[{"Name":"Introduction to Poynting Vector","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"1.2 Poynting Vector","Duration":"25m 24s","ChapterTopicVideoID":21379,"CourseChapterTopicPlaylistID":99493,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/21379.jpeg","UploadDate":"2020-04-06T22:39:13.1630000","DurationForVideoObject":"PT25M24S","Description":null,"MetaTitle":"1.2 Poynting Vector: Video + Workbook | Proprep","MetaDescription":"The Poynting Vector and the Energy Stored in Fields - Introduction to Poynting Vector. 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/the-poynting-vector-and-the-energy-stored-in-fields/introduction-to-poynting-vector/vid21469","VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.160","Text":"Hello, in this lesson,"},{"Start":"00:02.160 ","End":"00:05.960","Text":"we\u0027re going to be speaking about the pointing vector."},{"Start":"00:05.960 ","End":"00:09.270","Text":"First, let\u0027s write down the mathematical definition,"},{"Start":"00:09.270 ","End":"00:12.240","Text":"and then afterwards, we\u0027ll explain what\u0027s going on."},{"Start":"00:12.240 ","End":"00:18.690","Text":"The pointing vector is denoted by a capital S with a vector sign above."},{"Start":"00:18.690 ","End":"00:27.435","Text":"This is the pointing vector and it is equal to 1 divided by Mu naught."},{"Start":"00:27.435 ","End":"00:31.710","Text":"This is multiplied by E, the electric field,"},{"Start":"00:31.710 ","End":"00:37.600","Text":"cross multiplied with B, the magnetic field."},{"Start":"00:39.080 ","End":"00:43.680","Text":"Now let\u0027s start speaking about what this means."},{"Start":"00:43.680 ","End":"00:45.440","Text":"In the previous lesson,"},{"Start":"00:45.440 ","End":"00:50.720","Text":"we saw the energy needed to build a system"},{"Start":"00:50.720 ","End":"00:56.390","Text":"or the energy that is stored when there is a magnetic and electric field present."},{"Start":"00:56.390 ","End":"01:02.810","Text":"We saw that the energy is equal to the integral over a certain volume,"},{"Start":"01:02.810 ","End":"01:08.420","Text":"which was equal to Epsilon naught divided by 2 multiplied by"},{"Start":"01:08.420 ","End":"01:16.680","Text":"E^2 plus B squared divided by 2 Mu naught."},{"Start":"01:16.680 ","End":"01:18.955","Text":"All of this dV,"},{"Start":"01:18.955 ","End":"01:22.490","Text":"we\u0027re integrating with respect to volume."},{"Start":"01:22.490 ","End":"01:28.170","Text":"We saw that this section is called u_E."},{"Start":"01:28.370 ","End":"01:34.895","Text":"The energy density due to the magnetic field."},{"Start":"01:34.895 ","End":"01:38.480","Text":"This part was called u_B,"},{"Start":"01:38.480 ","End":"01:43.475","Text":"which was the energy density due to the magnetic field."},{"Start":"01:43.475 ","End":"01:47.630","Text":"Or we can look at everything inside these brackets,"},{"Start":"01:47.630 ","End":"01:49.940","Text":"which we called u_em,"},{"Start":"01:49.940 ","End":"01:57.490","Text":"which was the energy density due to the electromagnetic fields together."},{"Start":"01:57.490 ","End":"02:01.415","Text":"What we said is that we could integrate over"},{"Start":"02:01.415 ","End":"02:06.350","Text":"an entire volume to find the energy in that specific volume."},{"Start":"02:06.350 ","End":"02:07.970","Text":"Or in other words,"},{"Start":"02:07.970 ","End":"02:12.440","Text":"we could also take some unit volume."},{"Start":"02:12.440 ","End":"02:19.395","Text":"A single unit like so where its volume is dV."},{"Start":"02:19.395 ","End":"02:25.530","Text":"We could find that the energy inside of here is equal to u_em."},{"Start":"02:27.170 ","End":"02:30.740","Text":"Up until now, we\u0027ve been dealing with the energy."},{"Start":"02:30.740 ","End":"02:36.569","Text":"Now let\u0027s take a look at what S or the pointing vector means."},{"Start":"02:36.740 ","End":"02:42.184","Text":"The pointing vector S is the amount of energy"},{"Start":"02:42.184 ","End":"02:48.150","Text":"flowing through a unit of area in a unit of time."},{"Start":"02:48.150 ","End":"02:53.810","Text":"How much energy flows through a unit of area in a unit of time,"},{"Start":"02:53.810 ","End":"02:56.100","Text":"let\u0027s say per second."},{"Start":"02:56.330 ","End":"02:59.640","Text":"Let\u0027s try and explain this."},{"Start":"02:59.640 ","End":"03:02.045","Text":"Let\u0027s say that over here,"},{"Start":"03:02.045 ","End":"03:04.175","Text":"we have going through,"},{"Start":"03:04.175 ","End":"03:06.050","Text":"over here in this direction,"},{"Start":"03:06.050 ","End":"03:08.330","Text":"we have an electric field."},{"Start":"03:08.330 ","End":"03:10.520","Text":"Let\u0027s say that it\u0027s time-dependent,"},{"Start":"03:10.520 ","End":"03:14.210","Text":"and going in this perpendicular direction,"},{"Start":"03:14.210 ","End":"03:17.015","Text":"we have a magnetic field,"},{"Start":"03:17.015 ","End":"03:21.685","Text":"which is also time-dependent, like so."},{"Start":"03:21.685 ","End":"03:27.840","Text":"Now let\u0027s say that we\u0027re looking through this side of the cube."},{"Start":"03:27.840 ","End":"03:32.430","Text":"In other words, we\u0027re looking through this surface area."},{"Start":"03:32.430 ","End":"03:35.195","Text":"Or in other words, because this is dV,"},{"Start":"03:35.195 ","End":"03:40.025","Text":"this area will be dS so in a unit of area."},{"Start":"03:40.025 ","End":"03:45.830","Text":"How much of this energy u electromagnetic due to"},{"Start":"03:45.830 ","End":"03:53.150","Text":"this electric and magnetic field passes from this cube into another cube."},{"Start":"03:53.150 ","End":"03:57.530","Text":"Let\u0027s say that there is another cube over"},{"Start":"03:57.530 ","End":"04:04.900","Text":"here, like so."},{"Start":"04:04.900 ","End":"04:08.110","Text":"The energy is going to flow through here."},{"Start":"04:08.110 ","End":"04:14.520","Text":"Let\u0027s draw it in red in this direction."},{"Start":"04:14.520 ","End":"04:21.490","Text":"This flow of energy from this cube through this surface area in a unit of time,"},{"Start":"04:21.490 ","End":"04:28.520","Text":"let\u0027s say in 1 second is the S vector or the pointing vector."},{"Start":"04:29.080 ","End":"04:33.970","Text":"Here specifically with the axis that we randomly chose,"},{"Start":"04:33.970 ","End":"04:36.279","Text":"where the electric field is pointing"},{"Start":"04:36.279 ","End":"04:39.700","Text":"upwards and the magnetic field is pointing towards us."},{"Start":"04:39.700 ","End":"04:43.820","Text":"The pointing vector, which is equal to E cross B,"},{"Start":"04:43.820 ","End":"04:47.420","Text":"will be pointing in this rightwards direction."},{"Start":"04:47.420 ","End":"04:50.330","Text":"However, theoretically, if our E field was"},{"Start":"04:50.330 ","End":"04:53.390","Text":"pointing in a different direction and our B field was"},{"Start":"04:53.390 ","End":"04:59.090","Text":"pointing in a slightly different direction our E cross B could also be pointing upwards."},{"Start":"04:59.090 ","End":"05:03.200","Text":"Then energy would be transferred to a unit of volume which is"},{"Start":"05:03.200 ","End":"05:08.070","Text":"located above this original unit of volume."},{"Start":"05:09.680 ","End":"05:14.120","Text":"This gives us an idea of what the pointing vector is."},{"Start":"05:14.120 ","End":"05:21.909","Text":"But now let\u0027s give an example just a further explanation of why this can be useful."},{"Start":"05:21.909 ","End":"05:27.695","Text":"Now let\u0027s imagine that I have a much bigger box."},{"Start":"05:27.695 ","End":"05:31.350","Text":"I have this over here."},{"Start":"05:32.810 ","End":"05:39.230","Text":"Inside this box, I also have an electric and a magnetic field flowing through,"},{"Start":"05:39.230 ","End":"05:45.890","Text":"and I\u0027ve calculated the energy density or the total energy in this entire box."},{"Start":"05:45.890 ","End":"05:51.735","Text":"I\u0027ve integrated with the whole volume of this box and I\u0027ve found U,"},{"Start":"05:51.735 ","End":"05:54.520","Text":"so not the energy density u,"},{"Start":"05:54.520 ","End":"05:58.700","Text":"but the total energy inside this given volume."},{"Start":"05:58.700 ","End":"06:02.450","Text":"I know that the energy is U_em."},{"Start":"06:02.450 ","End":"06:08.820","Text":"The electromagnetic energy in this volume over here."},{"Start":"06:09.010 ","End":"06:12.650","Text":"Now what I want to do is I want to calculate"},{"Start":"06:12.650 ","End":"06:18.760","Text":"the amount of energy which is leaving this box."},{"Start":"06:18.760 ","End":"06:22.460","Text":"I\u0027m going to look at this point over here."},{"Start":"06:22.460 ","End":"06:28.840","Text":"Then I\u0027ll find the pointing vector over here for this area."},{"Start":"06:28.840 ","End":"06:35.480","Text":"Then I\u0027ll add up the energy or the pointing vector for this area,"},{"Start":"06:35.480 ","End":"06:37.835","Text":"and then for this area."},{"Start":"06:37.835 ","End":"06:41.540","Text":"I\u0027ll sum it all up until I calculated all of"},{"Start":"06:41.540 ","End":"06:46.910","Text":"the energy that has left this side of my box or of my cube."},{"Start":"06:46.910 ","End":"06:50.570","Text":"I\u0027m taking little surface areas and I\u0027m saying"},{"Start":"06:50.570 ","End":"06:56.155","Text":"how much energy is passing through that point per unit of time."},{"Start":"06:56.155 ","End":"07:01.520","Text":"That\u0027s my pointing vector and I add them all up for this entire side of the box."},{"Start":"07:01.520 ","End":"07:05.255","Text":"Then I do the exact same thing for over here,"},{"Start":"07:05.255 ","End":"07:11.510","Text":"I take a certain area and I calculate how much energy has exited from there."},{"Start":"07:11.510 ","End":"07:13.760","Text":"Then I take another surface area."},{"Start":"07:13.760 ","End":"07:17.660","Text":"Again, I calculate how much energy has come out of there."},{"Start":"07:17.660 ","End":"07:21.320","Text":"Then I add all of that up for this side of the cube."},{"Start":"07:21.320 ","End":"07:26.940","Text":"Then I carry on doing this for all the sides of the cube."},{"Start":"07:27.350 ","End":"07:36.150","Text":"Then I can see how much energy has left my box or how much energy has left my cube."},{"Start":"07:36.440 ","End":"07:41.060","Text":"The pointing vector gives us how much energy has left"},{"Start":"07:41.060 ","End":"07:45.975","Text":"a small surface area at a given time."},{"Start":"07:45.975 ","End":"07:49.065","Text":"Then when we sum all of them up,"},{"Start":"07:49.065 ","End":"07:50.975","Text":"all the pointing vectors up,"},{"Start":"07:50.975 ","End":"07:55.039","Text":"then we can find the total amount of energy that is left"},{"Start":"07:55.039 ","End":"08:01.630","Text":"our volume in a given unit of time."