k is equal to 1 divided by 4Pi Epsilon Naught."},{"Start":"14:22.535 ","End":"14:26.640","Text":"We could simply rearrange this and we\u0027d get that this is"},{"Start":"14:26.640 ","End":"14:33.630","Text":"equal to q divided by 4Pi R^2,"},{"Start":"14:33.630 ","End":"14:39.795","Text":"and that is simply the charge per surface area of our sphere."},{"Start":"14:39.795 ","End":"14:45.030","Text":"It\u0027s our charge Q divided by the total surface area of the spherical shell."},{"Start":"14:45.030 ","End":"14:49.790","Text":"That\u0027s exactly what we would have expected for our Sigma free, because it\u0027s just,"},{"Start":"14:49.790 ","End":"14:53.358","Text":"a Sigma free is in fact just as Sigma from what we learnt"},{"Start":"14:53.358 ","End":"14:57.900","Text":"earlier on in this topic or in this course,"},{"Start":"14:57.900 ","End":"15:01.425","Text":"which is simply the charge per unit area."},{"Start":"15:01.425 ","End":"15:03.990","Text":"That\u0027s exactly what we have over here."},{"Start":"15:03.990 ","End":"15:11.720","Text":"Now lastly, what if we were asked to find the potential in space?"},{"Start":"15:11.720 ","End":"15:14.315","Text":"Let\u0027s see what we would do."},{"Start":"15:14.315 ","End":"15:18.405","Text":"I\u0027m not going to solve it but if we\u0027re being asked to find the potential,"},{"Start":"15:18.405 ","End":"15:22.365","Text":"then refining the potential of the real field."},{"Start":"15:22.365 ","End":"15:29.115","Text":"What does that mean? That means that we have to integrate along our electric field."},{"Start":"15:29.115 ","End":"15:32.260","Text":"Here of course, our potential is going to be equal to 0 when"},{"Start":"15:32.260 ","End":"15:35.690","Text":"we\u0027re within our spherical shell."},{"Start":"15:35.690 ","End":"15:38.930","Text":"However, when we\u0027re outside of our spherical shell,"},{"Start":"15:38.930 ","End":"15:43.035","Text":"in the region between the spherical shell in our dielectric material, let\u0027s say."},{"Start":"15:43.035 ","End":"15:45.630","Text":"On which field are we going to be integrating?"},{"Start":"15:45.630 ","End":"15:47.805","Text":"Are we going to be integrating along E_0,"},{"Start":"15:47.805 ","End":"15:49.995","Text":"which is due to a spherical shell,"},{"Start":"15:49.995 ","End":"15:51.750","Text":"or along our real field."},{"Start":"15:51.750 ","End":"15:55.085","Text":"Which is taking into account the E field due to"},{"Start":"15:55.085 ","End":"15:58.850","Text":"a spherical shell and also due to our dielectric material."},{"Start":"15:58.850 ","End":"16:03.970","Text":"We\u0027re going to be integrating along this to find the potential our real field,"},{"Start":"16:03.970 ","End":"16:06.735","Text":"and that will find the real potential."},{"Start":"16:06.735 ","End":"16:09.825","Text":"Similarly, once we\u0027re outside of the dielectric,"},{"Start":"16:09.825 ","End":"16:13.650","Text":"we\u0027re also integrating along here are real field."},{"Start":"16:13.650 ","End":"16:20.040","Text":"Not the field due to the charged spherical shell."},{"Start":"16:20.040 ","End":"16:25.455","Text":"If we\u0027re dealing with a dielectric material in the question as well."},{"Start":"16:25.455 ","End":"16:27.810","Text":"That\u0027s really important to note."},{"Start":"16:27.810 ","End":"16:32.970","Text":"Now, let\u0027s give a slightly more intuitive explanation."},{"Start":"16:32.970 ","End":"16:37.300","Text":"Let\u0027s say I am putting a magnifying glass on this area"},{"Start":"16:37.300 ","End":"16:42.425","Text":"between the dielectric material and my charged spherical shell."},{"Start":"16:42.425 ","End":"16:44.880","Text":"Let\u0027s look at that over here."},{"Start":"16:44.880 ","End":"16:47.720","Text":"What we\u0027ll see is,"},{"Start":"16:47.720 ","End":"16:52.210","Text":"let\u0027s say my charged spherical shell is in this gray."},{"Start":"16:52.430 ","End":"16:55.505","Text":"This is my charged spherical shell."},{"Start":"16:55.505 ","End":"17:03.875","Text":"Then I have my dielectric material going around it, like so."},{"Start":"17:03.875 ","End":"17:07.170","Text":"There\u0027s always going to be some little space"},{"Start":"17:07.170 ","End":"17:10.085","Text":"between my charged spherical shell and my dielectric material."},{"Start":"17:10.085 ","End":"17:14.240","Text":"A little bit of air, something small."},{"Start":"17:14.240 ","End":"17:19.790","Text":"What I\u0027m doing is I can see that due to my charge spherical shell,"},{"Start":"17:19.790 ","End":"17:24.715","Text":"I have my Sigma free and due to my dielectric material,"},{"Start":"17:24.715 ","End":"17:27.830","Text":"I have my Sigma i over here."},{"Start":"17:27.830 ","End":"17:36.035","Text":"The first thing that I\u0027m going to do is in order to find my Sigma free,"},{"Start":"17:36.035 ","End":"17:39.825","Text":"I\u0027m going to measure the electric field here and here."},{"Start":"17:39.825 ","End":"17:43.175","Text":"I haven\u0027t yet gone into my dielectric material."},{"Start":"17:43.175 ","End":"17:52.015","Text":"I\u0027m measuring, sorry, the jump in the electric field just to my charged spherical shell."},{"Start":"17:52.015 ","End":"17:54.525","Text":"I\u0027m just doing this jump."},{"Start":"17:54.525 ","End":"17:58.040","Text":"Then when I\u0027m calculating my Sigma total,"},{"Start":"17:58.040 ","End":"18:03.510","Text":"then I\u0027m taking a measurement of my electric field right next to"},{"Start":"18:03.510 ","End":"18:07.775","Text":"my charged spherical shell and the jump in"},{"Start":"18:07.775 ","End":"18:13.234","Text":"the electric field when I\u0027m now inside my dielectric material."},{"Start":"18:13.234 ","End":"18:15.745","Text":"That\u0027s how I get my Sigma total."},{"Start":"18:15.745 ","End":"18:21.620","Text":"It\u0027s taking into account both the jump in the electric field due to jumping across"},{"Start":"18:21.620 ","End":"18:26.640","Text":"this charged surface and the jump in the electric field due to"},{"Start":"18:26.640 ","End":"18:32.280","Text":"jumping from there also into my dielectric material which has some E-field in it,"},{"Start":"18:32.280 ","End":"18:36.265","Text":"due to the lined up electric dipoles."},{"Start":"18:36.265 ","End":"18:45.015","Text":"I take the total jump and then I minus the mini jump from my charge surface."},{"Start":"18:45.015 ","End":"18:49.660","Text":"That\u0027s how I can find what my Sigma i is."},{"Start":"18:49.660 ","End":"18:53.715","Text":"If I jump from here to here,"},{"Start":"18:53.715 ","End":"19:00.365","Text":"I\u0027m finding my electric field just to my charged surface area,"},{"Start":"19:00.365 ","End":"19:03.085","Text":"and that means that I\u0027m going to get my Sigma free."},{"Start":"19:03.085 ","End":"19:06.860","Text":"My charge density on my charge surface area,"},{"Start":"19:06.860 ","End":"19:09.370","Text":"which is exactly what we can see over here."},{"Start":"19:09.370 ","End":"19:13.505","Text":"It\u0027s just the charge per unit area. We can see that."},{"Start":"19:13.505 ","End":"19:15.525","Text":"If I jump from here to here,"},{"Start":"19:15.525 ","End":"19:22.215","Text":"so from my charge surface area until I\u0027m inside my dielectric material."},{"Start":"19:22.215 ","End":"19:25.155","Text":"So what I\u0027m going to find is my real E-field,"},{"Start":"19:25.155 ","End":"19:29.970","Text":"and that\u0027s going to give me my Sigma total."},{"Start":"19:29.970 ","End":"19:38.820","Text":"That\u0027s the charge density due to both my charge surface area and my dielectric material."},{"Start":"19:38.820 ","End":"19:43.020","Text":"However, if I jump from just the space between"},{"Start":"19:43.020 ","End":"19:47.810","Text":"the charged surface area and my dielectric material into my dielectric material,"},{"Start":"19:47.810 ","End":"19:52.405","Text":"so what I\u0027m going to be doing is I\u0027m going to be"},{"Start":"19:52.405 ","End":"19:58.365","Text":"calculating the jump between these 2."},{"Start":"19:58.365 ","End":"20:03.765","Text":"It\u0027s going to be the jump between this over here and this over here."},{"Start":"20:03.765 ","End":"20:09.960","Text":"That jumping the electric field is given by this equation over here, here, and here."},{"Start":"20:09.960 ","End":"20:14.290","Text":"Because here I\u0027m at a radius of some R within"},{"Start":"20:14.290 ","End":"20:18.605","Text":"my dielectric material and here I met a capital"},{"Start":"20:18.605 ","End":"20:23.670","Text":"R. That is going to just give me my Sigma i."},{"Start":"20:23.670 ","End":"20:29.605","Text":"That\u0027s how we can find each one of these Sigmas separately by intuition,"},{"Start":"20:29.605 ","End":"20:32.025","Text":"between which points we\u0027re jumping,"},{"Start":"20:32.025 ","End":"20:34.935","Text":"in which electric fields jumps we\u0027re measuring."},{"Start":"20:34.935 ","End":"20:37.875","Text":"I hope that this explanation made it a little bit easier."},{"Start":"20:37.875 ","End":"20:43.785","Text":"If it didn\u0027t just do the tried and tested recipe for solving this type of question."},{"Start":"20:43.785 ","End":"20:46.660","Text":"That\u0027s the end of this lesson."}],"ID":21382},{"Watched":false,"Name":"Polarization Density","Duration":"14m 29s","ChapterTopicVideoID":21464,"CourseChapterTopicPlaylistID":99475,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.100","Text":"Hello. In this lesson,"},{"Start":"00:02.100 ","End":"00:05.550","Text":"we are going to be speaking about the polarization density."},{"Start":"00:05.550 ","End":"00:10.275","Text":"Other words for this are electric polarization or just polarization."},{"Start":"00:10.275 ","End":"00:16.110","Text":"It\u0027s denoted by the letter P and it\u0027s a vector field."},{"Start":"00:16.110 ","End":"00:18.180","Text":"Now in this lesson, we\u0027re going to be speaking"},{"Start":"00:18.180 ","End":"00:21.420","Text":"about this and seeing how we work with it."},{"Start":"00:21.420 ","End":"00:26.670","Text":"Let\u0027s imagine that here we have a polarized dielectric material."},{"Start":"00:26.670 ","End":"00:29.235","Text":"We\u0027re not going to deal with how it\u0027s become polarized."},{"Start":"00:29.235 ","End":"00:31.710","Text":"We just know that it is polarized."},{"Start":"00:31.710 ","End":"00:36.060","Text":"Now, there are some dielectric materials that can be polarized without"},{"Start":"00:36.060 ","End":"00:42.435","Text":"an external electric field and some are polarized via an external electric field."},{"Start":"00:42.435 ","End":"00:48.515","Text":"Here we have lots of dipoles that are pointing in the exact same direction."},{"Start":"00:48.515 ","End":"00:55.190","Text":"Which means that our dielectric material is going to have some net polarization."},{"Start":"00:55.190 ","End":"01:02.840","Text":"As we know, our arrows from our dipole go from our negative to our positive."},{"Start":"01:02.840 ","End":"01:10.145","Text":"These are represented by a small lowercase p. Let\u0027s call this p_1."},{"Start":"01:10.145 ","End":"01:11.855","Text":"Now, what is this p_1?"},{"Start":"01:11.855 ","End":"01:16.210","Text":"This is the dipole moment of this dipole over here."},{"Start":"01:16.210 ","End":"01:20.000","Text":"We can see that every single 1 of these dipoles and"},{"Start":"01:20.000 ","End":"01:24.170","Text":"the material is going to have a specific dipole moment."},{"Start":"01:24.170 ","End":"01:29.790","Text":"Dipole moments are all pointing from negative to positive."},{"Start":"01:30.830 ","End":"01:34.055","Text":"We have our lowercase p_1,"},{"Start":"01:34.055 ","End":"01:40.255","Text":"which is a single dipole moment within our dielectric material."},{"Start":"01:40.255 ","End":"01:43.400","Text":"Don\u0027t get confused between the lowercase p,"},{"Start":"01:43.400 ","End":"01:45.140","Text":"which is for a dipole moment,"},{"Start":"01:45.140 ","End":"01:55.110","Text":"and the capital P, which represents the polarization density."},{"Start":"01:55.110 ","End":"02:00.430","Text":"Lowercase p is given by the equation qd,"},{"Start":"02:00.430 ","End":"02:04.180","Text":"where q is the charge on the dipole."},{"Start":"02:04.180 ","End":"02:10.300","Text":"The absolute value of either the negative charge or the positive charge multiplied by d,"},{"Start":"02:10.300 ","End":"02:14.890","Text":"where d is the distance between the negative charge and the positive charge."},{"Start":"02:14.890 ","End":"02:18.280","Text":"Remember that the dipole moment can sometimes be represented"},{"Start":"02:18.280 ","End":"02:21.700","Text":"as a vector where our d is going to be"},{"Start":"02:21.700 ","End":"02:24.775","Text":"the size of the distance between the positive and negative charges"},{"Start":"02:24.775 ","End":"02:29.810","Text":"pointing in the direction from negative to positive."},{"Start":"02:30.260 ","End":"02:37.975","Text":"Now, let\u0027s define N. N is the number of dipoles in a given volume."},{"Start":"02:37.975 ","End":"02:41.945","Text":"The units for N is going to be per volume,"},{"Start":"02:41.945 ","End":"02:44.600","Text":"so 1 divided by meters cubed."},{"Start":"02:44.600 ","End":"02:52.430","Text":"That means if we look in some mini volume within our material,"},{"Start":"02:52.430 ","End":"02:56.285","Text":"so we can see how many dipoles,"},{"Start":"02:56.285 ","End":"03:00.770","Text":"so literally the numbers of dipoles within this volume."},{"Start":"03:00.770 ","End":"03:03.270","Text":"That\u0027s what N is."},{"Start":"03:03.710 ","End":"03:08.375","Text":"Now we can define our capital P vector,"},{"Start":"03:08.375 ","End":"03:12.440","Text":"which is our polarization density and it\u0027s equal to N,"},{"Start":"03:12.440 ","End":"03:18.650","Text":"the number of dipoles in a volume multiplied by our p_1 vector,"},{"Start":"03:18.650 ","End":"03:23.970","Text":"where p_1 vector is the dipole moment of a single dipole."},{"Start":"03:25.220 ","End":"03:32.170","Text":"As we can see, our polarization density is a vector,"},{"Start":"03:32.170 ","End":"03:38.705","Text":"and it\u0027s a vector which is dependent on its position in space."},{"Start":"03:38.705 ","End":"03:41.720","Text":"We can be within the same material,"},{"Start":"03:41.720 ","End":"03:45.965","Text":"the dielectric material and we can have the areas within the material where"},{"Start":"03:45.965 ","End":"03:48.530","Text":"our electric dipoles are pointing in"},{"Start":"03:48.530 ","End":"03:50.300","Text":"different directions or with"},{"Start":"03:50.300 ","End":"03:54.685","Text":"different magnitudes depending on where we are within the material."},{"Start":"03:54.685 ","End":"04:01.985","Text":"Polarization density is a vector as a function of its position in space."},{"Start":"04:01.985 ","End":"04:05.290","Text":"The units for P,"},{"Start":"04:05.290 ","End":"04:09.060","Text":"so we have the units of our p_1,"},{"Start":"04:09.060 ","End":"04:17.720","Text":"so that\u0027s going to be charge multiplied by meter divided by the units of our m,"},{"Start":"04:17.720 ","End":"04:24.710","Text":"which is 1 over m^3 so it\u0027s going to be divided by m^3."},{"Start":"04:24.710 ","End":"04:27.020","Text":"We can say that the units for"},{"Start":"04:27.020 ","End":"04:32.120","Text":"our polarization density is either dipole moment divided by meter cubed,"},{"Start":"04:32.120 ","End":"04:37.830","Text":"or the units for the dipole moment is cm divided by m^3."},{"Start":"04:38.420 ","End":"04:42.545","Text":"Now let\u0027s give a little bit of intuition."},{"Start":"04:42.545 ","End":"04:44.030","Text":"Earlier in the course,"},{"Start":"04:44.030 ","End":"04:51.545","Text":"we learned that the total charge that we\u0027ll have is equal to the integral of Rho dv,"},{"Start":"04:51.545 ","End":"04:57.510","Text":"where Rho is the charge density per unit volume."},{"Start":"04:58.040 ","End":"05:07.255","Text":"Here our polarization density denoted by capital P vector is also a type of density,"},{"Start":"05:07.255 ","End":"05:11.395","Text":"also given in the name polarization density and it\u0027s speaking about"},{"Start":"05:11.395 ","End":"05:19.010","Text":"how many dipole moments are in a given volume."},{"Start":"05:20.300 ","End":"05:23.790","Text":"Here we can see that our dp,"},{"Start":"05:23.790 ","End":"05:29.480","Text":"where this is a lowercase p so that means that we\u0027re speaking about our dipole moment,"},{"Start":"05:29.480 ","End":"05:34.165","Text":"is equal to capital P dV,"},{"Start":"05:34.165 ","End":"05:39.160","Text":"where capital P is our polarization density per unit volume."},{"Start":"05:39.160 ","End":"05:43.670","Text":"This is the dipole moment per unit volume."},{"Start":"05:43.670 ","End":"05:46.010","Text":"This is similar, okay,"},{"Start":"05:46.010 ","End":"05:48.830","Text":"if we\u0027re speaking about our q and our Rho dv."},{"Start":"05:48.830 ","End":"05:57.480","Text":"This is analogous to saying that our dq is equal to Rho dv."},{"Start":"05:58.310 ","End":"06:06.200","Text":"Our dq is how much charge is in per unit volume so our dp,"},{"Start":"06:06.200 ","End":"06:13.057","Text":"lowercase p is how many dipoles do I have in this unit of volume,"},{"Start":"06:13.057 ","End":"06:19.800","Text":"so how strong is this net vector field over here?"},{"Start":"06:19.940 ","End":"06:25.010","Text":"Then here, like Rho is talking about per unit volume,"},{"Start":"06:25.010 ","End":"06:27.127","Text":"so here, my capital P,"},{"Start":"06:27.127 ","End":"06:34.800","Text":"my polarization density is also speaking about the density of dipoles per unit volume."},{"Start":"06:36.080 ","End":"06:41.695","Text":"Now, if we want to find the total dipole moment in the material,"},{"Start":"06:41.695 ","End":"06:45.185","Text":"so the total dipole moment of all of this,"},{"Start":"06:45.185 ","End":"06:49.565","Text":"all we have to do is we have to integrate on our dp,"},{"Start":"06:49.565 ","End":"06:51.590","Text":"which means integrating on"},{"Start":"06:51.590 ","End":"06:57.690","Text":"our polarization density throughout the entire volume of the material."},{"Start":"06:58.340 ","End":"07:03.230","Text":"Now we can see that this looks rather similar to this over here."},{"Start":"07:03.230 ","End":"07:06.395","Text":"We can see that my polarization density,"},{"Start":"07:06.395 ","End":"07:08.375","Text":"our capital P vector,"},{"Start":"07:08.375 ","End":"07:16.330","Text":"is the density of electric dipoles per unit volume."},{"Start":"07:16.330 ","End":"07:20.050","Text":"Now this polarization density aside from being"},{"Start":"07:20.050 ","End":"07:24.370","Text":"useful in order to find the total dipole moment of the material,"},{"Start":"07:24.370 ","End":"07:32.390","Text":"it\u0027s also used in order to find our bound charge on the surface area of the material."},{"Start":"07:32.460 ","End":"07:35.605","Text":"As we learned a few lessons ago,"},{"Start":"07:35.605 ","End":"07:41.200","Text":"we have our bound charge on the surface of dielectric materials"},{"Start":"07:41.200 ","End":"07:47.620","Text":"which is often denoted Sigma_b for bound or Sigma_i for induced."},{"Start":"07:47.620 ","End":"07:57.655","Text":"Now this is equal to our polarization density dot product with our normal vector,"},{"Start":"07:57.655 ","End":"08:00.280","Text":"which is a unit vector pointing in"},{"Start":"08:00.280 ","End":"08:05.560","Text":"the perpendicular direction to the surface of our material."},{"Start":"08:05.560 ","End":"08:09.940","Text":"Now let\u0027s just remember what our Sigma_b or Sigma_i."},{"Start":"08:09.940 ","End":"08:15.110","Text":"This is also equal to Sigma_i, what this means."},{"Start":"08:16.380 ","End":"08:18.580","Text":"As we saw, we have"},{"Start":"08:18.580 ","End":"08:23.110","Text":"our dipole moments which is pointing from a negative to positive charge."},{"Start":"08:23.110 ","End":"08:28.330","Text":"Let\u0027s draw that in for everything."},{"Start":"08:28.330 ","End":"08:31.180","Text":"If negative, positive, negative, positive."},{"Start":"08:31.180 ","End":"08:32.500","Text":"Then on top of this,"},{"Start":"08:32.500 ","End":"08:40.000","Text":"we have some more arrows representing more dipole moments that"},{"Start":"08:40.000 ","End":"08:48.590","Text":"are stacked on top and they also have a negative and a positive in the same way."},{"Start":"08:51.540 ","End":"08:54.145","Text":"Now what we can see,"},{"Start":"08:54.145 ","End":"08:56.679","Text":"we spoke about this in a previous lesson,"},{"Start":"08:56.679 ","End":"09:01.255","Text":"is that when we stack all of these dipoles 1 on top of the other,"},{"Start":"09:01.255 ","End":"09:05.020","Text":"we can see that the inner charges here we have a plus"},{"Start":"09:05.020 ","End":"09:08.710","Text":"and then we have a minus on top and on and on and on until we get to the top."},{"Start":"09:08.710 ","End":"09:11.065","Text":"We will have negative, positive, negative, positive."},{"Start":"09:11.065 ","End":"09:15.430","Text":"Now the only areas where the positive and negative charges don\u0027t"},{"Start":"09:15.430 ","End":"09:20.320","Text":"balance each other out is right at the upper layer of the material."},{"Start":"09:20.320 ","End":"09:24.620","Text":"There\u0027s no above the upper dipole."},{"Start":"09:24.620 ","End":"09:27.375","Text":"There\u0027s no more dipoles above it."},{"Start":"09:27.375 ","End":"09:32.400","Text":"We can see that we\u0027re going to have some positive charge at the top."},{"Start":"09:32.400 ","End":"09:38.335","Text":"At the bottom, there\u0027s no more dipoles below the lower dipole."},{"Start":"09:38.335 ","End":"09:43.240","Text":"We can see that we\u0027re going to have a negative charge at the bottom."},{"Start":"09:43.240 ","End":"09:48.865","Text":"Now these charges are i Sigma_b, i Sigma bound."},{"Start":"09:48.865 ","End":"09:53.690","Text":"They\u0027re bound on the surface area of the material."},{"Start":"09:55.500 ","End":"09:58.210","Text":"We can see that our Sigma_b is"},{"Start":"09:58.210 ","End":"10:03.775","Text":"the bound charge and it\u0027s only on the surface of our material."},{"Start":"10:03.775 ","End":"10:05.710","Text":"Because as we can see in the middle,"},{"Start":"10:05.710 ","End":"10:10.720","Text":"we have our positive and negative charges which can cancel each other out."},{"Start":"10:10.720 ","End":"10:14.500","Text":"The only area where we don\u0027t have this canceling effect is"},{"Start":"10:14.500 ","End":"10:18.250","Text":"right at the bottom because there\u0027s no more dipoles below that have"},{"Start":"10:18.250 ","End":"10:19.720","Text":"a positive charge to cancel out"},{"Start":"10:19.720 ","End":"10:23.260","Text":"this negative charge and the layer on the top because there\u0027s"},{"Start":"10:23.260 ","End":"10:29.780","Text":"no more dipoles above that will have a negative charge to cancel out our positive charge."},{"Start":"10:30.930 ","End":"10:34.405","Text":"We can see that usually speaking,"},{"Start":"10:34.405 ","End":"10:39.370","Text":"we\u0027re not going to have any charge in the middle."},{"Start":"10:39.370 ","End":"10:44.575","Text":"Only on the surface areas of our material and not in the volume in-between."},{"Start":"10:44.575 ","End":"10:49.195","Text":"Now when will we get some charge in the volume in-between?"},{"Start":"10:49.195 ","End":"10:55.180","Text":"That will be if our dipole moments aren\u0027t uniform throughout the material."},{"Start":"10:55.180 ","End":"10:57.265","Text":"Let\u0027s say if we have"},{"Start":"10:57.265 ","End":"11:06.110","Text":"some different dipole distribution within the center of our dielectric material."},{"Start":"11:07.200 ","End":"11:14.830","Text":"Now we can see that our Rho_b is going to be the bound charge in the volume of"},{"Start":"11:14.830 ","End":"11:18.280","Text":"the dielectric material if there is some bound charge over"},{"Start":"11:18.280 ","End":"11:24.925","Text":"here if our plus and minus sections of our dipole moments don\u0027t cancel out."},{"Start":"11:24.925 ","End":"11:32.890","Text":"That is only when the dipole distribution is non-uniform."},{"Start":"11:32.890 ","End":"11:35.930","Text":"Then we\u0027ll get our Rho_b."},{"Start":"11:36.540 ","End":"11:40.855","Text":"Let\u0027s talk a little bit more about our Sigma_b."},{"Start":"11:40.855 ","End":"11:43.915","Text":"A bound charge on the surface is given by"},{"Start":"11:43.915 ","End":"11:49.090","Text":"our polarization density dot product with our normal vector."},{"Start":"11:49.090 ","End":"11:51.415","Text":"Our normal vector, as we said,"},{"Start":"11:51.415 ","End":"11:55.930","Text":"is going to always be in the direction outside of our material."},{"Start":"11:55.930 ","End":"11:59.230","Text":"If we\u0027re located on this surface,"},{"Start":"11:59.230 ","End":"12:01.990","Text":"our normal vector is going to be like this and this is"},{"Start":"12:01.990 ","End":"12:06.835","Text":"a unit vector and it\u0027s always perpendicular to the surface at that point."},{"Start":"12:06.835 ","End":"12:11.920","Text":"There will be a 90 degree angle between our normal vector and the surface."},{"Start":"12:11.920 ","End":"12:13.435","Text":"If we\u0027re looking over here,"},{"Start":"12:13.435 ","End":"12:15.610","Text":"our normal vector will be in this direction."},{"Start":"12:15.610 ","End":"12:17.140","Text":"If we\u0027re looking over here,"},{"Start":"12:17.140 ","End":"12:19.540","Text":"our normal vector will be in this direction."},{"Start":"12:19.540 ","End":"12:24.680","Text":"Over here, our normal vector will be in this direction."},{"Start":"12:25.320 ","End":"12:27.505","Text":"This is super easy."},{"Start":"12:27.505 ","End":"12:31.505","Text":"All I do is I get my polarization density function."},{"Start":"12:31.505 ","End":"12:36.180","Text":"Let\u0027s imagine that it\u0027s given to me and then I use the dot product"},{"Start":"12:36.180 ","End":"12:40.590","Text":"to multiply it with my normal vector and that\u0027s it,"},{"Start":"12:40.590 ","End":"12:43.270","Text":"that\u0027s my Sigma bound."},{"Start":"12:43.860 ","End":"12:48.580","Text":"Now what about our bound charge in the volume?"},{"Start":"12:48.580 ","End":"12:53.875","Text":"In order to get our bound charge, this is Rho,"},{"Start":"12:53.875 ","End":"12:57.279","Text":"we\u0027re going to have that this is equal to negative"},{"Start":"12:57.279 ","End":"13:01.705","Text":"our divergence of our polarization density."},{"Start":"13:01.705 ","End":"13:09.205","Text":"Let\u0027s imagine that our polarization density is given in Cartesian coordinates."},{"Start":"13:09.205 ","End":"13:12.460","Text":"The divergence of the Cartesian"},{"Start":"13:12.460 ","End":"13:16.660","Text":"coordinates is equal to negative from over here and then we\u0027ll"},{"Start":"13:16.660 ","End":"13:24.805","Text":"have d P_x by d_x plus d"},{"Start":"13:24.805 ","End":"13:33.790","Text":"P_y by d_y plus d P_z by d_z,"},{"Start":"13:33.790 ","End":"13:41.360","Text":"because this is a vector quantity and then we\u0027ll get our bound charge within our volume."},{"Start":"13:42.360 ","End":"13:46.825","Text":"Now we\u0027ve seen what the polarization density is."},{"Start":"13:46.825 ","End":"13:51.295","Text":"These 4 equations are important to note."},{"Start":"13:51.295 ","End":"13:55.540","Text":"This is what the polarization density is actually equal to,"},{"Start":"13:55.540 ","End":"13:58.750","Text":"this is how to work out the total dipole moment"},{"Start":"13:58.750 ","End":"14:02.185","Text":"in the material if we know the polarization density,"},{"Start":"14:02.185 ","End":"14:06.550","Text":"and of course the equations in order to know the bounds charges on"},{"Start":"14:06.550 ","End":"14:11.290","Text":"the surface and in the volume of the dielectric material."},{"Start":"14:11.290 ","End":"14:12.580","Text":"In future lessons,"},{"Start":"14:12.580 ","End":"14:17.230","Text":"we\u0027re going to learn how to use these equations and also how to get from"},{"Start":"14:17.230 ","End":"14:22.914","Text":"our polarization density our equation for the field produced by that."},{"Start":"14:22.914 ","End":"14:25.855","Text":"But that\u0027s going to be in future lessons."},{"Start":"14:25.855 ","End":"14:28.730","Text":"That\u0027s the end of this lesson."