Introduction to Resistors and Current Density
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[{"Name":"Introduction to Resistors and Current Density","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Calculating Resistance","Duration":"13m 16s","ChapterTopicVideoID":21515,"CourseChapterTopicPlaylistID":99491,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/21515.jpeg","UploadDate":"2020-04-21T15:13:54.4830000","DurationForVideoObject":"PT13M16S","Description":null,"MetaTitle":"Calculating Resistance: Video + Workbook | Proprep","MetaDescription":"Resistors, Current and Current Density - Introduction to Resistors and Current Density. 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/resistors%2c-current-and-current-density/introduction-to-resistors-and-current-density/vid22414","VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.680","Text":"Hello. In this lesson,"},{"Start":"00:01.680 ","End":"00:06.330","Text":"we\u0027re going to be speaking a bit about resistors and about current."},{"Start":"00:06.330 ","End":"00:08.685","Text":"The first thing that we have to remember about"},{"Start":"00:08.685 ","End":"00:12.775","Text":"a resistor is that it is in fact a conductor."},{"Start":"00:12.775 ","End":"00:16.485","Text":"In order for the resistor to resist the current,"},{"Start":"00:16.485 ","End":"00:19.935","Text":"it has to allow some current to pass through it,"},{"Start":"00:19.935 ","End":"00:23.860","Text":"which means therefore that it is a conductor."},{"Start":"00:23.860 ","End":"00:28.680","Text":"The equation for calculating the resistance of some resistor,"},{"Start":"00:28.680 ","End":"00:33.165","Text":"so R is equal to Rho,"},{"Start":"00:33.165 ","End":"00:35.250","Text":"which is resistivity,"},{"Start":"00:35.250 ","End":"00:38.615","Text":"multiplied by the length of the resistor"},{"Start":"00:38.615 ","End":"00:43.954","Text":"divided by the cross-sectional area of the resistor."},{"Start":"00:43.954 ","End":"00:50.185","Text":"Now, resistivity is a property of the material."},{"Start":"00:50.185 ","End":"00:56.825","Text":"What it will do is it will tell us how good of a conductor or a resistor the material is."},{"Start":"00:56.825 ","End":"01:00.005","Text":"Let\u0027s say we have a resistor made out of"},{"Start":"01:00.005 ","End":"01:03.755","Text":"copper and an exact same one with the same dimensions,"},{"Start":"01:03.755 ","End":"01:05.810","Text":"but made out of aluminum."},{"Start":"01:05.810 ","End":"01:09.365","Text":"Depending on copper\u0027s value for resistivity,"},{"Start":"01:09.365 ","End":"01:11.930","Text":"and aluminum\u0027s value for resistivity,"},{"Start":"01:11.930 ","End":"01:15.090","Text":"we\u0027ll know which one will be a better resistor."},{"Start":"01:15.090 ","End":"01:17.880","Text":"Which one has a higher resistance."},{"Start":"01:17.880 ","End":"01:22.370","Text":"Then we can know which is better as a conductor."},{"Start":"01:22.370 ","End":"01:24.050","Text":"Both are conductors."},{"Start":"01:24.050 ","End":"01:29.390","Text":"But let\u0027s say if we want to make wires for our electrical components,"},{"Start":"01:29.390 ","End":"01:32.690","Text":"we\u0027ll want something with a lower resistance."},{"Start":"01:32.690 ","End":"01:35.605","Text":"Something that is a better conductor."},{"Start":"01:35.605 ","End":"01:40.580","Text":"In that case, we\u0027ll want a lower value for resistivity."},{"Start":"01:40.580 ","End":"01:42.964","Text":"In our copper aluminum example,"},{"Start":"01:42.964 ","End":"01:45.590","Text":"copper has a lower value for resistivity,"},{"Start":"01:45.590 ","End":"01:51.390","Text":"and therefore, it would be a better conductor of current and electricity."},{"Start":"01:51.950 ","End":"01:58.680","Text":"Now, let\u0027s say that we have some box like so,"},{"Start":"01:58.680 ","End":"02:04.810","Text":"and it has dimensions like so."},{"Start":"02:04.810 ","End":"02:08.215","Text":"It\u0027s length is of length c over here."},{"Start":"02:08.215 ","End":"02:10.840","Text":"This side over here is a,"},{"Start":"02:10.840 ","End":"02:14.240","Text":"and this side over here is of length of b."},{"Start":"02:14.240 ","End":"02:21.040","Text":"First of all, in order to calculate the resistance of this type of shape,"},{"Start":"02:21.040 ","End":"02:22.659","Text":"this type of resistor,"},{"Start":"02:22.659 ","End":"02:25.060","Text":"the first thing we need to know is what is"},{"Start":"02:25.060 ","End":"02:28.525","Text":"the direction that the current is traveling in."},{"Start":"02:28.525 ","End":"02:33.205","Text":"Is the current traveling in this direction?"},{"Start":"02:33.205 ","End":"02:36.085","Text":"In which case, the cross-sectional area,"},{"Start":"02:36.085 ","End":"02:38.980","Text":"what we\u0027re dividing by will be a times b,"},{"Start":"02:38.980 ","End":"02:46.445","Text":"and the length of the resistor will be c. Or is the current traveling in this direction,"},{"Start":"02:46.445 ","End":"02:52.550","Text":"where therefore the cross-sectional area will be a times c?"