},{"Start":"08:02.090 ","End":"08:08.060","Text":"I wrote this down, so adding this up over the entire volume."},{"Start":"08:08.060 ","End":"08:15.455","Text":"We take this unit of area per unit time and we added up for the whole volume,"},{"Start":"08:15.455 ","End":"08:16.955","Text":"or in other words,"},{"Start":"08:16.955 ","End":"08:20.120","Text":"sum over the surface area that encases"},{"Start":"08:20.120 ","End":"08:26.210","Text":"the volume will give us the total energy lost from the volume,"},{"Start":"08:26.210 ","End":"08:30.840","Text":"from the entire volume in a given unit of time."},{"Start":"08:31.910 ","End":"08:40.413","Text":"Now what we want to do is we want to write this down in some kind of an equation."},{"Start":"08:40.413 ","End":"08:47.425","Text":"First of all, we know that we want to add this up over volume. Or i.e."},{"Start":"08:47.425 ","End":"08:52.750","Text":"what we\u0027re doing is we\u0027re summing over the surface area, encasing the volume."},{"Start":"08:52.750 ","End":"08:58.855","Text":"Whenever we\u0027re summing, what we want to do is we want to do an integral."},{"Start":"08:58.855 ","End":"09:06.920","Text":"We\u0027re summing the total energy lost from the volume in a unit of time."},{"Start":"09:06.990 ","End":"09:10.510","Text":"We\u0027re summing all of these Poynting vectors."},{"Start":"09:10.510 ","End":"09:14.740","Text":"We worked out the Poynting vector for this small surface area"},{"Start":"09:14.740 ","End":"09:19.960","Text":"over here and then we\u0027re adding it up across the entire surface area."},{"Start":"09:19.960 ","End":"09:23.380","Text":"We\u0027re integrating along S,"},{"Start":"09:23.380 ","End":"09:28.430","Text":"so this is the Poynting vector."},{"Start":"09:29.760 ","End":"09:35.620","Text":"It\u0027s a bit small, but this is the Poynting vector and then along the surface area,"},{"Start":"09:35.620 ","End":"09:39.550","Text":"so d lowercase s. Remember that this is"},{"Start":"09:39.550 ","End":"09:44.605","Text":"the surface area like we\u0027ve seen in all our other integrals dealing with surface area."},{"Start":"09:44.605 ","End":"09:50.035","Text":"This is surface area."},{"Start":"09:50.035 ","End":"09:56.275","Text":"Don\u0027t get confused between this capital S denoting the pointing vector"},{"Start":"09:56.275 ","End":"10:05.420","Text":"and this lowercase s ds denoting the surface area so this is ds."},{"Start":"10:07.290 ","End":"10:12.385","Text":"This, if we see how much energy has been lost,"},{"Start":"10:12.385 ","End":"10:17.335","Text":"that means how much energy has exited our volume."},{"Start":"10:17.335 ","End":"10:23.260","Text":"Just like when we were looking at Gauss\u0027s law and also Faraday\u0027s law,"},{"Start":"10:23.260 ","End":"10:26.770","Text":"we saw how much energy was exiting some shape."},{"Start":"10:26.770 ","End":"10:31.300","Text":"If we had a sphere and we were seeing not the energy,"},{"Start":"10:31.300 ","End":"10:36.310","Text":"but the electric field that was coming out of this charged sphere."},{"Start":"10:36.310 ","End":"10:39.265","Text":"Just like that, we had flux."},{"Start":"10:39.265 ","End":"10:43.884","Text":"Also over here, this is a calculation of flux."},{"Start":"10:43.884 ","End":"10:46.105","Text":"It\u0027s the Poynting flux."},{"Start":"10:46.105 ","End":"10:48.625","Text":"The flux of the pointing vector."},{"Start":"10:48.625 ","End":"10:53.650","Text":"Where the Gauss we were seeing the flux of the electric field."},{"Start":"10:53.650 ","End":"10:57.100","Text":"How much electric field is coming out of a shape?"},{"Start":"10:57.100 ","End":"11:01.405","Text":"Here we\u0027re looking at the Poynting vector flux."},{"Start":"11:01.405 ","End":"11:03.760","Text":"How much of this Poynting vector,"},{"Start":"11:03.760 ","End":"11:05.035","Text":"as we defined above,"},{"Start":"11:05.035 ","End":"11:08.065","Text":"is coming out of the shape."},{"Start":"11:08.065 ","End":"11:11.260","Text":"Now, of course, if we\u0027re seeing this,"},{"Start":"11:11.260 ","End":"11:17.245","Text":"the flux is how much energy was lost from our specific volume."},{"Start":"11:17.245 ","End":"11:19.540","Text":"How much was lost?"},{"Start":"11:19.540 ","End":"11:24.700","Text":"We started with a total energy of capital U,"},{"Start":"11:24.700 ","End":"11:27.670","Text":"don\u0027t get them confused with the lowercase u,"},{"Start":"11:27.670 ","End":"11:30.160","Text":"which is the energy density."},{"Start":"11:30.160 ","End":"11:33.100","Text":"This lowercase u is the energy density."},{"Start":"11:33.100 ","End":"11:37.345","Text":"The capital U is the total amount of energy."},{"Start":"11:37.345 ","End":"11:40.540","Text":"This is important, so don\u0027t get confused."},{"Start":"11:40.540 ","End":"11:45.025","Text":"We started off with an energy of capital U,"},{"Start":"11:45.025 ","End":"11:52.525","Text":"electromagnetic em, but now energy has exited our volume."},{"Start":"11:52.525 ","End":"11:55.060","Text":"There\u0027s been a change in energy."},{"Start":"11:55.060 ","End":"12:00.910","Text":"Of course, the whole point of the Poynting vector is that it\u0027s in a unit of time."},{"Start":"12:00.910 ","End":"12:03.790","Text":"Energy is lost per unit of time."},{"Start":"12:03.790 ","End":"12:10.405","Text":"Let\u0027s imagine, let\u0027s give an example that we started off with 20 joules."},{"Start":"12:10.405 ","End":"12:16.795","Text":"Inside this volume, 20 joules of energy and 5 joules exited."},{"Start":"12:16.795 ","End":"12:22.330","Text":"That means that the energy inside the volume has"},{"Start":"12:22.330 ","End":"12:28.345","Text":"gone from 20 and it\u0027s reduced to 15 joules."},{"Start":"12:28.345 ","End":"12:35.620","Text":"What we can see is that 5 joules of energy has gone out over some unit of time."},{"Start":"12:35.620 ","End":"12:42.910","Text":"That means that this whole equation is equal to the change in energy dU,"},{"Start":"12:42.910 ","End":"12:50.840","Text":"by dt, the change in energy per unit of time."},{"Start":"12:52.560 ","End":"12:56.635","Text":"Another thing, as we\u0027ve defined here,"},{"Start":"12:56.635 ","End":"13:00.625","Text":"so this flux will give the energy lost."},{"Start":"13:00.625 ","End":"13:05.095","Text":"That means that we\u0027re calculating how much energy was lost."},{"Start":"13:05.095 ","End":"13:13.630","Text":"This flux will be a positive value because we\u0027re just summing the energy that\u0027s left."},{"Start":"13:13.630 ","End":"13:16.930","Text":"However, our dU by dt,"},{"Start":"13:16.930 ","End":"13:20.200","Text":"this side of the equation will be going down."},{"Start":"13:20.200 ","End":"13:24.100","Text":"Because the energy inside is reducing."},{"Start":"13:24.100 ","End":"13:29.035","Text":"If we look at a graph of the,"},{"Start":"13:29.035 ","End":"13:31.970","Text":"sorry, this has to be capital U_em."},{"Start":"13:32.190 ","End":"13:38.440","Text":"The energy will start off somewhere over here and it will slowly go down."},{"Start":"13:38.440 ","End":"13:42.220","Text":"We have this negative gradient over here."},{"Start":"13:42.220 ","End":"13:46.015","Text":"But of course, if we defined this as how much energy was gained,"},{"Start":"13:46.015 ","End":"13:48.520","Text":"then it would have a positive gradient,"},{"Start":"13:48.520 ","End":"13:51.685","Text":"in which case here we would have a plus."},{"Start":"13:51.685 ","End":"13:55.060","Text":"Of course we could write this side without the"},{"Start":"13:55.060 ","End":"13:58.690","Text":"negative and just put the negative sign over here."},{"Start":"13:58.690 ","End":"14:00.610","Text":"It makes no difference."},{"Start":"14:00.610 ","End":"14:02.950","Text":"As 1 side is increasing,"},{"Start":"14:02.950 ","End":"14:04.465","Text":"the other is decreasing."},{"Start":"14:04.465 ","End":"14:10.430","Text":"If we\u0027re calculating how much energy was lost from the volume."},{"Start":"14:10.920 ","End":"14:14.875","Text":"The way that we\u0027ve defined it, energy is exiting."},{"Start":"14:14.875 ","End":"14:16.645","Text":"Here we\u0027ll have a positive,"},{"Start":"14:16.645 ","End":"14:20.320","Text":"which means that here we\u0027ll have a negative."},{"Start":"14:20.320 ","End":"14:22.959","Text":"The next thing, as previously discussed,"},{"Start":"14:22.959 ","End":"14:27.370","Text":"we\u0027re summing up over the surface area, encasing the volume."},{"Start":"14:27.370 ","End":"14:30.280","Text":"The entire surface area,"},{"Start":"14:30.280 ","End":"14:35.184","Text":"which means that this is going to be a closed loop integral."},{"Start":"14:35.184 ","End":"14:37.494","Text":"Just like what we did with Gauss,"},{"Start":"14:37.494 ","End":"14:43.210","Text":"where we integrated across an entire closed surface area."},{"Start":"14:43.210 ","End":"14:45.625","Text":"Just like that over here."},{"Start":"14:45.625 ","End":"14:47.680","Text":"Over all of the sides,"},{"Start":"14:47.680 ","End":"14:49.570","Text":"the total surface area,"},{"Start":"14:49.570 ","End":"14:52.760","Text":"so it\u0027s the closed loop integral."},{"Start":"14:54.900 ","End":"15:00.805","Text":"This is another equation that you should write in your equation sheets,"},{"Start":"15:00.805 ","End":"15:03.580","Text":"also this 1 and this 1."},{"Start":"15:03.580 ","End":"15:05.515","Text":"Let\u0027s just go over this again."},{"Start":"15:05.515 ","End":"15:09.130","Text":"This is the flux for the Poynting vector,"},{"Start":"15:09.130 ","End":"15:13.105","Text":"where the pointing vector is given over here."},{"Start":"15:13.105 ","End":"15:18.324","Text":"The flux of the pointing vector is given by the closed loop integral"},{"Start":"15:18.324 ","End":"15:24.475","Text":"of the pointing vector capital S.products with ds,"},{"Start":"15:24.475 ","End":"15:28.270","Text":"where this is a lowercase s denoting the surface area,"},{"Start":"15:28.270 ","End":"15:33.620","Text":"as we\u0027ve seen in other equations."},{"Start":"15:33.660 ","End":"15:41.875","Text":"This is equal to negative dU_em electromagnetic by dt."},{"Start":"15:41.875 ","End":"15:49.660","Text":"The negative change in electromagnetic energy per unit of time."},{"Start":"15:49.660 ","End":"15:53.140","Text":"This negative just comes from the fact that we defined it"},{"Start":"15:53.140 ","End":"15:56.485","Text":"as the energy exiting out of our"},{"Start":"15:56.485 ","End":"16:04.510","Text":"volume.Now let\u0027s rewrite this same equation"},{"Start":"16:04.510 ","End":"16:07.000","Text":"just in its differential form."},{"Start":"16:07.000 ","End":"16:10.150","Text":"A few lessons ago,"},{"Start":"16:10.150 ","End":"16:16.510","Text":"we had a lesson in a different chapter dealing with"},{"Start":"16:16.510 ","End":"16:20.845","Text":"Maxwell\u0027s equations where we saw that we can get from"},{"Start":"16:20.845 ","End":"16:25.825","Text":"different versions of the equations in the integral form to the differential form."