}],"ID":22312},{"Watched":false,"Name":"Exercise 2","Duration":"11m 19s","ChapterTopicVideoID":21465,"CourseChapterTopicPlaylistID":99475,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.740","Text":"Hello. In this question,"},{"Start":"00:01.740 ","End":"00:06.420","Text":"we have a box of the area A and width d. We\u0027re"},{"Start":"00:06.420 ","End":"00:12.000","Text":"being told that it\u0027s polarized with the polarization density of this."},{"Start":"00:12.000 ","End":"00:16.050","Text":"We\u0027re told that the origin of the box is in the center of the box."},{"Start":"00:16.050 ","End":"00:22.545","Text":"Now, Question 1 is to find the bound surface charge as well as the bound volume charge."},{"Start":"00:22.545 ","End":"00:25.470","Text":"The first thing that we have to do is we have to"},{"Start":"00:25.470 ","End":"00:29.880","Text":"realize that because our polarization density is this variable z,"},{"Start":"00:29.880 ","End":"00:33.630","Text":"and we\u0027re being told that the origin is at the center of the box,"},{"Start":"00:33.630 ","End":"00:37.845","Text":"we have to measure our z relative to the origin."},{"Start":"00:37.845 ","End":"00:40.760","Text":"This is important because if we don\u0027t do this right,"},{"Start":"00:40.760 ","End":"00:43.620","Text":"then our answer will be wrong."},{"Start":"00:43.620 ","End":"00:52.390","Text":"The upper side of the box is located at z=d over 2,"},{"Start":"00:52.390 ","End":"01:00.960","Text":"and the lower side of the box is located at z=negative d over 2."},{"Start":"01:02.270 ","End":"01:05.600","Text":"Now we can begin by answering our question."},{"Start":"01:05.600 ","End":"01:09.530","Text":"In order to find the bound surface charge at Sigma b,"},{"Start":"01:09.530 ","End":"01:16.435","Text":"and we know that that is equal to our polarization density dot our n-hat."},{"Start":"01:16.435 ","End":"01:18.980","Text":"Now what we\u0027re trying to do is,"},{"Start":"01:18.980 ","End":"01:27.900","Text":"let\u0027s work out the Sigma b at this side of the box."},{"Start":"01:28.970 ","End":"01:33.430","Text":"First of all, we can see that our normal vector,"},{"Start":"01:33.430 ","End":"01:40.205","Text":"our n-hat is simply pointing in this direction perpendicular to the surface."},{"Start":"01:40.205 ","End":"01:44.160","Text":"Here we can see that that is equal to z-hat."},{"Start":"01:45.890 ","End":"01:51.860","Text":"Our polarization density we\u0027re trying to find on the side of the box,"},{"Start":"01:51.860 ","End":"01:55.555","Text":"which is located at z=d over 2."},{"Start":"01:55.555 ","End":"01:57.960","Text":"We\u0027ll substitute that in."},{"Start":"01:57.960 ","End":"02:00.845","Text":"Then that\u0027s going to be a dot-product with our n-hat,"},{"Start":"02:00.845 ","End":"02:03.390","Text":"which we said is z-hat."},{"Start":"02:03.390 ","End":"02:05.040","Text":"Let\u0027s plug this in."},{"Start":"02:05.040 ","End":"02:10.800","Text":"That\u0027s going to be equal to P_naught multiplied by z divided by d,"},{"Start":"02:10.800 ","End":"02:17.459","Text":"so our z is d divided by 2 divided by d in the z-hat direction,"},{"Start":"02:17.459 ","End":"02:19.950","Text":"dot-product with our z-hat."},{"Start":"02:19.950 ","End":"02:23.100","Text":"First of all, we can see that our d\u0027s are going to cancel out,"},{"Start":"02:23.100 ","End":"02:26.775","Text":"and dot z-hat is simply equal to 1."},{"Start":"02:26.775 ","End":"02:33.510","Text":"We get that our final answer is equal to P_naught divided by 2."},{"Start":"02:34.300 ","End":"02:37.580","Text":"Now we can see that on this upper side,"},{"Start":"02:37.580 ","End":"02:43.470","Text":"I have lots of positive charges all along over here,"},{"Start":"02:43.470 ","End":"02:50.530","Text":"and the bound charge is equal to P_naught divided by 2."},{"Start":"02:50.530 ","End":"02:55.400","Text":"Now, let\u0027s see what we get for the bottom side."},{"Start":"02:55.400 ","End":"02:58.520","Text":"Here we\u0027re going to have lots of negative charges,"},{"Start":"02:58.520 ","End":"03:03.465","Text":"therefore, on our bottom side. Let\u0027s do that."},{"Start":"03:03.465 ","End":"03:10.950","Text":"Our Sigma b is equal to our p or polarization density."},{"Start":"03:10.950 ","End":"03:16.220","Text":"But this time, our z is equal to negative d divided by 2,"},{"Start":"03:16.220 ","End":"03:18.340","Text":"because we want our bottom edge."},{"Start":"03:18.340 ","End":"03:21.960","Text":"Then dot-product with our n-hat."},{"Start":"03:21.960 ","End":"03:24.540","Text":"Now, in which direction is our n-hats?"},{"Start":"03:24.540 ","End":"03:29.645","Text":"It\u0027s always going to be the unit vector perpendicular to the surface area,"},{"Start":"03:29.645 ","End":"03:31.835","Text":"and always coming out of the shape."},{"Start":"03:31.835 ","End":"03:33.770","Text":"This is now our n-hat."},{"Start":"03:33.770 ","End":"03:40.110","Text":"We can see that our n-hat is now in the negative z-hat direction."},{"Start":"03:40.280 ","End":"03:42.855","Text":"Now let\u0027s plug that in."},{"Start":"03:42.855 ","End":"03:46.970","Text":"It\u0027s pointing in the opposite direction to here in the negative side direction."},{"Start":"03:46.970 ","End":"03:51.530","Text":"This is going to be equal to P_naught multiplied by"},{"Start":"03:51.530 ","End":"03:58.565","Text":"negative d over 2 divided by d in the z-hat direction,"},{"Start":"03:58.565 ","End":"04:00.980","Text":"dot-product with our n-hat,"},{"Start":"04:00.980 ","End":"04:04.975","Text":"which is equal to negative z-hat."},{"Start":"04:04.975 ","End":"04:11.194","Text":"Now we can see that this negative and this negative cancel out to become positives."},{"Start":"04:11.194 ","End":"04:13.760","Text":"This d and this d cancels out,"},{"Start":"04:13.760 ","End":"04:17.865","Text":"and now z-hat.z-hat is equal to 1."},{"Start":"04:17.865 ","End":"04:20.925","Text":"Again, we\u0027ll get the same answer of P_naught"},{"Start":"04:20.925 ","End":"04:25.175","Text":"divided by 2 for our bound charge at the bottom side of the box."},{"Start":"04:25.175 ","End":"04:27.195","Text":"Now, wait one moment."},{"Start":"04:27.195 ","End":"04:30.480","Text":"We can see that our Sigma b is here, again,"},{"Start":"04:30.480 ","End":"04:34.340","Text":"equal to P_naught divided by 2, which is a positive."},{"Start":"04:34.340 ","End":"04:36.065","Text":"It\u0027s just like we had over here,"},{"Start":"04:36.065 ","End":"04:42.205","Text":"which means that these charges over here are also positive charges."},{"Start":"04:42.205 ","End":"04:45.155","Text":"At the beginning, we thought that they were"},{"Start":"04:45.155 ","End":"04:49.300","Text":"negative charge accumulation at the bottom over here."},{"Start":"04:49.300 ","End":"04:54.890","Text":"But now we can see that because we got the exact same answer as we did over here,"},{"Start":"04:54.890 ","End":"04:58.610","Text":"we can see that the charge accumulation at"},{"Start":"04:58.610 ","End":"05:03.050","Text":"the top side and at the bottom side are both positive."},{"Start":"05:03.050 ","End":"05:05.645","Text":"That\u0027s something to look out for."},{"Start":"05:05.645 ","End":"05:11.580","Text":"This also comes from the fact that our origin is located where it is."},{"Start":"05:13.370 ","End":"05:18.580","Text":"Now let\u0027s see what our Sigma b is on the sides."},{"Start":"05:18.580 ","End":"05:22.725","Text":"Now, we can see that our normal vector,"},{"Start":"05:22.725 ","End":"05:27.295","Text":"so our n-hat in the side is going to be in the,"},{"Start":"05:27.295 ","End":"05:28.540","Text":"depending where we\u0027re pointing,"},{"Start":"05:28.540 ","End":"05:33.355","Text":"in either the y-hat direction or the x-hat direction."},{"Start":"05:33.355 ","End":"05:37.150","Text":"When we do x-hat or y-hat.z-hat,"},{"Start":"05:37.150 ","End":"05:39.265","Text":"it\u0027s going to be equal to 0."},{"Start":"05:39.265 ","End":"05:44.020","Text":"We can see that we\u0027ll have no Sigma b on the sides which"},{"Start":"05:44.020 ","End":"05:50.155","Text":"are parallel to the direction of our polarization density."},{"Start":"05:50.155 ","End":"05:55.650","Text":"Only on the sides which are perpendicular,"},{"Start":"05:55.650 ","End":"05:59.175","Text":"which is just the top and the bottom."},{"Start":"05:59.175 ","End":"06:03.135","Text":"Now we can work out our Rho_b,"},{"Start":"06:03.135 ","End":"06:08.675","Text":"so our bound charge inside the internal volume."},{"Start":"06:08.675 ","End":"06:12.560","Text":"As we know, this equation is equal to negative"},{"Start":"06:12.560 ","End":"06:18.230","Text":"the divergence of our polarization density."},{"Start":"06:18.230 ","End":"06:22.475","Text":"As we can see, we\u0027re working with Cartesian coordinates."},{"Start":"06:22.475 ","End":"06:26.820","Text":"That means that we have to write dP_x by dx,"},{"Start":"06:26.820 ","End":"06:29.615","Text":"of course, a negative multiplies everything here,"},{"Start":"06:29.615 ","End":"06:33.350","Text":"plus dP_y by dy"},{"Start":"06:33.350 ","End":"06:41.070","Text":"plus dP_z by P_z."},{"Start":"06:41.070 ","End":"06:47.840","Text":"As we can see, we don\u0027t have any x or y components in our polarization density vector."},{"Start":"06:47.840 ","End":"06:50.525","Text":"That means that this is equal to 0,"},{"Start":"06:50.525 ","End":"06:52.450","Text":"and that this is equal to 0."},{"Start":"06:52.450 ","End":"07:00.010","Text":"All we have to do is have negative dP_z by dz,"},{"Start":"07:00.010 ","End":"07:10.125","Text":"which is simply going to be equal to negative P_0 divided by d. This is our answer."},{"Start":"07:10.125 ","End":"07:13.265","Text":"Because we can see that we have a negative sign over here,"},{"Start":"07:13.265 ","End":"07:18.380","Text":"which is different to what we saw with our Sigma bound,"},{"Start":"07:18.380 ","End":"07:23.000","Text":"so that means that inside the volume of our polarized box,"},{"Start":"07:23.000 ","End":"07:28.880","Text":"we\u0027re going to have some uniform charge,"},{"Start":"07:28.880 ","End":"07:34.400","Text":"which is negative because we have a negative sign over here."},{"Start":"07:34.400 ","End":"07:41.760","Text":"We have some uniform negative charge inside the volume of our box."},{"Start":"07:43.220 ","End":"07:48.220","Text":"We worked out the bound surface and volume charges."},{"Start":"07:48.220 ","End":"07:49.845","Text":"That was Question 1."},{"Start":"07:49.845 ","End":"07:52.230","Text":"Now let\u0027s look at question number 2."},{"Start":"07:52.230 ","End":"07:54.596","Text":"In question number 2, we\u0027re being asked,"},{"Start":"07:54.596 ","End":"08:00.960","Text":"what is the total bound charge in the entire box?"},{"Start":"08:02.540 ","End":"08:07.450","Text":"In general, the total bound charge is we have to sum"},{"Start":"08:07.450 ","End":"08:12.345","Text":"up on all of our bound charges multiplied by,"},{"Start":"08:12.345 ","End":"08:14.520","Text":"if they\u0027re Sigma, multiplied by a,"},{"Start":"08:14.520 ","End":"08:15.645","Text":"and if they\u0027re Rho,"},{"Start":"08:15.645 ","End":"08:20.605","Text":"then multiplied by v. Let\u0027s write this out in a mathematical way."},{"Start":"08:20.605 ","End":"08:27.855","Text":"We sum up all of our Sigma b multiplied by their a,"},{"Start":"08:27.855 ","End":"08:32.265","Text":"where i is equal from 1 until n,"},{"Start":"08:32.265 ","End":"08:34.890","Text":"along all of the sides of the shape."},{"Start":"08:34.890 ","End":"08:40.965","Text":"Then we add up from i is equal to 1 until n,"},{"Start":"08:40.965 ","End":"08:43.620","Text":"our bound charges,"},{"Start":"08:43.620 ","End":"08:49.035","Text":"and the volume multiplied by the volume."},{"Start":"08:49.035 ","End":"08:54.095","Text":"Here, specifically we only have 1 volume which we\u0027re talking about,"},{"Start":"08:54.095 ","End":"08:57.600","Text":"which is the total volume inside this box."},{"Start":"08:57.890 ","End":"09:00.645","Text":"Let\u0027s see what that is equal to."},{"Start":"09:00.645 ","End":"09:06.305","Text":"That means that we\u0027ll have our Sigma b for the top side of the box,"},{"Start":"09:06.305 ","End":"09:10.220","Text":"which was equal to P_naught divided by 2"},{"Start":"09:10.220 ","End":"09:15.500","Text":"multiplied by the area of that section, which is a."},{"Start":"09:15.500 ","End":"09:21.485","Text":"Then we\u0027ll add on our Sigma b for the bottom side of the box,"},{"Start":"09:21.485 ","End":"09:26.195","Text":"which we saw is also equal to P_0 divided by 2, and again,"},{"Start":"09:26.195 ","End":"09:30.320","Text":"multiplied by the area of the bottom side of the box,"},{"Start":"09:30.320 ","End":"09:33.005","Text":"which is also equal to a."},{"Start":"09:33.005 ","End":"09:37.050","Text":"Now we\u0027ll add on our Rho_b."},{"Start":"09:37.050 ","End":"09:42.210","Text":"We saw that our Rho_b is equal to negative P_0 divided by"},{"Start":"09:42.210 ","End":"09:47.180","Text":"d multiplied by the volume. What is the volume?"},{"Start":"09:47.180 ","End":"09:49.190","Text":"We have our surface area,"},{"Start":"09:49.190 ","End":"09:54.800","Text":"which is A multiplied by the depth or the width of the box,"},{"Start":"09:54.800 ","End":"09:59.250","Text":"which is d, and that is our volume."},{"Start":"10:02.540 ","End":"10:05.895","Text":"Let\u0027s see what our Q_b is equal to."},{"Start":"10:05.895 ","End":"10:09.015","Text":"We can see that this d and this d cancel out."},{"Start":"10:09.015 ","End":"10:14.070","Text":"Then we have P_naught over 2 multiplied by a plus P_naught over"},{"Start":"10:14.070 ","End":"10:19.515","Text":"2 multiplied by a minus P_naught multiplied by a."},{"Start":"10:19.515 ","End":"10:22.750","Text":"We\u0027re going to get that the total bound charge in"},{"Start":"10:22.750 ","End":"10:27.265","Text":"this entire box is equal to 0. Why is that?"},{"Start":"10:27.265 ","End":"10:31.645","Text":"Because we know that our box is polarized,"},{"Start":"10:31.645 ","End":"10:34.686","Text":"which means that it has lots of dipoles inside,"},{"Start":"10:34.686 ","End":"10:39.280","Text":"we can see that we\u0027re going to have just as many positive charges as we have"},{"Start":"10:39.280 ","End":"10:45.805","Text":"negative charges because each dipole is made out of a positive and a negative charge."},{"Start":"10:45.805 ","End":"10:48.515","Text":"All of these positive and negative charges,"},{"Start":"10:48.515 ","End":"10:52.930","Text":"when we look at them in total to see the total charge of our box,"},{"Start":"10:52.930 ","End":"10:55.825","Text":"we\u0027re going to cancel them out or cancel each other"},{"Start":"10:55.825 ","End":"10:59.780","Text":"out because each dipole has a positive and negative charge."},{"Start":"10:59.780 ","End":"11:03.800","Text":"In total, we have positive and negative cancel out,"},{"Start":"11:03.800 ","End":"11:07.970","Text":"and so on and so forth for all of the dipoles within the box."},{"Start":"11:07.970 ","End":"11:11.510","Text":"I don\u0027t know if that is a surprising answer or if you"},{"Start":"11:11.510 ","End":"11:15.275","Text":"could have predicted it because we have a box with lots of dipoles,"},{"Start":"11:15.275 ","End":"11:17.570","Text":"but there you have it."},{"Start":"11:17.570 ","End":"11:20.490","Text":"That\u0027s the end of this lesson."}],"ID":22313},{"Watched":false,"Name":"Deriving Bound Charge Density Part 1","Duration":"12m 42s","ChapterTopicVideoID":21466,"CourseChapterTopicPlaylistID":99475,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.380 ","End":"00:05.055","Text":"Hello. This is the first of 2 lessons dealing with"},{"Start":"00:05.055 ","End":"00:10.095","Text":"deriving the equation for the bound charge density."},{"Start":"00:10.095 ","End":"00:12.135","Text":"First of all, in this lesson,"},{"Start":"00:12.135 ","End":"00:15.165","Text":"we\u0027re going to see that when we take"},{"Start":"00:15.165 ","End":"00:21.300","Text":"the derivative or the gradient of the function 1 divided by"},{"Start":"00:21.300 ","End":"00:26.745","Text":"r its equal to r divided by r^3 or"},{"Start":"00:26.745 ","End":"00:32.865","Text":"r hat divided by r^2."},{"Start":"00:32.865 ","End":"00:39.685","Text":"Let\u0027s imagine that we have this point over here where we want to calculate the potential."},{"Start":"00:39.685 ","End":"00:43.790","Text":"Over here, we have some kind of material that has"},{"Start":"00:43.790 ","End":"00:48.995","Text":"some kind of charge density per unit volume Rho."},{"Start":"00:48.995 ","End":"00:52.085","Text":"What we can see is that we have"},{"Start":"00:52.085 ","End":"00:58.475","Text":"this vector r from the origin pointing to where we want to measure the potential,"},{"Start":"00:58.475 ","End":"01:03.800","Text":"and then we have this vector r tag which is pointing from"},{"Start":"01:03.800 ","End":"01:09.865","Text":"the origin to some piece or unit volume in this material,"},{"Start":"01:09.865 ","End":"01:12.525","Text":"which has some charge density,"},{"Start":"01:12.525 ","End":"01:18.725","Text":"and it\u0027s this piece in this material which is causing a potential over here."},{"Start":"01:18.725 ","End":"01:23.935","Text":"How do we describe the vector from here,"},{"Start":"01:23.935 ","End":"01:28.735","Text":"from this piece causing a potential to the place where we\u0027re measuring the potential?"},{"Start":"01:28.735 ","End":"01:30.865","Text":"That vector is called,"},{"Start":"01:30.865 ","End":"01:37.750","Text":"or we\u0027re defining it as r tilde vector and it is simply equal"},{"Start":"01:37.750 ","End":"01:45.650","Text":"to our r vector minus our r tag vector."},{"Start":"01:45.710 ","End":"01:48.580","Text":"What is this equal to?"},{"Start":"01:48.580 ","End":"01:53.710","Text":"If we say that our r vector is equal to, okay,"},{"Start":"01:53.710 ","End":"01:57.430","Text":"so from the origin to this point where we\u0027re measuring the potential,"},{"Start":"01:57.430 ","End":"02:00.175","Text":"so let\u0027s call that point x, y,"},{"Start":"02:00.175 ","End":"02:05.020","Text":"z and then our r tag vector, like so."},{"Start":"02:05.020 ","End":"02:07.205","Text":"Because it\u0027s a tag,"},{"Start":"02:07.205 ","End":"02:10.365","Text":"let\u0027s call it that it\u0027s at this place,"},{"Start":"02:10.365 ","End":"02:14.345","Text":"x tag, y tag, z tag."},{"Start":"02:14.345 ","End":"02:19.100","Text":"What we can see is in order to see what the potential is over here,"},{"Start":"02:19.100 ","End":"02:21.040","Text":"our r vector is constant,"},{"Start":"02:21.040 ","End":"02:25.415","Text":"it\u0027s always pointing from the origin to the same point in space."},{"Start":"02:25.415 ","End":"02:29.960","Text":"However, our r tag vector is changing because we\u0027re going to"},{"Start":"02:29.960 ","End":"02:34.835","Text":"sum up along all the unit volumes in this material,"},{"Start":"02:34.835 ","End":"02:37.325","Text":"as we can see over here."},{"Start":"02:37.325 ","End":"02:41.165","Text":"This is always going to be moving around,"},{"Start":"02:41.165 ","End":"02:43.730","Text":"so that means that our r tilde vector,"},{"Start":"02:43.730 ","End":"02:47.195","Text":"which is from this point over here to here,"},{"Start":"02:47.195 ","End":"02:49.580","Text":"so it\u0027s also going to be moving"},{"Start":"02:49.580 ","End":"02:53.855","Text":"around because if we\u0027re measuring a unit volume over here,"},{"Start":"02:53.855 ","End":"02:57.440","Text":"so our r tilde vector will be a different vector"},{"Start":"02:57.440 ","End":"03:02.495","Text":"and we can see that our r tilde vector as a function of our r tag vector."},{"Start":"03:02.495 ","End":"03:05.270","Text":"What is our r tilde vector equal to?"},{"Start":"03:05.270 ","End":"03:09.015","Text":"It\u0027s equal to x minus x tag,"},{"Start":"03:09.015 ","End":"03:12.675","Text":"y minus y tag,"},{"Start":"03:12.675 ","End":"03:16.540","Text":"and z minus z tag."},{"Start":"03:16.540 ","End":"03:20.240","Text":"Let\u0027s take a look at this piece of unit volume over"},{"Start":"03:20.240 ","End":"03:25.305","Text":"here and we know that it has a charge and what is its charge?"},{"Start":"03:25.305 ","End":"03:27.165","Text":"Its charge is dq,"},{"Start":"03:27.165 ","End":"03:36.100","Text":"and dq is equal to its charge density per unit volume Rho multiplied by its volume dv."},{"Start":"03:36.100 ","End":"03:45.710","Text":"Then we can say that the potential over here from this single piece over here,"},{"Start":"03:45.710 ","End":"03:55.295","Text":"so we can call this d Phi is equal to kdq divided by r magnitude."},{"Start":"03:55.295 ","End":"03:57.050","Text":"What is this r magnitude?"},{"Start":"03:57.050 ","End":"04:00.890","Text":"This is the distance from my piece to our potential,"},{"Start":"04:00.890 ","End":"04:06.390","Text":"which is this vector over here that we\u0027ve already defined, which is r tilde."},{"Start":"04:08.410 ","End":"04:14.330","Text":"What we have is that this is equal to kdq,"},{"Start":"04:14.330 ","End":"04:17.390","Text":"we can substitute that in a second divided,"},{"Start":"04:17.390 ","End":"04:22.810","Text":"by the magnitude of our r tilde vector."},{"Start":"04:22.810 ","End":"04:28.685","Text":"Let\u0027s write out what our magnitude of our r tilde vector is."},{"Start":"04:28.685 ","End":"04:31.110","Text":"Let\u0027s write it over here."},{"Start":"04:31.760 ","End":"04:34.535","Text":"This is going to be equal to,"},{"Start":"04:34.535 ","End":"04:36.470","Text":"like in Pythagoras;"},{"Start":"04:36.470 ","End":"04:39.020","Text":"so we have the x-component,"},{"Start":"04:39.020 ","End":"04:42.450","Text":"so x minus x tag^2,"},{"Start":"04:42.890 ","End":"04:49.775","Text":"plus y minus y tag^2 plus"},{"Start":"04:49.775 ","End":"04:58.330","Text":"z minus z tag^2 and then we take the square root of all of this."},{"Start":"04:59.680 ","End":"05:05.420","Text":"What we\u0027ll have to do is we\u0027ll have to sum up on the potential from"},{"Start":"05:05.420 ","End":"05:10.883","Text":"all of these pieces over here and then we\u0027ll find the total potential at this point."},{"Start":"05:10.883 ","End":"05:14.165","Text":"Then in order to find the bound charge density,"},{"Start":"05:14.165 ","End":"05:19.580","Text":"we\u0027re going to have to take the derivative or take the gradient of this function."},{"Start":"05:19.580 ","End":"05:26.285","Text":"We can see that kdq is going to be something that we\u0027ll see is constant,"},{"Start":"05:26.285 ","End":"05:29.900","Text":"but what we\u0027ll need to do is we\u0027ll have to take the derivative of this over here,"},{"Start":"05:29.900 ","End":"05:36.520","Text":"which is 1 divided by this magnitude of r tilde vector."},{"Start":"05:36.520 ","End":"05:41.180","Text":"What we want to see is that if we take the derivative or the gradient of this,"},{"Start":"05:41.180 ","End":"05:46.060","Text":"we\u0027re going to get this that is equal to this over here."},{"Start":"05:46.060 ","End":"05:49.215","Text":"If we\u0027re taking the derivative,"},{"Start":"05:49.215 ","End":"05:51.125","Text":"so let\u0027s take a look."},{"Start":"05:51.125 ","End":"05:55.400","Text":"What we can see is that our r vector is remaining constant"},{"Start":"05:55.400 ","End":"06:00.406","Text":"because our r tilde vector is made up of r vector minus r tag vector."},{"Start":"06:00.406 ","End":"06:02.870","Text":"Our r vector is remaining constant,"},{"Start":"06:02.870 ","End":"06:05.150","Text":"so if we take the gradient of our r vector,"},{"Start":"06:05.150 ","End":"06:06.830","Text":"it\u0027s going to be equal to 0."},{"Start":"06:06.830 ","End":"06:10.550","Text":"However, if we take the gradient of our r tag vector,"},{"Start":"06:10.550 ","End":"06:17.405","Text":"so we can see that it\u0027s constantly changing as we sum up along the entire material."},{"Start":"06:17.405 ","End":"06:25.400","Text":"Therefore, when we\u0027re taking the gradient of our r tilde vector, in actual fact,"},{"Start":"06:25.400 ","End":"06:29.315","Text":"we\u0027re taking the gradient with respect to r tag,"},{"Start":"06:29.315 ","End":"06:31.430","Text":"so that means with respect to x tag,"},{"Start":"06:31.430 ","End":"06:35.470","Text":"y tag, and z tag because those are variables."},{"Start":"06:35.470 ","End":"06:44.035","Text":"We put a tag here and we\u0027re taking the gradient of 1 divided by our r tilde vector."},{"Start":"06:44.035 ","End":"06:46.500","Text":"What does that mean?"},{"Start":"06:46.500 ","End":"06:50.030","Text":"The gradient tag is equal to, in the x-component,"},{"Start":"06:50.030 ","End":"06:52.880","Text":"we have d by dx tag,"},{"Start":"06:52.880 ","End":"06:56.270","Text":"in the y, we have d by dy tag,"},{"Start":"06:56.270 ","End":"06:57.815","Text":"and the z,"},{"Start":"06:57.815 ","End":"07:02.698","Text":"we have d by dz tag,"},{"Start":"07:02.698 ","End":"07:06.830","Text":"and all of this is working on our function for 1 divided"},{"Start":"07:06.830 ","End":"07:13.680","Text":"by the magnitude of our r tilde vector."},{"Start":"07:14.060 ","End":"07:18.145","Text":"If we put 1 divided by here,"},{"Start":"07:18.145 ","End":"07:22.195","Text":"we just have to add in a minus sign over here and it\u0027s equivalent."},{"Start":"07:22.195 ","End":"07:29.485","Text":"We\u0027re taking the gradient of x minus x tag ^2 plus"},{"Start":"07:29.485 ","End":"07:35.140","Text":"y minus y tag^2 plus z"},{"Start":"07:35.140 ","End":"07:42.670","Text":"minus z tag^2 and all of this to the power of negative 1/2."},{"Start":"07:42.670 ","End":"07:46.660","Text":"Now, let\u0027s see what this is equal to."},{"Start":"07:46.660 ","End":"07:51.410","Text":"We\u0027ll carry it on over here and let\u0027s scroll down."},{"Start":"07:51.410 ","End":"07:53.970","Text":"Let\u0027s do d by dx,"},{"Start":"07:53.970 ","End":"07:56.595","Text":"so this is the x-component."},{"Start":"07:56.595 ","End":"08:02.000","Text":"What we\u0027re going to have is we\u0027re using a combination of the product and the chain rule."},{"Start":"08:02.000 ","End":"08:04.460","Text":"We have over here, the derivative,"},{"Start":"08:04.460 ","End":"08:11.205","Text":"so we\u0027ll have negative 1/2 multiplied by everything in these brackets,"},{"Start":"08:11.205 ","End":"08:18.185","Text":"so x minus x tag^2 plus y minus y tag^2"},{"Start":"08:18.185 ","End":"08:27.395","Text":"plus z minus z tag^2 and then all of this to the power of negative 3 over 2."},{"Start":"08:27.395 ","End":"08:30.680","Text":"Then we have to multiply that by the derivative,"},{"Start":"08:30.680 ","End":"08:32.390","Text":"so here we\u0027re doing of x tag,"},{"Start":"08:32.390 ","End":"08:35.330","Text":"so here we have 2 multiplied by"},{"Start":"08:35.330 ","End":"08:41.830","Text":"x minus x tag and then multiplied by the inner derivative."},{"Start":"08:41.830 ","End":"08:45.040","Text":"So we\u0027re taking the derivative of negative x tag,"},{"Start":"08:45.040 ","End":"08:50.380","Text":"which is equal to negative 1 and this is in the x-direction."},{"Start":"08:50.380 ","End":"08:53.515","Text":"Now, in the y-direction,"},{"Start":"08:53.515 ","End":"08:56.380","Text":"we\u0027re going to have d by dy tag."},{"Start":"08:56.380 ","End":"08:58.105","Text":"Again, we have the same as here,"},{"Start":"08:58.105 ","End":"09:01.390","Text":"negative 1/2 multiplied by everything inside"},{"Start":"09:01.390 ","End":"09:06.880","Text":"these brackets and then multiplied by the inner derivatives,"},{"Start":"09:06.880 ","End":"09:08.920","Text":"so over here, okay,"},{"Start":"09:08.920 ","End":"09:15.495","Text":"so here we\u0027ll have multiplied by 2 y minus y tag,"},{"Start":"09:15.495 ","End":"09:21.275","Text":"and then multiplied by the derivative of negative y tag with respect to y tag."},{"Start":"09:21.275 ","End":"09:23.472","Text":"Then, again,"},{"Start":"09:23.472 ","End":"09:27.305","Text":"multiplied by negative 1 and this is in the y-direction."},{"Start":"09:27.305 ","End":"09:30.170","Text":"Then in the z-direction, again,"},{"Start":"09:30.170 ","End":"09:34.885","Text":"we have negative 1/2 multiplied by everything over here."},{"Start":"09:34.885 ","End":"09:37.170","Text":"Then, again,"},{"Start":"09:37.170 ","End":"09:42.825","Text":"d by z tag with respect to this over here,"},{"Start":"09:42.825 ","End":"09:51.495","Text":"so we have 2 multiplied by z minus z tag multiplied by d by dz tag of negative z tag,"},{"Start":"09:51.495 ","End":"09:55.370","Text":"which is multiplied by negative 1 in the z-direction."