},{"Start":"02:52.550 ","End":"02:54.705","Text":"This will be the cross-sectional area."},{"Start":"02:54.705 ","End":"02:59.630","Text":"The length of the resistor will be this length over here, which is b."},{"Start":"02:59.630 ","End":"03:02.990","Text":"In order to calculate the resistance,"},{"Start":"03:02.990 ","End":"03:06.445","Text":"we have to know the direction of the current."},{"Start":"03:06.445 ","End":"03:13.645","Text":"Therefore, we can define the length of the resistor as the path taken by the current."},{"Start":"03:13.645 ","End":"03:16.080","Text":"In this direction let\u0027s say,"},{"Start":"03:16.080 ","End":"03:20.180","Text":"and that the cross-sectional area is therefore the plane"},{"Start":"03:20.180 ","End":"03:25.040","Text":"which is perpendicular to the direction of current."},{"Start":"03:25.040 ","End":"03:30.440","Text":"In the case where our current is traveling in this direction,"},{"Start":"03:30.440 ","End":"03:34.950","Text":"so this would be the length,"},{"Start":"03:34.950 ","End":"03:40.100","Text":"and then this is the plane which is perpendicular to the direction of the current."},{"Start":"03:40.100 ","End":"03:43.830","Text":"Therefore, the cross-sectional area would be ab."},{"Start":"03:45.110 ","End":"03:49.835","Text":"A condition for using this equation as it is,"},{"Start":"03:49.835 ","End":"03:51.395","Text":"is that a,"},{"Start":"03:51.395 ","End":"03:54.325","Text":"the cross-sectional area and Rho,"},{"Start":"03:54.325 ","End":"03:58.425","Text":"the resistivity must be constants."},{"Start":"03:58.425 ","End":"04:04.335","Text":"Now let\u0027s look at an example where one of these isn\u0027t a constant."},{"Start":"04:04.335 ","End":"04:07.700","Text":"Let\u0027s see how we can adapt this equation in order to"},{"Start":"04:07.700 ","End":"04:12.085","Text":"still be able to use some version of it."},{"Start":"04:12.085 ","End":"04:18.380","Text":"Here\u0027s an example where we have a solid cylinder of radius r"},{"Start":"04:18.380 ","End":"04:26.030","Text":"and of height h. We can see that the z-axis is the axis of symmetry."},{"Start":"04:26.030 ","End":"04:32.450","Text":"The resistivity of this cylinder is Rho naught multiplied by z,"},{"Start":"04:32.450 ","End":"04:34.235","Text":"which is changing,"},{"Start":"04:34.235 ","End":"04:38.295","Text":"divided by h, the height of the cylinder."},{"Start":"04:38.295 ","End":"04:40.600","Text":"What we can see is that our Rho,"},{"Start":"04:40.600 ","End":"04:43.550","Text":"our resistivity isn\u0027t constant,"},{"Start":"04:43.550 ","End":"04:46.430","Text":"it isn\u0027t uniform throughout, it\u0027s changing."},{"Start":"04:46.430 ","End":"04:55.265","Text":"My question is, what is the total resistance of this cylindrical resistor."},{"Start":"04:55.265 ","End":"04:59.070","Text":"How do I solve this type of question?"},{"Start":"04:59.990 ","End":"05:07.270","Text":"First of all, we saw that our resistance is dependent on the direction of the current."},{"Start":"05:07.270 ","End":"05:12.995","Text":"Let\u0027s assume that we\u0027re being told that the current travels through the resistor,"},{"Start":"05:12.995 ","End":"05:17.335","Text":"like so. Along the z-axis."},{"Start":"05:17.335 ","End":"05:19.815","Text":"This is the direction of the current."},{"Start":"05:19.815 ","End":"05:22.295","Text":"What do we do in this type of case?"},{"Start":"05:22.295 ","End":"05:24.800","Text":"As we\u0027ve seen in many previous questions,"},{"Start":"05:24.800 ","End":"05:26.585","Text":"I\u0027m sure you can guess the technique."},{"Start":"05:26.585 ","End":"05:31.610","Text":"We split up this cylinder into lots and lots of discs."},{"Start":"05:31.610 ","End":"05:36.782","Text":"What we\u0027ll have is many discs."},{"Start":"05:36.782 ","End":"05:39.980","Text":"We\u0027ll calculate the resistance of each disc."},{"Start":"05:39.980 ","End":"05:44.070","Text":"Each disc has some height."},{"Start":"05:47.720 ","End":"05:51.285","Text":"It has some volume, this disc."},{"Start":"05:51.285 ","End":"05:58.020","Text":"Then we\u0027re going to sum up all of the discs that fit into this cylinder."},{"Start":"05:58.020 ","End":"06:02.600","Text":"First of all, we know that the radius of this disc is still r,"},{"Start":"06:02.600 ","End":"06:04.370","Text":"it\u0027s the same radius."},{"Start":"06:04.370 ","End":"06:08.710","Text":"What we\u0027ve just changed is this width over here."},{"Start":"06:08.710 ","End":"06:11.795","Text":"Instead of the height of the cylinder being h,"},{"Start":"06:11.795 ","End":"06:16.890","Text":"we have some very small height that we\u0027re going to call dz."