},{"Start":"16:25.825 ","End":"16:29.305","Text":"Now we\u0027re going to do basically that."},{"Start":"16:29.305 ","End":"16:34.360","Text":"We\u0027re just rewriting this over here."},{"Start":"16:34.360 ","End":"16:39.610","Text":"Through Gauss\u0027s law, we can rewrite this section of the"},{"Start":"16:39.610 ","End":"16:44.628","Text":"integral as instead of being a closed loop integral on S dS,"},{"Start":"16:44.628 ","End":"16:51.505","Text":"integrating along the surface from Gauss\u0027s law that we saw in Maxwell\u0027s equations,"},{"Start":"16:51.505 ","End":"16:57.370","Text":"we can convert from this integral on surface area to an integral volume by saying"},{"Start":"16:57.370 ","End":"17:03.774","Text":"that the vector field dot dS is the same as div,"},{"Start":"17:03.774 ","End":"17:10.110","Text":"the divergence, of S.dv."},{"Start":"17:10.110 ","End":"17:15.490","Text":"Here we\u0027re integrating along the volume."},{"Start":"17:15.920 ","End":"17:20.010","Text":"This comes from Gauss\u0027s law."},{"Start":"17:20.010 ","End":"17:29.800","Text":"That we can say that this is just equal to this and that is equal"},{"Start":"17:29.800 ","End":"17:40.045","Text":"to negative d. Let\u0027s just write this d by dt of U_em,"},{"Start":"17:40.045 ","End":"17:47.335","Text":"the total electromagnetic energy inside over here."},{"Start":"17:47.335 ","End":"17:49.870","Text":"Now, what is this em?"},{"Start":"17:49.870 ","End":"17:54.160","Text":"Let\u0027s write this again. It\u0027s negative d by dt."},{"Start":"17:54.160 ","End":"17:57.310","Text":"Our U_ em is,"},{"Start":"17:57.310 ","End":"18:01.360","Text":"as we saw over here,"},{"Start":"18:01.360 ","End":"18:07.210","Text":"is equal to the integral again with volume where we\u0027re"},{"Start":"18:07.210 ","End":"18:15.755","Text":"integrating along the exact same volume of our u_em,"},{"Start":"18:15.755 ","End":"18:22.855","Text":"our electromagnetic energy density, dv."},{"Start":"18:22.855 ","End":"18:25.410","Text":"I\u0027m just going to write this out again."},{"Start":"18:25.410 ","End":"18:31.480","Text":"Our u_em is our electromagnetic energy density,"},{"Start":"18:31.480 ","End":"18:36.190","Text":"which comprises of Epsilon_naught divided by 2 E"},{"Start":"18:36.190 ","End":"18:42.445","Text":"squared plus B squared divided by 2 Mu_naught,"},{"Start":"18:42.445 ","End":"18:45.670","Text":"where this is U_E,"},{"Start":"18:45.670 ","End":"18:51.055","Text":"the energy due to the electric field density,"},{"Start":"18:51.055 ","End":"18:57.655","Text":"and this is the energy due to the magnetic field density."},{"Start":"18:57.655 ","End":"19:06.070","Text":"Remember this, don\u0027t get confused between u and U."},{"Start":"19:06.070 ","End":"19:09.520","Text":"Then we have this, and now,"},{"Start":"19:09.520 ","End":"19:13.435","Text":"because we\u0027re assuming that the volume is constant,"},{"Start":"19:13.435 ","End":"19:15.779","Text":"it\u0027s not changing in time,"},{"Start":"19:15.779 ","End":"19:21.459","Text":"that means that we can insert this d by dt into this integral."},{"Start":"19:21.459 ","End":"19:30.580","Text":"We can say that this is just equal to negative the integral"},{"Start":"19:30.580 ","End":"19:40.750","Text":"along the volume of d u_em by the td dv."},{"Start":"19:40.750 ","End":"19:44.680","Text":"We can say that this is equal to this."},{"Start":"19:44.680 ","End":"19:50.605","Text":"In that case, what we have is this integral with respect to the same volume."},{"Start":"19:50.605 ","End":"19:55.540","Text":"That means that we can say that this over"},{"Start":"19:55.540 ","End":"20:04.040","Text":"here is equal to this over here."},{"Start":"20:05.910 ","End":"20:08.935","Text":"Sorry, there\u0027s no dot over here."},{"Start":"20:08.935 ","End":"20:18.590","Text":"In other words, we can say that the divergence of our pointing vector,"},{"Start":"20:18.950 ","End":"20:23.085","Text":"remember I took out the dot product over here,"},{"Start":"20:23.085 ","End":"20:28.695","Text":"is equal to negative d"},{"Start":"20:28.695 ","End":"20:35.050","Text":"u_em by dt."},{"Start":"20:36.210 ","End":"20:41.350","Text":"This is another equation to write in your equation sheets."},{"Start":"20:41.350 ","End":"20:44.050","Text":"Now let\u0027s just discuss this."},{"Start":"20:44.050 ","End":"20:48.640","Text":"Here we have the divergence of S. What does that mean?"},{"Start":"20:48.640 ","End":"20:50.650","Text":"Let\u0027s just write this out."},{"Start":"20:50.650 ","End":"21:01.000","Text":"Div S is equal to dS by dx in the x direction."},{"Start":"21:01.000 ","End":"21:11.725","Text":"We\u0027re taking the x-component of the pointing vector by dx plus dS_y by dy,"},{"Start":"21:11.725 ","End":"21:16.990","Text":"so the y component of the S vector dy"},{"Start":"21:16.990 ","End":"21:23.570","Text":"plus dS_z by dz."},{"Start":"21:23.570 ","End":"21:29.515","Text":"What we can see is the gradient of the S vector,"},{"Start":"21:29.515 ","End":"21:31.120","Text":"of the pointing vector,"},{"Start":"21:31.120 ","End":"21:34.050","Text":"in each of the axes,"},{"Start":"21:34.050 ","End":"21:35.620","Text":"so in the x-axis,"},{"Start":"21:35.620 ","End":"21:38.125","Text":"the y-axis, and the z-axis."},{"Start":"21:38.125 ","End":"21:40.180","Text":"Or in other words,"},{"Start":"21:40.180 ","End":"21:44.650","Text":"we can see the change of our pointing vector in the x,"},{"Start":"21:44.650 ","End":"21:47.810","Text":"y, and z directions."},{"Start":"21:47.970 ","End":"21:55.960","Text":"In other words, we can see the change of energy in each 1 of these directions added up,"},{"Start":"21:55.960 ","End":"21:59.575","Text":"which gives us the total change of directions."},{"Start":"21:59.575 ","End":"22:04.480","Text":"If the change in the x-direction plus the change in the energy in"},{"Start":"22:04.480 ","End":"22:10.210","Text":"the y direction plus the change of the energy in the z direction are added altogether,"},{"Start":"22:10.210 ","End":"22:12.805","Text":"it gives us the total change in energy."},{"Start":"22:12.805 ","End":"22:15.535","Text":"We\u0027re taking into account all of the directions."},{"Start":"22:15.535 ","End":"22:19.135","Text":"In that case, we have this total change of energy,"},{"Start":"22:19.135 ","End":"22:23.067","Text":"which is equal to this over here,"},{"Start":"22:23.067 ","End":"22:26.020","Text":"and as we said, S is per unit of time."},{"Start":"22:26.020 ","End":"22:29.920","Text":"Remember, the pointing vector is the amount of"},{"Start":"22:29.920 ","End":"22:34.960","Text":"energy flowing through a unit of area in a unit of time."},{"Start":"22:34.960 ","End":"22:39.475","Text":"We can see the total change per unit of time."},{"Start":"22:39.475 ","End":"22:41.530","Text":"Or in other words,"},{"Start":"22:41.530 ","End":"22:46.180","Text":"this is equal to the change in energy density."},{"Start":"22:46.180 ","End":"22:55.720","Text":"Because this is, remember we said that if we take a single unit of this volume,"},{"Start":"22:55.720 ","End":"22:59.125","Text":"this will be u_em."},{"Start":"22:59.125 ","End":"23:02.050","Text":"If we add all of this up, we get u."},{"Start":"23:02.050 ","End":"23:10.270","Text":"This will be equal to negative d u_em by dt."},{"Start":"23:10.270 ","End":"23:16.435","Text":"We get this change in the energy density per time."},{"Start":"23:16.435 ","End":"23:18.370","Text":"If we add all of that up,"},{"Start":"23:18.370 ","End":"23:25.220","Text":"we get the total change of energy throughout the entire large volume."},{"Start":"23:25.380 ","End":"23:31.540","Text":"We can see the total energy change in this small cube over here."},{"Start":"23:31.540 ","End":"23:33.475","Text":"That\u0027s why it makes sense."},{"Start":"23:33.475 ","End":"23:37.765","Text":"We add up the change in each of the directions x, y, and z."},{"Start":"23:37.765 ","End":"23:41.080","Text":"That gives us the total change in energy per"},{"Start":"23:41.080 ","End":"23:46.060","Text":"unit time because the pointing vector is also per unit time."},{"Start":"23:46.060 ","End":"23:50.920","Text":"The explanation of how to get from this version of the equation to"},{"Start":"23:50.920 ","End":"23:55.210","Text":"this version was done in an intuitive manner."},{"Start":"23:55.210 ","End":"24:02.110","Text":"But of course, there\u0027s also a mathematical proof that 1 can do to jump between the 2,"},{"Start":"24:02.110 ","End":"24:03.730","Text":"but it\u0027s more complicated."},{"Start":"24:03.730 ","End":"24:06.940","Text":"I think that using intuition right now,"},{"Start":"24:06.940 ","End":"24:12.055","Text":"just to understand it is what is relevant at the moment."},{"Start":"24:12.055 ","End":"24:18.520","Text":"Just 1 little thing to remember when using this equation is that"},{"Start":"24:18.520 ","End":"24:26.240","Text":"this equation and also this 1 is only correct when this is a vacuum."},{"Start":"24:26.940 ","End":"24:33.535","Text":"When there\u0027s no other material placed into here, because of course,"},{"Start":"24:33.535 ","End":"24:38.020","Text":"if we have energy as it passes through the other material,"},{"Start":"24:38.020 ","End":"24:40.510","Text":"then the energy will also be lost to heat,"},{"Start":"24:40.510 ","End":"24:46.240","Text":"as we saw with power when dealing with that topic."},{"Start":"24:46.240 ","End":"24:49.690","Text":"In that case, we\u0027d have to add different terms over here"},{"Start":"24:49.690 ","End":"24:52.645","Text":"to take into account energy lost to heat."},{"Start":"24:52.645 ","End":"24:55.630","Text":"However, if this is located in a vacuum,"},{"Start":"24:55.630 ","End":"24:58.540","Text":"then theoretically no energy is lost"},{"Start":"24:58.540 ","End":"25:02.995","Text":"because there\u0027s nothing for it to go through and turn to heat."},{"Start":"25:02.995 ","End":"25:08.845","Text":"That means that this equation is only correct when it\u0027s located inside a vacuum,"},{"Start":"25:08.845 ","End":"25:12.130","Text":"which is what\u0027s important for the meantime."},{"Start":"25:12.130 ","End":"25:15.081","Text":"Just remember that,"},{"Start":"25:15.081 ","End":"25:20.035","Text":"I\u0027ll also square that maybe make a note of that in your equation booklets."},{"Start":"25:20.035 ","End":"25:24.680","Text":"In the meantime, this is the end of this lesson."}],"ID":21469},{"Watched":false,"Name":"1.1 The Energy Required to Build a System within a Magnetic Field","Duration":"12m 5s","ChapterTopicVideoID":24817,"CourseChapterTopicPlaylistID":99493,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.920","Text":"Hello. In this lesson,"},{"Start":"00:01.920 ","End":"00:04.770","Text":"we are going to be speaking about the energy required to"},{"Start":"00:04.770 ","End":"00:07.845","Text":"build a system within a magnetic field."