},{"Start":"09:55.370 ","End":"10:00.573","Text":"Here we can see that this negative 1 cancels with this negative sign,"},{"Start":"10:00.573 ","End":"10:03.800","Text":"this negative 1 cancels with this negative sign,"},{"Start":"10:03.800 ","End":"10:06.755","Text":"and this negative 1 cancels with this negative sign."},{"Start":"10:06.755 ","End":"10:13.800","Text":"Then we can see we have 1/2 over here multiplied by 2 over here and the same over here,"},{"Start":"10:13.800 ","End":"10:19.150","Text":"1/2 multiplied by 2 and 1/2 multiplied by 2."},{"Start":"10:19.520 ","End":"10:24.290","Text":"Now, we can see that if we cube this,"},{"Start":"10:24.290 ","End":"10:30.605","Text":"so we have 1 divided r tilde vector^3, the magnitude,"},{"Start":"10:30.605 ","End":"10:37.365","Text":"so here we can just replace this 1 with 3,"},{"Start":"10:37.365 ","End":"10:40.820","Text":"so we have 3 over 2 and this is the equivalent to this,"},{"Start":"10:40.820 ","End":"10:43.685","Text":"negative 3 over 2."},{"Start":"10:43.685 ","End":"10:48.125","Text":"Now, we can carry all of this on over here."},{"Start":"10:48.125 ","End":"10:49.775","Text":"We can see in the x,"},{"Start":"10:49.775 ","End":"10:51.275","Text":"y, and z-direction,"},{"Start":"10:51.275 ","End":"10:54.230","Text":"we have this common multiple of x minus x"},{"Start":"10:54.230 ","End":"11:00.410","Text":"tag^2 plus y minus y tag^2 plus z minus z tag^2 to the power of negative 3 over 2."},{"Start":"11:00.410 ","End":"11:09.255","Text":"In other words, we have 1 divided by the magnitude of our r tilde vector^3."},{"Start":"11:09.255 ","End":"11:12.990","Text":"Then in the x-direction,"},{"Start":"11:12.990 ","End":"11:16.860","Text":"what we have is x minus x tag,"},{"Start":"11:16.860 ","End":"11:23.210","Text":"so x minus x tag in the x direction,"},{"Start":"11:23.210 ","End":"11:25.850","Text":"plus in the y-direction,"},{"Start":"11:25.850 ","End":"11:30.070","Text":"we have y minus y tag,"},{"Start":"11:30.070 ","End":"11:33.545","Text":"and then in the z-direction,"},{"Start":"11:33.545 ","End":"11:39.270","Text":"we have here z minus z tag."},{"Start":"11:41.270 ","End":"11:44.415","Text":"What is this equal to?"},{"Start":"11:44.415 ","End":"11:48.615","Text":"We can see that x minus x tag in the x direction,"},{"Start":"11:48.615 ","End":"11:51.615","Text":"y minus y tag in the y direction,"},{"Start":"11:51.615 ","End":"11:58.550","Text":"and z minus z tag in the z-direction is exactly our r tilde vector."},{"Start":"11:58.550 ","End":"12:04.730","Text":"In that case, we can just write all of this as r tilde vector divided by"},{"Start":"12:04.730 ","End":"12:11.310","Text":"the magnitude of r tilde vector^3 and as we know,"},{"Start":"12:11.310 ","End":"12:17.900","Text":"r vector or r tilde vector is equal to r tilde in the r hat direction,"},{"Start":"12:17.900 ","End":"12:23.765","Text":"so we can just write this in unit vector form as r tilde hat"},{"Start":"12:23.765 ","End":"12:31.210","Text":"divided by r tilde vector^2."},{"Start":"12:31.210 ","End":"12:33.350","Text":"Then if we look up,"},{"Start":"12:33.350 ","End":"12:36.755","Text":"this is exactly what we wanted to prove,"},{"Start":"12:36.755 ","End":"12:42.540","Text":"we got the exact same equation over here and that is the end of this lesson."}],"ID":22314},{"Watched":false,"Name":"Deriving Bound Charge Density Part 2","Duration":"5m 56s","ChapterTopicVideoID":21467,"CourseChapterTopicPlaylistID":99475,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.560","Text":"Hello. In this lesson,"},{"Start":"00:01.560 ","End":"00:07.545","Text":"we\u0027re going to be carrying on with showing how to derive the bound charge density."},{"Start":"00:07.545 ","End":"00:12.210","Text":"In this lesson, we\u0027re going to show how this,"},{"Start":"00:12.210 ","End":"00:20.475","Text":"so integrating along some volume of A which is some kind of vector function,"},{"Start":"00:20.475 ","End":"00:29.320","Text":"dot product with the gradient of a scalar function is equal to all of this."},{"Start":"00:30.260 ","End":"00:33.890","Text":"This is going to be equal to, again,"},{"Start":"00:33.890 ","End":"00:39.740","Text":"integrating along the volume of the divergence of A dot f."},{"Start":"00:39.740 ","End":"00:42.550","Text":"What we\u0027re going to have here is some kind of scalar"},{"Start":"00:42.550 ","End":"00:46.610","Text":"because we have a vector dot product with the scalar,"},{"Start":"00:46.610 ","End":"00:51.230","Text":"minus, again, integrating along the volume of"},{"Start":"00:51.230 ","End":"00:59.790","Text":"our scalar function over here multiplied by the divergence of our vector A."},{"Start":"01:00.950 ","End":"01:05.390","Text":"This is what you have to write down and remember,"},{"Start":"01:05.390 ","End":"01:08.210","Text":"and now we\u0027re going to show how we get here."},{"Start":"01:08.210 ","End":"01:13.900","Text":"First of all, we\u0027re going to start by writing out this over here."},{"Start":"01:13.900 ","End":"01:17.810","Text":"We have nabla,"},{"Start":"01:17.810 ","End":"01:26.825","Text":"and then we have in the brackets A dot f which is a function of x, y, and z."},{"Start":"01:26.825 ","End":"01:28.775","Text":"Now, we\u0027re going to open up these brackets."},{"Start":"01:28.775 ","End":"01:38.570","Text":"What we\u0027re going to have is nabla dot A multiplied by f as a function of x, y,"},{"Start":"01:38.570 ","End":"01:42.640","Text":"and z, and then we have plus,"},{"Start":"01:42.640 ","End":"01:52.710","Text":"and then we have A dot nabla of f as a function of x, y, z."},{"Start":"01:54.020 ","End":"02:01.455","Text":"Here we have the divergence of A multiplied by our function f plus we have"},{"Start":"02:01.455 ","End":"02:09.360","Text":"A dot product with the gradient of f. Now,"},{"Start":"02:09.360 ","End":"02:12.330","Text":"all we have to do is just move some stuff around,"},{"Start":"02:12.330 ","End":"02:17.673","Text":"so we just move this over here to this side of the equation,"},{"Start":"02:17.673 ","End":"02:19.075","Text":"and then we have,"},{"Start":"02:19.075 ","End":"02:22.569","Text":"if we put in the integration signs,"},{"Start":"02:22.569 ","End":"02:28.565","Text":"so then we\u0027ll have A dot grad f is equal to nabla of"},{"Start":"02:28.565 ","End":"02:35.665","Text":"A dot f minus the divergence of A multiplied by f, which is exactly this."},{"Start":"02:35.665 ","End":"02:38.600","Text":"We just have to put in the integration signs for"},{"Start":"02:38.600 ","End":"02:43.350","Text":"volume and move this term over to this side of the equation,"},{"Start":"02:43.350 ","End":"02:45.375","Text":"and that is it."},{"Start":"02:45.375 ","End":"02:49.565","Text":"I\u0027m going to show you that this does in fact"},{"Start":"02:49.565 ","End":"02:53.810","Text":"equal this as well just how you do it or so physically."},{"Start":"02:53.810 ","End":"02:59.105","Text":"What we have is the dot product between a vector and a scalar,"},{"Start":"02:59.105 ","End":"03:01.310","Text":"and then we take the divergence of it."},{"Start":"03:01.310 ","End":"03:03.380","Text":"How do we do this dot product?"},{"Start":"03:03.380 ","End":"03:05.315","Text":"What we do is we take"},{"Start":"03:05.315 ","End":"03:11.210","Text":"our scalar function f and then we multiply it by each component of A,"},{"Start":"03:11.210 ","End":"03:13.190","Text":"so we\u0027re going to have x, y,"},{"Start":"03:13.190 ","End":"03:15.530","Text":"z, for instance because we\u0027re dealing with volume."},{"Start":"03:15.530 ","End":"03:19.020","Text":"We have f multiplied by A_x,"},{"Start":"03:19.020 ","End":"03:23.310","Text":"and then we\u0027re going to have f multiplied by A_y,"},{"Start":"03:23.310 ","End":"03:27.970","Text":"and then we\u0027re going to have f multiplied by A_z."},{"Start":"03:28.130 ","End":"03:32.940","Text":"This, we can put it in brackets,"},{"Start":"03:32.940 ","End":"03:36.300","Text":"is this dot-product happening here,"},{"Start":"03:36.300 ","End":"03:38.130","Text":"and then we take the divergence."},{"Start":"03:38.130 ","End":"03:41.280","Text":"What does that mean? dx, dy, dz."},{"Start":"03:41.280 ","End":"03:50.460","Text":"We have d by dx of this plus d by dy of this,"},{"Start":"03:50.460 ","End":"03:56.385","Text":"plus d by dz of all of this."},{"Start":"03:56.385 ","End":"03:59.835","Text":"What is that equal to?"},{"Start":"03:59.835 ","End":"04:05.385","Text":"Now we have to use the product rule."},{"Start":"04:05.385 ","End":"04:12.395","Text":"First, we\u0027ll do df by dx,"},{"Start":"04:12.395 ","End":"04:21.810","Text":"and then this is going to be multiplied by A_x plus df by dy multiplied by A_y,"},{"Start":"04:22.450 ","End":"04:30.520","Text":"and then plus df by dz multiplied by A_z."},{"Start":"04:30.520 ","End":"04:38.540","Text":"What we\u0027re doing is we\u0027re differentiating our scalar function f according to x,"},{"Start":"04:38.540 ","End":"04:43.760","Text":"y and z, and now we\u0027re going to add on."},{"Start":"04:43.760 ","End":"04:49.200","Text":"Now we have to differentiate A_x, A_y, and A_z,"},{"Start":"04:49.200 ","End":"04:56.705","Text":"so now we have dA_x by dx plus"},{"Start":"04:56.705 ","End":"05:01.489","Text":"dA_y by dy plus"},{"Start":"05:01.489 ","End":"05:07.170","Text":"dA_z by dz,"},{"Start":"05:07.170 ","End":"05:15.630","Text":"and all of this is multiplied by our function f. Then what does that equal?"},{"Start":"05:15.630 ","End":"05:19.055","Text":"We have df by dx, this, this, this,"},{"Start":"05:19.055 ","End":"05:28.825","Text":"so what we have is the gradient of f dot product with our A vector,"},{"Start":"05:28.825 ","End":"05:31.515","Text":"and then we\u0027ll get what we have over here."},{"Start":"05:31.515 ","End":"05:36.005","Text":"Then what we have over here is, so plus,"},{"Start":"05:36.005 ","End":"05:46.420","Text":"and then we have the divergence of A,"},{"Start":"05:46.420 ","End":"05:56.290","Text":"we see that, and then just multiply it by our function f. That\u0027s the end of this lesson."}],"ID":22315},{"Watched":false,"Name":"Deriving Bound Charge Density Part 3","Duration":"15m 11s","ChapterTopicVideoID":21468,"CourseChapterTopicPlaylistID":99475,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.870","Text":"Hello. In this lesson,"},{"Start":"00:01.870 ","End":"00:03.580","Text":"we\u0027re going to continue deriving"},{"Start":"00:03.580 ","End":"00:08.575","Text":"the bound charge density for polarized dielectric materials."},{"Start":"00:08.575 ","End":"00:13.690","Text":"We\u0027re going to see how we can get from our polarization vector into"},{"Start":"00:13.690 ","End":"00:23.840","Text":"the bound charge densities for area and for volume in a dielectric material."},{"Start":"00:24.260 ","End":"00:31.000","Text":"Here we have some kind of polarized dielectric material and it"},{"Start":"00:31.000 ","End":"00:37.695","Text":"has polarization density or the polarization vector given like so,"},{"Start":"00:37.695 ","End":"00:40.345","Text":"and we have our r tag vector,"},{"Start":"00:40.345 ","End":"00:43.370","Text":"which is pointing from the origin until"},{"Start":"00:43.370 ","End":"00:48.979","Text":"some kind of point over here in the dielectric material."},{"Start":"00:48.979 ","End":"00:54.479","Text":"This point over here is some kind of unit volume."},{"Start":"00:55.370 ","End":"00:58.335","Text":"This has some kind of volume,"},{"Start":"00:58.335 ","End":"01:00.670","Text":"let\u0027s call it dv tag."},{"Start":"01:00.670 ","End":"01:09.580","Text":"The tag, everything that is inside the dielectric material will be having a tag on it."},{"Start":"01:09.580 ","End":"01:14.690","Text":"Now we want to know what its dipole moment is."},{"Start":"01:14.690 ","End":"01:19.415","Text":"Its dipole moment will be dp, a small p,"},{"Start":"01:19.415 ","End":"01:27.900","Text":"because it\u0027s a dipole moment and this is going to be equal to the polarization vector,"},{"Start":"01:27.900 ","End":"01:32.305","Text":"so that is this capital P vector,"},{"Start":"01:32.305 ","End":"01:35.050","Text":"which is as a function of r tag because of"},{"Start":"01:35.050 ","End":"01:38.110","Text":"course r tag is going to be constantly changing,"},{"Start":"01:38.110 ","End":"01:42.053","Text":"summing up along all of the dielectric material."},{"Start":"01:42.053 ","End":"01:48.700","Text":"Then, of course, multiplied by the volume so over here it\u0027s dv tag."},{"Start":"01:49.530 ","End":"01:51.970","Text":"Now just a quick note,"},{"Start":"01:51.970 ","End":"01:56.905","Text":"remember that the polarization vector gives the density of dipoles."},{"Start":"01:56.905 ","End":"02:02.650","Text":"We have the density of dipoles per unit volume multiplied by"},{"Start":"02:02.650 ","End":"02:09.930","Text":"the unit volume dv and that gives us the dipole moment of that volume over there."},{"Start":"02:10.730 ","End":"02:18.630","Text":"Now we can consider this little volume over here in red as some kind of dipole."},{"Start":"02:18.630 ","End":"02:23.170","Text":"What we want to do is we want to pick some kind of point in space over here"},{"Start":"02:23.170 ","End":"02:27.500","Text":"and we want to calculate the potential at that point."},{"Start":"02:27.500 ","End":"02:34.595","Text":"So the vector pointing from a dipole to this pointing space is called r tilde."},{"Start":"02:34.595 ","End":"02:41.060","Text":"Here we have a tilde and then we have a vector pointing from the origin until"},{"Start":"02:41.060 ","End":"02:44.090","Text":"this point in space to give us"},{"Start":"02:44.090 ","End":"02:47.644","Text":"the position of this point in space relative to the origin."},{"Start":"02:47.644 ","End":"02:54.930","Text":"The vector pointing from the origin till this point in space is called r vector."},{"Start":"02:56.360 ","End":"03:01.610","Text":"We\u0027ve chosen this point in space so we know what our r vector is and we can"},{"Start":"03:01.610 ","End":"03:06.410","Text":"also choose this point wherever it is in the dielectric material,"},{"Start":"03:06.410 ","End":"03:08.810","Text":"but we don\u0027t know what our r tilde is."},{"Start":"03:08.810 ","End":"03:10.355","Text":"What is our r tilde?"},{"Start":"03:10.355 ","End":"03:12.635","Text":"The vector pointing from"},{"Start":"03:12.635 ","End":"03:16.669","Text":"a dipole to the point in space where we want to measure the potential."},{"Start":"03:16.669 ","End":"03:23.825","Text":"We can see that our r tilde vector is simply going to be equal to,"},{"Start":"03:23.825 ","End":"03:25.760","Text":"we want to get to this point."},{"Start":"03:25.760 ","End":"03:29.210","Text":"What we can see is from our dv,"},{"Start":"03:29.210 ","End":"03:35.235","Text":"if we go down this r tag vector and up our r vector,"},{"Start":"03:35.235 ","End":"03:41.970","Text":"so we have r vector minus r tag vector,"},{"Start":"03:41.970 ","End":"03:44.290","Text":"and this is our r tilde."},{"Start":"03:45.470 ","End":"03:49.655","Text":"Then we can see that the magnitude,"},{"Start":"03:49.655 ","End":"03:53.000","Text":"or we can just write that the magnitude of r tilde,"},{"Start":"03:53.000 ","End":"03:54.875","Text":"let\u0027s just write it like this."},{"Start":"03:54.875 ","End":"04:03.779","Text":"It\u0027s easier to shorthand and this is just equal to the magnitude of our r tilde vector."},{"Start":"04:04.970 ","End":"04:08.150","Text":"We want to calculate the potential."},{"Start":"04:08.150 ","End":"04:13.355","Text":"We\u0027ve already seen that the equation for the potential due to"},{"Start":"04:13.355 ","End":"04:19.950","Text":"a dipole is equal to k multiplied by the dipole moment,"},{"Start":"04:19.950 ","End":"04:26.255","Text":"so this is a lowercase p dot r hat divided by r^2"},{"Start":"04:26.255 ","End":"04:29.150","Text":"where r is the distance between"},{"Start":"04:29.150 ","End":"04:32.930","Text":"the dipole to the point in space where we\u0027re measuring the potential."},{"Start":"04:32.930 ","End":"04:36.965","Text":"So over here it will be our r tilde vector and, of course,"},{"Start":"04:36.965 ","End":"04:42.230","Text":"this is the same as raising k multiplied by our dipole moment."},{"Start":"04:42.230 ","End":"04:50.280","Text":"So lowercase p dot r vector divided by r^3."},{"Start":"04:51.170 ","End":"04:54.180","Text":"Let\u0027s see what\u0027s happening in our case."},{"Start":"04:54.180 ","End":"04:57.490","Text":"In our case because we have dp,"},{"Start":"04:57.490 ","End":"05:00.798","Text":"so unit dipole moment,"},{"Start":"05:00.798 ","End":"05:04.760","Text":"we\u0027ll write that here we have d Phi,"},{"Start":"05:04.760 ","End":"05:07.995","Text":"the potential due to just this 1 dipole."},{"Start":"05:07.995 ","End":"05:10.715","Text":"Of course, then we\u0027re going to sum up along"},{"Start":"05:10.715 ","End":"05:14.840","Text":"all of the dipoles in this dielectric material."},{"Start":"05:14.840 ","End":"05:19.490","Text":"This is the potential over here due to 1 dipole."},{"Start":"05:19.490 ","End":"05:23.795","Text":"This is going to be equal to k multiplied by"},{"Start":"05:23.795 ","End":"05:31.264","Text":"dp and then dot-product with,"},{"Start":"05:31.264 ","End":"05:42.540","Text":"let\u0027s write our r tilde vector divided by r tilde^3."},{"Start":"05:43.670 ","End":"05:48.110","Text":"Now we can sub in what our dp is equal to."},{"Start":"05:48.110 ","End":"05:50.960","Text":"We have k multiplied by dp,"},{"Start":"05:50.960 ","End":"05:57.360","Text":"which is simply our polarization vector as a function of r tag vector."},{"Start":"05:57.360 ","End":"06:05.970","Text":"Then we have dot product with our r tilde vector and"},{"Start":"06:05.970 ","End":"06:15.220","Text":"then we have dv tag and all of this divided by r tilde^3."},{"Start":"06:15.800 ","End":"06:22.585","Text":"Now we can see that we have a dot product between our polarization vector and"},{"Start":"06:22.585 ","End":"06:30.340","Text":"our r tilde vector divided by r tilde^3."},{"Start":"06:30.950 ","End":"06:34.750","Text":"Now to find the total potential at this point"},{"Start":"06:34.750 ","End":"06:37.828","Text":"due to all of the dipoles and the dielectric material,"},{"Start":"06:37.828 ","End":"06:40.430","Text":"we\u0027re just going to integrate."},{"Start":"06:40.730 ","End":"06:45.190","Text":"This is going to be an integration on d Phi,"},{"Start":"06:45.190 ","End":"06:49.400","Text":"which is just equal to an integration on all of this."},{"Start":"06:50.870 ","End":"06:57.350","Text":"We can see that we\u0027re integrating along the tag."},{"Start":"06:57.930 ","End":"07:06.860","Text":"This r tilde vector is a constant when we\u0027re integrating."},{"Start":"07:08.870 ","End":"07:15.195","Text":"We saw 2 lessons ago that the gradient tag,"},{"Start":"07:15.195 ","End":"07:25.545","Text":"when we take the gradient with respect to r tag of 1 divided by r tilde."},{"Start":"07:25.545 ","End":"07:35.775","Text":"What we get is equal to our r tilde vector divided by r tilde^3."},{"Start":"07:35.775 ","End":"07:37.830","Text":"This we saw 2 lessons ago."},{"Start":"07:37.830 ","End":"07:40.470","Text":"If you don\u0027t remember how I did this,"},{"Start":"07:40.470 ","End":"07:43.230","Text":"please go back 2 lessons."},{"Start":"07:43.230 ","End":"07:44.865","Text":"A little note,"},{"Start":"07:44.865 ","End":"07:47.235","Text":"if this gradient didn\u0027t have a tag,"},{"Start":"07:47.235 ","End":"07:51.975","Text":"so if it was just the gradient of 1 divided by r,"},{"Start":"07:51.975 ","End":"08:00.720","Text":"so what we would get is negative r divided by r^3."},{"Start":"08:00.720 ","End":"08:04.170","Text":"There\u0027s no negative over here because we\u0027re taking the"},{"Start":"08:04.170 ","End":"08:08.950","Text":"derivative with respect to tag and here we aren\u0027t."},{"Start":"08:09.560 ","End":"08:12.870","Text":"What I\u0027m going to do is,"},{"Start":"08:12.870 ","End":"08:14.400","Text":"as we can see,"},{"Start":"08:14.400 ","End":"08:17.385","Text":"let\u0027s draw this in blue."},{"Start":"08:17.385 ","End":"08:23.865","Text":"This section of the equation is equal to this."},{"Start":"08:23.865 ","End":"08:26.985","Text":"I\u0027m going to switch this out,"},{"Start":"08:26.985 ","End":"08:33.330","Text":"r tilde vector divided by r tilde^3 by this over here,"},{"Start":"08:33.330 ","End":"08:35.835","Text":"the left side of the equation."},{"Start":"08:35.835 ","End":"08:44.415","Text":"Then what I\u0027m going to therefore get is that the potential is equal to the integral,"},{"Start":"08:44.415 ","End":"08:46.499","Text":"and we\u0027re integrating with volume,"},{"Start":"08:46.499 ","End":"08:52.650","Text":"of k and then our polarization density,"},{"Start":"08:52.650 ","End":"08:57.480","Text":"dot product with this over here."},{"Start":"08:57.480 ","End":"09:06.850","Text":"The gradient tag of 1 divided by r tilde dv tag."},{"Start":"09:07.520 ","End":"09:14.250","Text":"Now I\u0027m going to do another move which we saw actually in the previous lesson."},{"Start":"09:14.250 ","End":"09:19.005","Text":"I can rewrite this integral as this,"},{"Start":"09:19.005 ","End":"09:23.920","Text":"an integral with respect to v of k."},{"Start":"09:23.920 ","End":"09:33.270","Text":"Then gradient tag of our polarization vector,"},{"Start":"09:33.720 ","End":"09:44.625","Text":"divided by r tilde dv tag minus our integration"},{"Start":"09:44.625 ","End":"09:55.035","Text":"with respect to volume of k Nabla tag dot our polarization vector with respect to"},{"Start":"09:55.035 ","End":"09:58.440","Text":"r tag divided by"},{"Start":"09:58.440 ","End":"10:05.340","Text":"r tilde dv tag."},{"Start":"10:05.340 ","End":"10:06.975","Text":"This I showed in the previous lesson."},{"Start":"10:06.975 ","End":"10:11.880","Text":"If you can\u0027t remember how I can change this integral into these 2 integrals,"},{"Start":"10:11.880 ","End":"10:14.740","Text":"please go back to the previous lesson."},{"Start":"10:15.680 ","End":"10:18.480","Text":"Now using Stokes\u0027 law,"},{"Start":"10:18.480 ","End":"10:22.350","Text":"I\u0027m going to change this integral."},{"Start":"10:22.350 ","End":"10:24.390","Text":"If you don\u0027t remember Stokes\u0027 law,"},{"Start":"10:24.390 ","End":"10:26.475","Text":"please look it up on the Internet."},{"Start":"10:26.475 ","End":"10:31.155","Text":"I\u0027m going to change this integral from Stokes\u0027 law into,"},{"Start":"10:31.155 ","End":"10:33.165","Text":"instead of the integral and volume,"},{"Start":"10:33.165 ","End":"10:38.700","Text":"it\u0027s going to be a closed circuit integral on the surface area of"},{"Start":"10:38.700 ","End":"10:43.800","Text":"our dielectric material of k multiplied"},{"Start":"10:43.800 ","End":"10:49.320","Text":"by our polarization vector dot product with n,"},{"Start":"10:49.320 ","End":"10:51.375","Text":"where n is, of course, the normal."},{"Start":"10:51.375 ","End":"10:53.700","Text":"That means it is the vector which is"},{"Start":"10:53.700 ","End":"10:59.380","Text":"perpendicular to the surface of our dielectric material."},{"Start":"10:59.510 ","End":"11:04.410","Text":"This is going to be divided by r tilde."},{"Start":"11:04.410 ","End":"11:08.160","Text":"We\u0027re integrating along dsn tag."},{"Start":"11:08.160 ","End":"11:13.934","Text":"Because we said at the beginning that anything to do with our material itself,"},{"Start":"11:13.934 ","End":"11:17.170","Text":"we\u0027re going to write it out with a tag."},{"Start":"11:17.930 ","End":"11:22.020","Text":"Now, this has stayed the same. I just rewrote it."},{"Start":"11:22.020 ","End":"11:28.620","Text":"What we can see is that we have k multiplied by something divided by r,"},{"Start":"11:28.620 ","End":"11:32.655","Text":"which is an equation for our potential."},{"Start":"11:32.655 ","End":"11:37.230","Text":"Just like we saw if we\u0027re dealing with dq,"},{"Start":"11:37.230 ","End":"11:47.175","Text":"let\u0027s say that the potential will be some kdq divided by r. Here we have k,"},{"Start":"11:47.175 ","End":"11:52.570","Text":"some charge divided by r. Also here we have k,"},{"Start":"11:52.570 ","End":"11:58.685","Text":"some charge divided by r. Then we\u0027d of course have to integrate this,"},{"Start":"11:58.685 ","End":"12:02.250","Text":"which is what we have over here, we\u0027re integrating."},{"Start":"12:03.020 ","End":"12:06.494","Text":"We can see that these equations are very,"},{"Start":"12:06.494 ","End":"12:08.640","Text":"very similar, both of them."},{"Start":"12:08.640 ","End":"12:14.010","Text":"What we\u0027re going to do is we\u0027re going to take all of this and we\u0027re"},{"Start":"12:14.010 ","End":"12:19.275","Text":"going to say that this is equal to Sigma b."},{"Start":"12:19.275 ","End":"12:23.895","Text":"This is the bound charge on some kind of surface."},{"Start":"12:23.895 ","End":"12:28.350","Text":"If you imagine that we have some kind of plane,"},{"Start":"12:28.350 ","End":"12:29.880","Text":"some kind of surface,"},{"Start":"12:29.880 ","End":"12:34.200","Text":"and it has charged density per unit area of Sigma."},{"Start":"12:34.200 ","End":"12:37.350","Text":"Then what we would do is,"},{"Start":"12:37.350 ","End":"12:43.500","Text":"we would just integrate on k Sigma ds."},{"Start":"12:43.500 ","End":"12:48.850","Text":"If its area is s divided by"},{"Start":"12:48.980 ","End":"12:58.065","Text":"r. Then if we would integrate this,"},{"Start":"12:58.065 ","End":"13:03.375","Text":"we would get the potential due to this plane over here."},{"Start":"13:03.375 ","End":"13:05.490","Text":"This is Sigma b."},{"Start":"13:05.490 ","End":"13:11.505","Text":"Then let\u0027s circle this in blue over here."},{"Start":"13:11.505 ","End":"13:15.210","Text":"Let\u0027s just call this Rho b,"},{"Start":"13:15.210 ","End":"13:20.010","Text":"the bound charge density per unit volume."},{"Start":"13:20.010 ","End":"13:25.290","Text":"If you imagine that now this was some kind of cube,"},{"Start":"13:25.290 ","End":"13:29.205","Text":"with charge density Rho per unit volume."},{"Start":"13:29.205 ","End":"13:34.335","Text":"Then all we would do is we would write the integral exactly the same,"},{"Start":"13:34.335 ","End":"13:43.455","Text":"k Rho dv divided by r. In other words,"},{"Start":"13:43.455 ","End":"13:46.920","Text":"we can just write this out again."},{"Start":"13:46.920 ","End":"13:50.730","Text":"We\u0027re going to have an integral along s,"},{"Start":"13:50.730 ","End":"13:53.325","Text":"the surface area of"},{"Start":"13:53.325 ","End":"14:01.980","Text":"k Sigma b ds tag divided by r tilde."},{"Start":"14:01.980 ","End":"14:05.655","Text":"Let\u0027s write this a plus."},{"Start":"14:05.655 ","End":"14:11.625","Text":"Let\u0027s make this negative Rho b"},{"Start":"14:11.625 ","End":"14:19.320","Text":"plus the integral along the volume of k. Then we have a minus and a minus,"},{"Start":"14:19.320 ","End":"14:30.100","Text":"so it\u0027s a plus, so Rho bdv tag divided by r tilde."},{"Start":"14:31.310 ","End":"14:35.495","Text":"The potential, it can just be written like so."},{"Start":"14:35.495 ","End":"14:38.570","Text":"Then we can see that our Sigma b,"},{"Start":"14:38.570 ","End":"14:43.030","Text":"or let\u0027s scroll up because we already have this, our Sigma b,"},{"Start":"14:43.030 ","End":"14:50.765","Text":"our bound charge per unit area is our polarization density dot our normal vector,"},{"Start":"14:50.765 ","End":"14:52.775","Text":"which is exactly what we have here."},{"Start":"14:52.775 ","End":"14:57.860","Text":"Our bound charge per unit volume is equal to"},{"Start":"14:57.860 ","End":"15:04.530","Text":"negative Nabla dot our polarization density."},{"Start":"15:04.530 ","End":"15:08.735","Text":"That is where these equations come from."},{"Start":"15:08.735 ","End":"15:11.850","Text":"That is the end of this lesson."}],"ID":22316},{"Watched":false,"Name":"Exercise 3","Duration":"12m 50s","ChapterTopicVideoID":21469,"CourseChapterTopicPlaylistID":99475,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.115","Text":"Hello. In this lesson,"},{"Start":"00:02.115 ","End":"00:04.455","Text":"we\u0027re going to be answering the following question."