},{"Start":"06:18.010 ","End":"06:26.030","Text":"Because this height or the width of this disc is so small, it\u0027s approaching 0,"},{"Start":"06:26.030 ","End":"06:33.770","Text":"we can assume that the resistivity at both the bottom over here,"},{"Start":"06:33.770 ","End":"06:40.415","Text":"this point and at this point is the same because the change in z is so small."},{"Start":"06:40.415 ","End":"06:45.745","Text":"We can assume therefore that the resistivity is constant."},{"Start":"06:45.745 ","End":"06:51.550","Text":"Therefore, let\u0027s calculate the resistance of this small cylinder."},{"Start":"06:51.550 ","End":"06:53.530","Text":"Let\u0027s call this dR,"},{"Start":"06:53.530 ","End":"06:55.490","Text":"the resistance of the small cylinder."},{"Start":"06:55.490 ","End":"07:00.600","Text":"Then later we\u0027ll sum up all of the tiny cylinders."},{"Start":"07:01.100 ","End":"07:03.500","Text":"We have our resistivity,"},{"Start":"07:03.500 ","End":"07:04.850","Text":"which now is constant,"},{"Start":"07:04.850 ","End":"07:09.110","Text":"which is equal to Rho naught multiplied by z."},{"Start":"07:09.110 ","End":"07:12.620","Text":"Here we are at some height z divided by h,"},{"Start":"07:12.620 ","End":"07:15.905","Text":"the total height of the cylinder."},{"Start":"07:15.905 ","End":"07:19.895","Text":"Then all of this is multiplied by L,"},{"Start":"07:19.895 ","End":"07:22.435","Text":"the length of the resistor."},{"Start":"07:22.435 ","End":"07:27.395","Text":"If our current is traveling in this direction along the z axis,"},{"Start":"07:27.395 ","End":"07:31.385","Text":"so the current will be traveling like so"},{"Start":"07:31.385 ","End":"07:37.320","Text":"also through this many cylinder or this thick disc."},{"Start":"07:37.320 ","End":"07:41.060","Text":"That means the length of the resistor is the path taken"},{"Start":"07:41.060 ","End":"07:44.209","Text":"by the current in this specific resistor,"},{"Start":"07:44.209 ","End":"07:46.505","Text":"which is this over here,"},{"Start":"07:46.505 ","End":"07:48.735","Text":"which is a path of dz."},{"Start":"07:48.735 ","End":"07:52.950","Text":"It\u0027s just the height of this thick disc."},{"Start":"07:52.950 ","End":"07:56.565","Text":"Dz, that\u0027s the length of the resistor."},{"Start":"07:56.565 ","End":"07:58.925","Text":"Then it\u0027s divided by the cross-sectional area,"},{"Start":"07:58.925 ","End":"08:02.450","Text":"which is the plane perpendicular to the direction of current."},{"Start":"08:02.450 ","End":"08:05.615","Text":"What is the plane perpendicular to the direction of current?"},{"Start":"08:05.615 ","End":"08:09.530","Text":"Is just this surface area over here."},{"Start":"08:09.530 ","End":"08:13.190","Text":"This is the cross-sectional surface area. What is that?"},{"Start":"08:13.190 ","End":"08:17.060","Text":"It\u0027s just the cross-sectional area of a circle, which as we know,"},{"Start":"08:17.060 ","End":"08:22.260","Text":"is equal to Pi multiplied by the radius of the circle squared."},{"Start":"08:22.910 ","End":"08:26.520","Text":"Now we want to sum this up."},{"Start":"08:26.520 ","End":"08:31.620","Text":"We\u0027re just going to be adding these tiny resistors up."},{"Start":"08:31.970 ","End":"08:34.430","Text":"Before we add this up,"},{"Start":"08:34.430 ","End":"08:36.290","Text":"we have to know how to add them up."},{"Start":"08:36.290 ","End":"08:42.095","Text":"Are the resistors connected to one another in series or in parallel?"},{"Start":"08:42.095 ","End":"08:46.040","Text":"If we draw another one of these discs,"},{"Start":"08:46.040 ","End":"08:48.915","Text":"like so, it will be easier."},{"Start":"08:48.915 ","End":"08:52.070","Text":"Now we can imagine that this is connected by a wire."},{"Start":"08:52.070 ","End":"08:56.870","Text":"Then we can see that we have a resistor and then a wire and a resistor."},{"Start":"08:56.870 ","End":"09:01.640","Text":"It\u0027s as if we have a case like so."},{"Start":"09:01.640 ","End":"09:08.150","Text":"These two cases are exactly the same. What does that mean?"},{"Start":"09:08.150 ","End":"09:12.720","Text":"That means that these resistors are all connected in series."},{"Start":"09:12.720 ","End":"09:14.585","Text":"This is series."},{"Start":"09:14.585 ","End":"09:19.700","Text":"The total resistance is simply going to be"},{"Start":"09:19.700 ","End":"09:25.700","Text":"how we add all of these tiny resistors when they\u0027re connected in series."},{"Start":"09:25.700 ","End":"09:33.025","Text":"In series, we just sum each of their resistances."},{"Start":"09:33.025 ","End":"09:37.669","Text":"Because we\u0027re dealing with infinitesimal numbers,"},{"Start":"09:37.669 ","End":"09:42.505","Text":"so this summation sign becomes the integral."},{"Start":"09:42.505 ","End":"09:49.155","Text":"What we\u0027re integrating is along dR. Now we can plug that in."},{"Start":"09:49.155 ","End":"09:53.310","Text":"We\u0027re integrating along Rho naught zdz"},{"Start":"09:53.310 ","End":"10:00.460","Text":", divided by hPir^2."