},{"Start":"00:07.845 ","End":"00:11.955","Text":"Now, previously, when speaking about electrostatics,"},{"Start":"00:11.955 ","End":"00:16.650","Text":"we saw a similar type of a situation where we also"},{"Start":"00:16.650 ","End":"00:21.705","Text":"looked at the energy required to build a system of a number of charges."},{"Start":"00:21.705 ","End":"00:25.050","Text":"We saw that the equation was this."},{"Start":"00:25.050 ","End":"00:30.890","Text":"U, for the energy which was equal to half"},{"Start":"00:30.890 ","End":"00:36.140","Text":"multiplied by the sum of all of the charges."},{"Start":"00:36.140 ","End":"00:40.640","Text":"So q_i multiplied by Phi_i,"},{"Start":"00:40.640 ","End":"00:42.740","Text":"which is their potential,"},{"Start":"00:42.740 ","End":"00:44.345","Text":"which we also saw."},{"Start":"00:44.345 ","End":"00:46.040","Text":"This was if we had point charges,"},{"Start":"00:46.040 ","End":"00:49.565","Text":"but if we had some charge density,"},{"Start":"00:49.565 ","End":"00:55.260","Text":"then it was the integral of Epsilon naught divided by"},{"Start":"00:55.260 ","End":"01:04.670","Text":"2 multiplied by E^2dV over the whole volume."},{"Start":"01:04.670 ","End":"01:09.395","Text":"So here if we had some point charges q_1,"},{"Start":"01:09.395 ","End":"01:12.090","Text":"q_2, q_3,"},{"Start":"01:12.090 ","End":"01:16.590","Text":"or we can have some sphere with charge density Rho."},{"Start":"01:16.590 ","End":"01:21.110","Text":"Then we could figure out the work done or the energy required in"},{"Start":"01:21.110 ","End":"01:25.490","Text":"order to build such a system via using these equations."},{"Start":"01:25.490 ","End":"01:32.120","Text":"This is what we\u0027ve seen up until now in electrostatics."},{"Start":"01:32.120 ","End":"01:34.745","Text":"All of this we spoke about,"},{"Start":"01:34.745 ","End":"01:36.650","Text":"and if you can\u0027t remember this,"},{"Start":"01:36.650 ","End":"01:41.345","Text":"please go back to the relevant chapter in electrostatics."},{"Start":"01:41.345 ","End":"01:46.000","Text":"All of this is correct in electrostatics."},{"Start":"01:46.000 ","End":"01:48.950","Text":"So why am I highlighting that?"},{"Start":"01:48.950 ","End":"01:52.595","Text":"This is because all of the charges are static."},{"Start":"01:52.595 ","End":"01:54.020","Text":"They\u0027re not moving."},{"Start":"01:54.020 ","End":"01:55.745","Text":"Or in other words,"},{"Start":"01:55.745 ","End":"01:57.485","Text":"when the charges aren\u0027t moving,"},{"Start":"01:57.485 ","End":"02:01.355","Text":"that means that there is no current."},{"Start":"02:01.355 ","End":"02:06.245","Text":"All of these equations are correct when there is no current."},{"Start":"02:06.245 ","End":"02:09.990","Text":"But what happens when there is current?"},{"Start":"02:10.400 ","End":"02:14.190","Text":"What do we do if there is current?"},{"Start":"02:14.190 ","End":"02:17.710","Text":"If there is current, I,"},{"Start":"02:17.710 ","End":"02:22.390","Text":"then that means that there\u0027s going to be a magnetic field."},{"Start":"02:22.390 ","End":"02:25.335","Text":"In which case, our equation for, U,"},{"Start":"02:25.335 ","End":"02:30.995","Text":"for the energy required is going to be the integral of the same thing over here."},{"Start":"02:30.995 ","End":"02:37.405","Text":"So Epsilon naught divided by 2 multiplied by E^2."},{"Start":"02:37.405 ","End":"02:39.890","Text":"We\u0027ll put this in brackets."},{"Start":"02:39.890 ","End":"02:41.045","Text":"Of course, when we square it,"},{"Start":"02:41.045 ","End":"02:43.220","Text":"we get rid of the vector symbol."},{"Start":"02:43.220 ","End":"02:45.140","Text":"But then on top of that,"},{"Start":"02:45.140 ","End":"02:50.705","Text":"we\u0027re going to add in the element dealing with the current or with the magnetic field,"},{"Start":"02:50.705 ","End":"02:58.770","Text":"which is B^2 divided by 2 Mu naught,"},{"Start":"02:58.770 ","End":"03:01.482","Text":"and then all of this dv."},{"Start":"03:01.482 ","End":"03:07.940","Text":"Again, we\u0027re integrating with respect to the entire region all the volume in the region."},{"Start":"03:07.940 ","End":"03:09.350","Text":"But when we have current,"},{"Start":"03:09.350 ","End":"03:15.674","Text":"we have to add in this element over here of magnetic field."},{"Start":"03:15.674 ","End":"03:20.270","Text":"This is an equation to write in your equation sheets."},{"Start":"03:20.270 ","End":"03:26.540","Text":"Remember that this integral you do on the entire region from 0"},{"Start":"03:26.540 ","End":"03:29.270","Text":"to infinity or from negative infinity to"},{"Start":"03:29.270 ","End":"03:33.490","Text":"infinity depending on how you\u0027re defining the question."},{"Start":"03:33.490 ","End":"03:39.790","Text":"Now just like if we\u0027re trying to calculate the charge Q,"},{"Start":"03:39.790 ","End":"03:43.900","Text":"so we have an integral of Rho dv,"},{"Start":"03:43.900 ","End":"03:49.570","Text":"where Rho over here is the density of the charge."},{"Start":"03:49.570 ","End":"03:53.930","Text":"So the charge density."},{"Start":"03:53.930 ","End":"03:58.224","Text":"Just like that here we have U,"},{"Start":"03:58.224 ","End":"04:04.840","Text":"the energy also here is equal to the integral of something."},{"Start":"04:04.840 ","End":"04:08.960","Text":"Here\u0027s the something and again, dv."},{"Start":"04:09.360 ","End":"04:20.030","Text":"This something is the density of the energy or the energy density."},{"Start":"04:20.810 ","End":"04:25.865","Text":"Over here, we can have that the energy density,"},{"Start":"04:25.865 ","End":"04:29.525","Text":"which is symbolized by either Mu_E."},{"Start":"04:29.525 ","End":"04:32.105","Text":"Here are the electric density, sorry,"},{"Start":"04:32.105 ","End":"04:41.078","Text":"E for electric because here we have the electric field, E field."},{"Start":"04:41.078 ","End":"04:47.105","Text":"This is equal to or so sometimes just it\u0027s called a lowercase u_E,"},{"Start":"04:47.105 ","End":"04:50.255","Text":"again, denoting the electric field,"},{"Start":"04:50.255 ","End":"04:53.000","Text":"not energy electric field,"},{"Start":"04:53.000 ","End":"05:02.380","Text":"which is equal to Epsilon naught multiplied by E^2 divided by 2."},{"Start":"05:02.380 ","End":"05:06.170","Text":"Then similarly, this over here."},{"Start":"05:06.170 ","End":"05:08.360","Text":"So let\u0027s make it in blue."},{"Start":"05:08.360 ","End":"05:16.560","Text":"This over here is often called Mu_B."},{"Start":"05:16.560 ","End":"05:23.580","Text":"The electric density from the magnetic field."},{"Start":"05:23.630 ","End":"05:27.970","Text":"This is equal to"},{"Start":"05:31.190 ","End":"05:35.610","Text":"B^2 divided by 2 Mu naught."},{"Start":"05:35.610 ","End":"05:43.200","Text":"This is also called lowercase u_B from the magnetic field."},{"Start":"05:43.200 ","End":"05:48.690","Text":"We can have that, the lowercase u_em."},{"Start":"05:48.690 ","End":"05:56.600","Text":"The energy density of the electromagnetic field."},{"Start":"05:56.600 ","End":"05:59.360","Text":"So that\u0027s taking into account this,"},{"Start":"05:59.360 ","End":"06:09.630","Text":"the electric field and this the magnetic field is equal to u_E plus u_B."},{"Start":"06:09.630 ","End":"06:14.150","Text":"This plus this, or in other words,"},{"Start":"06:14.150 ","End":"06:23.109","Text":"all of this from here until here is called u_em."},{"Start":"06:23.109 ","End":"06:32.130","Text":"The energy density of the electromagnetic fields, electromagnetic."},{"Start":"06:32.900 ","End":"06:35.630","Text":"What does this give us?"},{"Start":"06:35.630 ","End":"06:42.605","Text":"Let\u0027s say here we have this charged sphere and as we know,"},{"Start":"06:42.605 ","End":"06:46.310","Text":"we have electric and magnetic field lines."},{"Start":"06:46.310 ","End":"06:51.330","Text":"Let\u0027s imagine that there\u0027s also magnetic field over here because the charges are moving."},{"Start":"06:51.330 ","End":"06:58.927","Text":"We have an electric and a magnetic field over here coming out."},{"Start":"06:58.927 ","End":"07:03.800","Text":"Now what we can do is we can look at,"},{"Start":"07:03.800 ","End":"07:08.105","Text":"instead of looking at the electric field and the magnetic field as"},{"Start":"07:08.105 ","End":"07:14.990","Text":"the entity or something that connects between each of the particles,"},{"Start":"07:14.990 ","End":"07:16.640","Text":"between each of the charges."},{"Start":"07:16.640 ","End":"07:21.680","Text":"Now we\u0027re looking at the electric and the magnetic field as separate entities."},{"Start":"07:21.680 ","End":"07:24.155","Text":"They are just entities over here."},{"Start":"07:24.155 ","End":"07:31.380","Text":"That means that if I look at some unit of volume,"},{"Start":"07:31.380 ","End":"07:38.935","Text":"so let\u0027s say over here I\u0027m looking at this unit of volume that has a volume of dv,"},{"Start":"07:38.935 ","End":"07:45.200","Text":"small volume, this little unit of volume is going to have energy."},{"Start":"07:45.200 ","End":"07:48.680","Text":"There\u0027s going to be energy inside."},{"Start":"07:48.680 ","End":"07:50.900","Text":"What is the energy inside?"},{"Start":"07:50.900 ","End":"07:56.620","Text":"First of all, the energy will be dU because it\u0027s the energy just in this little box."},{"Start":"07:56.620 ","End":"07:58.725","Text":"What it\u0027s equal to,"},{"Start":"07:58.725 ","End":"08:01.680","Text":"it\u0027s equal to u_em."},{"Start":"08:01.680 ","End":"08:07.700","Text":"The energy density of the electromagnetic field."},{"Start":"08:07.700 ","End":"08:13.395","Text":"Then of course, multiplied by dv, its volume."},{"Start":"08:13.395 ","End":"08:21.450","Text":"Because the units of this u is energy divided by density."},{"Start":"08:21.450 ","End":"08:23.860","Text":"In order to get energy, dU,"},{"Start":"08:23.860 ","End":"08:29.298","Text":"we have to multiply it by density to get units of energy."},{"Start":"08:29.298 ","End":"08:32.250","Text":"That\u0027s what this means."},{"Start":"08:32.250 ","End":"08:36.620","Text":"If we have, some were located in an area where there\u0027s an"},{"Start":"08:36.620 ","End":"08:42.295","Text":"electric and a magnetic fields present or if there\u0027s just an electric field,"},{"Start":"08:42.295 ","End":"08:49.900","Text":"then dU will just be eu dv and if there\u0027s just a magnetic field,"},{"Start":"08:49.900 ","End":"08:55.395","Text":"then dU will be uB multiplied by dv."},{"Start":"08:55.395 ","End":"09:01.885","Text":"If there\u0027s both, then we take into account both multiplied by the volume."