},{"Start":"00:04.455 ","End":"00:09.225","Text":"Sphere of radius R is polarized with polarization density,"},{"Start":"00:09.225 ","End":"00:12.035","Text":"which is equal to Ar vector."},{"Start":"00:12.035 ","End":"00:17.590","Text":"A is a constant and r vector is a vector from the center of the sphere."},{"Start":"00:17.590 ","End":"00:22.155","Text":"Question Number 1 is to calculate the bound charge densities,"},{"Start":"00:22.155 ","End":"00:28.815","Text":"so that means the bound charge density per unit area and per unit volume."},{"Start":"00:28.815 ","End":"00:31.965","Text":"Let\u0027s begin answering Question Number 1."},{"Start":"00:31.965 ","End":"00:38.715","Text":"First of all, our polarization density is given as Ar vector."},{"Start":"00:38.715 ","End":"00:46.635","Text":"Another way of writing this as A multiplied by r in the r hat direction."},{"Start":"00:46.635 ","End":"00:53.675","Text":"Therefore, we can say that the radial component of our polarization density,"},{"Start":"00:53.675 ","End":"01:01.710","Text":"so P_r is simply equal to Ar."},{"Start":"01:01.710 ","End":"01:04.310","Text":"Let\u0027s just write over here from a previous lesson,"},{"Start":"01:04.310 ","End":"01:10.340","Text":"we saw that Rho b is equal to"},{"Start":"01:10.340 ","End":"01:18.540","Text":"the negative Nabla dot p and that our sigma b is"},{"Start":"01:18.540 ","End":"01:27.975","Text":"equal to P dot n. Let\u0027s begin with our Rho."},{"Start":"01:27.975 ","End":"01:32.285","Text":"This is our bound charge per unit volume so Rho b,"},{"Start":"01:32.285 ","End":"01:33.890","Text":"and so as we said,"},{"Start":"01:33.890 ","End":"01:40.130","Text":"this is equal to negative Nabla dot p. First of all,"},{"Start":"01:40.130 ","End":"01:43.790","Text":"we can see that our P only has a radial vector,"},{"Start":"01:43.790 ","End":"01:53.790","Text":"we don\u0027t have a Theta or a z component to this or a Theta and a Phi component."},{"Start":"01:53.790 ","End":"02:00.150","Text":"Therefore, we can write this as the negative Nabla,"},{"Start":"02:00.150 ","End":"02:02.450","Text":"so what we\u0027ll have is negative,"},{"Start":"02:02.450 ","End":"02:06.140","Text":"then we\u0027re dealing with spherical coordinates so we\u0027re going to"},{"Start":"02:06.140 ","End":"02:10.215","Text":"have negative 1 divided by r^2."},{"Start":"02:10.215 ","End":"02:13.430","Text":"Then we take the derivative with respect to r,"},{"Start":"02:13.430 ","End":"02:17.930","Text":"so d by dr and then when we\u0027re using spherical coordinates,"},{"Start":"02:17.930 ","End":"02:23.030","Text":"don\u0027t forget, you have to first multiply it by r^2."},{"Start":"02:23.030 ","End":"02:25.430","Text":"The r component of our p,"},{"Start":"02:25.430 ","End":"02:29.990","Text":"we multiply by r^2 and only then we can take the derivative with respect to"},{"Start":"02:29.990 ","End":"02:35.850","Text":"r and then we divide by r^2 and multiply by negative 1."},{"Start":"02:36.080 ","End":"02:39.680","Text":"As we just said, the other components are equal to 0"},{"Start":"02:39.680 ","End":"02:43.940","Text":"because our p vector only has a radial component."},{"Start":"02:43.940 ","End":"02:48.975","Text":"We can just write plus 0 plus 0 just so that you remember that."},{"Start":"02:48.975 ","End":"02:54.390","Text":"This is going to be equal to negative 1 divided by r^2 d"},{"Start":"02:54.390 ","End":"02:59.865","Text":"by dr of r^2 multiplied by P_r which is Ar,"},{"Start":"02:59.865 ","End":"03:05.070","Text":"so of r^2 multiplied by Ar, which is,"},{"Start":"03:05.070 ","End":"03:12.640","Text":"in other words, A multiplied by r^3."},{"Start":"03:12.740 ","End":"03:15.990","Text":"That is, of course,"},{"Start":"03:15.990 ","End":"03:22.470","Text":"going to be equal to negative 1 divided by r^2 of d by dr"},{"Start":"03:22.470 ","End":"03:30.960","Text":"of Ar^3 is simply going to be 3Ar^2."},{"Start":"03:30.960 ","End":"03:37.680","Text":"Therefore, we can say that our Rho b is equal to,"},{"Start":"03:37.680 ","End":"03:43.450","Text":"this r^2 will cancel with this r^2 so it\u0027s equal to negative 3A."},{"Start":"03:44.510 ","End":"03:50.365","Text":"Now let\u0027s calculate our Sigma b. Sigma b, as we saw,"},{"Start":"03:50.365 ","End":"03:55.630","Text":"is equal our p vector dot n hat,"},{"Start":"03:55.630 ","End":"03:59.390","Text":"the normal unit vector."},{"Start":"04:00.020 ","End":"04:03.015","Text":"What is our normal vector over here?"},{"Start":"04:03.015 ","End":"04:05.115","Text":"We\u0027re dealing with a sphere,"},{"Start":"04:05.115 ","End":"04:08.680","Text":"so why don\u0027t we have a sphere the normal vector to"},{"Start":"04:08.680 ","End":"04:12.640","Text":"the surface of a sphere is simply going to be the r vector."},{"Start":"04:12.640 ","End":"04:18.030","Text":"In our example we have our p vector dot r hat."},{"Start":"04:18.030 ","End":"04:23.160","Text":"Remember that this is n hats so we\u0027re dealing with r-hat if it was n vector,"},{"Start":"04:23.160 ","End":"04:25.810","Text":"we\u0027d have r vector."},{"Start":"04:26.120 ","End":"04:29.990","Text":"In other words, we have our p,"},{"Start":"04:29.990 ","End":"04:35.800","Text":"which is Ar multiplied by r hat,"},{"Start":"04:35.800 ","End":"04:41.520","Text":"remember we wrote that out over here dot product with r hat."},{"Start":"04:41.520 ","End":"04:44.165","Text":"We have r hat dot r hat,"},{"Start":"04:44.165 ","End":"04:47.220","Text":"which is simply equal to 1."},{"Start":"04:47.690 ","End":"04:54.875","Text":"Therefore, this is going to be equal to A multiplied by r. Now,"},{"Start":"04:54.875 ","End":"05:04.625","Text":"here we have to remember that our Sigma b is the bound charge per unit area."},{"Start":"05:04.625 ","End":"05:07.370","Text":"Where is the surface area?"},{"Start":"05:07.370 ","End":"05:12.165","Text":"The surface area is of course around the sphere,"},{"Start":"05:12.165 ","End":"05:13.770","Text":"on the outskirts of the sphere,"},{"Start":"05:13.770 ","End":"05:15.405","Text":"it\u0027s not inside the sphere."},{"Start":"05:15.405 ","End":"05:22.880","Text":"So we have to remember that r is equal to the radius of the sphere."},{"Start":"05:22.880 ","End":"05:26.885","Text":"The surface area is the area that surrounds the sphere"},{"Start":"05:26.885 ","End":"05:30.980","Text":"so it\u0027s a radius R. Therefore this is equal"},{"Start":"05:30.980 ","End":"05:38.850","Text":"to A multiplied by R. That\u0027s the answer to Question Number 1."},{"Start":"05:38.850 ","End":"05:41.185","Text":"Now let\u0027s take a look at Question Number 2,"},{"Start":"05:41.185 ","End":"05:46.880","Text":"which is to calculate the electric field inside and outside of the sphere."},{"Start":"05:46.880 ","End":"05:52.100","Text":"What we have over here is some sphere where on the inside of"},{"Start":"05:52.100 ","End":"05:57.764","Text":"the sphere we have the bound charge per unit volume."},{"Start":"05:57.764 ","End":"05:59.730","Text":"We have Rho b,"},{"Start":"05:59.730 ","End":"06:03.765","Text":"which we just calculated is equal to negative 3A."},{"Start":"06:03.765 ","End":"06:07.145","Text":"Then on the outskirts of the sphere,"},{"Start":"06:07.145 ","End":"06:09.755","Text":"we have some envelope,"},{"Start":"06:09.755 ","End":"06:17.195","Text":"which is just the surface area of the sphere and we know that the charge over there,"},{"Start":"06:17.195 ","End":"06:23.435","Text":"so Sigma b, the bound surface charge is equal,"},{"Start":"06:23.435 ","End":"06:27.115","Text":"we just calculated it, Ar."},{"Start":"06:27.115 ","End":"06:31.145","Text":"Now we\u0027re just going to calculate the electric field."},{"Start":"06:31.145 ","End":"06:41.010","Text":"Let\u0027s look at when r is greater than capital R. Now we\u0027re located outside of our sphere."},{"Start":"06:41.010 ","End":"06:48.320","Text":"Using Gauss\u0027s law, so we have that E multiplied by the surface area,"},{"Start":"06:48.320 ","End":"06:51.560","Text":"which of course is equal to 4 Pi r^2,"},{"Start":"06:51.560 ","End":"06:53.105","Text":"the surface area of a sphere,"},{"Start":"06:53.105 ","End":"07:00.945","Text":"which is equal to the charge inside divided by Epsilon_0."},{"Start":"07:00.945 ","End":"07:05.645","Text":"First of all, let\u0027s see what our charge inside is."},{"Start":"07:05.645 ","End":"07:07.790","Text":"What is our Qin."},{"Start":"07:07.790 ","End":"07:13.545","Text":"Qin is equal to the total bound charge."},{"Start":"07:13.545 ","End":"07:21.710","Text":"Q bound total, the total bound charge located inside."},{"Start":"07:21.710 ","End":"07:26.380","Text":"That includes our Rho b and our Sigma b."},{"Start":"07:26.380 ","End":"07:28.785","Text":"Let\u0027s calculate this."},{"Start":"07:28.785 ","End":"07:31.055","Text":"We have sigma b,"},{"Start":"07:31.055 ","End":"07:34.190","Text":"which is our charged density per unit area"},{"Start":"07:34.190 ","End":"07:38.210","Text":"and then that means we have to multiply it by the surface area so"},{"Start":"07:38.210 ","End":"07:42.800","Text":"the surface area of the sphere is"},{"Start":"07:42.800 ","End":"07:49.100","Text":"simply going to be equal to 4 Pi R^2."},{"Start":"07:49.100 ","End":"07:52.355","Text":"Because the spheres of radius R,"},{"Start":"07:52.355 ","End":"07:56.975","Text":"and here we\u0027re using r where we\u0027re located somewhere here,"},{"Start":"07:56.975 ","End":"08:05.125","Text":"this is r. Then we have to add on our Rho."},{"Start":"08:05.125 ","End":"08:11.900","Text":"Our Rho b is our bound charge density per unit volume so then we"},{"Start":"08:11.900 ","End":"08:18.350","Text":"have to multiply it by the volume of a sphere so the volume of a sphere is of course,"},{"Start":"08:18.350 ","End":"08:21.430","Text":"4/3 Pi r^3,"},{"Start":"08:21.430 ","End":"08:23.900","Text":"where of course we\u0027re using capital R because we are"},{"Start":"08:23.900 ","End":"08:27.330","Text":"taking the volume of the entire sphere."},{"Start":"08:27.620 ","End":"08:35.435","Text":"Now let\u0027s sub in our values for Sigma b. Sigma b we know is Ar."},{"Start":"08:35.435 ","End":"08:40.620","Text":"We have 4 Pi A and then we have"},{"Start":"08:40.620 ","End":"08:46.875","Text":"r multiplied by r^2 so we have r^3 plus our Rho b."},{"Start":"08:46.875 ","End":"08:50.805","Text":"Our Rho b is negative 3A so we have"},{"Start":"08:50.805 ","End":"08:59.565","Text":"negative 3A and then it\u0027s multiplied by 4 divided by 3 s we can just cancel that out."},{"Start":"08:59.565 ","End":"09:09.940","Text":"Let\u0027s put negative 4 Pi A multiplied by r^3."},{"Start":"09:12.620 ","End":"09:19.620","Text":"What we can see is that this is equal to 0."},{"Start":"09:19.620 ","End":"09:22.405","Text":"Why is this equal to 0?"},{"Start":"09:22.405 ","End":"09:24.590","Text":"Because we remember that all of this is"},{"Start":"09:24.590 ","End":"09:31.320","Text":"a dielectric material so we have dipoles. What is a dipole?"},{"Start":"09:31.320 ","End":"09:36.470","Text":"It\u0027s some positive charge coming together with a negative charge."},{"Start":"09:36.470 ","End":"09:42.170","Text":"We know that we have an equal number of positive and negative charges,"},{"Start":"09:42.170 ","End":"09:51.060","Text":"so that means that the net total charge inside this Gaussian surface,"},{"Start":"09:51.060 ","End":"09:54.650","Text":"the gray dots is going to be equal to 0 because all of"},{"Start":"09:54.650 ","End":"09:58.920","Text":"the pluses cancel out with all of the minuses."},{"Start":"09:59.810 ","End":"10:05.630","Text":"Therefore, we know that the electric field"},{"Start":"10:05.630 ","End":"10:11.655","Text":"at this position where r is greater than R is equal to 0."},{"Start":"10:11.655 ","End":"10:17.810","Text":"Now let\u0027s look at the electric field in the region where r is smaller than"},{"Start":"10:17.810 ","End":"10:24.890","Text":"R. Now our Gaussian surface"},{"Start":"10:24.890 ","End":"10:29.765","Text":"is somewhere located inside the sphere itself."},{"Start":"10:29.765 ","End":"10:35.480","Text":"Again, we have E multiplied by the surface area."},{"Start":"10:35.480 ","End":"10:39.030","Text":"This is our r this time."},{"Start":"10:41.230 ","End":"10:47.610","Text":"The surface area is going to be 4 Pi r^2 and"},{"Start":"10:47.610 ","End":"10:53.955","Text":"this is again equal to Qin divided by Epsilon_0."},{"Start":"10:53.955 ","End":"10:58.785","Text":"Now we have to see what Qin is equal to."},{"Start":"10:58.785 ","End":"11:01.355","Text":"We can see that we\u0027re located inside the sphere,"},{"Start":"11:01.355 ","End":"11:08.145","Text":"which means that our Sigma b isn\u0027t affecting what\u0027s going on inside here."},{"Start":"11:08.145 ","End":"11:14.360","Text":"Our Gaussian surface doesn\u0027t include our Sigma b so we just have to look at our Rho b."},{"Start":"11:14.360 ","End":"11:20.615","Text":"We have, our Rho b multiplied by the volume because Rho is per unit volume,"},{"Start":"11:20.615 ","End":"11:28.590","Text":"where of course the volume is 4/3 Pi and this time it\u0027s r^3 again,"},{"Start":"11:28.590 ","End":"11:32.300","Text":"it\u0027s a lowercase r because"},{"Start":"11:32.300 ","End":"11:36.725","Text":"not the entire sphere is incorporated in this Gaussian surface."},{"Start":"11:36.725 ","End":"11:41.330","Text":"Rho b, we already saw is equal to negative 3A,"},{"Start":"11:41.330 ","End":"11:44.390","Text":"so we have negative and then the 3 and the 3 cancel out,"},{"Start":"11:44.390 ","End":"11:51.610","Text":"so negative 4 Pi Ar^3."},{"Start":"11:51.610 ","End":"11:55.185","Text":"This is our Qin so let\u0027s scroll down."},{"Start":"11:55.185 ","End":"12:02.820","Text":"Then we have E multiplied by 4 Pi r^2 is equal to Qin,"},{"Start":"12:02.820 ","End":"12:09.990","Text":"which is equal to negative 4 Pi Ar^3 divided by Epsilon_0."},{"Start":"12:09.990 ","End":"12:13.665","Text":"We can cancel out the 4,"},{"Start":"12:13.665 ","End":"12:16.935","Text":"the Pi and the r^2."},{"Start":"12:16.935 ","End":"12:23.495","Text":"Then therefore we\u0027re left with our electric field inside the sphere,"},{"Start":"12:23.495 ","End":"12:32.490","Text":"which is equal to negative Ar divided by Epsilon_0."},{"Start":"12:32.490 ","End":"12:35.554","Text":"If we want to add on a vector quantity,"},{"Start":"12:35.554 ","End":"12:39.080","Text":"this is a vector so we\u0027re adding in our direction,"},{"Start":"12:39.080 ","End":"12:43.410","Text":"we know that this is of course in the radial direction."},{"Start":"12:43.600 ","End":"12:50.670","Text":"This is the answer to question Number 2 and that is the end of our lesson."}],"ID":22317},{"Watched":false,"Name":"Exercise 4","Duration":"14m 5s","ChapterTopicVideoID":21303,"CourseChapterTopicPlaylistID":99475,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.055","Text":"Hello, in this lesson,"},{"Start":"00:02.055 ","End":"00:04.560","Text":"we\u0027re going to be answering the following question."},{"Start":"00:04.560 ","End":"00:11.700","Text":"A cylinder of radius R and length L is uniformly polarized along its axis of symmetry."},{"Start":"00:11.700 ","End":"00:14.670","Text":"Calculate the bound charge densities and draw"},{"Start":"00:14.670 ","End":"00:18.090","Text":"the field lines under the following circumstances."},{"Start":"00:18.090 ","End":"00:20.205","Text":"Soon we\u0027re going to look at this,"},{"Start":"00:20.205 ","End":"00:23.580","Text":"let\u0027s just first write out the basics."},{"Start":"00:23.580 ","End":"00:27.465","Text":"First of all, we can see that the base of the cylinder,"},{"Start":"00:27.465 ","End":"00:32.775","Text":"it has 1 base in this direction and 1 base in this direction."},{"Start":"00:32.775 ","End":"00:36.375","Text":"The normal vectors to each of these bases,"},{"Start":"00:36.375 ","End":"00:38.420","Text":"we can see it\u0027s along the z-axis."},{"Start":"00:38.420 ","End":"00:42.590","Text":"Here we can see the normal vector is in the positive z direction."},{"Start":"00:42.590 ","End":"00:49.370","Text":"And here we can see that the normal vector is in the negative z direction."},{"Start":"00:49.370 ","End":"00:56.014","Text":"Now what we can write is we can write the polarization vector,"},{"Start":"00:56.014 ","End":"01:02.300","Text":"P is equal to some constant, it\u0027s uniformly polarized,"},{"Start":"01:02.300 ","End":"01:07.350","Text":"so let\u0027s call it p zero along its axis of symmetry,"},{"Start":"01:07.350 ","End":"01:09.900","Text":"its axis of symmetry is z,"},{"Start":"01:09.900 ","End":"01:14.620","Text":"so in the z direction."},{"Start":"01:14.860 ","End":"01:20.000","Text":"Let\u0027s write out our Sigma b first of all,"},{"Start":"01:20.000 ","End":"01:23.650","Text":"our balance charge per unit area, Sigma b."},{"Start":"01:23.650 ","End":"01:32.545","Text":"As we know, this is equal to our polarization vector dot product with our normal."},{"Start":"01:32.545 ","End":"01:37.330","Text":"Of course, our normal vector along the cylinder,"},{"Start":"01:37.330 ","End":"01:46.218","Text":"along the shape of the cylinder and not at its base is in the radial direction."},{"Start":"01:46.218 ","End":"01:56.205","Text":"Now let us carry on this is going to be equal to we have on this base over here,"},{"Start":"01:56.205 ","End":"02:04.470","Text":"we\u0027ll have p_0 dot-product with our z hat."},{"Start":"02:04.470 ","End":"02:08.775","Text":"Sorry, p naught z hat is our p vector,"},{"Start":"02:08.775 ","End":"02:11.085","Text":"dot product with our normal."},{"Start":"02:11.085 ","End":"02:19.210","Text":"Our normal over here at the bottom base is equal to z hat, positive z hat."},{"Start":"02:19.210 ","End":"02:23.414","Text":"This z hat dot z hat is equal to 1,"},{"Start":"02:23.414 ","End":"02:28.780","Text":"what we\u0027re going to have is that this is equal to p_0."},{"Start":"02:28.780 ","End":"02:36.774","Text":"Then our next option is this base over here what we\u0027re going to have is a p vector,"},{"Start":"02:36.774 ","End":"02:42.665","Text":"so p naught z hat multiplied or a dot product with the normal over here,"},{"Start":"02:42.665 ","End":"02:45.240","Text":"which is negative z hat."},{"Start":"02:45.240 ","End":"02:50.635","Text":"Z hat dot negative z hat is just equal to negative 1."},{"Start":"02:50.635 ","End":"02:55.460","Text":"So here it will equal to negative p_0."},{"Start":"02:55.460 ","End":"03:00.180","Text":"And then around the cylinder we\u0027ll"},{"Start":"03:00.180 ","End":"03:07.315","Text":"have p_0 z hat dot product with our r hat."},{"Start":"03:07.315 ","End":"03:13.415","Text":"Which of course we know is equal to zero because z hat and r hat,"},{"Start":"03:13.415 ","End":"03:18.920","Text":"we can also see from the diagram are perpendicular to 1 another all the time."},{"Start":"03:18.920 ","End":"03:29.070","Text":"These are our 3 options for our balanced charge density per unit area."},{"Start":"03:29.540 ","End":"03:34.470","Text":"Now what about our row b."},{"Start":"03:34.470 ","End":"03:37.670","Text":"Our bound charge density per unit volume."},{"Start":"03:37.670 ","End":"03:44.475","Text":"As we know that this is equal to negative Nabla dot our polarization vector"},{"Start":"03:44.475 ","End":"03:53.090","Text":"p. We\u0027re being told that p is a constant."},{"Start":"03:53.090 ","End":"03:55.670","Text":"We\u0027re being told that the radius is uniformly"},{"Start":"03:55.670 ","End":"03:59.975","Text":"polarized which means that the polarization density is a constant."},{"Start":"03:59.975 ","End":"04:02.900","Text":"Which means that if we take the divergence,"},{"Start":"04:02.900 ","End":"04:06.130","Text":"it is going to be equal to zero."},{"Start":"04:06.130 ","End":"04:10.325","Text":"We can also see that if you try and take"},{"Start":"04:10.325 ","End":"04:16.205","Text":"the Nabla dot product of p 0 z hat, you will get 0."},{"Start":"04:16.205 ","End":"04:21.065","Text":"We have no volumetric bound charge density."},{"Start":"04:21.065 ","End":"04:25.357","Text":"Now let\u0027s look at question number 1,"},{"Start":"04:25.357 ","End":"04:27.545","Text":"now that we have a charge densities."},{"Start":"04:27.545 ","End":"04:30.800","Text":"Question number 1 is saying that r is"},{"Start":"04:30.800 ","End":"04:35.500","Text":"significantly smaller than l. That means that our radius is much - much"},{"Start":"04:35.500 ","End":"04:39.320","Text":"smaller than l. That means that our cylinder will look"},{"Start":"04:39.320 ","End":"04:46.350","Text":"something like this, long and thin."},{"Start":"04:47.230 ","End":"04:56.204","Text":"What we saw is that we only have some charge density on the bases."},{"Start":"04:56.204 ","End":"05:00.357","Text":"Where I\u0027ve drawn the green,"},{"Start":"05:00.357 ","End":"05:03.575","Text":"here, at z hat,"},{"Start":"05:03.575 ","End":"05:07.640","Text":"we have a positive charge and a negative z hat,"},{"Start":"05:07.640 ","End":"05:09.620","Text":"where the normal is negative z hat,"},{"Start":"05:09.620 ","End":"05:12.575","Text":"so that\u0027s this base we have a negative charge."},{"Start":"05:12.575 ","End":"05:18.875","Text":"In question 1 we can see that on this side over here,"},{"Start":"05:18.875 ","End":"05:22.340","Text":"we\u0027re going to have this negative charge and then we\u0027re going to"},{"Start":"05:22.340 ","End":"05:27.335","Text":"have like a thin wire or some hose."},{"Start":"05:27.335 ","End":"05:30.605","Text":"Over here on this end,"},{"Start":"05:30.605 ","End":"05:37.055","Text":"we\u0027re going to have a positive charge so we have a dipole essentially."},{"Start":"05:37.055 ","End":"05:40.760","Text":"I just put a square around this and"},{"Start":"05:40.760 ","End":"05:43.700","Text":"this because that\u0027s the answer to the charge densities,"},{"Start":"05:43.700 ","End":"05:46.235","Text":"and now we\u0027re talking about the field lines."},{"Start":"05:46.235 ","End":"05:50.330","Text":"First, in order to draw the field lines,"},{"Start":"05:50.330 ","End":"05:54.350","Text":"we\u0027re going to write out our charge q."},{"Start":"05:54.350 ","End":"05:57.260","Text":"On this end, over here,"},{"Start":"05:57.260 ","End":"05:58.475","Text":"on the negative end,"},{"Start":"05:58.475 ","End":"06:06.230","Text":"we\u0027re going to have that q is equal to Sigma multiplied by the total area."},{"Start":"06:06.230 ","End":"06:12.155","Text":"We have that our Sigma over here is equal to negative p,"},{"Start":"06:12.155 ","End":"06:16.745","Text":"o and our area is equal to the area of a circle,"},{"Start":"06:16.745 ","End":"06:21.575","Text":"so that is Pi r^2."},{"Start":"06:21.575 ","End":"06:26.210","Text":"Then over here, our q is going to be equal"},{"Start":"06:26.210 ","End":"06:31.070","Text":"to sigma multiplied by the area and here our sigma is positive,"},{"Start":"06:31.070 ","End":"06:38.930","Text":"so we have p o multiplied by Pi r^2."},{"Start":"06:38.930 ","End":"06:46.085","Text":"Now our electric field lines are going to look like the electric field lines of a dipole."},{"Start":"06:46.085 ","End":"06:49.800","Text":"The electric field for a dipole."},{"Start":"06:49.800 ","End":"06:53.915","Text":"First of all, we have our dipole moment so p,"},{"Start":"06:53.915 ","End":"06:57.710","Text":"which is equal to our charge q,"},{"Start":"06:57.710 ","End":"07:05.734","Text":"of course, the positive version multiplied by the vector from negative to positive."},{"Start":"07:05.734 ","End":"07:07.700","Text":"That\u0027s how it goes in dipoles."},{"Start":"07:07.700 ","End":"07:10.435","Text":"Remember always from negative to positive in a dipole."},{"Start":"07:10.435 ","End":"07:16.750","Text":"And of course we know that this rightwards direction is the z direction."},{"Start":"07:18.710 ","End":"07:22.845","Text":"The distance between the 2 is L of course."},{"Start":"07:22.845 ","End":"07:27.420","Text":"We can see that it\u0027s in the positive direction, so we can write this as q,"},{"Start":"07:27.420 ","End":"07:29.860","Text":"which is equal to P0piR^2"},{"Start":"07:30.690 ","End":"07:38.190","Text":"multiplied by the length or the distance between the 2 charges,"},{"Start":"07:38.190 ","End":"07:46.300","Text":"which is just the length of this cylinder and then going in the z direction."},{"Start":"07:47.630 ","End":"07:52.350","Text":"Then the electric field is just the electric field of a dipole moment,"},{"Start":"07:52.350 ","End":"07:55.990","Text":"where this is the dipole moment."},{"Start":"07:57.290 ","End":"08:01.055","Text":"Now let\u0027s look at number 2."},{"Start":"08:01.055 ","End":"08:05.370","Text":"Number 2 is where L is much smaller than R,"},{"Start":"08:05.370 ","End":"08:13.720","Text":"so the length is much smaller R. That means that we have something like so."},{"Start":"08:14.630 ","End":"08:18.750","Text":"It looks more like a coin or maybe."},{"Start":"08:18.750 ","End":"08:23.470","Text":"The radius is much larger than the length."},{"Start":"08:23.900 ","End":"08:30.735","Text":"What we have is 2 circular charged plates."},{"Start":"08:30.735 ","End":"08:35.760","Text":"Of course here we have the negatively charged plate,"},{"Start":"08:35.760 ","End":"08:41.735","Text":"and over here we have the positively charged plate."},{"Start":"08:41.735 ","End":"08:45.120","Text":"This is working like a circular capacitor."},{"Start":"08:45.120 ","End":"08:50.064","Text":"If you\u0027re trying to work out the electric field with a capacitor,"},{"Start":"08:50.064 ","End":"08:53.260","Text":"you can just work it out as if it\u0027s the electric field of"},{"Start":"08:53.260 ","End":"08:57.440","Text":"2 infinite planes that are close together."},{"Start":"08:57.600 ","End":"09:01.090","Text":"Number 1, what we can see is that"},{"Start":"09:01.090 ","End":"09:05.295","Text":"the electric field from the positive plate is going in this direction,"},{"Start":"09:05.295 ","End":"09:11.021","Text":"and the electric field from the negative plate is also going in this direction,"},{"Start":"09:11.021 ","End":"09:17.540","Text":"so we multiply the electric field of 2 infinite planes by 2."},{"Start":"09:17.540 ","End":"09:21.730","Text":"Therefore we can write that the electric field is equal"},{"Start":"09:21.730 ","End":"09:28.347","Text":"to 2 multiplied by Sigma divided by 2 Epsilon naughts,"},{"Start":"09:28.347 ","End":"09:32.490","Text":"where sigma divided by 2 Epsilon_0 is the electric field of an infinite plane,"},{"Start":"09:32.490 ","End":"09:34.925","Text":"and then we\u0027re multiplying it by 2."},{"Start":"09:34.925 ","End":"09:43.975","Text":"That is just going to be equal to some direction."},{"Start":"09:43.975 ","End":"09:50.400","Text":"Over here what we\u0027re going to have is Sigma which is simply equal"},{"Start":"09:50.400 ","End":"09:57.305","Text":"to this P_o divided by Epsilon_0."},{"Start":"09:57.305 ","End":"10:06.783","Text":"Of course this is going in the negative z direction,"},{"Start":"10:06.783 ","End":"10:10.855","Text":"because we can see that the arrows are pointing in the leftward direction,"},{"Start":"10:10.855 ","End":"10:15.047","Text":"but the positive z direction is in the rightwards direction,"},{"Start":"10:15.047 ","End":"10:17.470","Text":"so here we have a negative."},{"Start":"10:18.020 ","End":"10:21.930","Text":"This is the electric field, of course,"},{"Start":"10:21.930 ","End":"10:26.300","Text":"when r is smaller than L, r,"},{"Start":"10:26.300 ","End":"10:31.275","Text":"so we\u0027re located somewhere inside the cylinder."},{"Start":"10:31.275 ","End":"10:35.847","Text":"If we\u0027re located above or below the 2 bases,"},{"Start":"10:35.847 ","End":"10:40.818","Text":"so here I did where r is between negative L over 2 and positive L over 2,"},{"Start":"10:40.818 ","End":"10:45.470","Text":"so when we\u0027re located somewhere in here,"},{"Start":"10:45.470 ","End":"10:48.670","Text":"as in not outside of the cylinder,"},{"Start":"10:48.670 ","End":"10:50.360","Text":"if this is our 0 point."},{"Start":"10:50.360 ","End":"10:55.030","Text":"Over here, if we\u0027re located outside of the cylinder,"},{"Start":"10:55.030 ","End":"10:58.320","Text":"so here or here,"},{"Start":"10:58.320 ","End":"11:00.835","Text":"so if I position it\u0027s greater than L over 2,"},{"Start":"11:00.835 ","End":"11:02.140","Text":"so that\u0027s somewhere here,"},{"Start":"11:02.140 ","End":"11:04.284","Text":"or less than negative L over 2,"},{"Start":"11:04.284 ","End":"11:06.080","Text":"so that\u0027s somewhere over here,"},{"Start":"11:06.080 ","End":"11:09.815","Text":"then the electric field is simply equal to 0."