},{"Start":"10:00.460 ","End":"10:05.150","Text":"The bounds of this integral are from this height over here, which is 0,"},{"Start":"10:05.150 ","End":"10:08.315","Text":"until the top of the cylindrical resistor,"},{"Start":"10:08.315 ","End":"10:12.035","Text":"which is h. Now,"},{"Start":"10:12.035 ","End":"10:14.190","Text":"all we have to do is do this integrals."},{"Start":"10:14.190 ","End":"10:19.460","Text":"What we\u0027ll have is that this is equal to Rho divided by"},{"Start":"10:19.460 ","End":"10:26.745","Text":"Pir^2 multiplied by h^2 divided by 2h."},{"Start":"10:26.745 ","End":"10:29.100","Text":"This will cancel out with this."},{"Start":"10:29.100 ","End":"10:38.820","Text":"What we\u0027ll have is Rho_0 h divided by 2Pir^2."},{"Start":"10:38.820 ","End":"10:43.540","Text":"This is the total resistance of this cylindrical resistor when we were told"},{"Start":"10:43.540 ","End":"10:48.625","Text":"at the beginning that its resistivity is not uniform throughout."},{"Start":"10:48.625 ","End":"10:52.435","Text":"The trick that we used in order to calculate the resistance"},{"Start":"10:52.435 ","End":"10:56.740","Text":"is we split up our total resistor so that"},{"Start":"10:56.740 ","End":"11:00.400","Text":"each disk would be so thin that we could"},{"Start":"11:00.400 ","End":"11:05.715","Text":"consider the resistivity across it as being uniform."},{"Start":"11:05.715 ","End":"11:08.925","Text":"Then we applied this equation."},{"Start":"11:08.925 ","End":"11:12.400","Text":"Then we saw how each of these resistors was"},{"Start":"11:12.400 ","End":"11:16.715","Text":"joined up to all of the others that make up the rest of the resistor."},{"Start":"11:16.715 ","End":"11:22.155","Text":"Then that gave us the way to sum this all up."},{"Start":"11:22.155 ","End":"11:23.870","Text":"If it\u0027s in series or in parallel,"},{"Start":"11:23.870 ","End":"11:26.465","Text":"there\u0027s different ways for summing up the resistors."},{"Start":"11:26.465 ","End":"11:31.115","Text":"Then we just integrated and found the total resistance."},{"Start":"11:31.115 ","End":"11:36.725","Text":"Then if this is a regular resistor that abides by Ohm\u0027s law,"},{"Start":"11:36.725 ","End":"11:41.900","Text":"then if we\u0027re told that the resistor is then connected"},{"Start":"11:41.900 ","End":"11:48.225","Text":"to some voltage source like so,"},{"Start":"11:48.225 ","End":"11:50.505","Text":"voltage source V naught."},{"Start":"11:50.505 ","End":"11:52.740","Text":"Then from the equation,"},{"Start":"11:52.740 ","End":"11:55.215","Text":"V is equal to IR,"},{"Start":"11:55.215 ","End":"11:57.285","Text":"we could calculate the current."},{"Start":"11:57.285 ","End":"12:04.770","Text":"We can say that the current is just equal to the voltage divided by the resistance,"},{"Start":"12:04.770 ","End":"12:07.545","Text":"which is what we just calculated."},{"Start":"12:07.545 ","End":"12:10.790","Text":"From just these details that we had at the beginning,"},{"Start":"12:10.790 ","End":"12:17.790","Text":"we can calculate both the resistance and the current passing through this resistor."},{"Start":"12:17.790 ","End":"12:21.575","Text":"One final thing before we finish this lesson,"},{"Start":"12:21.575 ","End":"12:23.870","Text":"I\u0027m just going to square this as well,"},{"Start":"12:23.870 ","End":"12:29.165","Text":"just so you remember that we can use this equation to calculate the current."},{"Start":"12:29.165 ","End":"12:34.039","Text":"The last thing that there is to remember is that there\u0027s something called Sigma,"},{"Start":"12:34.039 ","End":"12:38.930","Text":"which is conductivity, which is just the reciprocal of the resistivity."},{"Start":"12:38.930 ","End":"12:41.420","Text":"It\u0027s 1 divided by Rho,"},{"Start":"12:41.420 ","End":"12:44.275","Text":"where Rho is the resistivity."},{"Start":"12:44.275 ","End":"12:47.490","Text":"They basically mean the same thing,"},{"Start":"12:47.490 ","End":"12:50.250","Text":"but Rho is resistivity."},{"Start":"12:50.250 ","End":"12:53.090","Text":"The higher it is, the better the resistor it is."},{"Start":"12:53.090 ","End":"12:56.045","Text":"Then Sigma, which is the reciprocal of resistivity,"},{"Start":"12:56.045 ","End":"12:59.780","Text":"shows us how good a conductor a material is."},{"Start":"12:59.780 ","End":"13:03.845","Text":"Obviously, if the resistivity is very low,"},{"Start":"13:03.845 ","End":"13:06.740","Text":"the conductivity will be very high,"},{"Start":"13:06.740 ","End":"13:09.883","Text":"and therefore the resistance will be low,"},{"Start":"13:09.883 ","End":"13:13.490","Text":"and therefore it will be a better conductor."},{"Start":"13:13.490 ","End":"13:16.650","Text":"That\u0027s the end of this lesson."