},{"Start":"09:01.885 ","End":"09:08.330","Text":"That means that there\u0027s just energy located at this point over here in space."},{"Start":"09:09.290 ","End":"09:12.430","Text":"Now, another thing."},{"Start":"09:12.430 ","End":"09:16.180","Text":"Later on, you\u0027re going to see that the electric field and"},{"Start":"09:16.180 ","End":"09:20.635","Text":"the magnetic field are in fact, the same thing."},{"Start":"09:20.635 ","End":"09:23.350","Text":"If you\u0027re just looking at,"},{"Start":"09:23.350 ","End":"09:27.305","Text":"if you are located in the S frame of reference,"},{"Start":"09:27.305 ","End":"09:33.290","Text":"then your electric field will be a magnetic field in the S tag frame of reference."},{"Start":"09:33.290 ","End":"09:35.510","Text":"But this I\u0027m not going to go into now,"},{"Start":"09:35.510 ","End":"09:40.639","Text":"but just know that these fields are connected and there in fact,"},{"Start":"09:40.639 ","End":"09:46.590","Text":"the same thing if you\u0027re looking through different frames of reference."},{"Start":"09:47.140 ","End":"09:50.240","Text":"This, you\u0027ll learn about more."},{"Start":"09:50.240 ","End":"09:53.810","Text":"Now, another thing to get to this a B field,"},{"Start":"09:53.810 ","End":"09:56.850","Text":"so we said that it comes from current."},{"Start":"09:56.850 ","End":"10:05.685","Text":"What we can see for the equation for the energy stored in a coil."},{"Start":"10:05.685 ","End":"10:12.675","Text":"This is equal to 1/2 LI^2."},{"Start":"10:12.675 ","End":"10:19.415","Text":"Then there\u0027s a mathematical jump that we can do from this equation to this equation."},{"Start":"10:19.415 ","End":"10:25.695","Text":"B^2 divided by 2 Mu naught dv."},{"Start":"10:25.695 ","End":"10:32.995","Text":"This is this, so you could also write this in here without the dv."},{"Start":"10:32.995 ","End":"10:41.480","Text":"But what we have over here in blue is just the energy stored in a coil."},{"Start":"10:42.440 ","End":"10:47.510","Text":"We saw a similar jump when we were dealing with electrostatics."},{"Start":"10:47.510 ","End":"10:50.720","Text":"We have a very similar equation here."},{"Start":"10:50.720 ","End":"10:53.375","Text":"Half times the sum,"},{"Start":"10:53.375 ","End":"10:58.160","Text":"which then turns into this integral with respect to volume."},{"Start":"10:58.160 ","End":"11:05.650","Text":"It\u0027s a similar step over here to get the magnetic field from the current."},{"Start":"11:05.650 ","End":"11:11.120","Text":"Just like here, the electric field gives us the or"},{"Start":"11:11.120 ","End":"11:15.964","Text":"helps us to get the energy storage due to the electric charges."},{"Start":"11:15.964 ","End":"11:24.060","Text":"This magnetic field gives us the energy stored due to the current."},{"Start":"11:24.890 ","End":"11:29.360","Text":"I\u0027m not going to show the mathematical jumps,"},{"Start":"11:29.360 ","End":"11:33.380","Text":"but what is important to note is this equation over here,"},{"Start":"11:33.380 ","End":"11:39.448","Text":"for the energy required to build a system within a magnetic field."},{"Start":"11:39.448 ","End":"11:44.150","Text":"We have to take into account both the electric and the magnetic field."},{"Start":"11:44.150 ","End":"11:47.030","Text":"We have 2 terms, and this is important."},{"Start":"11:47.030 ","End":"11:49.445","Text":"This is the equation when there is current."},{"Start":"11:49.445 ","End":"11:50.945","Text":"If there is no current,"},{"Start":"11:50.945 ","End":"11:53.870","Text":"that means the charges aren\u0027t moving or in other words,"},{"Start":"11:53.870 ","End":"11:57.825","Text":"the charges are static."},{"Start":"11:57.825 ","End":"12:02.420","Text":"Then we have this equation over here for electrostatics."},{"Start":"12:02.420 ","End":"12:05.969","Text":"That is the end of this lesson."}],"ID":25730},{"Watched":false,"Name":"1.3 Exercise - Parallel Plate Capacitor and Voltage","Duration":"31m 39s","ChapterTopicVideoID":24818,"CourseChapterTopicPlaylistID":99493,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.965","Text":"Hello. In this lesson,"},{"Start":"00:01.965 ","End":"00:04.815","Text":"we\u0027re going to be answering the following question."},{"Start":"00:04.815 ","End":"00:07.800","Text":"We have a parallel plate capacitor,"},{"Start":"00:07.800 ","End":"00:11.354","Text":"which consists of 2 circular plates,"},{"Start":"00:11.354 ","End":"00:16.019","Text":"each of radius a located a distance d from one another,"},{"Start":"00:16.019 ","End":"00:18.630","Text":"where d is much smaller than a."},{"Start":"00:18.630 ","End":"00:21.989","Text":"Of course, the picture isn\u0027t drawn to scale."},{"Start":"00:21.989 ","End":"00:28.125","Text":"I have just done this gap bigger so that it\u0027s easier for us to draw on the diagram."},{"Start":"00:28.125 ","End":"00:32.999","Text":"The capacitor is connected to a voltage source over here,"},{"Start":"00:32.999 ","End":"00:34.784","Text":"V as a function of t,"},{"Start":"00:34.784 ","End":"00:36.930","Text":"which is equal to at,"},{"Start":"00:36.930 ","End":"00:38.939","Text":"where a is a constant and t,"},{"Start":"00:38.939 ","End":"00:41.505","Text":"of course, represents time."},{"Start":"00:41.505 ","End":"00:48.330","Text":"Question number 1 is to calculate the electric field inside the capacitor."},{"Start":"00:49.190 ","End":"00:53.374","Text":"First of all, we can see that we have this electric circuit"},{"Start":"00:53.374 ","End":"00:58.730","Text":"consisting just of the voltage source and of the capacitor."},{"Start":"00:58.730 ","End":"01:05.525","Text":"Where this is the positive and this is the negative sides of the voltage source."},{"Start":"01:05.525 ","End":"01:10.115","Text":"As we know the voltage drops or sometimes increases across"},{"Start":"01:10.115 ","End":"01:15.139","Text":"each component in an electric circuit."},{"Start":"01:15.139 ","End":"01:19.205","Text":"Here, we only have this capacitor in the electric circuit,"},{"Start":"01:19.205 ","End":"01:26.014","Text":"which means that the voltage of the source is equal to the voltage across the capacitor"},{"Start":"01:26.014 ","End":"01:33.425","Text":"because there\u0027s no jumps in voltage anywhere because there\u0027s only this other component."},{"Start":"01:33.425 ","End":"01:37.785","Text":"We can say that the voltage of"},{"Start":"01:37.785 ","End":"01:44.630","Text":"the source is equal to the voltage across the capacitor."},{"Start":"01:44.630 ","End":"01:50.509","Text":"That was given to us as a multiplied by t. We"},{"Start":"01:50.509 ","End":"01:56.520","Text":"can see that the voltage is increasing linearly as a function of time."},{"Start":"01:57.020 ","End":"01:59.705","Text":"If this is the positive side,"},{"Start":"01:59.705 ","End":"02:02.374","Text":"this is the positive capacitor plate."},{"Start":"02:02.374 ","End":"02:03.950","Text":"This is the negative side,"},{"Start":"02:03.950 ","End":"02:06.784","Text":"so this is the negative capacitor plate,"},{"Start":"02:06.784 ","End":"02:11.600","Text":"and so this is the direction of the voltage."},{"Start":"02:11.600 ","End":"02:17.320","Text":"Let\u0027s define this as the positive z direction."},{"Start":"02:17.320 ","End":"02:23.779","Text":"Now, we know that the electric field inside a capacitor E is equal"},{"Start":"02:23.779 ","End":"02:30.890","Text":"to the voltage across the capacitor divided by the distance between the 2 plates."},{"Start":"02:30.890 ","End":"02:35.119","Text":"The voltage across the capacitor is At,"},{"Start":"02:35.119 ","End":"02:40.084","Text":"the distance is d. In that case,"},{"Start":"02:40.084 ","End":"02:44.255","Text":"this is the answer to question number 1."},{"Start":"02:44.255 ","End":"02:47.180","Text":"Of course, we can make this a vector with the direction,"},{"Start":"02:47.180 ","End":"02:51.810","Text":"which is going in the z-direction."},{"Start":"02:52.790 ","End":"02:56.064","Text":"Now, let\u0027s answer question 2."},{"Start":"02:56.064 ","End":"03:02.640","Text":"Calculate the magnetic field inside and outside of the capacitor."},{"Start":"03:02.810 ","End":"03:07.204","Text":"First of all, because we saw that we have an electric field,"},{"Start":"03:07.204 ","End":"03:10.585","Text":"we know that we\u0027re going to have a magnetic field."},{"Start":"03:10.585 ","End":"03:13.220","Text":"In order to calculate the magnetic fields,"},{"Start":"03:13.220 ","End":"03:17.629","Text":"we\u0027re going to be using one of Maxwell\u0027s equations where we have"},{"Start":"03:17.629 ","End":"03:23.209","Text":"the closed loop integral of B dot dl is equal"},{"Start":"03:23.209 ","End":"03:31.985","Text":"to Mu naught multiplied by J_d\u0027s plus"},{"Start":"03:31.985 ","End":"03:42.630","Text":"Mu naught multiplied by Epsilon naught dE by dt ds."},{"Start":"03:44.990 ","End":"03:51.000","Text":"This J is the current density."},{"Start":"03:51.000 ","End":"03:53.285","Text":"Then when we multiply it by the area,"},{"Start":"03:53.285 ","End":"03:56.045","Text":"we just get the current per area."},{"Start":"03:56.045 ","End":"04:01.870","Text":"Let\u0027s just call this In the meantime, the real current."},{"Start":"04:02.360 ","End":"04:09.030","Text":"Over here, all of this is J_d,"},{"Start":"04:09.030 ","End":"04:15.730","Text":"so the displacement current."},{"Start":"04:16.130 ","End":"04:22.265","Text":"Now, this J we know is going to be equal to 0."},{"Start":"04:22.265 ","End":"04:24.485","Text":"J, the real current,"},{"Start":"04:24.485 ","End":"04:26.989","Text":"is equal to 0. Why is that?"},{"Start":"04:26.989 ","End":"04:30.935","Text":"Because over here between the capacitor plates,"},{"Start":"04:30.935 ","End":"04:33.949","Text":"we just have a vacuum or we have air."},{"Start":"04:33.949 ","End":"04:38.420","Text":"We know that no charges are passing through which means that"},{"Start":"04:38.420 ","End":"04:43.039","Text":"we have no real current passing through."},{"Start":"04:43.039 ","End":"04:47.496","Text":"That\u0027s why J is equal to 0 over here,"},{"Start":"04:47.496 ","End":"04:49.669","Text":"so this is equal to 0."},{"Start":"04:49.669 ","End":"04:57.119","Text":"Now, J_d displacement current isn\u0027t a real current,"},{"Start":"04:57.119 ","End":"05:00.710","Text":"it\u0027s something else because, again,"},{"Start":"05:00.710 ","End":"05:04.745","Text":"as we said, there\u0027s no charges passing between the 2 plates."},{"Start":"05:04.745 ","End":"05:10.250","Text":"But what we see is it\u0027s something we just call it a displacement current,"},{"Start":"05:10.250 ","End":"05:12.335","Text":"but it isn\u0027t really a current."},{"Start":"05:12.335 ","End":"05:20.290","Text":"It\u0027s equal to Epsilon naught multiplied by the time derivative of the electric field."