},{"Start":"11:09.815 ","End":"11:13.305","Text":"Why is that? Because over here,"},{"Start":"11:13.305 ","End":"11:16.405","Text":"from the negative charges we have"},{"Start":"11:16.405 ","End":"11:21.455","Text":"the electric field going like so and from the positively charged plate,"},{"Start":"11:21.455 ","End":"11:25.355","Text":"we have the electric field going like so and then just together,"},{"Start":"11:25.355 ","End":"11:27.940","Text":"they just cancel each other out."},{"Start":"11:28.080 ","End":"11:31.905","Text":"Now let\u0027s look at question number 3."},{"Start":"11:31.905 ","End":"11:39.260","Text":"This is where the radius is approximately equal to the length."},{"Start":"11:39.770 ","End":"11:42.160","Text":"In an example like this,"},{"Start":"11:42.160 ","End":"11:45.960","Text":"it just means that we have the regular old cylinder"},{"Start":"11:45.960 ","End":"11:49.620","Text":"where there\u0027s nothing special about it."},{"Start":"11:49.620 ","End":"11:51.570","Text":"It\u0027s completely in proportion."},{"Start":"11:51.570 ","End":"11:53.190","Text":"First of all, again,"},{"Start":"11:53.190 ","End":"11:55.910","Text":"we know that this is the negatively charged plate,"},{"Start":"11:55.910 ","End":"11:59.495","Text":"and this is the positively charged plate and"},{"Start":"11:59.495 ","End":"12:03.720","Text":"there\u0027s some distance L away from one another."},{"Start":"12:03.720 ","End":"12:07.515","Text":"We don\u0027t know how to find the electric field everywhere,"},{"Start":"12:07.515 ","End":"12:12.580","Text":"but we can find the electric field along the axis of symmetry."},{"Start":"12:12.580 ","End":"12:17.240","Text":"The electric field in this case is just a superposition of"},{"Start":"12:17.240 ","End":"12:22.875","Text":"the negative plate causing an electric field at some point,"},{"Start":"12:22.875 ","End":"12:26.680","Text":"and the positive plate causing an electric field at"},{"Start":"12:26.680 ","End":"12:31.540","Text":"that point and we just superimpose those 2 values."},{"Start":"12:31.910 ","End":"12:34.200","Text":"Quickly before we finish,"},{"Start":"12:34.200 ","End":"12:38.085","Text":"let\u0027s just draw the field lines because we didn\u0027t do that."},{"Start":"12:38.085 ","End":"12:41.460","Text":"The field lines for this over here,"},{"Start":"12:41.460 ","End":"12:43.895","Text":"case number 1 is of a dipole."},{"Start":"12:43.895 ","End":"12:49.305","Text":"We have field eyes going from positive to negative."},{"Start":"12:49.305 ","End":"12:51.830","Text":"Like so."},{"Start":"12:51.830 ","End":"12:58.450","Text":"Just going like this from positive to negative,"},{"Start":"12:58.450 ","End":"13:03.330","Text":"so it\u0027s going in this direction."},{"Start":"13:03.330 ","End":"13:07.010","Text":"Then over here, the electric field lines."},{"Start":"13:07.010 ","End":"13:10.720","Text":"Within this coin shape,"},{"Start":"13:10.720 ","End":"13:13.980","Text":"the electric field lines are just going straight."},{"Start":"13:13.980 ","End":"13:16.650","Text":"Then around the edge,"},{"Start":"13:16.650 ","End":"13:20.392","Text":"we have a little bit of this going along,"},{"Start":"13:20.392 ","End":"13:24.105","Text":"again, from positive to negative."},{"Start":"13:24.105 ","End":"13:30.165","Text":"We have these shapes like so all around the edge."},{"Start":"13:30.165 ","End":"13:31.635","Text":"In case number 3,"},{"Start":"13:31.635 ","End":"13:33.810","Text":"where we have the irregular old cylinder,"},{"Start":"13:33.810 ","End":"13:39.575","Text":"we have a combination of both."},{"Start":"13:39.575 ","End":"13:44.565","Text":"We\u0027re also going to have electric field lines going straight,"},{"Start":"13:44.565 ","End":"13:47.803","Text":"between the 2 disks or the 2 bases,"},{"Start":"13:47.803 ","End":"13:49.608","Text":"from positive to negative,"},{"Start":"13:49.608 ","End":"13:52.545","Text":"and we\u0027re also going to get in the dipole,"},{"Start":"13:52.545 ","End":"13:59.445","Text":"these electric field lines that are looping around,"},{"Start":"13:59.445 ","End":"14:03.060","Text":"from positive to negative."},{"Start":"14:03.060 ","End":"14:06.260","Text":"That is the end of this lesson."}],"ID":21383},{"Watched":false,"Name":"Exercise 5","Duration":"28m 35s","ChapterTopicVideoID":21470,"CourseChapterTopicPlaylistID":99475,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.100","Text":"Hello. In this lesson,"},{"Start":"00:02.100 ","End":"00:08.255","Text":"we\u0027re going to see how to calculate the electric field of a uniformly polarized sphere."},{"Start":"00:08.255 ","End":"00:12.930","Text":"Here we have a uniformly polarized sphere of radius R,"},{"Start":"00:12.930 ","End":"00:24.040","Text":"where its vector of polarization P is equal to P_0 in the Z direction."},{"Start":"00:24.080 ","End":"00:28.260","Text":"Let\u0027s work out the bound charge distributions."},{"Start":"00:28.260 ","End":"00:32.565","Text":"The bound charge distribution per unit volume."},{"Start":"00:32.565 ","End":"00:38.415","Text":"We know that this is going to be equal to 0 because this is equal to negative"},{"Start":"00:38.415 ","End":"00:44.375","Text":"Nabla dot P. Because P is a constant,"},{"Start":"00:44.375 ","End":"00:46.160","Text":"when we take the divergence of it,"},{"Start":"00:46.160 ","End":"00:48.935","Text":"that\u0027s going to be equal to 0."},{"Start":"00:48.935 ","End":"00:54.305","Text":"Then if we look at the bound charge per unit area,"},{"Start":"00:54.305 ","End":"01:02.580","Text":"so that we know is equal to P dot n. Our p over here is p_0 in"},{"Start":"01:02.580 ","End":"01:06.860","Text":"the Z direction so"},{"Start":"01:06.860 ","End":"01:11.930","Text":"we can see that the normal vector of a sphere is always going to be r hat."},{"Start":"01:12.800 ","End":"01:16.380","Text":"What is z dot r hat?"},{"Start":"01:16.380 ","End":"01:22.365","Text":"If we have a vector z in this direction,"},{"Start":"01:22.365 ","End":"01:29.435","Text":"this is z hat and our vector r hat is in some radial direction."},{"Start":"01:29.435 ","End":"01:34.670","Text":"Let\u0027s say that the angle between them is feet."},{"Start":"01:34.670 ","End":"01:41.240","Text":"As we know, the dot-product between z hat and r hat is equal to the magnitude of"},{"Start":"01:41.240 ","End":"01:44.345","Text":"z hat multiplied by the magnitude of"},{"Start":"01:44.345 ","End":"01:48.745","Text":"r hat multiplied by cosine of the angle between the 2."},{"Start":"01:48.745 ","End":"01:53.900","Text":"The magnitude of any unit vector be it z hat or r-hat is 1."},{"Start":"01:53.900 ","End":"01:56.480","Text":"We have p_0 multiplied by 1,"},{"Start":"01:56.480 ","End":"02:02.640","Text":"multiplied by cosine of the angle between the 2, which is Phi."},{"Start":"02:04.070 ","End":"02:09.770","Text":"That means that the charges on the sphere look something like this."},{"Start":"02:09.770 ","End":"02:14.035","Text":"If we draw over here that this is our axis,"},{"Start":"02:14.035 ","End":"02:21.540","Text":"and over here the angle is 0 and over here the angle is 90 degrees."},{"Start":"02:21.890 ","End":"02:26.945","Text":"We can see that at Phi is equal to 0."},{"Start":"02:26.945 ","End":"02:30.155","Text":"Cosine of Phi is equal to 1."},{"Start":"02:30.155 ","End":"02:36.200","Text":"We have our p_0 polarization density over here."},{"Start":"02:36.200 ","End":"02:39.005","Text":"The greatest amount of polarization density."},{"Start":"02:39.005 ","End":"02:43.130","Text":"As our angle gets larger and larger,"},{"Start":"02:43.130 ","End":"02:45.650","Text":"cosine of the angle become smaller."},{"Start":"02:45.650 ","End":"02:51.530","Text":"That means that our charged distribution is less dense."},{"Start":"02:51.530 ","End":"02:54.230","Text":"Then when we get to 90 degrees,"},{"Start":"02:54.230 ","End":"02:57.710","Text":"so cosine of 90 is equal to 0 and then we have"},{"Start":"02:57.710 ","End":"03:02.600","Text":"a 0 charge density over here and also over here."},{"Start":"03:02.600 ","End":"03:07.160","Text":"Then as we go into the negative area, so again,"},{"Start":"03:07.160 ","End":"03:11.965","Text":"we have the same pattern where as the angle reaches a 180,"},{"Start":"03:11.965 ","End":"03:16.160","Text":"so cosine of 180 degrees is going to be 1 so then we"},{"Start":"03:16.160 ","End":"03:20.300","Text":"have the greatest charged distribution but negative charge distribution."},{"Start":"03:20.300 ","End":"03:23.405","Text":"Then again, when we reach 270,"},{"Start":"03:23.405 ","End":"03:26.610","Text":"we get a 0 charge distribution."},{"Start":"03:27.440 ","End":"03:29.620","Text":"Now our question is,"},{"Start":"03:29.620 ","End":"03:33.490","Text":"how to calculate the electric field in this type of case."},{"Start":"03:33.490 ","End":"03:35.440","Text":"In order to do this,"},{"Start":"03:35.440 ","End":"03:43.400","Text":"there\u0027s a trick where we take 2 spheres and we overlap them slightly."},{"Start":"03:43.610 ","End":"03:46.254","Text":"Here we have 2 spheres,"},{"Start":"03:46.254 ","End":"03:51.820","Text":"1 sphere has charged density per unit volume plus Rho,"},{"Start":"03:51.820 ","End":"03:56.980","Text":"and the other sphere has charged density per unit volume negative Rho."},{"Start":"03:56.980 ","End":"03:59.260","Text":"Then we overlap them,"},{"Start":"03:59.260 ","End":"04:02.380","Text":"but their centers aren\u0027t overlapping."},{"Start":"04:02.380 ","End":"04:07.270","Text":"We can see that we get this type of situation over here,"},{"Start":"04:07.270 ","End":"04:09.390","Text":"where the distance between"},{"Start":"04:09.390 ","End":"04:16.110","Text":"each sphere\u0027s center is a distance d away from the sphere\u0027s center."},{"Start":"04:17.480 ","End":"04:19.805","Text":"There\u0027s this distance d,"},{"Start":"04:19.805 ","End":"04:22.415","Text":"and we have to set a condition that"},{"Start":"04:22.415 ","End":"04:31.860","Text":"this distance d is much smaller than the radius of our sphere."},{"Start":"04:32.500 ","End":"04:37.114","Text":"We can see that these 2 spheres have the same charge density,"},{"Start":"04:37.114 ","End":"04:39.635","Text":"just 1 in plus and 1 in minus."},{"Start":"04:39.635 ","End":"04:41.990","Text":"In this area of overlap,"},{"Start":"04:41.990 ","End":"04:46.045","Text":"so all of this area over here,"},{"Start":"04:46.045 ","End":"04:50.585","Text":"the charges are going to cancel each other out."},{"Start":"04:50.585 ","End":"04:54.350","Text":"All of that will be left are these positive charges"},{"Start":"04:54.350 ","End":"04:58.415","Text":"over here on the outskirts that have no negative charges to cancel out with."},{"Start":"04:58.415 ","End":"05:05.165","Text":"These negative charges on this outskirt that have no positive charges to cancel out with."},{"Start":"05:05.165 ","End":"05:13.035","Text":"We can see that there\u0027s no charges at this overlap."},{"Start":"05:13.035 ","End":"05:18.870","Text":"We can see that the charge density at these points is equal to 0."},{"Start":"05:20.240 ","End":"05:23.075","Text":"This is the diagram that we\u0027re left with."},{"Start":"05:23.075 ","End":"05:25.670","Text":"The charges inside have canceled out,"},{"Start":"05:25.670 ","End":"05:30.330","Text":"and we\u0027re left with these positive and negative charges over here."},{"Start":"05:31.610 ","End":"05:36.460","Text":"At these points over here, and over here,"},{"Start":"05:36.460 ","End":"05:41.080","Text":"we can see that there\u0027s more volume of the sphere left."},{"Start":"05:41.080 ","End":"05:47.020","Text":"That represents a higher charge distribution or a larger charge density"},{"Start":"05:47.020 ","End":"05:53.870","Text":"over here and over here than we have at these tips over here."},{"Start":"05:54.950 ","End":"05:58.665","Text":"Now we\u0027re going to do our trick."},{"Start":"05:58.665 ","End":"06:05.505","Text":"Here is our origin for this question and let\u0027s choose some volume over here."},{"Start":"06:05.505 ","End":"06:08.160","Text":"Let\u0027s call this dq."},{"Start":"06:08.160 ","End":"06:14.800","Text":"We have this vector over here pointing to dq."},{"Start":"06:14.800 ","End":"06:21.370","Text":"What we can see, let\u0027s write it over here is that dq is equal"},{"Start":"06:21.370 ","End":"06:29.140","Text":"to Rho multiplied by the unit volume dv."},{"Start":"06:29.140 ","End":"06:35.080","Text":"However, this we\u0027re saying is equal to this example over here,"},{"Start":"06:35.080 ","End":"06:39.725","Text":"where all we have is this surface charge,"},{"Start":"06:39.725 ","End":"06:46.910","Text":"so that means that this is equal to Sigma multiplied by the unit area ds."},{"Start":"06:47.520 ","End":"06:54.385","Text":"Because we\u0027re using spherical coordinates because we have a sphere. Let\u0027s write this out."},{"Start":"06:54.385 ","End":"06:58.850","Text":"We have Rho and then dv in spherical coordinates is equal"},{"Start":"06:58.850 ","End":"07:04.065","Text":"to R squared multiplied by the sine of Phi"},{"Start":"07:04.065 ","End":"07:09.870","Text":"multiplied by dr d Theta d Phi"},{"Start":"07:09.870 ","End":"07:14.839","Text":"and this is equal to Sigma multiplied by ds in spherical coordinates,"},{"Start":"07:14.839 ","End":"07:20.570","Text":"where ds is equal to R squared multiplied by the sine of"},{"Start":"07:20.570 ","End":"07:27.520","Text":"Phi multiplied by d Theta d Phi."},{"Start":"07:28.220 ","End":"07:33.650","Text":"In the dv, I substitute it in this capital R,"},{"Start":"07:33.650 ","End":"07:37.775","Text":"the radius of the sphere and that\u0027s because I\u0027m trying to find"},{"Start":"07:37.775 ","End":"07:42.830","Text":"the charge dq along the outskirts,"},{"Start":"07:42.830 ","End":"07:44.345","Text":"on the surface of the sphere."},{"Start":"07:44.345 ","End":"07:48.530","Text":"Because obviously, dv could be anywhere inside these spheres,"},{"Start":"07:48.530 ","End":"07:49.730","Text":"okay, anywhere in the volume."},{"Start":"07:49.730 ","End":"07:54.140","Text":"But I\u0027m looking specifically on the edge because that\u0027s what I actually have over here."},{"Start":"07:54.140 ","End":"07:58.580","Text":"I have a bound charge along the surface area."},{"Start":"07:58.580 ","End":"08:01.919","Text":"That\u0027s why I substituted in the radius."},{"Start":"08:02.750 ","End":"08:07.609","Text":"Now we can cancel R squared on both sides."},{"Start":"08:07.609 ","End":"08:14.165","Text":"We can cancel sine of Phi on both sides and d Theta d Phi on both sides."},{"Start":"08:14.165 ","End":"08:21.330","Text":"Now what we have is that Rho dr is equal to Sigma."},{"Start":"08:21.560 ","End":"08:31.310","Text":"As we know, our Sigma is equal to p_0 multiplied by the cosine of Phi."},{"Start":"08:31.640 ","End":"08:36.055","Text":"What I want to find out here is what my Rho is equal to."},{"Start":"08:36.055 ","End":"08:40.435","Text":"Because then I will know my relationship between my Rho and my Sigma."},{"Start":"08:40.435 ","End":"08:48.260","Text":"My Rho represents the charge density in this parallel problem with 2 overlapping spheres."},{"Start":"08:48.260 ","End":"08:50.075","Text":"If I figure out my Rho,"},{"Start":"08:50.075 ","End":"08:52.930","Text":"then it\u0027s going to help me find the relationship with"},{"Start":"08:52.930 ","End":"08:58.345","Text":"my Sigma in the actual problem and then I can calculate my electric field."},{"Start":"08:58.345 ","End":"09:03.914","Text":"My problem right now is that I\u0027m stuck with this dr over here."},{"Start":"09:03.914 ","End":"09:09.840","Text":"What I\u0027m going to do is I\u0027m going to try and work out what my dr is equal to."},{"Start":"09:10.520 ","End":"09:12.875","Text":"Let\u0027s look at this picture."},{"Start":"09:12.875 ","End":"09:15.755","Text":"If this distance over here is d,"},{"Start":"09:15.755 ","End":"09:23.445","Text":"then we can say that this distance over here is also d and black."},{"Start":"09:23.445 ","End":"09:33.761","Text":"If here\u0027s our origin and we have some radius over here,"},{"Start":"09:33.761 ","End":"09:43.513","Text":"then this over here is going to be our dr."},{"Start":"09:43.513 ","End":"09:48.745","Text":"Here we\u0027ve enlarged this."},{"Start":"09:48.745 ","End":"09:53.860","Text":"This over here is the curve representing"},{"Start":"09:53.860 ","End":"09:59.605","Text":"the sphere over here of negative charge density and up here,"},{"Start":"09:59.605 ","End":"10:06.580","Text":"we have our sphere with positive charge density."},{"Start":"10:06.580 ","End":"10:08.890","Text":"Here we have our positive charges."},{"Start":"10:08.890 ","End":"10:10.840","Text":"Here we can see that this is D,"},{"Start":"10:10.840 ","End":"10:13.795","Text":"this is the separation of the 2 spheres."},{"Start":"10:13.795 ","End":"10:15.400","Text":"What we saw over here,"},{"Start":"10:15.400 ","End":"10:19.160","Text":"a line going straight up in the Z direction."},{"Start":"10:19.220 ","End":"10:22.530","Text":"Then our dr for our radius,"},{"Start":"10:22.530 ","End":"10:24.225","Text":"which is going out to here."},{"Start":"10:24.225 ","End":"10:27.210","Text":"It\u0027s going to be at some angle."},{"Start":"10:27.210 ","End":"10:31.455","Text":"That is a close-up over here."},{"Start":"10:31.455 ","End":"10:37.135","Text":"We already saw that if we look at this over here,"},{"Start":"10:37.135 ","End":"10:41.050","Text":"this angle is phi."},{"Start":"10:41.050 ","End":"10:44.110","Text":"Just like what we saw over here,"},{"Start":"10:44.110 ","End":"10:45.910","Text":"we\u0027re looking at the exact same problem,"},{"Start":"10:45.910 ","End":"10:51.820","Text":"the exact same angle phi so that means that\u0027s the angle between D,"},{"Start":"10:51.820 ","End":"10:54.355","Text":"the Z direction, and our"},{"Start":"10:54.355 ","End":"11:01.030","Text":"R. That\u0027s the exact same angle phi over here between d and our dr"},{"Start":"11:01.030 ","End":"11:04.060","Text":"where of course dr is in the same direction as"},{"Start":"11:04.060 ","End":"11:11.350","Text":"r. Because we said that d is much smaller than r. Then,"},{"Start":"11:11.350 ","End":"11:15.025","Text":"of course, dr is going to be very small."},{"Start":"11:15.025 ","End":"11:21.190","Text":"We can say that we can complete this over"},{"Start":"11:21.190 ","End":"11:27.580","Text":"here into a right-angled triangle."},{"Start":"11:27.580 ","End":"11:31.524","Text":"Well, here we have a right angle and so, therefore,"},{"Start":"11:31.524 ","End":"11:37.375","Text":"we can say that dr is equal to"},{"Start":"11:37.375 ","End":"11:44.230","Text":"d multiplied by the cosine of phi."},{"Start":"11:44.230 ","End":"11:48.940","Text":"This is our dr from this over"},{"Start":"11:48.940 ","End":"11:54.070","Text":"here and it works just because we\u0027re dealing with tiny lengths."},{"Start":"11:54.070 ","End":"11:56.305","Text":"That\u0027s why we can use this."},{"Start":"11:56.305 ","End":"12:00.217","Text":"Therefore I\u0027m going to substitute dr in over here."},{"Start":"12:00.217 ","End":"12:02.170","Text":"What I\u0027ll get is rho dr,"},{"Start":"12:02.170 ","End":"12:07.960","Text":"which has rho multiplied by d cosine of phi,"},{"Start":"12:07.960 ","End":"12:09.895","Text":"is equal to Sigma,"},{"Start":"12:09.895 ","End":"12:13.825","Text":"which is p_0 multiplied by the cosine of phi."},{"Start":"12:13.825 ","End":"12:18.145","Text":"Then I can cancel out the cosine of phi and divide both sides by it,"},{"Start":"12:18.145 ","End":"12:25.280","Text":"and therefore, I get that Rho d is equal to p_0."},{"Start":"12:27.320 ","End":"12:36.840","Text":"Now we\u0027ve calculated our rho charge density per unit volume for the parallel problem."},{"Start":"12:36.840 ","End":"12:39.240","Text":"Now what I\u0027m going to do using the Rho,"},{"Start":"12:39.240 ","End":"12:41.880","Text":"I\u0027m going to calculate the electric field from"},{"Start":"12:41.880 ","End":"12:45.180","Text":"these 2 overlapping spheres and that is going"},{"Start":"12:45.180 ","End":"12:51.550","Text":"to be parallel to this sphere with charge density per unit area."},{"Start":"12:51.550 ","End":"12:57.920","Text":"We\u0027re going to calculate the electric field for the 2 overlapping spheres."},{"Start":"12:58.350 ","End":"13:02.245","Text":"Just a reminder of how to calculate the electric field,"},{"Start":"13:02.245 ","End":"13:04.960","Text":"we just use Gauss\u0027s law."},{"Start":"13:04.960 ","End":"13:08.770","Text":"When r is smaller than the radius,"},{"Start":"13:08.770 ","End":"13:15.895","Text":"that means that we\u0027re calculating the electric field inside the sphere."},{"Start":"13:15.895 ","End":"13:22.180","Text":"From Gauss\u0027s law, what we have is E multiplied by the surface area."},{"Start":"13:22.180 ","End":"13:26.530","Text":"So that\u0027s equal to 4 Pi r squared is equal"},{"Start":"13:26.530 ","End":"13:31.090","Text":"to 1 divided by Epsilon naught multiplied by the charge density,"},{"Start":"13:31.090 ","End":"13:34.390","Text":"which is Rho multiplied by the volume."},{"Start":"13:34.390 ","End":"13:39.910","Text":"The volume over here is 4/3 pi r cubed,"},{"Start":"13:39.910 ","End":"13:44.230","Text":"and that\u0027s the lowercase r because we\u0027re just dealing"},{"Start":"13:44.230 ","End":"13:49.270","Text":"with the lowercase r section of the volume of the sphere."},{"Start":"13:49.270 ","End":"13:54.010","Text":"Then what we can do is we can isolate out our E,"},{"Start":"13:54.010 ","End":"13:56.350","Text":"so we can divide both sides by 4,"},{"Start":"13:56.350 ","End":"14:00.130","Text":"by pi by r squared."},{"Start":"14:00.130 ","End":"14:09.520","Text":"What we get is that E is equal to rho divided by Epsilon naught r in the r hat direction,"},{"Start":"14:09.520 ","End":"14:12.805","Text":"the radial direction or we can just write this,"},{"Start":"14:12.805 ","End":"14:15.970","Text":"r in the r hat direction is just equal to our vector."},{"Start":"14:15.970 ","End":"14:21.040","Text":"We can write this as rho r vector divided by Epsilon naught."},{"Start":"14:21.040 ","End":"14:24.925","Text":"What we can see is that the E field in"},{"Start":"14:24.925 ","End":"14:29.995","Text":"this area inside the sphere is dependent on our r vector."},{"Start":"14:29.995 ","End":"14:35.590","Text":"What happens when r is bigger than the radius of the sphere?"},{"Start":"14:35.590 ","End":"14:39.985","Text":"Now we\u0027re located outside of the sphere."},{"Start":"14:39.985 ","End":"14:46.725","Text":"Again, what we have is that e dot 4 pi r"},{"Start":"14:46.725 ","End":"14:49.620","Text":"squared is equal to 1 divided by"},{"Start":"14:49.620 ","End":"14:54.075","Text":"Epsilon naught multiplied by rho multiplied by the volume."},{"Start":"14:54.075 ","End":"14:57.975","Text":"What is the volume? It\u0027s 4/3 Pi."},{"Start":"14:57.975 ","End":"15:04.180","Text":"Then all our charge density stops at the maximum radius of our sphere,"},{"Start":"15:04.180 ","End":"15:10.435","Text":"which has capital R. There\u0027s no charge density between capital R and lowercase r,"},{"Start":"15:10.435 ","End":"15:15.865","Text":"between the edge of the sphere and our Gaussian envelope."},{"Start":"15:15.865 ","End":"15:18.940","Text":"So this is going to be multiplied by r cubed,"},{"Start":"15:18.940 ","End":"15:20.995","Text":"with capital r cubed."},{"Start":"15:20.995 ","End":"15:29.150","Text":"Again, we can isolate out our E so we can divide both sides by 4 and pi."},{"Start":"15:29.580 ","End":"15:33.400","Text":"Sorry, over here, I forgot the 3."},{"Start":"15:33.400 ","End":"15:35.740","Text":"We have a 3 over here."},{"Start":"15:35.740 ","End":"15:38.095","Text":"Now I can isolate out my E,"},{"Start":"15:38.095 ","End":"15:46.300","Text":"so I have that my electric field is equal to rho multiplied by r cubed divided"},{"Start":"15:46.300 ","End":"15:57.110","Text":"by 3 epsilon naught r squared in the radial direction."},{"Start":"15:58.080 ","End":"16:01.660","Text":"Of course, we know that the electric field outside of"},{"Start":"16:01.660 ","End":"16:06.880","Text":"a charged sphere is equal to the electric field of a point charge."},{"Start":"16:06.880 ","End":"16:10.090","Text":"Because if we look at our charged sphere from far away,"},{"Start":"16:10.090 ","End":"16:12.190","Text":"it looks like a point charge."},{"Start":"16:12.190 ","End":"16:15.805","Text":"The electric field of a point charge is simply equal to"},{"Start":"16:15.805 ","End":"16:20.410","Text":"k multiplied by the charge of the point charge or in this case,"},{"Start":"16:20.410 ","End":"16:24.070","Text":"it\u0027s the total charge of the sphere and it\u0027s"},{"Start":"16:24.070 ","End":"16:29.900","Text":"divided by r squared in the radial direction."},{"Start":"16:30.720 ","End":"16:37.465","Text":"That\u0027s just a reminder of the electric field of 1 sphere with charge density Rho."},{"Start":"16:37.465 ","End":"16:41.660","Text":"What happens when we have 2 spheres?"},{"Start":"16:43.230 ","End":"16:47.935","Text":"First of all, let\u0027s look inside the 2 spheres."},{"Start":"16:47.935 ","End":"16:51.025","Text":"Right now I want to measure, let\u0027s say,"},{"Start":"16:51.025 ","End":"16:55.465","Text":"the electric field at this point over here."},{"Start":"16:55.465 ","End":"17:01.615","Text":"This point over here is located inside the upper sphere or the red sphere,"},{"Start":"17:01.615 ","End":"17:06.430","Text":"and it\u0027s inside the lowest sphere or the blue sphere."},{"Start":"17:06.430 ","End":"17:11.110","Text":"The electric field at this point over here is going to be a superposition of"},{"Start":"17:11.110 ","End":"17:17.480","Text":"the electric field inside the red sphere and inside the blue sphere."},{"Start":"17:19.290 ","End":"17:21.820","Text":"Let\u0027s take a look."},{"Start":"17:21.820 ","End":"17:27.130","Text":"We have this red dot in the center of the red sphere."},{"Start":"17:27.130 ","End":"17:31.270","Text":"The radius from the center of the red sphere to this point."},{"Start":"17:31.270 ","End":"17:34.075","Text":"Let\u0027s call this r plus."},{"Start":"17:34.075 ","End":"17:38.650","Text":"Then there\u0027s a blue dot at the center of the blue sphere."},{"Start":"17:38.650 ","End":"17:41.470","Text":"The radius from the center of the blue sphere"},{"Start":"17:41.470 ","End":"17:44.620","Text":"to this point over here where we\u0027re measuring the electric field."},{"Start":"17:44.620 ","End":"17:52.369","Text":"Let\u0027s call it r minus and these are, of course, vectors."},{"Start":"17:52.369 ","End":"17:57.983","Text":"Right now we\u0027re in the region where we\u0027re inside the 2 spheres."},{"Start":"17:57.983 ","End":"18:05.220","Text":"R is smaller than the radius of the spheres so the electric field is, as we just said,"},{"Start":"18:05.220 ","End":"18:10.770","Text":"going to be a superposition of the electric field in the positively charged sphere in"},{"Start":"18:10.770 ","End":"18:18.400","Text":"the red sphere plus the electric field in the negatively charged blue sphere."},{"Start":"18:18.400 ","End":"18:20.695","Text":"If we look over here,"},{"Start":"18:20.695 ","End":"18:30.550","Text":"we can see that we\u0027re going to have Rho divided by 3 epsilon naught multiplied by"},{"Start":"18:30.550 ","End":"18:34.900","Text":"I vector so what we\u0027re going to have is Rho divided by"},{"Start":"18:34.900 ","End":"18:45.860","Text":"3 epsilon_naughts and then we have r_plus vector plus r_minus vector."},{"Start":"18:46.590 ","End":"18:53.770","Text":"Sorry, this is meant to be minus because our E_minus the electric field"},{"Start":"18:53.770 ","End":"18:56.170","Text":"from the negatively charged sphere is obviously going to"},{"Start":"18:56.170 ","End":"19:01.040","Text":"be a negative electric field so that\u0027s why it\u0027s minus."},{"Start":"19:01.620 ","End":"19:04.000","Text":"If we open up these brackets,"},{"Start":"19:04.000 ","End":"19:08.035","Text":"we have Rho divided by 3 epsilon_naught r_plus,"},{"Start":"19:08.035 ","End":"19:12.370","Text":"and that\u0027s the electric field for the positively charged sphere and then we"},{"Start":"19:12.370 ","End":"19:17.020","Text":"have minus Rho divided by 3 epsilon_naught r_minus,"},{"Start":"19:17.020 ","End":"19:22.765","Text":"which is the electric field for the negatively charged sphere."