}],"ID":22414},{"Watched":false,"Name":"Current Density and Electric Field in Resistor","Duration":"17m 28s","ChapterTopicVideoID":21516,"CourseChapterTopicPlaylistID":99491,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.235","Text":"Hello. In the previous lesson,"},{"Start":"00:02.235 ","End":"00:08.190","Text":"we looked at a resistor with this type of shape and we spoke about"},{"Start":"00:08.190 ","End":"00:14.340","Text":"how we can calculate the resistance of the resistor with the help of our equation,"},{"Start":"00:14.340 ","End":"00:18.810","Text":"which was like so that the resistance is equal to the resistivity"},{"Start":"00:18.810 ","End":"00:21.495","Text":"multiplied by the length of the resistor"},{"Start":"00:21.495 ","End":"00:25.350","Text":"divided by the cross-sectional area of the resistor."},{"Start":"00:25.350 ","End":"00:29.250","Text":"We also spoke about how the length and the area is"},{"Start":"00:29.250 ","End":"00:33.180","Text":"dependent on the direction that the current is traveling."},{"Start":"00:33.180 ","End":"00:38.040","Text":"For instance, if we have a current which is traveling in this direction,"},{"Start":"00:38.040 ","End":"00:42.945","Text":"so it will go through here and out through here."},{"Start":"00:42.945 ","End":"00:48.460","Text":"In that case, L would be equal to this length over here,"},{"Start":"00:48.460 ","End":"00:52.375","Text":"which is C and the cross-sectional area"},{"Start":"00:52.375 ","End":"00:56.845","Text":"A would be equal to the area of this plane over here,"},{"Start":"00:56.845 ","End":"01:00.155","Text":"which is equal to a times b."},{"Start":"01:00.155 ","End":"01:06.280","Text":"What we\u0027re going to do in this lesson is we\u0027re going to learn about current density,"},{"Start":"01:06.280 ","End":"01:10.520","Text":"which is denoted by the letter J."},{"Start":"01:10.980 ","End":"01:17.680","Text":"Current density is how much current is passing through a unit of area."},{"Start":"01:17.680 ","End":"01:20.860","Text":"In this lesson, we\u0027re going to speak a little bit more"},{"Start":"01:20.860 ","End":"01:25.125","Text":"about what that means and the importance of a current density."},{"Start":"01:25.125 ","End":"01:29.250","Text":"But first let\u0027s speak about what current actually is."},{"Start":"01:29.660 ","End":"01:32.340","Text":"Let\u0027s speak about current."},{"Start":"01:32.340 ","End":"01:40.765","Text":"Let\u0027s say that we have current which is now traveling like so in this direction."},{"Start":"01:40.765 ","End":"01:43.710","Text":"This is our current passing through."},{"Start":"01:43.710 ","End":"01:47.180","Text":"We said that in order to use this equation over here,"},{"Start":"01:47.180 ","End":"01:51.460","Text":"we have to have a constant Rho,"},{"Start":"01:51.460 ","End":"01:55.820","Text":"constant resistivity throughout and a constant cross-sectional area."},{"Start":"01:55.820 ","End":"02:01.520","Text":"If a certain amount of current is passing through this cross-sectional area,"},{"Start":"02:01.520 ","End":"02:06.836","Text":"then the same amount of current is going to be passing through here"},{"Start":"02:06.836 ","End":"02:14.125","Text":"because this is the same cross-sectional area as this."},{"Start":"02:14.125 ","End":"02:23.210","Text":"Current is defined to be the number of charges passing through an area in a unit of time."},{"Start":"02:23.210 ","End":"02:29.485","Text":"Let\u0027s say how many charges pass through this cross-sectional area in 1 second?"},{"Start":"02:29.485 ","End":"02:32.170","Text":"That is current."},{"Start":"02:32.270 ","End":"02:36.485","Text":"If we wrote out this statement as an equation,"},{"Start":"02:36.485 ","End":"02:41.360","Text":"so that would mean that the current is equal to the number of charges"},{"Start":"02:41.360 ","End":"02:47.740","Text":"passing so dq in a unit of time, dt."},{"Start":"02:47.840 ","End":"02:50.265","Text":"This is this."},{"Start":"02:50.265 ","End":"02:56.814","Text":"If we had someone standing over here and"},{"Start":"02:56.814 ","End":"03:04.525","Text":"she\u0027s watching how many charges pass through this cross-sectional area in 1 second."},{"Start":"03:04.525 ","End":"03:10.000","Text":"What she\u0027ll do is she\u0027ll count the number of charges."},{"Start":"03:10.000 ","End":"03:16.295","Text":"What we can see is that current is a scalar product."},{"Start":"03:16.295 ","End":"03:18.640","Text":"Or this is a scalar function."},{"Start":"03:18.640 ","End":"03:20.155","Text":"This isn\u0027t a vector."},{"Start":"03:20.155 ","End":"03:21.790","Text":"We don\u0027t have direction,"},{"Start":"03:21.790 ","End":"03:25.240","Text":"we just have size or magnitude."},{"Start":"03:25.240 ","End":"03:30.634","Text":"Current is a scalar product."},{"Start":"03:30.634 ","End":"03:34.230","Text":"Well, or rather a scalar quantity."},{"Start":"03:34.230 ","End":"03:40.180","Text":"Now if I look at this cross-sectional area, again over here."},{"Start":"03:40.180 ","End":"03:43.910","Text":"This is the same cross-sectional area."},{"Start":"03:45.080 ","End":"03:48.345","Text":"What is current density?"