},{"Start":"05:20.290 ","End":"05:25.020","Text":"We see that our electric field is dependent on time."},{"Start":"05:25.280 ","End":"05:33.065","Text":"In that case, we can see that our J_d over here is equal to, as we said,"},{"Start":"05:33.065 ","End":"05:37.174","Text":"Epsilon naught multiplied by the time derivative of the electric field,"},{"Start":"05:37.174 ","End":"05:47.910","Text":"which will just be a divided by d. It is in the z-direction."},{"Start":"05:49.550 ","End":"05:56.405","Text":"Now, what we want to do is we want to calculate the magnetic field."},{"Start":"05:56.405 ","End":"06:03.885","Text":"What we have is we have this J_d over here going in this direction."},{"Start":"06:03.885 ","End":"06:13.875","Text":"This is our J_d in the positive z direction and this is mimicking current."},{"Start":"06:13.875 ","End":"06:15.695","Text":"From the right-hand rule,"},{"Start":"06:15.695 ","End":"06:19.460","Text":"if the current is pointing in the z direction,"},{"Start":"06:19.460 ","End":"06:24.304","Text":"then the thumb points in the z direction and the fingers"},{"Start":"06:24.304 ","End":"06:31.710","Text":"curl in the Theta direction around the thumb."},{"Start":"06:31.750 ","End":"06:35.985","Text":"We can see that this direction,"},{"Start":"06:35.985 ","End":"06:41.815","Text":"like so, is the positive Theta direction."},{"Start":"06:41.815 ","End":"06:46.255","Text":"That will be the direction of the magnetic field."},{"Start":"06:46.255 ","End":"06:51.849","Text":"We know that the magnetic field is going to be in the Theta direction."},{"Start":"06:51.849 ","End":"06:56.395","Text":"Which we\u0027ve defined this direction as being the positive direction."},{"Start":"06:56.395 ","End":"07:03.159","Text":"What we\u0027re going to do is we\u0027re going to take a little loop like"},{"Start":"07:03.159 ","End":"07:10.840","Text":"so that I\u0027m drawing in gray and we\u0027re saying that this is of radius r. First,"},{"Start":"07:10.840 ","End":"07:14.634","Text":"we\u0027re going to calculate the magnetic field inside the capacitor"},{"Start":"07:14.634 ","End":"07:19.445","Text":"and then we\u0027ll calculate the magnetic field outside of the capacitor."},{"Start":"07:19.445 ","End":"07:23.835","Text":"Now with this, we can write out this integral."},{"Start":"07:23.835 ","End":"07:32.480","Text":"We\u0027re doing the closed loop integral of B.dl."},{"Start":"07:32.480 ","End":"07:36.100","Text":"This is simply going to become,"},{"Start":"07:36.100 ","End":"07:40.045","Text":"so we can see that the magnetic field is going to be constant."},{"Start":"07:40.045 ","End":"07:42.910","Text":"We can just multiply it by the loop,"},{"Start":"07:42.910 ","End":"07:47.710","Text":"which is just the circumference of this r radius,"},{"Start":"07:47.710 ","End":"07:49.640","Text":"which of course is changing."},{"Start":"07:49.640 ","End":"07:54.010","Text":"However, we\u0027re calculating for every r inside the capacitor."},{"Start":"07:54.010 ","End":"07:58.960","Text":"This is going to be equal to b multiplied by the circumference of our loop,"},{"Start":"07:58.960 ","End":"08:00.490","Text":"just like in Gauss\u0027s law,"},{"Start":"08:00.490 ","End":"08:03.235","Text":"a case we\u0027ve seen this done before."},{"Start":"08:03.235 ","End":"08:09.979","Text":"That is 2Pi r. This is equal to this over here,"},{"Start":"08:09.979 ","End":"08:12.870","Text":"which is 0 plus this."},{"Start":"08:12.870 ","End":"08:23.430","Text":"This is equal to Mu naught multiplied by the integral of this over here so J_d,"},{"Start":"08:24.070 ","End":"08:28.805","Text":"dot ds, the surface area."},{"Start":"08:28.805 ","End":"08:35.255","Text":"Because our J_d is constant with respect to area,"},{"Start":"08:35.255 ","End":"08:37.350","Text":"which is what we\u0027re integrating."},{"Start":"08:37.350 ","End":"08:43.774","Text":"We can say that this is equal to Mu naught Epsilon naught multiplied"},{"Start":"08:43.774 ","End":"08:51.890","Text":"by a divided by d,"},{"Start":"08:51.890 ","End":"08:55.290","Text":"and then multiplied by the surface area."},{"Start":"08:55.290 ","End":"09:02.610","Text":"The surface area of this ring over here is Pi r^2."},{"Start":"09:03.290 ","End":"09:06.210","Text":"Now we can cancel out."},{"Start":"09:06.210 ","End":"09:11.494","Text":"What we have is b multiplied by 2Pi r,"},{"Start":"09:11.494 ","End":"09:20.320","Text":"which is equal to Mu naught Epsilon naught a divided by d multiplied by Pi r^2."},{"Start":"09:20.320 ","End":"09:25.225","Text":"We can cancel out Pi and r. Pi and one of the r\u0027s over here."},{"Start":"09:25.225 ","End":"09:30.769","Text":"Therefore, what we get is that B is equal to Mu naught,"},{"Start":"09:30.769 ","End":"09:36.930","Text":"Epsilon naught Ar divided by 2d."},{"Start":"09:37.610 ","End":"09:42.150","Text":"This is for when r is smaller than a."},{"Start":"09:42.150 ","End":"09:45.589","Text":"In other words, this is the magnetic field and,"},{"Start":"09:45.589 ","End":"09:48.499","Text":"of course, we can add in its direction."},{"Start":"09:48.499 ","End":"09:51.005","Text":"We already said it\u0027s in the Theta direction."},{"Start":"09:51.005 ","End":"09:57.620","Text":"This is the magnetic field when we\u0027re located inside of the capacitor."},{"Start":"09:58.910 ","End":"10:03.690","Text":"Now, let\u0027s look outside of the capacitor."},{"Start":"10:03.690 ","End":"10:11.690","Text":"Now, we\u0027re taking a loop that is still a radius r but where r is greater than a."},{"Start":"10:11.960 ","End":"10:17.680","Text":"What we can see is that this side of the equation will have the same thing,"},{"Start":"10:17.680 ","End":"10:25.345","Text":"b multiplied by 2Pi r. Nothing has changed."},{"Start":"10:25.345 ","End":"10:29.424","Text":"The magnetic field is still in the Theta direction,"},{"Start":"10:29.424 ","End":"10:33.235","Text":"and it\u0027s still going along this line,"},{"Start":"10:33.235 ","End":"10:41.125","Text":"which has this loop around the capacitor of radius r. This is equal 2."},{"Start":"10:41.125 ","End":"10:43.315","Text":"Now we\u0027re looking over here,"},{"Start":"10:43.315 ","End":"10:45.325","Text":"here are real current,"},{"Start":"10:45.325 ","End":"10:50.060","Text":"Rj is still 0, but our J_d is still some things."},{"Start":"10:50.060 ","End":"10:55.044","Text":"We have this is equal to Mu naught."},{"Start":"10:55.044 ","End":"11:00.785","Text":"Again, this is constant with respect to the area."},{"Start":"11:00.785 ","End":"11:03.049","Text":"We don\u0027t have to integrate."},{"Start":"11:03.049 ","End":"11:07.295","Text":"We can just multiply Mu naught multiplied by J_d,"},{"Start":"11:07.295 ","End":"11:15.019","Text":"which is Epsilon naught a divided by d. Then again,"},{"Start":"11:15.019 ","End":"11:20.325","Text":"this is multiplied by the surface area."},{"Start":"11:20.325 ","End":"11:26.135","Text":"Now the surface area is up until the radius of a."},{"Start":"11:26.135 ","End":"11:28.265","Text":"This is important to note."},{"Start":"11:28.265 ","End":"11:29.615","Text":"As we can see,"},{"Start":"11:29.615 ","End":"11:33.619","Text":"we have our J_d is only"},{"Start":"11:33.619 ","End":"11:42.140","Text":"acting on the surface area that corresponds to our plates of the capacitor."},{"Start":"11:42.140 ","End":"11:45.110","Text":"There is no J_d outside of here."},{"Start":"11:45.110 ","End":"11:48.272","Text":"J_d, our displacement current,"},{"Start":"11:48.272 ","End":"11:51.245","Text":"only occurs within the capacitor."},{"Start":"11:51.245 ","End":"11:53.420","Text":"When we\u0027re out of the capacitor,"},{"Start":"11:53.420 ","End":"11:58.200","Text":"there\u0027s no plate for the J_d to flow through."},{"Start":"11:58.800 ","End":"12:03.685","Text":"That means that when we\u0027re multiplying this by the surface area,"},{"Start":"12:03.685 ","End":"12:08.905","Text":"the surface area is just the area of the disk."},{"Start":"12:08.905 ","End":"12:11.065","Text":"That\u0027s of radius a."},{"Start":"12:11.065 ","End":"12:17.080","Text":"That means that we multiply this by Pi a^2."},{"Start":"12:17.080 ","End":"12:22.945","Text":"I\u0027m just going to highlight this because this is important to remember that it\u0027s a."},{"Start":"12:22.945 ","End":"12:27.280","Text":"Our displacement current only flows through the plate."},{"Start":"12:27.280 ","End":"12:32.785","Text":"When we\u0027re calculating the maximal surface area over here,"},{"Start":"12:32.785 ","End":"12:35.649","Text":"it can only be the surface area of the plate because"},{"Start":"12:35.649 ","End":"12:42.040","Text":"our J_d over here isn\u0027t flowing outside of the plate."},{"Start":"12:42.040 ","End":"12:46.045","Text":"Now again, we can cancel out the Pi."},{"Start":"12:46.045 ","End":"12:52.044","Text":"What we get is that B is equal to Mu naught Epsilon naught"},{"Start":"12:52.044 ","End":"13:00.010","Text":"A a^2 divided by 2rd."},{"Start":"13:00.010 ","End":"13:03.700","Text":"Again, we can make this a vector."},{"Start":"13:03.700 ","End":"13:11.380","Text":"Again, it\u0027s in the Theta direction and this is when r is greater than a."},{"Start":"13:11.380 ","End":"13:15.760","Text":"Now what happens if our radius is equal to a?"},{"Start":"13:15.760 ","End":"13:21.580","Text":"We\u0027re located right on the edge of our parallel plate capacitor."},{"Start":"13:21.580 ","End":"13:29.499","Text":"What we can see is that if we substitute na for r into this equation,"},{"Start":"13:29.499 ","End":"13:34.614","Text":"for r is less than a, or into here,"},{"Start":"13:34.614 ","End":"13:39.430","Text":"where r is greater than a,"},{"Start":"13:39.430 ","End":"13:41.770","Text":"we substitute a over here,"},{"Start":"13:41.770 ","End":"13:45.760","Text":"then we get the exact same magnetic field."},{"Start":"13:45.760 ","End":"13:49.915","Text":"We can see that this magnetic field is continuous."},{"Start":"13:49.915 ","End":"13:57.040","Text":"We can say r is less than or equal to a or r is greater than or equal to a."},{"Start":"13:57.040 ","End":"14:01.150","Text":"Now let\u0027s answer question number 3."},{"Start":"14:01.150 ","End":"14:05.485","Text":"Before we start that, I\u0027m just going to rub out the board."},{"Start":"14:05.485 ","End":"14:09.790","Text":"Question number 3 is to calculate the energy stored within"},{"Start":"14:09.790 ","End":"14:14.230","Text":"the surface area of the capacitor."},{"Start":"14:14.230 ","End":"14:24.990","Text":"If we remember, the energy is equal to the integral of Epsilon"},{"Start":"14:24.990 ","End":"14:32.545","Text":"naught divided by 2 multiplied by the electric field squared dv plus"},{"Start":"14:32.545 ","End":"14:40.544","Text":"the integral of 1 divided by 2 Mu naught B^2,"},{"Start":"14:40.544 ","End":"14:44.080","Text":"where B^2 is the magnetic field dV."