},{"Start":"19:22.765 ","End":"19:30.490","Text":"Now, let\u0027s see what these vectors are equal to."},{"Start":"19:30.490 ","End":"19:36.370","Text":"If we just use our vector addition, which we know,"},{"Start":"19:36.370 ","End":"19:40.525","Text":"we can see that if we add our d vector"},{"Start":"19:40.525 ","End":"19:47.470","Text":"plus our r_plus vector so we go up here,"},{"Start":"19:47.470 ","End":"19:53.065","Text":"up d, and then up r_plus then that\u0027s the same as going from this blue point,"},{"Start":"19:53.065 ","End":"19:59.455","Text":"where i minus and d start at this blue point and just going up r_minus."},{"Start":"19:59.455 ","End":"20:06.080","Text":"We can see that r minus vector is equal to d vector plus r_plus vector."},{"Start":"20:07.050 ","End":"20:13.750","Text":"Then I can just use algebra and I can subtract r_minus from"},{"Start":"20:13.750 ","End":"20:22.029","Text":"both sides and subtract d from both sides and what I\u0027ll get is that r_plus vector minus,"},{"Start":"20:22.029 ","End":"20:31.765","Text":"r_minus vector is equal to negative d vector and this is exactly what we have over here."},{"Start":"20:31.765 ","End":"20:37.060","Text":"Therefore, I can say that my E field is equal to Rho divided by"},{"Start":"20:37.060 ","End":"20:43.730","Text":"3 epsilon_naught multiplied by negative d vector."},{"Start":"20:45.480 ","End":"20:48.145","Text":"I can see that my d vector,"},{"Start":"20:48.145 ","End":"20:49.930","Text":"let\u0027s write it up over here,"},{"Start":"20:49.930 ","End":"20:55.990","Text":"is equal to some length d in this upwards direction,"},{"Start":"20:55.990 ","End":"21:03.100","Text":"which if we remember from the original drawing that this was the z direction,"},{"Start":"21:03.100 ","End":"21:05.905","Text":"and it\u0027s the same direction as"},{"Start":"21:05.905 ","End":"21:15.310","Text":"our polarization vector so that means that our d vector can be written as dz hat,"},{"Start":"21:15.310 ","End":"21:17.920","Text":"d in the z direction."},{"Start":"21:17.920 ","End":"21:20.230","Text":"Therefore, I can substitute that in here,"},{"Start":"21:20.230 ","End":"21:22.630","Text":"so I can write, so I have negative d,"},{"Start":"21:22.630 ","End":"21:27.160","Text":"so I have negative Rho multiplied by the d vector,"},{"Start":"21:27.160 ","End":"21:31.450","Text":"which is d in the z hat direction,"},{"Start":"21:31.450 ","End":"21:36.230","Text":"divided by 3 epsilon_naught."},{"Start":"21:37.080 ","End":"21:40.975","Text":"What is Rho d?"},{"Start":"21:40.975 ","End":"21:44.500","Text":"Rho d if we just go up,"},{"Start":"21:44.500 ","End":"21:45.970","Text":"we saw this before,"},{"Start":"21:45.970 ","End":"21:50.995","Text":"Rho d is equal to P naught,"},{"Start":"21:50.995 ","End":"21:57.070","Text":"the constant from our polarization vector that we just calculated."},{"Start":"21:57.070 ","End":"22:00.100","Text":"Therefore, I can substitute that in."},{"Start":"22:00.100 ","End":"22:02.230","Text":"I have negative Rho d,"},{"Start":"22:02.230 ","End":"22:05.440","Text":"which is equal to P naught divided by"},{"Start":"22:05.440 ","End":"22:13.540","Text":"3 epsilon_naughts and this is in the z direction."},{"Start":"22:13.540 ","End":"22:16.840","Text":"This is the electric field inside"},{"Start":"22:16.840 ","End":"22:21.790","Text":"our original sphere with"},{"Start":"22:21.790 ","End":"22:29.300","Text":"our charge density or charge distribution that looked something like this."},{"Start":"22:32.520 ","End":"22:38.200","Text":"Inside the sphere, we have an electric field which is"},{"Start":"22:38.200 ","End":"22:43.210","Text":"going in the negative z direction so in"},{"Start":"22:43.210 ","End":"22:48.130","Text":"the opposite direction to our polarization vector and"},{"Start":"22:48.130 ","End":"22:55.585","Text":"its magnitude is P_0 divided by 3 epsilon_naughts."},{"Start":"22:55.585 ","End":"23:00.970","Text":"Now let\u0027s look in the region where r is greater than r,"},{"Start":"23:00.970 ","End":"23:07.180","Text":"so we\u0027re located outside."},{"Start":"23:07.180 ","End":"23:12.910","Text":"That means that we\u0027re calculating the electric field at this point over here."},{"Start":"23:12.910 ","End":"23:24.050","Text":"Again, this will be our r_plus vector and this over here will be our r_minus vector."},{"Start":"23:24.690 ","End":"23:32.005","Text":"In this case, we\u0027re located outside of the 2 spheres and we saw that the electric field"},{"Start":"23:32.005 ","End":"23:38.935","Text":"outside of a sphere can be considered the same electric field as of a point charge."},{"Start":"23:38.935 ","End":"23:44.815","Text":"What we have in actual fact is that we can imagine that at the center of the red sphere,"},{"Start":"23:44.815 ","End":"23:47.860","Text":"that we\u0027re considering the red sphere as a point charge with"},{"Start":"23:47.860 ","End":"23:53.530","Text":"some positive charge because it\u0027s got a positive charge distribution and we can say that"},{"Start":"23:53.530 ","End":"24:02.320","Text":"the blue sphere can be considered a point charge as well with a charge q minus well,"},{"Start":"24:02.320 ","End":"24:03.985","Text":"both of their spheres,"},{"Start":"24:03.985 ","End":"24:09.520","Text":"or both of these point charges are at distance d from 1 another so,"},{"Start":"24:09.520 ","End":"24:11.800","Text":"therefore, we have a positive and a negative charge,"},{"Start":"24:11.800 ","End":"24:14.350","Text":"a distance d away from 1 another."},{"Start":"24:14.350 ","End":"24:16.900","Text":"We can see that the electric field that will come from"},{"Start":"24:16.900 ","End":"24:20.605","Text":"here is the same that will be measured over here,"},{"Start":"24:20.605 ","End":"24:24.110","Text":"is going to be the electric field of a dipole."},{"Start":"24:24.450 ","End":"24:29.740","Text":"First of all, our dipole moment is equal"},{"Start":"24:29.740 ","End":"24:34.675","Text":"to the charge multiplied by the distance between the 2."},{"Start":"24:34.675 ","End":"24:36.520","Text":"What is our q?"},{"Start":"24:36.520 ","End":"24:41.650","Text":"Our q is simply going to be equal to,"},{"Start":"24:41.650 ","End":"24:48.655","Text":"we can backtrack over here so it\u0027s simply going to be equal to the volume of the sphere,"},{"Start":"24:48.655 ","End":"24:54.400","Text":"which is 4/3 Pir^3,"},{"Start":"24:54.400 ","End":"24:57.100","Text":"multiplied by the charge density,"},{"Start":"24:57.100 ","End":"24:58.585","Text":"which is Rho,"},{"Start":"24:58.585 ","End":"25:02.600","Text":"and then multiplied by our d vector,"},{"Start":"25:02.600 ","End":"25:05.155","Text":"so our d vector."},{"Start":"25:05.155 ","End":"25:15.475","Text":"This is equal to 4/3 Pir^3 Rho and then our d vector,"},{"Start":"25:15.475 ","End":"25:20.365","Text":"as we said, is d in the z-direction."},{"Start":"25:20.365 ","End":"25:31.420","Text":"Then we can see that Rho d is equal to P_0 so we can write this as"},{"Start":"25:31.420 ","End":"25:39.620","Text":"4/3 Pir^3 P_0"},{"Start":"25:39.620 ","End":"25:44.180","Text":"in the z-direction."},{"Start":"25:45.360 ","End":"25:50.380","Text":"This is our dipole moment and I\u0027m reminding you that the radius of"},{"Start":"25:50.380 ","End":"25:55.645","Text":"the spheres is much larger than the distance d that they\u0027re separated."},{"Start":"25:55.645 ","End":"26:01.930","Text":"Therefore, I can use the equation for the electric field of a dipole when we\u0027re located"},{"Start":"26:01.930 ","End":"26:04.600","Text":"very far away from the dipole because right now we\u0027re"},{"Start":"26:04.600 ","End":"26:08.050","Text":"located at a distance greater than I,"},{"Start":"26:08.050 ","End":"26:13.600","Text":"which means that it\u0027s obviously going to be much greater than d. Now"},{"Start":"26:13.600 ","End":"26:21.250","Text":"let\u0027s go down and then I can write that my electric field is equal to like so."},{"Start":"26:21.250 ","End":"26:27.130","Text":"I have negative P_0 divided by 3 epsilon_naught in"},{"Start":"26:27.130 ","End":"26:31.720","Text":"the z direction when I\u0027m located inside"},{"Start":"26:31.720 ","End":"26:37.270","Text":"the sphere and then when I\u0027m located outside of the sphere,"},{"Start":"26:37.270 ","End":"26:39.640","Text":"the electric field is of a dipole,"},{"Start":"26:39.640 ","End":"26:46.705","Text":"so it\u0027s k multiplied by 3 multiplied by the dipole moment dot product with"},{"Start":"26:46.705 ","End":"26:55.725","Text":"the r vector dot product multiplied in the r hat direction minus"},{"Start":"26:55.725 ","End":"27:02.970","Text":"the dipole moment and all of this is divided by r^3 and this is of course"},{"Start":"27:02.970 ","End":"27:10.420","Text":"when r is greater than R. This is the onset to the question."},{"Start":"27:10.420 ","End":"27:14.410","Text":"Now, just a little note we can see that our dipole moment is equal"},{"Start":"27:14.410 ","End":"27:20.230","Text":"to P_0 multiplied by the volume of the sphere,"},{"Start":"27:20.230 ","End":"27:25.210","Text":"4/3 Pir^3 because that makes sense as we know,"},{"Start":"27:25.210 ","End":"27:28.159","Text":"that our"},{"Start":"27:36.450 ","End":"27:38.530","Text":"dipole moment"},{"Start":"27:38.530 ","End":"27:43.210","Text":"is equal to the integral of our vector of polarization"},{"Start":"27:43.210 ","End":"27:49.165","Text":"dv multiplied by the volume."},{"Start":"27:49.165 ","End":"27:50.350","Text":"If we integrate it,"},{"Start":"27:50.350 ","End":"27:59.280","Text":"if we had a sphere and we were told that we have P_0 in 1 unit of volume,"},{"Start":"27:59.280 ","End":"28:03.735","Text":"so if we would sum up all of the volumes in the entire sphere"},{"Start":"28:03.735 ","End":"28:08.055","Text":"then what we would have is P_0 multiplied by the volume of the sphere,"},{"Start":"28:08.055 ","End":"28:11.920","Text":"which is 4/3 Pir^3."},{"Start":"28:11.920 ","End":"28:16.085","Text":"Then if we\u0027re looking very far away from the sphere,"},{"Start":"28:16.085 ","End":"28:19.590","Text":"so then it won\u0027t anymore look like a sphere,"},{"Start":"28:19.590 ","End":"28:22.289","Text":"but it will just look like a dipole,"},{"Start":"28:22.289 ","End":"28:27.330","Text":"just 1 point of dipole and then that makes sense so then we get that"},{"Start":"28:27.330 ","End":"28:32.730","Text":"our dipole moment is equal to P_0 multiplied by the volume."},{"Start":"28:32.730 ","End":"28:35.680","Text":"That\u0027s the end of the lesson."}],"ID":22318},{"Watched":false,"Name":"Gauss_s Law and Displacement Vector","Duration":"7m 8s","ChapterTopicVideoID":21471,"CourseChapterTopicPlaylistID":99475,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.035","Text":"Hello. In this lesson,"},{"Start":"00:02.035 ","End":"00:04.860","Text":"we\u0027re going to be speaking about Gauss\u0027s law and how"},{"Start":"00:04.860 ","End":"00:08.130","Text":"to use it when dealing with dielectric materials."},{"Start":"00:08.130 ","End":"00:10.965","Text":"We\u0027re going to be defining a few terms."},{"Start":"00:10.965 ","End":"00:18.610","Text":"One of which will be important is the term for the displacement vector."},{"Start":"00:18.920 ","End":"00:24.405","Text":"Up until now, we\u0027ve seen the bound charge densities."},{"Start":"00:24.405 ","End":"00:29.640","Text":"We\u0027ve had our bound charge density per unit area,"},{"Start":"00:29.640 ","End":"00:32.370","Text":"which was Sigma B, and per unit volume,"},{"Start":"00:32.370 ","End":"00:34.395","Text":"which was Rho B."},{"Start":"00:34.395 ","End":"00:41.610","Text":"These bound charge densities were due to some polarization in the material,"},{"Start":"00:41.610 ","End":"00:46.880","Text":"so it\u0027s just how the dipoles are arranged within the material."},{"Start":"00:46.880 ","End":"00:53.340","Text":"Now we\u0027re going to define something new which is called Rho free,"},{"Start":"00:53.340 ","End":"00:58.305","Text":"or will also call it Rho f for short."},{"Start":"00:58.305 ","End":"01:03.330","Text":"This is all of the free charges."},{"Start":"01:03.330 ","End":"01:08.060","Text":"All of the charge density which isn\u0027t bound,"},{"Start":"01:08.060 ","End":"01:13.505","Text":"which isn\u0027t as a result of some polarization,"},{"Start":"01:13.505 ","End":"01:21.675","Text":"so that is Rho f. Now let\u0027s define Rho total."},{"Start":"01:21.675 ","End":"01:26.545","Text":"This is the total charge density per unit volume."},{"Start":"01:26.545 ","End":"01:32.080","Text":"This is equal to the bound charge density per unit volume,"},{"Start":"01:32.080 ","End":"01:35.590","Text":"plus the charge density per unit volume,"},{"Start":"01:35.590 ","End":"01:37.885","Text":"which isn\u0027t due to polarization,"},{"Start":"01:37.885 ","End":"01:41.785","Text":"so Rho f. Let\u0027s give an example."},{"Start":"01:41.785 ","End":"01:45.540","Text":"Let\u0027s say we have some material over here,"},{"Start":"01:45.540 ","End":"01:49.509","Text":"and then inside we have a dipole,"},{"Start":"01:49.509 ","End":"01:51.790","Text":"so a plus and a minus,"},{"Start":"01:51.790 ","End":"01:55.030","Text":"and the whole material is just a bunch of dipoles plus minus,"},{"Start":"01:55.030 ","End":"01:57.420","Text":"so that is our Rho b,"},{"Start":"01:57.420 ","End":"02:06.330","Text":"and then into that we add in some positive point charge."},{"Start":"02:06.330 ","End":"02:08.310","Text":"This is our Rho f,"},{"Start":"02:08.310 ","End":"02:11.304","Text":"the charge density not due to the polarization."},{"Start":"02:11.304 ","End":"02:18.200","Text":"The total charge density"},{"Start":"02:18.200 ","End":"02:26.400","Text":"per unit volume is going to be the bound charges plus the free charges."},{"Start":"02:27.890 ","End":"02:31.385","Text":"According to Gauss\u0027s law,"},{"Start":"02:31.385 ","End":"02:33.665","Text":"if this is the total charge,"},{"Start":"02:33.665 ","End":"02:39.215","Text":"so that means that Rho total has to be equal to"},{"Start":"02:39.215 ","End":"02:45.780","Text":"Epsilon naught multiplied by the divergence of the electric field,"},{"Start":"02:45.780 ","End":"02:47.925","Text":"so this is from Gauss\u0027s law,"},{"Start":"02:47.925 ","End":"02:56.640","Text":"that means that this is equal to Rho p plus Rho f. What is Rho b?"},{"Start":"02:56.640 ","End":"03:05.265","Text":"We remember that Rho b is equal to the negative divergence of the polarization vector,"},{"Start":"03:05.265 ","End":"03:08.235","Text":"and then plus Rho f,"},{"Start":"03:08.235 ","End":"03:11.800","Text":"which is just this charge over here."},{"Start":"03:11.900 ","End":"03:16.080","Text":"What is this Rho free, this Rho f?"},{"Start":"03:16.080 ","End":"03:19.255","Text":"What we can do is we can just isolate it out."},{"Start":"03:19.255 ","End":"03:26.275","Text":"A common multiple, if you will is this Nabla taking the divergence,"},{"Start":"03:26.275 ","End":"03:35.035","Text":"so we can just rewrite this as the divergence of epsilon naught"},{"Start":"03:35.035 ","End":"03:44.750","Text":"multiplied by the E-field plus the polarization vector is equal to Rho free."},{"Start":"03:44.750 ","End":"03:48.740","Text":"In this, you\u0027re actually allowed to do."},{"Start":"03:51.660 ","End":"03:55.705","Text":"If you open up these brackets,"},{"Start":"03:55.705 ","End":"04:02.500","Text":"then you will get this minus or plus this."},{"Start":"04:02.990 ","End":"04:07.705","Text":"This is where we define the displacement vector,"},{"Start":"04:07.705 ","End":"04:11.560","Text":"so what we have over here inside the brackets is"},{"Start":"04:11.560 ","End":"04:17.995","Text":"the displacement vector and this is called D. Let\u0027s write it out."},{"Start":"04:17.995 ","End":"04:20.815","Text":"This is what is important to remember,"},{"Start":"04:20.815 ","End":"04:26.440","Text":"So D the displacement vector is equal to Epsilon naught multiplied"},{"Start":"04:26.440 ","End":"04:33.040","Text":"by the E-field plus the polarization vector."},{"Start":"04:34.160 ","End":"04:37.515","Text":"Then once we substitute this N,"},{"Start":"04:37.515 ","End":"04:44.630","Text":"so we can see that if we take the divergence of our displacement vector D,"},{"Start":"04:44.630 ","End":"04:48.820","Text":"we get that this is equal to a Rho free."},{"Start":"04:48.820 ","End":"04:53.080","Text":"Why do I want to do this?"},{"Start":"04:53.210 ","End":"04:57.499","Text":"The reason is because this lux,"},{"Start":"04:57.499 ","End":"05:03.630","Text":"if it\u0027s divided by the factor the constant epsilon naught,"},{"Start":"05:03.630 ","End":"05:09.540","Text":"so we can see that this is Gauss\u0027s Law for the free charges."},{"Start":"05:10.190 ","End":"05:12.960","Text":"What does that mean?"},{"Start":"05:12.960 ","End":"05:16.625","Text":"Therefore, and this arrow means it works both ways."},{"Start":"05:16.625 ","End":"05:25.155","Text":"If we take the closed loop integral of d/ds,"},{"Start":"05:25.155 ","End":"05:31.610","Text":"what we\u0027ll get is q in the internal charges,"},{"Start":"05:31.610 ","End":"05:33.965","Text":"but the internal free charges."},{"Start":"05:33.965 ","End":"05:38.665","Text":"Because it\u0027s Rho f, so we get the free internal charges."},{"Start":"05:38.665 ","End":"05:42.305","Text":"In other words, if we look at this example over here,"},{"Start":"05:42.305 ","End":"05:45.590","Text":"if I would know what D is over here,"},{"Start":"05:45.590 ","End":"05:48.274","Text":"then I could do a Gaussian surface,"},{"Start":"05:48.274 ","End":"05:51.155","Text":"some envelope around the shape,"},{"Start":"05:51.155 ","End":"05:54.985","Text":"and then I could integrate along d/ds,"},{"Start":"05:54.985 ","End":"06:00.860","Text":"and I\u0027ll get all the free charges inside this shape or inside this materials."},{"Start":"06:00.860 ","End":"06:04.465","Text":"Here it would be this positive charge over here."},{"Start":"06:04.465 ","End":"06:07.295","Text":"Whenever we\u0027re dealing with dielectric materials,"},{"Start":"06:07.295 ","End":"06:09.455","Text":"we use this equation."},{"Start":"06:09.455 ","End":"06:15.230","Text":"The reason for this is because with my free charges, I control that."},{"Start":"06:15.230 ","End":"06:19.340","Text":"I can add charge in and take charge out and I control"},{"Start":"06:19.340 ","End":"06:24.320","Text":"exactly what\u0027s going on if the overall net charge is positive or negative."},{"Start":"06:24.320 ","End":"06:26.540","Text":"With the bound charges,"},{"Start":"06:26.540 ","End":"06:27.920","Text":"I don\u0027t control this,,"},{"Start":"06:27.920 ","End":"06:34.054","Text":"the second I add in some free charges so I can cause some polarization in the material."},{"Start":"06:34.054 ","End":"06:36.485","Text":"Maybe the material was polarized before,"},{"Start":"06:36.485 ","End":"06:39.320","Text":"and it\u0027s something that I can\u0027t really control."},{"Start":"06:39.320 ","End":"06:42.365","Text":"However, the free charges to put in or out,"},{"Start":"06:42.365 ","End":"06:45.695","Text":"that\u0027s something that I can control and that is why we use"},{"Start":"06:45.695 ","End":"06:51.130","Text":"this equation and that is why we have the displacement vector."},{"Start":"06:51.130 ","End":"06:56.615","Text":"Whenever you use dielectric materials and you want to use Gauss\u0027s law,"},{"Start":"06:56.615 ","End":"07:00.080","Text":"you have to use the version with the displacement vector."},{"Start":"07:00.080 ","End":"07:01.595","Text":"In the next few lessons,"},{"Start":"07:01.595 ","End":"07:05.090","Text":"we\u0027re going to see some examples of how to use this."},{"Start":"07:05.090 ","End":"07:08.580","Text":"That\u0027s the end of this explanation."}],"ID":22319},{"Watched":false,"Name":"Exercise 6","Duration":"18m 15s","ChapterTopicVideoID":21472,"CourseChapterTopicPlaylistID":99475,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.135","Text":"Hello. In this lesson we\u0027re going to be answering the following question."},{"Start":"00:04.135 ","End":"00:06.670","Text":"A spherical shell of inner radius"},{"Start":"00:06.670 ","End":"00:11.620","Text":"a and outer radius b is made of a dielectric material of"},{"Start":"00:11.620 ","End":"00:15.925","Text":"polarization density given by P as"},{"Start":"00:15.925 ","End":"00:22.140","Text":"a vector of r is equal to A divided by r in the radial direction."},{"Start":"00:22.140 ","End":"00:27.430","Text":"A is a constant and r is the distance from the center of the shell."},{"Start":"00:27.430 ","End":"00:32.335","Text":"Calculate the electric field throughout once by making use"},{"Start":"00:32.335 ","End":"00:37.805","Text":"of the bound charges and again by making use of the displacement vector."},{"Start":"00:37.805 ","End":"00:41.170","Text":"There\u0027s 2 methods that we can use in order to calculate"},{"Start":"00:41.170 ","End":"00:45.830","Text":"the electric field when dealing with dielectric materials."},{"Start":"00:45.830 ","End":"00:48.770","Text":"One is what we\u0027ve practiced quite a few times which is"},{"Start":"00:48.770 ","End":"00:52.145","Text":"by finding out the bound charges and utilizing that,"},{"Start":"00:52.145 ","End":"00:55.070","Text":"and the other way is by using"},{"Start":"00:55.070 ","End":"00:59.015","Text":"the displacement vector which we\u0027ll practice in this question."},{"Start":"00:59.015 ","End":"01:03.850","Text":"Let\u0027s start from the bound charges."},{"Start":"01:06.290 ","End":"01:09.165","Text":"Let\u0027s start with method number 1."},{"Start":"01:09.165 ","End":"01:14.265","Text":"Our polarization vector as a function of r"},{"Start":"01:14.265 ","End":"01:20.025","Text":"is equal to A divided by r in the radial direction."},{"Start":"01:20.025 ","End":"01:21.800","Text":"We can see that we only have"},{"Start":"01:21.800 ","End":"01:24.920","Text":"a radial component and we\u0027re"},{"Start":"01:24.920 ","End":"01:28.585","Text":"working in spherical coordinates because this is a spherical shell."},{"Start":"01:28.585 ","End":"01:33.395","Text":"Let\u0027s start from our bound charge per unit volume."},{"Start":"01:33.395 ","End":"01:38.405","Text":"This is equal to the negative divergence of"},{"Start":"01:38.405 ","End":"01:43.909","Text":"our vector P. Because we only have a radial component,"},{"Start":"01:43.909 ","End":"01:47.840","Text":"we\u0027re only going to find"},{"Start":"01:47.840 ","End":"01:50.240","Text":"the divergence of the radial component if there"},{"Start":"01:50.240 ","End":"01:52.805","Text":"are other components then we would use that as well,"},{"Start":"01:52.805 ","End":"01:58.190","Text":"so this is going to be equal to negative 1 divided by r^2 d by"},{"Start":"01:58.190 ","End":"02:05.415","Text":"d r of r^2 and our radial component of our P vector."},{"Start":"02:05.415 ","End":"02:11.130","Text":"This is equal to negative 1 divided by r^2 d by"},{"Start":"02:11.130 ","End":"02:16.850","Text":"d r of r^2 multiplied by a radial component of P,"},{"Start":"02:16.850 ","End":"02:20.325","Text":"which is just a divided by r. So r^2"},{"Start":"02:20.325 ","End":"02:25.935","Text":"multiplied by A divided by r is equal to r multiplied by A."},{"Start":"02:25.935 ","End":"02:29.265","Text":"When we do d by d r of r A,"},{"Start":"02:29.265 ","End":"02:31.005","Text":"we\u0027re just left with A."},{"Start":"02:31.005 ","End":"02:38.410","Text":"So then we\u0027re left with negative A divided by r^2."},{"Start":"02:40.130 ","End":"02:43.130","Text":"The next thing that we\u0027re going to do is we\u0027re going to"},{"Start":"02:43.130 ","End":"02:45.950","Text":"find our bound charge per unit area."},{"Start":"02:45.950 ","End":"02:52.380","Text":"As we know, this occurs on the outskirts, on the surface."},{"Start":"02:52.380 ","End":"02:55.875","Text":"What we can see is we have 2 surfaces,"},{"Start":"02:55.875 ","End":"02:57.990","Text":"one at a radius b,"},{"Start":"02:57.990 ","End":"02:59.825","Text":"and another at a radius a,"},{"Start":"02:59.825 ","End":"03:02.275","Text":"because it\u0027s a spherical shell."},{"Start":"03:02.275 ","End":"03:07.970","Text":"Let\u0027s first find the bound charges on our surface b."},{"Start":"03:07.970 ","End":"03:11.825","Text":"So we\u0027re substituting that r is equal to b."},{"Start":"03:11.825 ","End":"03:13.670","Text":"This is, as we know,"},{"Start":"03:13.670 ","End":"03:23.955","Text":"equal to a polarization vector dot product with our unit normal vector."},{"Start":"03:23.955 ","End":"03:26.745","Text":"That means that we have A,"},{"Start":"03:26.745 ","End":"03:30.335","Text":"and of course over here,"},{"Start":"03:30.335 ","End":"03:31.550","Text":"r is equal to b,"},{"Start":"03:31.550 ","End":"03:35.475","Text":"so A divided by b,"},{"Start":"03:35.475 ","End":"03:38.790","Text":"and this is in the r direction,"},{"Start":"03:38.790 ","End":"03:41.090","Text":"dot-product with our normal vector."},{"Start":"03:41.090 ","End":"03:44.360","Text":"Our normal vector is also in the radial direction,"},{"Start":"03:44.360 ","End":"03:48.905","Text":"because here we have a spherical shell of radius a,"},{"Start":"03:48.905 ","End":"03:52.550","Text":"and here we have another spherical shell of radius b."},{"Start":"03:52.550 ","End":"03:55.315","Text":"That\u0027s what it really looks like."},{"Start":"03:55.315 ","End":"04:01.365","Text":"So r hat dot r hat is of course equal to 1."},{"Start":"04:01.365 ","End":"04:05.660","Text":"Therefore this is equal to A divided by b."},{"Start":"04:05.660 ","End":"04:10.790","Text":"The bound charge density at r is equal to"},{"Start":"04:10.790 ","End":"04:18.590","Text":"a of this inner surface area of our thick spherical shell."},{"Start":"04:18.590 ","End":"04:20.645","Text":"It\u0027s again P dot n,"},{"Start":"04:20.645 ","End":"04:23.495","Text":"which is equal to A divided by r,"},{"Start":"04:23.495 ","End":"04:28.880","Text":"where r is equal to A in the r direction dot the normal vector,"},{"Start":"04:28.880 ","End":"04:32.705","Text":"which is again r hat dot r hat as 1."},{"Start":"04:32.705 ","End":"04:38.100","Text":"Here we have capital A divided by lowercase a."},{"Start":"04:39.010 ","End":"04:42.364","Text":"Now of course here I made a mistake."},{"Start":"04:42.364 ","End":"04:44.795","Text":"Let\u0027s just go over it."},{"Start":"04:44.795 ","End":"04:49.880","Text":"Here my normal vector at b is in this direction."},{"Start":"04:49.880 ","End":"04:56.885","Text":"This is normal to my surface at radius b. That\u0027s great."},{"Start":"04:56.885 ","End":"05:03.770","Text":"However, my normal vector to surface area a is going to be in this direction."},{"Start":"05:03.770 ","End":"05:07.775","Text":"Because over here I have a dielectric material,"},{"Start":"05:07.775 ","End":"05:11.660","Text":"so my free empty space is in this direction."},{"Start":"05:11.660 ","End":"05:14.045","Text":"This is of course,"},{"Start":"05:14.045 ","End":"05:19.220","Text":"as we can see in the negative radial direction."},{"Start":"05:19.220 ","End":"05:23.460","Text":"I have to add in a minus over here."},{"Start":"05:24.530 ","End":"05:27.990","Text":"It\u0027s important to look at that."},{"Start":"05:27.990 ","End":"05:34.535","Text":"If you need help or if it\u0027s harder for you to visualize or you can forget,"},{"Start":"05:34.535 ","End":"05:35.975","Text":"like I did draw,"},{"Start":"05:35.975 ","End":"05:39.920","Text":"here\u0027s your surface and the closest empty space is over here."},{"Start":"05:39.920 ","End":"05:42.670","Text":"The normal is in this direction,"},{"Start":"05:42.670 ","End":"05:46.440","Text":"and here we have this surface,"},{"Start":"05:46.440 ","End":"05:52.040","Text":"and the closest empty space is in this direction."},{"Start":"05:52.040 ","End":"05:56.120","Text":"So here we have in the negative r direction,"},{"Start":"05:56.120 ","End":"05:59.609","Text":"and here we have in the positive r direction."},{"Start":"06:00.980 ","End":"06:05.165","Text":"Now we can work out the electric field."},{"Start":"06:05.165 ","End":"06:07.760","Text":"Let\u0027s split ourselves into regions."