},{"Start":"03:48.345 ","End":"03:50.385","Text":"With current density,"},{"Start":"03:50.385 ","End":"03:54.270","Text":"we\u0027ll take a small area from here."},{"Start":"03:54.270 ","End":"03:56.385","Text":"Let\u0027s say this."},{"Start":"03:56.385 ","End":"04:00.735","Text":"This is one point in the cross-sectional area."},{"Start":"04:00.735 ","End":"04:04.269","Text":"Let\u0027s call the area of this square ds."},{"Start":"04:04.269 ","End":"04:07.160","Text":"It\u0027s a very, very small area."},{"Start":"04:07.160 ","End":"04:13.915","Text":"What we\u0027ll do is we\u0027ll see how many charges are passing through this specific point."},{"Start":"04:13.915 ","End":"04:18.980","Text":"If we just have one charge which is passing through,"},{"Start":"04:18.980 ","End":"04:20.780","Text":"so that\u0027s what that means."},{"Start":"04:20.780 ","End":"04:26.310","Text":"We\u0027re counting how many charges pass through a point on the cross-sectional area."},{"Start":"04:26.310 ","End":"04:28.880","Text":"J or current density,"},{"Start":"04:28.880 ","End":"04:31.625","Text":"is how much current is passing through this point,"},{"Start":"04:31.625 ","End":"04:34.535","Text":"or through this point, or this point,"},{"Start":"04:34.535 ","End":"04:39.630","Text":"and so on and so forth in this cross-sectional area."},{"Start":"04:39.920 ","End":"04:46.595","Text":"Now what we want to do is find the connection between current and current density."},{"Start":"04:46.595 ","End":"04:54.350","Text":"Current is equal to the integral along J, the current density,"},{"Start":"04:54.350 ","End":"04:58.310","Text":"which is a vector quantity because we\u0027re looking at"},{"Start":"04:58.310 ","End":"05:05.375","Text":"the direction that the charges are traveling in dot product with ds,"},{"Start":"05:05.375 ","End":"05:11.720","Text":"where ds is this small area over here and ds,"},{"Start":"05:11.720 ","End":"05:18.330","Text":"the direction of ds is of course normal to the cross-sectional area."},{"Start":"05:19.060 ","End":"05:23.820","Text":"We can see that over here we have vectors."},{"Start":"05:23.820 ","End":"05:25.340","Text":"But our current of course,"},{"Start":"05:25.340 ","End":"05:26.915","Text":"is still a scalar."},{"Start":"05:26.915 ","End":"05:31.280","Text":"What we want to do is we want to take the component of J,"},{"Start":"05:31.280 ","End":"05:33.425","Text":"which is traveling through"},{"Start":"05:33.425 ","End":"05:39.320","Text":"our cross-sectional area that isn\u0027t traveling parallel to our cross-sectional area."},{"Start":"05:39.320 ","End":"05:43.565","Text":"Because we know that the current has to travel through here."},{"Start":"05:43.565 ","End":"05:46.280","Text":"We take the component of J,"},{"Start":"05:46.280 ","End":"05:53.660","Text":"which is perpendicular to the cross sectional area and then when we sum up on our J ds,"},{"Start":"05:53.660 ","End":"05:55.190","Text":"that means we\u0027re summing up on all of"},{"Start":"05:55.190 ","End":"05:58.100","Text":"the charges that are passing through every single point"},{"Start":"05:58.100 ","End":"06:05.340","Text":"in this cross-sectional area and of course that leads us to the definition of current."},{"Start":"06:05.780 ","End":"06:11.420","Text":"Write down these 2 equations in your equation sheets."},{"Start":"06:11.420 ","End":"06:17.945","Text":"If J is a constant throughout as in an every point along this cross-sectional area,"},{"Start":"06:17.945 ","End":"06:21.700","Text":"the same amount of charges are passing."},{"Start":"06:21.700 ","End":"06:24.455","Text":"If J is equal to a constant,"},{"Start":"06:24.455 ","End":"06:32.460","Text":"then this equation becomes I is equal to J.S."},{"Start":"06:32.460 ","End":"06:37.599","Text":"This is only if J is a constant and then the integral just becomes"},{"Start":"06:37.599 ","End":"06:43.415","Text":"J multiplied by S. Of course later on in this chapter,"},{"Start":"06:43.415 ","End":"06:48.970","Text":"we\u0027re going to learn different equations for calculating our J."},{"Start":"06:48.970 ","End":"06:51.710","Text":"But I\u0027m not going to speak about that now."},{"Start":"06:51.710 ","End":"06:57.575","Text":"What we\u0027re going to do for the rest of this lesson is we\u0027re going to look at an example."},{"Start":"06:57.575 ","End":"06:59.510","Text":"In the previous lesson,"},{"Start":"06:59.510 ","End":"07:04.370","Text":"we saw this solid cylinder of radius r and of height h,"},{"Start":"07:04.370 ","End":"07:11.750","Text":"where we had this non uniform value for resistivity and we calculated the resistance"},{"Start":"07:11.750 ","End":"07:20.340","Text":"of this resistor over here and we got that it is equal to Rho_0 h divided by 2 Pi r^2."},{"Start":"07:20.340 ","End":"07:26.180","Text":"Then we said what would happen if we connected this resistor to a voltage source."