},{"Start":"14:44.080 ","End":"14:48.220","Text":"We can also put all of this under one integral sign,"},{"Start":"14:48.220 ","End":"14:50.240","Text":"it doesn\u0027t really matter."},{"Start":"14:50.940 ","End":"14:55.719","Text":"Now let\u0027s start substituting in values."},{"Start":"14:55.719 ","End":"15:04.795","Text":"We have over here the integral of Epsilon naught divided by 2 multiplied by E^2 dV,"},{"Start":"15:04.795 ","End":"15:07.940","Text":"our magnetic field is this."},{"Start":"15:09.420 ","End":"15:19.570","Text":"The electric field squared is A^2 t^2 divided by d^2."},{"Start":"15:19.570 ","End":"15:24.430","Text":"All of this is dV plus the integral of"},{"Start":"15:24.430 ","End":"15:30.070","Text":"1 divided by 2 Mu naught of our magnetic field squared."},{"Start":"15:30.070 ","End":"15:35.634","Text":"Because we want to find the magnetic field within the surface area of the capacitor,"},{"Start":"15:35.634 ","End":"15:38.110","Text":"we\u0027re going to use this equation over"},{"Start":"15:38.110 ","End":"15:41.574","Text":"here because this is where the radius is less than a,"},{"Start":"15:41.574 ","End":"15:45.745","Text":"which means that we\u0027re inside the capacitor."},{"Start":"15:45.745 ","End":"15:49.557","Text":"We have our magnetic field squared."},{"Start":"15:49.557 ","End":"15:53.979","Text":"That means we have all of this B^2,"},{"Start":"15:53.979 ","End":"15:59.379","Text":"we have Mu naught squared Epsilon naught squared"},{"Start":"15:59.379 ","End":"16:07.799","Text":"A^2 r^2"},{"Start":"16:07.799 ","End":"16:10.885","Text":"divided by 2^2,"},{"Start":"16:10.885 ","End":"16:19.600","Text":"which is 4, multiplied by d^2 dV."},{"Start":"16:19.600 ","End":"16:25.900","Text":"Now, what we can do is we can take all of our constants out of the integral sign."},{"Start":"16:25.900 ","End":"16:30.399","Text":"First of all, we can see here that all of"},{"Start":"16:30.399 ","End":"16:35.230","Text":"what we have in this section in this integral is all a constant."},{"Start":"16:35.230 ","End":"16:38.529","Text":"That means that we\u0027re just multiplying all of"},{"Start":"16:38.529 ","End":"16:42.685","Text":"these constants by the volume of the capacitor."},{"Start":"16:42.685 ","End":"16:47.900","Text":"The volume of the capacitor is just the volume of this cylinder over here."},{"Start":"16:48.600 ","End":"16:53.185","Text":"Over here we\u0027ll have Epsilon naught"},{"Start":"16:53.185 ","End":"17:01.749","Text":"A^2 t^2 divided by 2d^2,"},{"Start":"17:01.749 ","End":"17:05.710","Text":"and then we\u0027ll multiply that by d"},{"Start":"17:05.710 ","End":"17:12.970","Text":"multiplied by Pi a^2."},{"Start":"17:12.970 ","End":"17:16.660","Text":"This d with this d can cancel out."},{"Start":"17:16.660 ","End":"17:18.520","Text":"Then we have plus,"},{"Start":"17:18.520 ","End":"17:22.330","Text":"now we can take out all of the constants from here."},{"Start":"17:22.330 ","End":"17:24.898","Text":"First of all, let\u0027s just cancel some stuff out."},{"Start":"17:24.898 ","End":"17:28.390","Text":"This Mu naught will cancel out with this."},{"Start":"17:28.390 ","End":"17:30.989","Text":"Then we can take this out."},{"Start":"17:30.989 ","End":"17:38.530","Text":"We have Mu naught Epsilon naught squared A^2 divided by 2 times 4,"},{"Start":"17:38.530 ","End":"17:46.360","Text":"which is 8d^2 multiplied by the integral of r squared dV,"},{"Start":"17:46.360 ","End":"17:52.390","Text":"where dV in cylindrical coordinates is simply equal"},{"Start":"17:52.390 ","End":"18:01.450","Text":"to rdr dTheta dz."},{"Start":"18:01.450 ","End":"18:06.250","Text":"Of course, this is a triple integral."},{"Start":"18:06.250 ","End":"18:11.170","Text":"For our r, we\u0027re going from 0 up until a."},{"Start":"18:11.170 ","End":"18:12.730","Text":"For our Theta,"},{"Start":"18:12.730 ","End":"18:16.315","Text":"we\u0027re going from 0 up until 2Pi full circle."},{"Start":"18:16.315 ","End":"18:25.840","Text":"For our z, we\u0027re going from 0 up until d. Then,"},{"Start":"18:25.840 ","End":"18:34.614","Text":"what we\u0027re going to have is Epsilon naught A^2 t^2 Pi a^2 divided by"},{"Start":"18:34.614 ","End":"18:38.680","Text":"2d plus Epsilon naught"},{"Start":"18:38.680 ","End":"18:45.745","Text":"Mu naught A^2 divided by 8d^2."},{"Start":"18:45.745 ","End":"18:51.709","Text":"Here we can see that we don\u0027t have any variables of Theta and z,"},{"Start":"18:51.709 ","End":"18:56.964","Text":"so we can just multiply this by 2Pi and"},{"Start":"18:56.964 ","End":"19:04.150","Text":"by d. Then over here we have r^2 times r, which is r^3dr."},{"Start":"19:04.150 ","End":"19:12.910","Text":"We have therefore multiplied by r^4 divided by 4."},{"Start":"19:12.910 ","End":"19:19.870","Text":"Then, we plug in over here for r between 0 and a."},{"Start":"19:19.870 ","End":"19:27.290","Text":"In other words, we can just write this as a^4."},{"Start":"19:28.260 ","End":"19:33.310","Text":"Now we can cancel out this d with one of these d\u0027s and"},{"Start":"19:33.310 ","End":"19:38.185","Text":"this 2 over here with this 4 making this 2 over here."},{"Start":"19:38.185 ","End":"19:42.049","Text":"Now, let\u0027s rewrite all of this."},{"Start":"19:42.049 ","End":"19:50.409","Text":"We could also write with like terms and then we get this as the answer to question 3;"},{"Start":"19:50.409 ","End":"19:56.799","Text":"the energy U is equal to Epsilon naught a^2 Pia^2 divided by"},{"Start":"19:56.799 ","End":"20:05.560","Text":"2d multiplied by t^2 plus Mu naught Epsilon naught a^2 divided by 8."},{"Start":"20:05.560 ","End":"20:08.905","Text":"Now let\u0027s answer question number 4,"},{"Start":"20:08.905 ","End":"20:13.045","Text":"which is to calculate the pointing vector for question number 3."},{"Start":"20:13.045 ","End":"20:15.489","Text":"The pointing vector, of course,"},{"Start":"20:15.489 ","End":"20:20.140","Text":"is a vector at the surface of the capacitor,"},{"Start":"20:20.140 ","End":"20:24.129","Text":"which as we learned in one of the previous lessons,"},{"Start":"20:24.129 ","End":"20:31.840","Text":"is the amount of energy exchange per unit of area per unit of time."},{"Start":"20:31.840 ","End":"20:34.270","Text":"If we\u0027re looking at per unit of area,"},{"Start":"20:34.270 ","End":"20:39.340","Text":"we\u0027re of course looking at the surface area of the capacitor."},{"Start":"20:39.340 ","End":"20:45.298","Text":"Our surface, our radius is the only thing that we\u0027re looking at,"},{"Start":"20:45.298 ","End":"20:50.170","Text":"so I\u0027ll just remind you of the equation for the pointing vector S. It\u0027s 1 divided"},{"Start":"20:50.170 ","End":"20:55.660","Text":"by Mu naught E cross B."},{"Start":"20:55.660 ","End":"20:58.609","Text":"We\u0027re using these equations."},{"Start":"20:58.609 ","End":"21:00.820","Text":"We can see that our variable is the radius,"},{"Start":"21:00.820 ","End":"21:05.319","Text":"so the radius of the surface area of the capacitor is a."},{"Start":"21:05.319 ","End":"21:10.645","Text":"Because only at a we\u0027re the surface,"},{"Start":"21:10.645 ","End":"21:14.710","Text":"the outskirts of the capacitor. Let\u0027s write this out."},{"Start":"21:14.710 ","End":"21:17.500","Text":"This is equal to 1 divided by Mu naught,"},{"Start":"21:17.500 ","End":"21:19.630","Text":"and then we have the electric field,"},{"Start":"21:19.630 ","End":"21:27.250","Text":"so we have At divided by d in the z direction cross multiplied with the magnetic field."},{"Start":"21:27.250 ","End":"21:32.245","Text":"Of course, we\u0027re looking at the edge at radius r is equal to a."},{"Start":"21:32.245 ","End":"21:36.970","Text":"As we saw, because this equation for the magnetic field is continuous,"},{"Start":"21:36.970 ","End":"21:41.005","Text":"we can use either version of the equation, they\u0027re both equal."},{"Start":"21:41.005 ","End":"21:43.105","Text":"Let\u0027s just use the first one."},{"Start":"21:43.105 ","End":"21:46.840","Text":"We have Mu naught Epsilon naught A,"},{"Start":"21:46.840 ","End":"21:54.310","Text":"and then the radius we\u0027re substituting in a divided by 2d,"},{"Start":"21:54.310 ","End":"21:57.710","Text":"and this is in the Theta direction."},{"Start":"21:58.650 ","End":"22:05.470","Text":"Over here first of all this Mu naught can cancel with this Mu naught"},{"Start":"22:05.470 ","End":"22:13.600","Text":"and then we just multiply all of the constants."},{"Start":"22:13.600 ","End":"22:23.820","Text":"We have At divided by d multiplied by Epsilon naught multiplied by another A,"},{"Start":"22:23.820 ","End":"22:28.710","Text":"multiplied by a and then here we have another 2d,"},{"Start":"22:28.710 ","End":"22:31.959","Text":"so here it\u0027s 2d^2."},{"Start":"22:31.959 ","End":"22:35.650","Text":"Then, when we cross multiply the z and the Theta,"},{"Start":"22:35.650 ","End":"22:41.515","Text":"we get here that this is in the negative r direction."},{"Start":"22:41.515 ","End":"22:44.680","Text":"This was the answer to Question 4."},{"Start":"22:44.680 ","End":"22:47.934","Text":"I\u0027ll just remind you that here we\u0027re looking for"},{"Start":"22:47.934 ","End":"22:52.037","Text":"the electric and magnetic fields at r is equal to a,"},{"Start":"22:52.037 ","End":"23:01.075","Text":"because we\u0027re looking at what\u0027s going on over here on the surface of the capacitor."},{"Start":"23:01.075 ","End":"23:06.489","Text":"Question number 5 is to calculate the pointing flux"},{"Start":"23:06.489 ","End":"23:12.340","Text":"and show that it is equal to negative the energy change we got in 3."},{"Start":"23:12.340 ","End":"23:14.635","Text":"First of all,"},{"Start":"23:14.635 ","End":"23:21.295","Text":"we can see that our pointing vector is pointing in the negative radial direction,"},{"Start":"23:21.295 ","End":"23:23.589","Text":"and it\u0027s always at r is equal to a."},{"Start":"23:23.589 ","End":"23:30.219","Text":"So it\u0027s always on the surface and pointing inwards like so."},{"Start":"23:30.219 ","End":"23:37.089","Text":"This is our S vector pointing into the radial direction inside to the capacitor."},{"Start":"23:37.089 ","End":"23:41.680","Text":"We can also see that throughout the surface area,"},{"Start":"23:41.680 ","End":"23:44.049","Text":"so whether we\u0027re at the bottom of the capacitor,"},{"Start":"23:44.049 ","End":"23:45.905","Text":"the middle or the top,"},{"Start":"23:45.905 ","End":"23:48.160","Text":"this is going to be constant,"},{"Start":"23:48.160 ","End":"23:52.250","Text":"the pointing vector is constant."},{"Start":"23:52.590 ","End":"23:57.100","Text":"So 4, 5,"},{"Start":"23:57.100 ","End":"24:01.675","Text":"we\u0027re looking at the flux of the pointing vector."},{"Start":"24:01.675 ","End":"24:04.240","Text":"This is exactly like what we saw with Gauss."