},{"Start":"06:07.760 ","End":"06:09.410","Text":"Our first region is when we\u0027re"},{"Start":"06:09.410 ","End":"06:15.810","Text":"located between the center of the sphere and our surface area a."},{"Start":"06:16.550 ","End":"06:22.910","Text":"We\u0027re located somewhere here between the inner shell and the center."},{"Start":"06:22.910 ","End":"06:25.115","Text":"Here we can see that we don\u0027t have"},{"Start":"06:25.115 ","End":"06:29.735","Text":"any dielectric material or anything we just have empty space."},{"Start":"06:29.735 ","End":"06:38.310","Text":"Over here, we can see that the electric field is equal to 0."},{"Start":"06:39.520 ","End":"06:43.055","Text":"Now we\u0027re looking in this region."},{"Start":"06:43.055 ","End":"06:49.125","Text":"This is the region between the surface areas."},{"Start":"06:49.125 ","End":"06:51.950","Text":"We\u0027re located inside the shell."},{"Start":"06:51.950 ","End":"06:56.500","Text":"We\u0027re located between a and b."},{"Start":"06:56.500 ","End":"07:01.860","Text":"Here we have that E multiplied by 4 pi r^2"},{"Start":"07:01.860 ","End":"07:07.785","Text":"is equal to Q_in divided by Epsilon naught."},{"Start":"07:07.785 ","End":"07:12.755","Text":"Now what we have to do is we have to work out what Q_in is equal to."},{"Start":"07:12.755 ","End":"07:19.080","Text":"Q_in is going to be equal to."},{"Start":"07:19.820 ","End":"07:26.635","Text":"Let\u0027s take a look. We have our surface charge at our Sigma at A,"},{"Start":"07:26.635 ","End":"07:29.515","Text":"which has some charge,"},{"Start":"07:29.515 ","End":"07:33.425","Text":"as we saw this over here."},{"Start":"07:33.425 ","End":"07:39.460","Text":"Then we also have the charges in this space over here,"},{"Start":"07:39.460 ","End":"07:41.410","Text":"which is a volume."},{"Start":"07:41.410 ","End":"07:48.570","Text":"This is going to be equal to our Sigma_b at radius a,"},{"Start":"07:48.570 ","End":"07:56.170","Text":"plus and now we\u0027re going to integrate from a until our radius r. Where of course"},{"Start":"07:56.170 ","End":"08:01.440","Text":"our radius r is this of"},{"Start":"08:01.440 ","End":"08:07.710","Text":"Rho_b d v, because it\u0027s a volume."},{"Start":"08:07.710 ","End":"08:14.365","Text":"Our Sigma_b at a is simply equal to,"},{"Start":"08:14.365 ","End":"08:20.585","Text":"of course, this is the charge density per unit area multiplied by the area."},{"Start":"08:20.585 ","End":"08:25.105","Text":"So 4 Pi multiplied by the radius squared."},{"Start":"08:25.105 ","End":"08:29.330","Text":"Where the radius of Sigma_b at a is of course a,"},{"Start":"08:29.330 ","End":"08:30.920","Text":"so multiplied by a^2."},{"Start":"08:30.920 ","End":"08:40.580","Text":"Sigma_b b at A is equal to negative A divided by a multiplied by 4 Pi a^2."},{"Start":"08:40.580 ","End":"08:44.405","Text":"We\u0027ll cancel out those things soon,"},{"Start":"08:44.405 ","End":"08:50.855","Text":"plus the integral from a up until r of Rho b."},{"Start":"08:50.855 ","End":"08:58.310","Text":"So that was negative A divided by r^2 dV."},{"Start":"09:01.230 ","End":"09:05.095","Text":"What is our dV equal to?"},{"Start":"09:05.095 ","End":"09:10.585","Text":"We can see that we just have the surface area of a circle,"},{"Start":"09:10.585 ","End":"09:13.960","Text":"which is 4 pi r^2."},{"Start":"09:13.960 ","End":"09:18.565","Text":"Then the radius is just changing as we move out."},{"Start":"09:18.565 ","End":"09:21.280","Text":"We have this circle, then,"},{"Start":"09:21.280 ","End":"09:24.625","Text":"this surface area plus this,"},{"Start":"09:24.625 ","End":"09:27.100","Text":"and we\u0027re just adding along that."},{"Start":"09:27.100 ","End":"09:28.870","Text":"It\u0027s going to be equal to"},{"Start":"09:28.870 ","End":"09:37.580","Text":"4 pi r^2 dr because it\u0027s the radius that is changing the amount of volume."},{"Start":"09:37.940 ","End":"09:42.475","Text":"Now let\u0027s scroll down."},{"Start":"09:42.475 ","End":"09:46.960","Text":"What we have is that Q_in is equal to."},{"Start":"09:46.960 ","End":"09:48.910","Text":"Here we can cross this out."},{"Start":"09:48.910 ","End":"09:54.070","Text":"We have negative 4 pi,"},{"Start":"09:54.070 ","End":"10:02.215","Text":"a multiplied by a plus the integral of a until r of"},{"Start":"10:02.215 ","End":"10:12.770","Text":"negative a divided by r^2 multiplied by 4 pi r^2 dr."},{"Start":"10:12.770 ","End":"10:16.650","Text":"This r^2 will cancel out with this r^2."},{"Start":"10:16.650 ","End":"10:22.800","Text":"Then what we\u0027ll get is that this is equal to negative 4 pi a,"},{"Start":"10:22.800 ","End":"10:28.600","Text":"a plus or minus, rather."},{"Start":"10:28.600 ","End":"10:33.730","Text":"Minus and then we have 4 pi a and then in brackets,"},{"Start":"10:33.730 ","End":"10:38.920","Text":"r minus a,4 pi a,"},{"Start":"10:38.920 ","End":"10:46.190","Text":"r minus a, where r minus a are the bounds of the integration."},{"Start":"10:46.650 ","End":"10:49.615","Text":"Now we can plug this back."},{"Start":"10:49.615 ","End":"10:59.710","Text":"We have that e multiplied by 4 Pi r^2 is equal to Q_in,"},{"Start":"10:59.710 ","End":"11:05.815","Text":"which is equal to negative 4 Pi A."},{"Start":"11:05.815 ","End":"11:08.800","Text":"Then we have"},{"Start":"11:08.800 ","End":"11:13.645","Text":"a plus r"},{"Start":"11:13.645 ","End":"11:20.470","Text":"minus a,"},{"Start":"11:20.470 ","End":"11:24.815","Text":"divided by epsilon naught."},{"Start":"11:24.815 ","End":"11:30.330","Text":"This is equal to negative 4 Pi a."},{"Start":"11:30.330 ","End":"11:32.190","Text":"Then we have a minus a,"},{"Start":"11:32.190 ","End":"11:38.890","Text":"so that cancels out multiplied by r divided by epsilon naught."},{"Start":"11:38.890 ","End":"11:42.850","Text":"Now I can cancel both sides by 4 Pi."},{"Start":"11:42.850 ","End":"11:46.960","Text":"4 Pi, cancel out one of the r\u0027s."},{"Start":"11:46.960 ","End":"11:53.470","Text":"Therefore, I will get that my electric field is equal to simply"},{"Start":"11:53.470 ","End":"12:03.025","Text":"negative a divided by epsilon naught r. Sorry,"},{"Start":"12:03.025 ","End":"12:06.310","Text":"over here, these will switch sides."},{"Start":"12:06.310 ","End":"12:09.010","Text":"Then we\u0027ll get a negative and a negative,"},{"Start":"12:09.010 ","End":"12:11.035","Text":"so this is a positive."},{"Start":"12:11.035 ","End":"12:14.390","Text":"It\u0027s a positive a."},{"Start":"12:14.850 ","End":"12:17.830","Text":"Of course if we put this in a vector,"},{"Start":"12:17.830 ","End":"12:20.870","Text":"it\u0027s in the radial direction."},{"Start":"12:22.080 ","End":"12:27.085","Text":"Our next region is when we\u0027re located over here,"},{"Start":"12:27.085 ","End":"12:29.740","Text":"outside of the sphere completely."},{"Start":"12:29.740 ","End":"12:34.240","Text":"We know that the Q_in,"},{"Start":"12:34.240 ","End":"12:36.640","Text":"over here in the volume,"},{"Start":"12:36.640 ","End":"12:40.525","Text":"between the two surface areas of the spherical shell."},{"Start":"12:40.525 ","End":"12:43.675","Text":"Because we have polarization over here,"},{"Start":"12:43.675 ","End":"12:49.150","Text":"we know that the total charge is going to be equal to 0 in-between."},{"Start":"12:49.150 ","End":"12:53.800","Text":"If you want to check it, you can do this integration between a and"},{"Start":"12:53.800 ","End":"12:59.780","Text":"b of rho_(b)dV and you will get that it is equal to 0."},{"Start":"13:00.360 ","End":"13:09.595","Text":"What we\u0027re going to get is that Q_in is simply the total charge, we can write it out."},{"Start":"13:09.595 ","End":"13:17.170","Text":"That is going to be equal to the integral from a to b of Rho_(b)dV"},{"Start":"13:17.170 ","End":"13:25.645","Text":"plus our Sigma b (a) multiplied by the surface area of a,"},{"Start":"13:25.645 ","End":"13:34.165","Text":"4 pi a squared plus our Sigma b (b) multiplied by the surface area of b,"},{"Start":"13:34.165 ","End":"13:37.915","Text":"which is 4 pi b squared."},{"Start":"13:37.915 ","End":"13:40.810","Text":"Once we do all of these calculations,"},{"Start":"13:40.810 ","End":"13:44.030","Text":"we\u0027ll see that this is equal to 0."},{"Start":"13:44.840 ","End":"13:53.560","Text":"Therefore, we can say that our electric field over here is also going to be equal to 0."},{"Start":"13:55.370 ","End":"13:59.385","Text":"This is the electric field throughout all of the regions."},{"Start":"13:59.385 ","End":"14:02.745","Text":"Here we\u0027ve used the idea of our bound charges."},{"Start":"14:02.745 ","End":"14:06.645","Text":"Now what we\u0027re going to do is we\u0027re going to find out the electric field,"},{"Start":"14:06.645 ","End":"14:10.690","Text":"but we\u0027re going to be using the displacement vector."},{"Start":"14:11.760 ","End":"14:15.115","Text":"When we\u0027re using the displacement vector,"},{"Start":"14:15.115 ","End":"14:19.195","Text":"we have to know what our rho_3 is equal to."},{"Start":"14:19.195 ","End":"14:23.695","Text":"That is how many charges do we have that are free?"},{"Start":"14:23.695 ","End":"14:27.415","Text":"Or in other words, charges that aren\u0027t bound charges."},{"Start":"14:27.415 ","End":"14:32.365","Text":"What we have is a dielectric material and it has a polarization vector,"},{"Start":"14:32.365 ","End":"14:38.635","Text":"which means that we have lots and lots of dipoles or polarized charges."},{"Start":"14:38.635 ","End":"14:45.745","Text":"We can see that we haven\u0027t put into the system some point charge or anything like that."},{"Start":"14:45.745 ","End":"14:50.245","Text":"Therefore, we know that all of our charges are bound charges,"},{"Start":"14:50.245 ","End":"14:54.205","Text":"so we have 0 free charges."},{"Start":"14:54.205 ","End":"15:01.130","Text":"That means that our value D is equal to 0."},{"Start":"15:02.340 ","End":"15:07.690","Text":"In that case, let\u0027s look in the region of r is smaller than a."},{"Start":"15:07.690 ","End":"15:11.530","Text":"Here we know that we don\u0027t have polarization,"},{"Start":"15:11.530 ","End":"15:13.480","Text":"because this is empty space."},{"Start":"15:13.480 ","End":"15:15.955","Text":"There\u0027s no charges over here."},{"Start":"15:15.955 ","End":"15:20.215","Text":"Therefore, if we write out our equation for the displacement vector,"},{"Start":"15:20.215 ","End":"15:23.245","Text":"which D is equal to epsilon naught"},{"Start":"15:23.245 ","End":"15:27.925","Text":"multiplied by the E field plus our polarization vector."},{"Start":"15:27.925 ","End":"15:29.965","Text":"Here we can see."},{"Start":"15:29.965 ","End":"15:32.185","Text":"We know that d is equal to 0."},{"Start":"15:32.185 ","End":"15:35.650","Text":"We saw, we know that P in this region,"},{"Start":"15:35.650 ","End":"15:39.730","Text":"specifically over here because it\u0027s empty space."},{"Start":"15:39.730 ","End":"15:43.375","Text":"Over here, P is equal to 0."},{"Start":"15:43.375 ","End":"15:47.800","Text":"Therefore, we know that epsilon naught multiplied by"},{"Start":"15:47.800 ","End":"15:52.990","Text":"the E-field has to be therefore equal to 0 if we play around with this."},{"Start":"15:52.990 ","End":"15:58.900","Text":"Therefore, we get that our E field in this region is equal to 0."},{"Start":"15:58.900 ","End":"16:03.610","Text":"Now what about in the region between a and b."},{"Start":"16:03.610 ","End":"16:07.100","Text":"We\u0027re located inside the spherical shell."},{"Start":"16:07.110 ","End":"16:09.730","Text":"We have our D vector,"},{"Start":"16:09.730 ","End":"16:11.275","Text":"which we know is equal to 0,"},{"Start":"16:11.275 ","End":"16:18.685","Text":"and this is equal to epsilon naught multiplied by E plus our P vector."},{"Start":"16:18.685 ","End":"16:25.210","Text":"Therefore, what we get is that our E field is equal"},{"Start":"16:25.210 ","End":"16:32.425","Text":"to negative of polarization vector divided by epsilon naught."},{"Start":"16:32.425 ","End":"16:36.310","Text":"I\u0027ve just rearranged this equation because it\u0027s equal to 0."},{"Start":"16:36.310 ","End":"16:40.540","Text":"My polarization vector, negative of it."},{"Start":"16:40.540 ","End":"16:46.045","Text":"I have negative A divided by r in the r direction,"},{"Start":"16:46.045 ","End":"16:49.645","Text":"divided by epsilon naught."},{"Start":"16:49.645 ","End":"16:55.030","Text":"This is just equal to negative A divided by epsilon naught r,"},{"Start":"16:55.030 ","End":"16:57.505","Text":"in the radial direction,"},{"Start":"16:57.505 ","End":"16:59.995","Text":"and that is the electric field over here."},{"Start":"16:59.995 ","End":"17:10.850","Text":"Then let\u0027s look in the region where we\u0027re outside of our spherical shell."},{"Start":"17:11.250 ","End":"17:14.185","Text":"We\u0027re located somewhere over here."},{"Start":"17:14.185 ","End":"17:19.390","Text":"Here again, I don\u0027t have polarization because I\u0027m an empty space."},{"Start":"17:19.390 ","End":"17:25.540","Text":"The only area where I have polarization is over here,"},{"Start":"17:25.540 ","End":"17:27.640","Text":"in the middle of the spherical shell,"},{"Start":"17:27.640 ","End":"17:30.669","Text":"because that\u0027s where I have dielectric material."},{"Start":"17:30.669 ","End":"17:35.950","Text":"Here I know that my d is equal to 0 and this is equal to"},{"Start":"17:35.950 ","End":"17:43.000","Text":"Epsilon naught e plus my polarization vector in this region over here and empty space,"},{"Start":"17:43.000 ","End":"17:45.205","Text":"which is also equal to 0."},{"Start":"17:45.205 ","End":"17:51.115","Text":"Therefore, I get that my E field or so over here is equal to 0."},{"Start":"17:51.115 ","End":"17:52.870","Text":"We got these answers,"},{"Start":"17:52.870 ","End":"17:54.715","Text":"and if you look back in the video,"},{"Start":"17:54.715 ","End":"17:59.980","Text":"these are the exact answers that we got when we made use of the bound charges."},{"Start":"17:59.980 ","End":"18:02.735","Text":"We can see that using the displacement vector is"},{"Start":"18:02.735 ","End":"18:07.175","Text":"a much quicker and easier calculation to make,"},{"Start":"18:07.175 ","End":"18:12.930","Text":"rather than doing lots and lots of different calculations using the bound charges."},{"Start":"18:12.930 ","End":"18:15.920","Text":"That\u0027s the end of this lesson."}],"ID":22320},{"Watched":false,"Name":"Electric Field Vs Displacement Vector","Duration":"15m 43s","ChapterTopicVideoID":21473,"CourseChapterTopicPlaylistID":99475,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.725","Text":"Hello, in this lesson,"},{"Start":"00:01.725 ","End":"00:03.930","Text":"we\u0027re going to be speaking about D,"},{"Start":"00:03.930 ","End":"00:05.985","Text":"which is our displacement vector,"},{"Start":"00:05.985 ","End":"00:11.010","Text":"and how it is similar to the field caused by the free charges."},{"Start":"00:11.010 ","End":"00:17.010","Text":"But this is something very confusing that lots of people can sometimes misunderstand,"},{"Start":"00:17.010 ","End":"00:19.815","Text":"so this is a very important lesson."},{"Start":"00:19.815 ","End":"00:24.074","Text":"The question that we\u0027re going to be asking is if D,"},{"Start":"00:24.074 ","End":"00:28.710","Text":"our displacement vector is a type of electric field which is"},{"Start":"00:28.710 ","End":"00:34.620","Text":"formed just from our free charged particles."},{"Start":"00:34.620 ","End":"00:39.725","Text":"We\u0027ve seen that D, our displacement vector is equal to"},{"Start":"00:39.725 ","End":"00:47.370","Text":"epsilon naught multiplied by E naught so what is E naught?"},{"Start":"00:47.370 ","End":"00:52.865","Text":"E naught as the electric field formed just by the free particles."},{"Start":"00:52.865 ","End":"00:59.970","Text":"If we imagine that I had some dielectric material,"},{"Start":"00:59.970 ","End":"01:03.630","Text":"like so in some box,"},{"Start":"01:03.630 ","End":"01:10.560","Text":"and it\u0027s polarized, remember the charge goes from minus to plus,"},{"Start":"01:10.560 ","End":"01:12.780","Text":"so this is the polarization,"},{"Start":"01:12.780 ","End":"01:17.915","Text":"and then let\u0027s imagine that I have some charged particles over here,"},{"Start":"01:17.915 ","End":"01:26.930","Text":"rho f. Now let\u0027s say that I choose a Gaussian formula somewhere in space."},{"Start":"01:26.930 ","End":"01:35.335","Text":"I include my free particles and part of my polarized dielectric material."},{"Start":"01:35.335 ","End":"01:39.615","Text":"If I use Gauss\u0027s law on D,"},{"Start":"01:39.615 ","End":"01:42.300","Text":"this is Gauss\u0027s law on D,"},{"Start":"01:42.300 ","End":"01:46.110","Text":"then I know that this is going to be the answer."},{"Start":"01:46.110 ","End":"01:50.900","Text":"E naught is the electric field,"},{"Start":"01:50.900 ","End":"01:57.690","Text":"as we just said, just due to these free charges or free charge particles."},{"Start":"01:57.690 ","End":"02:04.590","Text":"If I were to choose some point be at in the dielectric material or outside of it,"},{"Start":"02:04.590 ","End":"02:07.730","Text":"so what I would do is like what we learned in Coulomb\u0027s law."},{"Start":"02:07.730 ","End":"02:11.855","Text":"I would find the electric field at this point or this point wherever,"},{"Start":"02:11.855 ","End":"02:15.800","Text":"due to all of these free charges."},{"Start":"02:15.800 ","End":"02:20.345","Text":"Then I would find the electric field at this point,"},{"Start":"02:20.345 ","End":"02:24.650","Text":"and this would be my E naught electric field."},{"Start":"02:24.650 ","End":"02:30.320","Text":"We\u0027re ignoring all of the bound charges in the dielectric material."},{"Start":"02:30.320 ","End":"02:34.115","Text":"What we can see is that our displacement vector"},{"Start":"02:34.115 ","End":"02:38.975","Text":"is equal to epsilon naught multiplied by E naught."},{"Start":"02:38.975 ","End":"02:44.720","Text":"But what is important to note is that D isn\u0027t always"},{"Start":"02:44.720 ","End":"02:51.620","Text":"equal to epsilon naught multiplied by E naught so this only happens in specific cases,"},{"Start":"02:51.620 ","End":"02:55.565","Text":"and we\u0027re going to speak about which cases this happens in."},{"Start":"02:55.565 ","End":"02:58.445","Text":"This is very important to remember."},{"Start":"02:58.445 ","End":"03:04.715","Text":"The displacement vector isn\u0027t always equal to our E naught."},{"Start":"03:04.715 ","End":"03:12.170","Text":"D isn\u0027t exactly a type of field formed only by the free charges."},{"Start":"03:12.170 ","End":"03:16.220","Text":"It isn\u0027t exactly right to equate these two,"},{"Start":"03:16.220 ","End":"03:21.320","Text":"but it\u0027s better to look at D as the displacement vector as"},{"Start":"03:21.320 ","End":"03:25.640","Text":"some mathematical tool that we can use in order to"},{"Start":"03:25.640 ","End":"03:31.250","Text":"find the electric field due to the free particles."},{"Start":"03:31.250 ","End":"03:36.275","Text":"Why isn\u0027t D a type of field?"},{"Start":"03:36.275 ","End":"03:42.140","Text":"Yes, it does adhere to Gauss\u0027s law for the free particles."},{"Start":"03:42.140 ","End":"03:44.075","Text":"this is what we get over here,"},{"Start":"03:44.075 ","End":"03:47.150","Text":"we\u0027re doing Gauss over here and we get the free particles."},{"Start":"03:47.150 ","End":"03:51.265","Text":"However, in order to define some field,"},{"Start":"03:51.265 ","End":"03:55.940","Text":"we also need to know the rotor of the field."},{"Start":"03:55.940 ","End":"04:01.010","Text":"The rotor of the electric field,"},{"Start":"04:01.010 ","End":"04:03.335","Text":"for example, what is the rotor?"},{"Start":"04:03.335 ","End":"04:08.015","Text":"It\u0027s nabla cross by the vector field so in this example,"},{"Start":"04:08.015 ","End":"04:09.650","Text":"we\u0027re looking at the E field."},{"Start":"04:09.650 ","End":"04:13.670","Text":"We know that the rotor of the electric field is always equal"},{"Start":"04:13.670 ","End":"04:18.500","Text":"to 0 because the electric field is conservative,"},{"Start":"04:18.500 ","End":"04:23.840","Text":"and this we\u0027ve seen is always equal to 0."},{"Start":"04:23.840 ","End":"04:28.648","Text":"However, and we\u0027ll look at some examples soon,"},{"Start":"04:28.648 ","End":"04:36.000","Text":"the rotor of the displacement vector D is not always equal to 0."},{"Start":"04:36.000 ","End":"04:42.690","Text":"Our displacement vector D isn\u0027t always a conservative field,"},{"Start":"04:42.690 ","End":"04:47.600","Text":"and we know that that means that it can\u0027t be the field due to"},{"Start":"04:47.600 ","End":"04:50.300","Text":"a charged particles because the field due to charge"},{"Start":"04:50.300 ","End":"04:54.240","Text":"particles is always a conservative field."},{"Start":"04:55.130 ","End":"05:01.565","Text":"In other words, Coulomb\u0027s law doesn\u0027t work for D the displacement vector."},{"Start":"05:01.565 ","End":"05:04.910","Text":"If I were to add up"},{"Start":"05:04.910 ","End":"05:12.260","Text":"the Coulomb\u0027s law to find the force due to each one of these free particles,"},{"Start":"05:12.260 ","End":"05:14.050","Text":"so doing Coulomb\u0027s law,"},{"Start":"05:14.050 ","End":"05:20.480","Text":"I wouldn\u0027t get D. What do we can see is that only"},{"Start":"05:20.480 ","End":"05:26.975","Text":"Gauss\u0027s law works on D. When we\u0027re thinking of D as some field,"},{"Start":"05:26.975 ","End":"05:29.210","Text":"only Gauss\u0027s law works with D,"},{"Start":"05:29.210 ","End":"05:34.880","Text":"but our rotor doesn\u0027t equal to 0 when we\u0027re dealing with D,"},{"Start":"05:34.880 ","End":"05:40.260","Text":"and Coulomb\u0027s law doesn\u0027t work for D either."},{"Start":"05:41.420 ","End":"05:45.350","Text":"Let\u0027s take a look why the rotor of D,"},{"Start":"05:45.350 ","End":"05:47.195","Text":"it doesn\u0027t equal to 0."},{"Start":"05:47.195 ","End":"05:49.970","Text":"First, we have to remember that E,"},{"Start":"05:49.970 ","End":"05:57.965","Text":"the total electric field which comes from both the bound charges and the free charges,"},{"Start":"05:57.965 ","End":"06:00.590","Text":"does not equal to E naught."},{"Start":"06:00.590 ","End":"06:03.080","Text":"Where E naught, as we said,"},{"Start":"06:03.080 ","End":"06:08.460","Text":"was the electric fields due to just the free charges."},{"Start":"06:09.140 ","End":"06:15.515","Text":"Now let\u0027s be reminded by the equation for our displacement vector."},{"Start":"06:15.515 ","End":"06:20.750","Text":"D is equal to epsilon naught multiplied by E,"},{"Start":"06:20.750 ","End":"06:29.375","Text":"the E field from bound and free charges plus our polarization vector."},{"Start":"06:29.375 ","End":"06:34.835","Text":"Now when we take the rotor of D,"},{"Start":"06:34.835 ","End":"06:42.395","Text":"that means that we\u0027re taking the rotor of each one of these separately so of"},{"Start":"06:42.395 ","End":"06:52.260","Text":"epsilon naught E plus the rotor on our vector of polarization."},{"Start":"06:53.990 ","End":"06:57.350","Text":"Epsilon naught, we know is a constant,"},{"Start":"06:57.350 ","End":"06:59.660","Text":"so the rotor of a constant is always equal to 0,"},{"Start":"06:59.660 ","End":"07:01.859","Text":"it doesn\u0027t make a difference."},{"Start":"07:03.000 ","End":"07:06.205","Text":"The rotor of our E field,"},{"Start":"07:06.205 ","End":"07:08.650","Text":"we know that the E field is a conservative field,"},{"Start":"07:08.650 ","End":"07:11.080","Text":"so it\u0027s always going to be equal to 0."},{"Start":"07:11.080 ","End":"07:12.520","Text":"This we saw over here."},{"Start":"07:12.520 ","End":"07:15.280","Text":"Then we just have 0 multiplied by epsilon naught,"},{"Start":"07:15.280 ","End":"07:16.690","Text":"which is a constant."},{"Start":"07:16.690 ","End":"07:19.615","Text":"That\u0027s equal to 0."},{"Start":"07:19.615 ","End":"07:23.410","Text":"Let\u0027s write that over here."},{"Start":"07:23.410 ","End":"07:28.645","Text":"What we\u0027re actually doing is we can take epsilon naught out to the side,"},{"Start":"07:28.645 ","End":"07:30.520","Text":"and we\u0027re doing a rotor on E,"},{"Start":"07:30.520 ","End":"07:34.540","Text":"which is equal to 0, and then we\u0027re multiplying by a constant."},{"Start":"07:34.540 ","End":"07:39.980","Text":"Well, now, let\u0027s look at the rotor of the polarization vector."},{"Start":"07:40.050 ","End":"07:48.190","Text":"Now, the rotor of P doesn\u0027t have to be equal to 0."},{"Start":"07:48.190 ","End":"07:53.360","Text":"Now, we\u0027re going to look at this via an easy example."},{"Start":"07:54.060 ","End":"07:56.830","Text":"Here\u0027s an example."},{"Start":"07:56.830 ","End":"08:04.150","Text":"I have some box which has some uniform polarization throughout,"},{"Start":"08:04.150 ","End":"08:09.685","Text":"where the polarization is pointing in the x-direction."},{"Start":"08:09.685 ","End":"08:15.265","Text":"Now, what I\u0027m going to do is I\u0027m going to use Stokes theorem."},{"Start":"08:15.265 ","End":"08:22.030","Text":"That says that if I take an integral of the rotor of some vector."},{"Start":"08:22.030 ","End":"08:24.610","Text":"In this case, we\u0027ll do P,"},{"Start":"08:24.610 ","End":"08:30.055","Text":"our polarization vector ds along some area,"},{"Start":"08:30.055 ","End":"08:37.960","Text":"that is equal to the closed circuit integral of that same vector field."},{"Start":"08:37.960 ","End":"08:42.710","Text":"Let\u0027s say P in this example, dot dl."},{"Start":"08:44.640 ","End":"08:50.845","Text":"We\u0027ve seen this type of equation when we were dealing with conservative forces."},{"Start":"08:50.845 ","End":"08:58.045","Text":"We saw that if we took the integral of the rotor of a force and it was equal to 0,"},{"Start":"08:58.045 ","End":"09:01.240","Text":"then that force was conservative."},{"Start":"09:01.240 ","End":"09:06.430","Text":"Then the closed circuit integral on the work."},{"Start":"09:06.430 ","End":"09:10.435","Text":"This is exactly the same equation as we would have for work,"},{"Start":"09:10.435 ","End":"09:13.805","Text":"some vector dl or dr."},{"Start":"09:13.805 ","End":"09:17.740","Text":"The closed circuit integral for the work will"},{"Start":"09:17.740 ","End":"09:23.060","Text":"also be equal to 0 if we\u0027re dealing with a conservative force."},{"Start":"09:24.030 ","End":"09:29.110","Text":"That means that if we take any closed circuit loop,"},{"Start":"09:29.110 ","End":"09:32.920","Text":"then we should be getting 0 if this is"},{"Start":"09:32.920 ","End":"09:40.255","Text":"a conservative force or a conservative function or field."},{"Start":"09:40.255 ","End":"09:46.640","Text":"Therefore, the rotor of P of this field will be equal to 0."},{"Start":"09:46.710 ","End":"09:49.585","Text":"Let\u0027s take some loop,"},{"Start":"09:49.585 ","End":"09:51.100","Text":"we can take any loop."},{"Start":"09:51.100 ","End":"09:54.280","Text":"Let\u0027s choose this loop over here."},{"Start":"09:54.280 ","End":"10:00.115","Text":"That part of it goes inside the material and part of it is outside of the material."},{"Start":"10:00.115 ","End":"10:02.185","Text":"We can see in this region,"},{"Start":"10:02.185 ","End":"10:05.290","Text":"we get something from the fact that we\u0027re"},{"Start":"10:05.290 ","End":"10:09.055","Text":"going in the direction of the field and we\u0027re inside the material."},{"Start":"10:09.055 ","End":"10:14.290","Text":"However, when we carry on with this loop outside,"},{"Start":"10:14.290 ","End":"10:18.220","Text":"we see that we\u0027re not getting any help from this P field,"},{"Start":"10:18.220 ","End":"10:21.490","Text":"this polarization vector inside."},{"Start":"10:21.490 ","End":"10:29.080","Text":"What we can see is that we\u0027re only getting our work if you want to call it that,"},{"Start":"10:29.080 ","End":"10:32.425","Text":"isn\u0027t being canceled out on this closed loop."},{"Start":"10:32.425 ","End":"10:35.290","Text":"We\u0027re adding work and then nothing, nothing,"},{"Start":"10:35.290 ","End":"10:39.355","Text":"nothing, and then we\u0027ll get back to the start and adding nothing."},{"Start":"10:39.355 ","End":"10:41.890","Text":"The work is in canceling out."