},{"Start":"07:26.180 ","End":"07:31.745","Text":"We said that as long as this resistor acts like a regular ohmic resistor,"},{"Start":"07:31.745 ","End":"07:36.125","Text":"then we can use Ohm\u0027s law and we can say therefore that V_0,"},{"Start":"07:36.125 ","End":"07:38.930","Text":"the voltage of the voltage source,"},{"Start":"07:38.930 ","End":"07:44.525","Text":"is equal to our current multiplied by our total resistance,"},{"Start":"07:44.525 ","End":"07:48.815","Text":"which we already calculated and this we know from the question."},{"Start":"07:48.815 ","End":"07:50.510","Text":"Therefore from this equation,"},{"Start":"07:50.510 ","End":"07:53.840","Text":"we could calculate the current flowing through the resistor as"},{"Start":"07:53.840 ","End":"07:59.260","Text":"being V note divided by R total."},{"Start":"08:00.650 ","End":"08:05.515","Text":"Now we\u0027re going to calculate our J, so what is our J?"},{"Start":"08:05.515 ","End":"08:07.285","Text":"Our current density."},{"Start":"08:07.285 ","End":"08:10.775","Text":"Current is passing through here."},{"Start":"08:10.775 ","End":"08:18.040","Text":"The same current that is passing through this area over here is going to"},{"Start":"08:18.040 ","End":"08:25.610","Text":"be the same current that will pass through this cross-sectional area over here."},{"Start":"08:25.610 ","End":"08:31.540","Text":"Because our cross-sectional area in this specific question is uniform."},{"Start":"08:31.540 ","End":"08:32.980","Text":"What does that mean?"},{"Start":"08:32.980 ","End":"08:38.334","Text":"That means that our charge density is also going to be uniform."},{"Start":"08:38.334 ","End":"08:41.920","Text":"That means that the same amount of charges that are passing"},{"Start":"08:41.920 ","End":"08:46.730","Text":"through this disk over here is going to"},{"Start":"08:46.730 ","End":"08:50.150","Text":"be the exact same number of charges"},{"Start":"08:50.150 ","End":"08:54.740","Text":"passing through every point or so on this disk over here."},{"Start":"08:54.740 ","End":"09:01.370","Text":"You can just take this disk and copy and paste it throughout the entire cylinder."},{"Start":"09:01.370 ","End":"09:03.800","Text":"In that case, if we want to calculate J,"},{"Start":"09:03.800 ","End":"09:08.485","Text":"we can assume in these types of questions that J is going to be a constant."},{"Start":"09:08.485 ","End":"09:10.970","Text":"What is J going to be equal to?"},{"Start":"09:10.970 ","End":"09:13.969","Text":"J is equal to from this equation,"},{"Start":"09:13.969 ","End":"09:17.960","Text":"our current divided by our cross-sectional area."},{"Start":"09:17.960 ","End":"09:21.860","Text":"Current divided by our cross-sectional area,"},{"Start":"09:21.860 ","End":"09:27.785","Text":"which is going to be equal to V_0 divided by the total resistance."},{"Start":"09:27.785 ","End":"09:32.120","Text":"This is our current, divided by the cross-sectional area of a circle,"},{"Start":"09:32.120 ","End":"09:36.240","Text":"which we know is equal to Pi r^2."},{"Start":"09:36.860 ","End":"09:40.325","Text":"Once you substitute in our total,"},{"Start":"09:40.325 ","End":"09:42.710","Text":"this will be your value for J,"},{"Start":"09:42.710 ","End":"09:46.985","Text":"2 multiplied by the voltage of the voltage source divided by"},{"Start":"09:46.985 ","End":"09:53.025","Text":"Rho_0 h. This is the answer for J."},{"Start":"09:53.025 ","End":"09:58.655","Text":"Now another equation for calculating J, the current density,"},{"Start":"09:58.655 ","End":"10:02.825","Text":"is by saying that J,"},{"Start":"10:02.825 ","End":"10:05.640","Text":"I\u0027ll write it over here."},{"Start":"10:05.870 ","End":"10:11.670","Text":"That J is equal to Sigma."},{"Start":"10:11.670 ","End":"10:18.150","Text":"Remember where we said that Sigma is equal to the conductivity."},{"Start":"10:18.870 ","End":"10:22.360","Text":"Our conductivity if we remember,"},{"Start":"10:22.360 ","End":"10:27.940","Text":"Sigma is equal to 1 divided by our resistivity."},{"Start":"10:27.940 ","End":"10:30.400","Text":"This we also saw in the previous lesson."},{"Start":"10:30.400 ","End":"10:34.600","Text":"J is equal to the conductivity or the reciprocal of"},{"Start":"10:34.600 ","End":"10:40.990","Text":"resistivity multiplied by the electric field."},{"Start":"10:40.990 ","End":"10:45.250","Text":"We can see that the current density is connected to"},{"Start":"10:45.250 ","End":"10:49.765","Text":"the electric field inside our resistor."},{"Start":"10:49.765 ","End":"10:52.210","Text":"Now, what is weird in the previous lesson,"},{"Start":"10:52.210 ","End":"10:56.785","Text":"we spoke about how a resistor is a conductor."},{"Start":"10:56.785 ","End":"10:59.200","Text":"It\u0027s a poor conductor,"},{"Start":"10:59.200 ","End":"11:05.170","Text":"but it still has to be a type of conductor because in order to resist current,"},{"Start":"11:05.170 ","End":"11:07.960","Text":"it has to be able to allow some current through,"},{"Start":"11:07.960 ","End":"11:10.405","Text":"meaning it\u0027s a resistor."