},{"Start":"24:04.240 ","End":"24:09.490","Text":"We have the integral on S our pointing vector.ds,"},{"Start":"24:09.490 ","End":"24:12.909","Text":"where of course this denotes surface area."},{"Start":"24:12.909 ","End":"24:15.639","Text":"Now, because as we\u0027ve already said,"},{"Start":"24:15.639 ","End":"24:18.595","Text":"our pointing vector is constant,"},{"Start":"24:18.595 ","End":"24:21.295","Text":"we don\u0027t have any variables in here."},{"Start":"24:21.295 ","End":"24:24.534","Text":"We can use like what we did in Gauss\u0027s law,"},{"Start":"24:24.534 ","End":"24:32.020","Text":"where we just take the magnitude of our pointing vector dot product with S,"},{"Start":"24:32.020 ","End":"24:35.454","Text":"where S is, of course, surface area."},{"Start":"24:35.454 ","End":"24:39.550","Text":"We\u0027re just multiplying our pointing vector"},{"Start":"24:39.550 ","End":"24:44.155","Text":"multiplied by the surface area of our capacitor."},{"Start":"24:44.155 ","End":"24:47.979","Text":"We\u0027re just taking the magnitude,"},{"Start":"24:47.979 ","End":"24:52.305","Text":"so we won\u0027t look at the direction."},{"Start":"24:52.305 ","End":"24:57.265","Text":"We just have Epsilon naught A^2 ta"},{"Start":"24:57.265 ","End":"25:03.595","Text":"divided by 2d^2 and all of this is multiplied by the surface area."},{"Start":"25:03.595 ","End":"25:05.560","Text":"We take the circumference over here,"},{"Start":"25:05.560 ","End":"25:08.035","Text":"the surface area of what I\u0027ve drawn in blue."},{"Start":"25:08.035 ","End":"25:13.330","Text":"The circumference is 2 Pi multiplied by the radius,"},{"Start":"25:13.330 ","End":"25:17.275","Text":"so a and then multiplied by this length over here,"},{"Start":"25:17.275 ","End":"25:25.360","Text":"d. This is the surface area over here of the sides."},{"Start":"25:25.360 ","End":"25:28.750","Text":"Then this 2 can cancel out,"},{"Start":"25:28.750 ","End":"25:31.795","Text":"this d can cancel out over here,"},{"Start":"25:31.795 ","End":"25:35.530","Text":"and then we\u0027ll just multiply this afterwards."},{"Start":"25:35.530 ","End":"25:44.358","Text":"What we see is that we get that the flux is equal to Epsilon naught A^2 t,"},{"Start":"25:44.358 ","End":"25:47.470","Text":"and then we have a multiplied by a,"},{"Start":"25:47.470 ","End":"25:53.110","Text":"so a^2 multiplied by Pi and all of this is"},{"Start":"25:53.110 ","End":"25:59.469","Text":"divided by d. This is the pointing flux."},{"Start":"25:59.469 ","End":"26:01.150","Text":"Now just a little explanation,"},{"Start":"26:01.150 ","End":"26:04.269","Text":"here we can see that we\u0027re doing the dot product between"},{"Start":"26:04.269 ","End":"26:11.110","Text":"the pointing vector and the vector denoting the surface area."},{"Start":"26:11.110 ","End":"26:14.199","Text":"Now, as we can see over here,"},{"Start":"26:14.199 ","End":"26:19.330","Text":"when we\u0027re looking at the vector denoting the surface area on the side,"},{"Start":"26:19.330 ","End":"26:23.424","Text":"so on the cylinder encasing the sides of the capacitor,"},{"Start":"26:23.424 ","End":"26:26.980","Text":"we can see that the 2 vectors are parallel."},{"Start":"26:26.980 ","End":"26:29.200","Text":"When we do the dot product between them,"},{"Start":"26:29.200 ","End":"26:33.035","Text":"they don\u0027t cancel out and that\u0027s why we get this value over here."},{"Start":"26:33.035 ","End":"26:39.621","Text":"However, if we were to try and see the pointing vector on the base of the capacitor,"},{"Start":"26:39.621 ","End":"26:41.897","Text":"so on the top plate and on the bottom plate,"},{"Start":"26:41.897 ","End":"26:45.630","Text":"we can see that the vector denoting the surface area is"},{"Start":"26:45.630 ","End":"26:49.769","Text":"pointing in this direction, upwards or downwards,"},{"Start":"26:49.769 ","End":"26:52.425","Text":"but on this axis over here,"},{"Start":"26:52.425 ","End":"26:54.180","Text":"in the z direction,"},{"Start":"26:54.180 ","End":"27:00.120","Text":"and then we can see that our pointing vector is in the radial direction."},{"Start":"27:00.120 ","End":"27:04.769","Text":"Then when we do the dot product between the radial direction and the z direction,"},{"Start":"27:04.769 ","End":"27:09.369","Text":"the dot product means that they\u0027ll be equal to 0."},{"Start":"27:09.369 ","End":"27:12.310","Text":"Then it will cancel out, and we\u0027ll get that the pointing flux"},{"Start":"27:12.310 ","End":"27:17.064","Text":"on the top plate and the bottom plate is equal to 0."},{"Start":"27:17.064 ","End":"27:22.299","Text":"We only have this pointing flux on the cylinder encasing the capacitor,"},{"Start":"27:22.299 ","End":"27:26.150","Text":"but not on its plates."},{"Start":"27:26.910 ","End":"27:32.785","Text":"We have that the flux through the basis is equal to 0,"},{"Start":"27:32.785 ","End":"27:40.914","Text":"as we said because the pointing vector is perpendicular to the vector for the base."},{"Start":"27:40.914 ","End":"27:42.640","Text":"Just a little note over here,"},{"Start":"27:42.640 ","End":"27:46.375","Text":"because we took the size or the magnitude of the pointing vector,"},{"Start":"27:46.375 ","End":"27:48.249","Text":"we canceled out the minus,"},{"Start":"27:48.249 ","End":"27:52.765","Text":"but we actually have to add in the minus again."},{"Start":"27:52.765 ","End":"27:54.189","Text":"We got the magnitude,"},{"Start":"27:54.189 ","End":"27:57.910","Text":"but we know that the flux is in the opposite direction."},{"Start":"27:57.910 ","End":"28:00.999","Text":"Usually, we\u0027re speaking about the pointing vector,"},{"Start":"28:00.999 ","End":"28:04.029","Text":"pointing out of our body,"},{"Start":"28:04.029 ","End":"28:05.619","Text":"as energy is going out."},{"Start":"28:05.619 ","End":"28:09.009","Text":"But here we can see that energy is actually going in,"},{"Start":"28:09.009 ","End":"28:11.259","Text":"so the pointing vector is pointing in"},{"Start":"28:11.259 ","End":"28:13.750","Text":"the opposite direction to the direction that it usually"},{"Start":"28:13.750 ","End":"28:19.195","Text":"is and that\u0027s according to how we defined it at least in the previous lesson."},{"Start":"28:19.195 ","End":"28:23.750","Text":"Remember that there is a minus over here."},{"Start":"28:24.150 ","End":"28:29.710","Text":"Now, what we want to do is we want to do the final stage of question number 5,"},{"Start":"28:29.710 ","End":"28:35.545","Text":"show that it\u0027s equal to the negative energy change from question number 3."},{"Start":"28:35.545 ","End":"28:39.709","Text":"What we\u0027re doing is Poynting\u0027s law."},{"Start":"28:39.900 ","End":"28:42.115","Text":"This is Poynting\u0027s law,"},{"Start":"28:42.115 ","End":"28:47.140","Text":"it\u0027s also may be worthwhile to write this in your equation sheets."},{"Start":"28:47.140 ","End":"28:51.940","Text":"We have the close loop integral of SDS,"},{"Start":"28:51.940 ","End":"28:53.379","Text":"or in other words,"},{"Start":"28:53.379 ","End":"28:56.320","Text":"the flux over here."},{"Start":"28:56.320 ","End":"29:03.384","Text":"Poynting\u0027s laws says that this is equal to negative dU_em by dt,"},{"Start":"29:03.384 ","End":"29:09.895","Text":"so the negative time derivative of the electromagnetic energy."},{"Start":"29:09.895 ","End":"29:11.950","Text":"This is of course,"},{"Start":"29:11.950 ","End":"29:18.460","Text":"the flux as we saw over here, Poynting\u0027s flux."},{"Start":"29:18.460 ","End":"29:21.730","Text":"I\u0027ll just write it out as P flux, Poynting\u0027s flux."},{"Start":"29:21.730 ","End":"29:26.499","Text":"Over here, we got that the flux was equal to, as we said,"},{"Start":"29:26.499 ","End":"29:32.140","Text":"negative because it\u0027s going inwards instead of outwards as we define the direction."},{"Start":"29:32.140 ","End":"29:39.415","Text":"Epsilon naught A^2t a^2 Pi divided by d,"},{"Start":"29:39.415 ","End":"29:45.595","Text":"and it\u0027s equal to negative the time derivative of the electromagnetic energy."},{"Start":"29:45.595 ","End":"29:47.979","Text":"This was our electromagnetic energy,"},{"Start":"29:47.979 ","End":"29:50.230","Text":"and now we\u0027re taking the negative time derivatives."},{"Start":"29:50.230 ","End":"29:54.865","Text":"First of all, we\u0027ll add a minus over here and now the time derivative of this."},{"Start":"29:54.865 ","End":"29:57.325","Text":"In the brackets, we have 2 terms."},{"Start":"29:57.325 ","End":"29:59.560","Text":"This term over here,"},{"Start":"29:59.560 ","End":"30:05.380","Text":"we can see when multiplied with this is independent of time,"},{"Start":"30:05.380 ","End":"30:09.400","Text":"we have no t variable over here,"},{"Start":"30:09.400 ","End":"30:11.005","Text":"so this cancels out."},{"Start":"30:11.005 ","End":"30:16.419","Text":"What we\u0027re left with is what\u0027s outside of the brackets multiplied by t^2."},{"Start":"30:16.419 ","End":"30:18.775","Text":"Then if we take the derivative of this,"},{"Start":"30:18.775 ","End":"30:23.352","Text":"so 1 of the t\u0027s cancels out due to taking the derivative,"},{"Start":"30:23.352 ","End":"30:24.550","Text":"not exactly cancels out,"},{"Start":"30:24.550 ","End":"30:26.335","Text":"but you know what I mean."},{"Start":"30:26.335 ","End":"30:33.085","Text":"We\u0027re taking the time derivative of this with the t^2 over here."},{"Start":"30:33.085 ","End":"30:36.565","Text":"What we get is that this is equal to Epsilon naught"},{"Start":"30:36.565 ","End":"30:46.645","Text":"A^2 Pi a^2t divided by 2d,"},{"Start":"30:46.645 ","End":"30:51.084","Text":"but then we multiply it by 2 from the time derivative."},{"Start":"30:51.084 ","End":"30:57.092","Text":"We have divided by 2 multiplied by 2 over here so that cancels out,"},{"Start":"30:57.092 ","End":"30:59.379","Text":"and we just have d. We have"},{"Start":"30:59.379 ","End":"31:02.919","Text":"the exact same thing written over here just in a slightly different order,"},{"Start":"31:02.919 ","End":"31:04.979","Text":"switch the Pi and the t,"},{"Start":"31:04.979 ","End":"31:07.455","Text":"and you\u0027ll see that this is exactly the same."},{"Start":"31:07.455 ","End":"31:12.839","Text":"Now we\u0027ve shown that the Poynting flux is equal to"},{"Start":"31:12.839 ","End":"31:21.509","Text":"the negative derivative of the electromagnetic energy."},{"Start":"31:21.509 ","End":"31:24.349","Text":"Or we showed Poynting\u0027s law."},{"Start":"31:24.840 ","End":"31:28.210","Text":"This is the answer to Question 5."},{"Start":"31:28.210 ","End":"31:32.740","Text":"This is the flux through the surface area,"},{"Start":"31:32.740 ","End":"31:40.010","Text":"and here we\u0027ve shown Poynting\u0027s law and that is the end of this lesson."}],"ID":25731}],"Thumbnail":null,"ID":99493}]

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