},{"Start":"10:41.890 ","End":"10:44.965","Text":"This side of the equation isn\u0027t canceling out,"},{"Start":"10:44.965 ","End":"10:49.360","Text":"and therefore this side of the equation also is in canceling out,"},{"Start":"10:49.360 ","End":"10:55.910","Text":"and therefore, the rotor(P) cannot be equal to 0."},{"Start":"10:57.600 ","End":"11:02.330","Text":"In that case, we can see that the"},{"Start":"11:03.270 ","End":"11:10.735","Text":"rotor(D) is equal to 0 plus the rotor(P)."},{"Start":"11:10.735 ","End":"11:15.220","Text":"It\u0027s just equal to the rotor(P),"},{"Start":"11:15.220 ","End":"11:17.140","Text":"and of course in some cases,"},{"Start":"11:17.140 ","End":"11:22.070","Text":"it will be equal to 0, but it doesn\u0027t have to be equal to 0."},{"Start":"11:23.400 ","End":"11:28.300","Text":"If D is in a conservative vector field,"},{"Start":"11:28.300 ","End":"11:33.685","Text":"then D also cannot be the field due to charged particles."},{"Start":"11:33.685 ","End":"11:35.710","Text":"Also, D doesn\u0027t have"},{"Start":"11:35.710 ","End":"11:42.290","Text":"the same potential function as that of a field with the charged particles."},{"Start":"11:43.950 ","End":"11:47.830","Text":"In that case, how can we work with D?"},{"Start":"11:47.830 ","End":"11:51.080","Text":"How can we work with the displacement vector?"},{"Start":"11:51.780 ","End":"11:55.285","Text":"When can I use my D?"},{"Start":"11:55.285 ","End":"12:01.015","Text":"I can work with Gauss\u0027s law when I have cases with some symmetry."},{"Start":"12:01.015 ","End":"12:09.640","Text":"If I have some wire or some cylinder like so,"},{"Start":"12:09.640 ","End":"12:13.750","Text":"where I have some symmetry, or a sphere."},{"Start":"12:13.750 ","End":"12:20.905","Text":"Or let\u0027s say if I have some infinite plane like so,"},{"Start":"12:20.905 ","End":"12:26.305","Text":"and all of these have some Rho free or Sigma free."},{"Start":"12:26.305 ","End":"12:28.870","Text":"Here it would actually be Sigma free."},{"Start":"12:28.870 ","End":"12:31.285","Text":"Here we\u0027ll have Rho free,"},{"Start":"12:31.285 ","End":"12:35.185","Text":"and then if I have some dielectric material on top."},{"Start":"12:35.185 ","End":"12:41.185","Text":"I can work out the electric field here by using Gauss\u0027s law."},{"Start":"12:41.185 ","End":"12:44.080","Text":"I can just calculate my D,"},{"Start":"12:44.080 ","End":"12:53.570","Text":"my displacement vector from Gauss\u0027s law by using the free charges or the Sigma free."},{"Start":"12:54.270 ","End":"12:59.605","Text":"Sorry, I can calculate through Gauss\u0027s law on D,"},{"Start":"12:59.605 ","End":"13:04.345","Text":"my Sigma free, and then my free charges."},{"Start":"13:04.345 ","End":"13:11.660","Text":"From there, I can find out what my E field and my polarization vectors are."},{"Start":"13:12.750 ","End":"13:18.415","Text":"What do we do if we\u0027re dealing with a case where we don\u0027t have symmetry."},{"Start":"13:18.415 ","End":"13:23.810","Text":"That means that we have to use the boundary conditions."},{"Start":"13:24.780 ","End":"13:28.660","Text":"What are the boundary conditions?"},{"Start":"13:28.660 ","End":"13:35.125","Text":"First of all, if we find the jump in our displacement vector,"},{"Start":"13:35.125 ","End":"13:38.890","Text":"which is perpendicular to our surface,"},{"Start":"13:38.890 ","End":"13:42.730","Text":"that is only due to our Sigma free,"},{"Start":"13:42.730 ","End":"13:44.995","Text":"our free surface charges."},{"Start":"13:44.995 ","End":"13:49.690","Text":"If here I have the surface of some material,"},{"Start":"13:49.690 ","End":"13:52.555","Text":"and here I have my perpendicular D here,"},{"Start":"13:52.555 ","End":"13:55.630","Text":"and here my perpendicular D,"},{"Start":"13:55.630 ","End":"14:02.170","Text":"there\u0027s a jump in my value for D is only due to the Sigma free."},{"Start":"14:02.170 ","End":"14:10.430","Text":"Also, I can take the jump in my parallel D vector."},{"Start":"14:11.280 ","End":"14:14.780","Text":"This is without the vector sign also here."},{"Start":"14:14.780 ","End":"14:24.000","Text":"The parallel D, and that is equal to Delta P parallel."},{"Start":"14:24.900 ","End":"14:29.275","Text":"This is my change in the polarization."},{"Start":"14:29.275 ","End":"14:33.745","Text":"This was D perpendicular and D perpendicular,"},{"Start":"14:33.745 ","End":"14:39.850","Text":"and this is D parallel and this is my other D parallel."},{"Start":"14:39.850 ","End":"14:42.485","Text":"I look at the difference between the two."},{"Start":"14:42.485 ","End":"14:45.710","Text":"Whereas when we\u0027re dealing with the E field,"},{"Start":"14:45.710 ","End":"14:49.070","Text":"what I have is Epsilon naught and then if I look at"},{"Start":"14:49.070 ","End":"14:52.550","Text":"the change in my perpendicular E field,"},{"Start":"14:52.550 ","End":"14:56.113","Text":"this is equal to my Sigma total,"},{"Start":"14:56.113 ","End":"15:00.980","Text":"so all of my charges both bound and free."},{"Start":"15:00.980 ","End":"15:12.070","Text":"If I look at the change in my parallel E fields."},{"Start":"15:12.070 ","End":"15:14.170","Text":"This will always be equal to 0,"},{"Start":"15:14.170 ","End":"15:19.170","Text":"and that is because the electric field is always conservative."},{"Start":"15:20.310 ","End":"15:22.760","Text":"That is the end of the lesson."},{"Start":"15:22.760 ","End":"15:27.770","Text":"What\u0027s important to take from this is that if we have a case of symmetry,"},{"Start":"15:27.770 ","End":"15:30.800","Text":"then we use D with Gauss\u0027s law,"},{"Start":"15:30.800 ","End":"15:34.020","Text":"and if we don\u0027t have symmetry in the problem,"},{"Start":"15:34.020 ","End":"15:37.360","Text":"then we have to use the boundary conditions."},{"Start":"15:37.360 ","End":"15:41.230","Text":"For that, you should remember these equations."},{"Start":"15:41.230 ","End":"15:44.030","Text":"That\u0027s the end of this lesson."}],"ID":22321},{"Watched":false,"Name":"Linear Dielectric Materials","Duration":"10m 22s","ChapterTopicVideoID":21474,"CourseChapterTopicPlaylistID":99475,"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 lesson,"},{"Start":"00:02.145 ","End":"00:06.405","Text":"we\u0027re going to be speaking about linear dielectric materials."},{"Start":"00:06.405 ","End":"00:13.095","Text":"If we have some dielectric material, like so,"},{"Start":"00:13.095 ","End":"00:17.520","Text":"and we put it in an E field, so,"},{"Start":"00:17.520 ","End":"00:22.623","Text":"let\u0027s say that the E field is pointing in this direction,"},{"Start":"00:22.623 ","End":"00:25.910","Text":"we know that the charges within the dielectric material"},{"Start":"00:25.910 ","End":"00:29.875","Text":"are going to line up with the E field,"},{"Start":"00:29.875 ","End":"00:36.995","Text":"such that they\u0027re linear to the E field and the equation for our polarization vector is"},{"Start":"00:36.995 ","End":"00:44.620","Text":"equal to Epsilon naught multiplied by Chi_e where Chi_e is the Greek letter Chi_e,"},{"Start":"00:44.620 ","End":"00:52.590","Text":"which means the susceptibility of the material multiplied by E, the electric field."},{"Start":"00:53.030 ","End":"00:55.640","Text":"What does susceptibility mean?"},{"Start":"00:55.640 ","End":"00:57.145","Text":"What is this Chi_e?"},{"Start":"00:57.145 ","End":"01:05.470","Text":"It tells us how strong the polarization will be given some electric field."},{"Start":"01:05.470 ","End":"01:09.350","Text":"But what\u0027s important to know is that the polarization will"},{"Start":"01:09.350 ","End":"01:13.505","Text":"be linear according to our electric field."},{"Start":"01:13.505 ","End":"01:17.165","Text":"Something that is important to note is that"},{"Start":"01:17.165 ","End":"01:22.925","Text":"our polarization vector is equal to Epsilon naught Chi_e multiplied by E,"},{"Start":"01:22.925 ","End":"01:32.189","Text":"but it is not equal to Epsilon naught Chi_e multiplied by E naught."},{"Start":"01:32.189 ","End":"01:33.555","Text":"Why I\u0027m reminding you,"},{"Start":"01:33.555 ","End":"01:40.290","Text":"E naught represents the electric field due to the free particles alone,"},{"Start":"01:40.670 ","End":"01:44.590","Text":"or rather the free charges."},{"Start":"01:44.810 ","End":"01:49.910","Text":"Why does this not equal all of this multiplied by E naught?"},{"Start":"01:49.910 ","End":"01:54.065","Text":"Let\u0027s imagine that I have some surface over here"},{"Start":"01:54.065 ","End":"01:59.905","Text":"with some charges over here and these are my free charges."},{"Start":"01:59.905 ","End":"02:02.945","Text":"I have Sigma_f over here."},{"Start":"02:02.945 ","End":"02:06.830","Text":"Then that means that I\u0027ll have in my dielectric materials,"},{"Start":"02:06.830 ","End":"02:10.670","Text":"some electric field like so E naught."},{"Start":"02:10.670 ","End":"02:17.705","Text":"Why can\u0027t I find my polarization density due to this E naught,"},{"Start":"02:17.705 ","End":"02:21.690","Text":"this electric field due to these charges?"},{"Start":"02:21.710 ","End":"02:27.305","Text":"The reason is because in this material I have lots of dipoles,"},{"Start":"02:27.305 ","End":"02:31.860","Text":"and then due to this electric field E naught,"},{"Start":"02:31.860 ","End":"02:37.055","Text":"my dipoles are going to further be polarized."},{"Start":"02:37.055 ","End":"02:40.715","Text":"Then by this change in polarization,"},{"Start":"02:40.715 ","End":"02:46.555","Text":"that is in itself going to change the electric field"},{"Start":"02:46.555 ","End":"02:49.640","Text":"and this change in electric field is going to"},{"Start":"02:49.640 ","End":"02:54.290","Text":"change again the polarization of the dipoles."},{"Start":"02:54.290 ","End":"02:56.855","Text":"Then their polarization will change again,"},{"Start":"02:56.855 ","End":"02:58.505","Text":"again, changing the E field,"},{"Start":"02:58.505 ","End":"03:00.380","Text":"again, changing their polarization,"},{"Start":"03:00.380 ","End":"03:02.420","Text":"and so on and so forth."},{"Start":"03:02.420 ","End":"03:06.860","Text":"What we\u0027re getting into is a never ending cycle."},{"Start":"03:06.860 ","End":"03:10.715","Text":"Therefore we see that our polarization vector isn\u0027t linear"},{"Start":"03:10.715 ","End":"03:15.065","Text":"according to our E naught electric field,"},{"Start":"03:15.065 ","End":"03:19.130","Text":"the electric field formed by our free particles but it is"},{"Start":"03:19.130 ","End":"03:23.405","Text":"linear due to the total electric field,"},{"Start":"03:23.405 ","End":"03:27.829","Text":"which of course includes the electric field due to our free particles"},{"Start":"03:27.829 ","End":"03:33.000","Text":"and due to our dipoles inside the material."},{"Start":"03:33.000 ","End":"03:36.360","Text":"This is the final electric field that we"},{"Start":"03:36.360 ","End":"03:40.610","Text":"get once everything is settled down and then we\u0027ll"},{"Start":"03:40.610 ","End":"03:48.515","Text":"see that P is linear to this E field that includes the free charges,"},{"Start":"03:48.515 ","End":"03:51.390","Text":"and the bound charges."},{"Start":"03:53.000 ","End":"04:00.680","Text":"Now let\u0027s write out our equation for d. We know that our displacement vector is equal to"},{"Start":"04:00.680 ","End":"04:08.840","Text":"Epsilon naught multiplied by E plus P. But we know that P is equal to this,"},{"Start":"04:08.840 ","End":"04:10.340","Text":"so let\u0027s write that out."},{"Start":"04:10.340 ","End":"04:14.225","Text":"We have Epsilon naught multiplied by E plus P,"},{"Start":"04:14.225 ","End":"04:23.945","Text":"which is equal to Epsilon naught Chi_e multiplied by E. Now I can take out"},{"Start":"04:23.945 ","End":"04:30.200","Text":"all of my common factors over here so I have Epsilon naught and then I have"},{"Start":"04:30.200 ","End":"04:36.505","Text":"1+Chi_e multiplied by E."},{"Start":"04:36.505 ","End":"04:41.253","Text":"Then what I can do is I can call all of this,"},{"Start":"04:41.253 ","End":"04:45.650","Text":"let\u0027s give it a short name and let\u0027s call it Epsilon."},{"Start":"04:45.650 ","End":"04:48.710","Text":"Therefore I can say that all that,"},{"Start":"04:48.710 ","End":"04:56.190","Text":"D is simply equal to Epsilon multiplied by my E field."},{"Start":"04:57.730 ","End":"05:05.180","Text":"This Epsilon, which is equal to this is called the permeability."},{"Start":"05:05.970 ","End":"05:10.610","Text":"Permeability is very similar to susceptibility."},{"Start":"05:10.610 ","End":"05:15.550","Text":"1, both are constants and we can see that if we have the susceptibility,"},{"Start":"05:15.550 ","End":"05:18.640","Text":"then we know what the permeability is because we know what"},{"Start":"05:18.640 ","End":"05:22.105","Text":"Epsilon naught is and vice versa."},{"Start":"05:22.105 ","End":"05:27.535","Text":"These 2 susceptibility and permeability are interchangeable."},{"Start":"05:27.535 ","End":"05:29.920","Text":"You can work with both of them."},{"Start":"05:29.920 ","End":"05:31.300","Text":"They both mean the same thing."},{"Start":"05:31.300 ","End":"05:34.360","Text":"Just obviously, if you\u0027re working out P,"},{"Start":"05:34.360 ","End":"05:39.925","Text":"then you have to use susceptibility and if you\u0027re working out D,"},{"Start":"05:39.925 ","End":"05:42.950","Text":"then you will get permeability,"},{"Start":"05:42.950 ","End":"05:47.220","Text":"but it includes the susceptibility in it."},{"Start":"05:47.990 ","End":"05:52.040","Text":"Another term that we can use is this k,"},{"Start":"05:52.040 ","End":"05:56.210","Text":"which is the dielectric constant and that\u0027s sometimes symbolized by"},{"Start":"05:56.210 ","End":"06:00.680","Text":"Epsilon r and it\u0027s just equal to Epsilon divided by Epsilon naught."},{"Start":"06:00.680 ","End":"06:07.735","Text":"We\u0027re getting rid of the units of our Epsilon and it\u0027s just equal to 1+Chi_e."},{"Start":"06:07.735 ","End":"06:15.120","Text":"We can use any 1 of these 3 terms in these equations."},{"Start":"06:15.120 ","End":"06:20.340","Text":"We just have to work around with the algebra in order to use them."},{"Start":"06:20.660 ","End":"06:25.280","Text":"What we can see is that P and D are linear"},{"Start":"06:25.280 ","End":"06:29.705","Text":"to our E field and it\u0027s important to remember which E field."},{"Start":"06:29.705 ","End":"06:35.670","Text":"It\u0027s the total E field due to both the bound and free charges."},{"Start":"06:35.930 ","End":"06:41.300","Text":"What\u0027s important to say over here is that D over here still"},{"Start":"06:41.300 ","End":"06:46.370","Text":"isn\u0027t necessarily equal to Epsilon naught E naught even"},{"Start":"06:46.370 ","End":"06:50.480","Text":"for linear dielectric materials and that\u0027s because the rotor of"},{"Start":"06:50.480 ","End":"06:57.010","Text":"D or the rotor of P over here will not be equal to 0."},{"Start":"06:57.860 ","End":"07:05.315","Text":"The 1 case where I can say that the rotor of D is equal to 0,"},{"Start":"07:05.315 ","End":"07:08.165","Text":"is in a very specific case,"},{"Start":"07:08.165 ","End":"07:17.585","Text":"where my entire region is filled with a uniform dielectric material."},{"Start":"07:17.585 ","End":"07:27.630","Text":"In that case, I can say that my rotor for my displacement vector is equal to 0."},{"Start":"07:28.520 ","End":"07:31.400","Text":"Only in this specific case where"},{"Start":"07:31.400 ","End":"07:34.849","Text":"the entire region is filled with a uniform dielectric material,"},{"Start":"07:34.849 ","End":"07:39.855","Text":"will the rotor of D and the rotor of P be equal to 0."},{"Start":"07:39.855 ","End":"07:41.870","Text":"Then, in that case,"},{"Start":"07:41.870 ","End":"07:50.985","Text":"we can actually write that D is equal to Epsilon naught multiplied by E naught."},{"Start":"07:50.985 ","End":"07:58.595","Text":"In that case, we can say that our displacement vector is equal to"},{"Start":"07:58.595 ","End":"08:07.470","Text":"the electric field formed by the free charged particles."},{"Start":"08:08.180 ","End":"08:13.895","Text":"This is what we have in the case of a capacitor."},{"Start":"08:13.895 ","End":"08:17.660","Text":"In a lot of cases where we\u0027re dealing with dielectric materials,"},{"Start":"08:17.660 ","End":"08:20.210","Text":"we\u0027re actually going to be dealing with capacitors,"},{"Start":"08:20.210 ","End":"08:23.949","Text":"in which case we can use this equation."},{"Start":"08:23.949 ","End":"08:29.195","Text":"Let\u0027s imagine that we have some capacitor like so."},{"Start":"08:29.195 ","End":"08:34.260","Text":"It has its upper sheet and"},{"Start":"08:34.260 ","End":"08:41.495","Text":"its lower sheets and in the middle between these 2 plates,"},{"Start":"08:41.495 ","End":"08:48.625","Text":"we have this dielectric material, like so."},{"Start":"08:48.625 ","End":"08:51.515","Text":"Here we have a dielectric material."},{"Start":"08:51.515 ","End":"08:53.390","Text":"Now as we know with a capacitor,"},{"Start":"08:53.390 ","End":"08:57.770","Text":"the E field outside of the capacitor is equal to 0."},{"Start":"08:57.770 ","End":"09:02.190","Text":"The E field above and below is equal to 0."},{"Start":"09:02.240 ","End":"09:07.580","Text":"Then, in this case, it doesn\u0027t matter that the entire region isn\u0027t filled with"},{"Start":"09:07.580 ","End":"09:10.204","Text":"this uniform dielectric material"},{"Start":"09:10.204 ","End":"09:14.615","Text":"because the electric field over in this region is equal to 0,"},{"Start":"09:14.615 ","End":"09:18.860","Text":"D=0 and P=0 so it doesn\u0027t bother me."},{"Start":"09:18.860 ","End":"09:24.510","Text":"What\u0027s interesting to me is just what\u0027s going on in the middle over here."},{"Start":"09:24.620 ","End":"09:29.570","Text":"What does interest me is the only region where I have an electric field here,"},{"Start":"09:29.570 ","End":"09:35.120","Text":"which is between the capacitor plates and between the capacitor plates,"},{"Start":"09:35.120 ","End":"09:41.720","Text":"that whole area is filled with a uniform dielectric material."},{"Start":"09:41.720 ","End":"09:44.030","Text":"We\u0027re working with this statement over"},{"Start":"09:44.030 ","End":"09:48.025","Text":"here and that means that the rotor of D and P is equal to 0."},{"Start":"09:48.025 ","End":"09:50.140","Text":"Therefore in this region,"},{"Start":"09:50.140 ","End":"09:55.755","Text":"I can say that D is equal to Epsilon naught multiplied by E naught."},{"Start":"09:55.755 ","End":"09:58.415","Text":"In this region, inside the capacitor,"},{"Start":"09:58.415 ","End":"10:02.495","Text":"I can say that my displacement vector is"},{"Start":"10:02.495 ","End":"10:08.890","Text":"linear to my electric field caused by my free charged particles."},{"Start":"10:08.890 ","End":"10:19.340","Text":"I can say that D is basically equal to the free charges on the capacitor plates."},{"Start":"10:19.340 ","End":"10:22.320","Text":"That\u0027s the end of this lesson."}],"ID":22322},{"Watched":false,"Name":"Exercise 7","Duration":"7m 14s","ChapterTopicVideoID":21475,"CourseChapterTopicPlaylistID":99475,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.710","Text":"Hello. In this lesson,"},{"Start":"00:01.710 ","End":"00:04.065","Text":"we\u0027re going to be answering the following question."},{"Start":"00:04.065 ","End":"00:10.290","Text":"A light ray is made up of an electric and magnetic fields propagating in space."},{"Start":"00:10.290 ","End":"00:15.210","Text":"Show that if the ray passes from one dielectric material of"},{"Start":"00:15.210 ","End":"00:21.825","Text":"constant Epsilon 1 and into another dielectric material of constant Epsilon 2,"},{"Start":"00:21.825 ","End":"00:24.120","Text":"then we get Snell\u0027s law."},{"Start":"00:24.120 ","End":"00:28.390","Text":"Here is a reminder of what Snell\u0027s law is."},{"Start":"00:29.240 ","End":"00:34.140","Text":"For this question, we can ignore"},{"Start":"00:34.140 ","End":"00:36.060","Text":"the magnetic fields and we can imagine that"},{"Start":"00:36.060 ","End":"00:39.125","Text":"a light ray is just made up of an electric field."},{"Start":"00:39.125 ","End":"00:46.250","Text":"When a ray passes through two different materials or two different dielectric materials,"},{"Start":"00:46.250 ","End":"00:53.190","Text":"then the light ray bends or refracts a little bit."},{"Start":"00:53.390 ","End":"00:57.740","Text":"What do we have over here is that the light ray travels"},{"Start":"00:57.740 ","End":"01:01.745","Text":"through the first dielectric material and it comes in,"},{"Start":"01:01.745 ","End":"01:07.715","Text":"it\u0027s incident on the first dielectric material at an angle of Theta 1."},{"Start":"01:07.715 ","End":"01:11.990","Text":"Then at the interface between the two dielectric materials,"},{"Start":"01:11.990 ","End":"01:18.140","Text":"the light ray will refract or will bend a little bit depending on the refractive index,"},{"Start":"01:18.140 ","End":"01:22.280","Text":"Epsilon 2, or the difference in the two refractive indices"},{"Start":"01:22.280 ","End":"01:30.875","Text":"and it will bend and the ray will now have a new angle,"},{"Start":"01:30.875 ","End":"01:35.280","Text":"which will be given as Theta 2."},{"Start":"01:35.540 ","End":"01:40.475","Text":"Snell\u0027s law finds the relationship between Theta 1, Theta 2,"},{"Start":"01:40.475 ","End":"01:46.575","Text":"and Epsilon 1 and Epsilon 2. Let\u0027s begin."},{"Start":"01:46.575 ","End":"01:51.915","Text":"First of all, what is tan of Theta 1?"},{"Start":"01:51.915 ","End":"01:58.980","Text":"Tan of Theta 1 is tan of this angle over here."},{"Start":"01:58.980 ","End":"02:00.770","Text":"What does that equal to?"},{"Start":"02:00.770 ","End":"02:05.800","Text":"That is equal to the electric field,"},{"Start":"02:05.800 ","End":"02:11.135","Text":"so the parallel component in Region number 1,"},{"Start":"02:11.135 ","End":"02:18.275","Text":"divided by the perpendicular component still in Region number 1,"},{"Start":"02:18.275 ","End":"02:21.960","Text":"where this over here,"},{"Start":"02:21.960 ","End":"02:23.983","Text":"our parallel component,"},{"Start":"02:23.983 ","End":"02:29.010","Text":"and this over here is our perpendicular component."},{"Start":"02:30.080 ","End":"02:32.525","Text":"A few lessons ago,"},{"Start":"02:32.525 ","End":"02:36.095","Text":"we learned that the rotor of"},{"Start":"02:36.095 ","End":"02:41.030","Text":"an electric field due to all of"},{"Start":"02:41.030 ","End":"02:47.255","Text":"the charges is equal to 0 because the electric field is a conservative field."},{"Start":"02:47.255 ","End":"02:51.350","Text":"Therefore, we know that the jump in"},{"Start":"02:51.350 ","End":"02:57.980","Text":"the parallel components of the electric field has to be equal to 0."},{"Start":"02:57.980 ","End":"03:04.505","Text":"Therefore, we know that the parallel component in"},{"Start":"03:04.505 ","End":"03:13.620","Text":"the first material is equal to the parallel component in the second material."},{"Start":"03:13.910 ","End":"03:17.060","Text":"Then from a boundary conditions,"},{"Start":"03:17.060 ","End":"03:22.025","Text":"we know that the change in the perpendicular component"},{"Start":"03:22.025 ","End":"03:27.815","Text":"of our displacement vector is equal to our Sigma free, our free charges."},{"Start":"03:27.815 ","End":"03:29.480","Text":"Now, over here,"},{"Start":"03:29.480 ","End":"03:31.985","Text":"we\u0027re dealing with dielectric materials."},{"Start":"03:31.985 ","End":"03:33.800","Text":"We might have bound charges,"},{"Start":"03:33.800 ","End":"03:36.965","Text":"but we\u0027re not told of any free charges added."},{"Start":"03:36.965 ","End":"03:38.990","Text":"In this example over here,"},{"Start":"03:38.990 ","End":"03:42.100","Text":"this is equal to 0."},{"Start":"03:43.160 ","End":"03:52.070","Text":"I\u0027ll just remind you that our perpendicular components of our D-field are over here."},{"Start":"03:52.070 ","End":"03:55.590","Text":"This is in Region 1 and this is in Region 2."},{"Start":"03:55.590 ","End":"04:01.305","Text":"Our jump in a perpendicular component of D is equal to 0."},{"Start":"04:01.305 ","End":"04:06.820","Text":"Therefore, the perpendicular component in Region 2"},{"Start":"04:06.820 ","End":"04:12.581","Text":"minus the perpendicular component in Region 1 is of course equal to 0."},{"Start":"04:12.581 ","End":"04:16.230","Text":"Therefore, we know that D is equal to,"},{"Start":"04:16.230 ","End":"04:19.185","Text":"in Region 2,"},{"Start":"04:19.185 ","End":"04:24.200","Text":"Epsilon 2 multiplied by the perpendicular component of"},{"Start":"04:24.200 ","End":"04:30.240","Text":"the E-field there minus the perpendicular component D in Region 1,"},{"Start":"04:30.240 ","End":"04:32.290","Text":"which is equal to Epsilon 1,"},{"Start":"04:32.290 ","End":"04:38.750","Text":"multiplied by the perpendicular component of the E-field in Region 1."},{"Start":"04:38.750 ","End":"04:43.240","Text":"This is of course, all equal to 0."},{"Start":"04:43.880 ","End":"04:46.535","Text":"Let\u0027s move that up to here."},{"Start":"04:46.535 ","End":"04:48.275","Text":"This is equal to 0."},{"Start":"04:48.275 ","End":"04:51.365","Text":"Therefore, I can say,"},{"Start":"04:51.365 ","End":"04:57.440","Text":"if I just rearrange this onto this side and divide both sides by Epsilon 1,"},{"Start":"04:57.440 ","End":"04:59.885","Text":"so what I will get is that"},{"Start":"04:59.885 ","End":"05:06.060","Text":"the perpendicular component of the E-field in Region 1 is equal"},{"Start":"05:06.060 ","End":"05:09.910","Text":"to Epsilon 2 divided by Epsilon 1"},{"Start":"05:09.910 ","End":"05:15.630","Text":"multiplied by the perpendicular component of my E-field in Region 2."},{"Start":"05:16.160 ","End":"05:20.150","Text":"Now a quick note that I didn\u0027t mention is that"},{"Start":"05:20.150 ","End":"05:23.180","Text":"the reason I can substitute in these equations is"},{"Start":"05:23.180 ","End":"05:25.340","Text":"because we\u0027re assuming that"},{"Start":"05:25.340 ","End":"05:30.355","Text":"these two dielectric materials are linear dielectric materials."},{"Start":"05:30.355 ","End":"05:32.780","Text":"That\u0027s why it works. If you don\u0027t understand that,"},{"Start":"05:32.780 ","End":"05:36.090","Text":"please go back to the previous lesson."},{"Start":"05:38.200 ","End":"05:42.095","Text":"Now, let\u0027s plug this all into this equation."},{"Start":"05:42.095 ","End":"05:46.195","Text":"I\u0027m back to my tan of Theta 1,"},{"Start":"05:46.195 ","End":"05:55.875","Text":"which is equal to my parallel E component divided by my perpendicular E component."},{"Start":"05:55.875 ","End":"05:59.795","Text":"My parallel E component in Region 1, I know,"},{"Start":"05:59.795 ","End":"06:05.555","Text":"is also equal from over here to my parallel E component in Region 2,"},{"Start":"06:05.555 ","End":"06:13.530","Text":"and I know that my perpendicular component in Region 1 is equal to this over here,"},{"Start":"06:13.530 ","End":"06:17.540","Text":"Epsilon 2 divided by Epsilon 1 multiplied"},{"Start":"06:17.540 ","End":"06:23.190","Text":"by my perpendicular component of my E-field in Region 2."},{"Start":"06:24.110 ","End":"06:29.150","Text":"My parallel component of the E-field in Region 2 divided by"},{"Start":"06:29.150 ","End":"06:35.500","Text":"my perpendicular component in Region 2 is similar to this."},{"Start":"06:35.500 ","End":"06:38.595","Text":"It\u0027s going to be equal to tan of Theta 2."},{"Start":"06:38.595 ","End":"06:43.455","Text":"But then I have Epsilon 2 divided by Epsilon 1 in the denominator."},{"Start":"06:43.455 ","End":"06:48.200","Text":"This is going to become Epsilon 1 divided by Epsilon 2,"},{"Start":"06:48.200 ","End":"06:54.590","Text":"and then this over here is equal to tan of Theta 2,"},{"Start":"06:54.590 ","End":"07:01.025","Text":"so multiplied by tan of Theta 2."},{"Start":"07:01.025 ","End":"07:04.010","Text":"That\u0027s it. Now we see we got to Snell\u0027s law;"},{"Start":"07:04.010 ","End":"07:11.465","Text":"tan of Theta 1 is equal to Epsilon 1 divided by Epsilon 2 of tan of Theta 2."},{"Start":"07:11.465 ","End":"07:14.580","Text":"That is the end of this lesson.R"}],"ID":22323}],"Thumbnail":null,"ID":99475}]

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