},{"Start":"11:10.405 ","End":"11:12.850","Text":"In one of the previous chapters,"},{"Start":"11:12.850 ","End":"11:16.480","Text":"we spoke about how if we\u0027re dealing with a conductor,"},{"Start":"11:16.480 ","End":"11:21.340","Text":"then the electric field inside the conductor is going to be equal to 0."},{"Start":"11:21.340 ","End":"11:24.910","Text":"How can we have this equation over here,"},{"Start":"11:24.910 ","End":"11:29.015","Text":"which is connected to the electric field inside a conductor?"},{"Start":"11:29.015 ","End":"11:32.115","Text":"The electric field inside a conductor is"},{"Start":"11:32.115 ","End":"11:36.370","Text":"equal to 0 when we\u0027re dealing with electrostatics."},{"Start":"11:36.920 ","End":"11:45.250","Text":"The E field is equal to 0 in a conductor when we were dealing with electrostatics."},{"Start":"11:45.250 ","End":"11:48.205","Text":"But here we\u0027re talking about charges moving,"},{"Start":"11:48.205 ","End":"11:50.545","Text":"that\u0027s why we have a current."},{"Start":"11:50.545 ","End":"11:53.845","Text":"It means that there are charges moving."},{"Start":"11:53.845 ","End":"11:57.370","Text":"In that case, if charges are moving and we have a current,"},{"Start":"11:57.370 ","End":"12:00.805","Text":"that means that we have electrodynamics."},{"Start":"12:00.805 ","End":"12:07.430","Text":"Then electrodynamics, the E field does not equal to 0 in a conductor."},{"Start":"12:08.130 ","End":"12:10.975","Text":"Just remember if there\u0027s a current,"},{"Start":"12:10.975 ","End":"12:12.970","Text":"that means that charges are moving,"},{"Start":"12:12.970 ","End":"12:16.870","Text":"that means that we\u0027re dealing with electrodynamics and that means that the E field in"},{"Start":"12:16.870 ","End":"12:22.645","Text":"a conductor is some value which is non-zero."},{"Start":"12:22.645 ","End":"12:29.630","Text":"Therefore, we can see that we will get some value for the current density."},{"Start":"12:31.470 ","End":"12:37.030","Text":"Also, write this down in your equation sheets."},{"Start":"12:37.030 ","End":"12:43.180","Text":"The current density is equal to the conductivity multiplied by the E field."},{"Start":"12:43.180 ","End":"12:45.280","Text":"Now, given that,"},{"Start":"12:45.280 ","End":"12:49.840","Text":"let\u0027s calculate the E field in this resistor over here."},{"Start":"12:49.840 ","End":"12:54.310","Text":"From this equation, we can see that our E field is"},{"Start":"12:54.310 ","End":"12:59.900","Text":"going to be equal to J divided by Sigma."},{"Start":"13:00.000 ","End":"13:08.140","Text":"We have over here that this is 1 divided by Sigma multiplied by J."},{"Start":"13:08.140 ","End":"13:09.940","Text":"What\u0027s 1 divided by Sigma?"},{"Start":"13:09.940 ","End":"13:11.815","Text":"From this equation we can see over here,"},{"Start":"13:11.815 ","End":"13:19.165","Text":"it\u0027s equal to Rho or the resistivity multiplied by J."},{"Start":"13:19.165 ","End":"13:27.700","Text":"In that case, we\u0027ll get that our E field within the resistor is equal to our resistivity,"},{"Start":"13:27.700 ","End":"13:29.395","Text":"which we were given in the question,"},{"Start":"13:29.395 ","End":"13:37.255","Text":"is equal to Rho_0 z divided by h multiplied by J,"},{"Start":"13:37.255 ","End":"13:43.240","Text":"where our J is equal to 2V_0 divided by"},{"Start":"13:43.240 ","End":"13:51.325","Text":"Rho_0 h. The direction is of course the direction of the current,"},{"Start":"13:51.325 ","End":"13:55.570","Text":"which in this case over here is in the z direction."},{"Start":"13:55.570 ","End":"14:03.160","Text":"We\u0027ll get that this is the electric field within our resistor."},{"Start":"14:03.160 ","End":"14:10.930","Text":"Then of course we know that if we integrate this E field along the route,"},{"Start":"14:10.930 ","End":"14:13.885","Text":"so along the z-axis,"},{"Start":"14:13.885 ","End":"14:19.090","Text":"we will get the potential difference of our voltage source."},{"Start":"14:19.090 ","End":"14:21.820","Text":"We will get what V_0 is equal to."},{"Start":"14:21.820 ","End":"14:29.150","Text":"V_0 is equal to the integral of E dot dz,"},{"Start":"14:29.250 ","End":"14:33.040","Text":"just as would be expected,"},{"Start":"14:33.040 ","End":"14:37.150","Text":"and we will get the same value for V_0."},{"Start":"14:37.150 ","End":"14:40.960","Text":"You can pause the video and try it now."},{"Start":"14:40.960 ","End":"14:47.090","Text":"Of course, there are other equations for our current density."},{"Start":"14:47.670 ","End":"14:53.740","Text":"The first new equation for current density that we\u0027re going to look at is this."},{"Start":"14:53.740 ","End":"14:56.500","Text":"Now, please be careful over here."},{"Start":"14:56.500 ","End":"14:59.485","Text":"I put a star, exclamation marks."},{"Start":"14:59.485 ","End":"15:02.635","Text":"Over here the current density is equal to Rho,"},{"Start":"15:02.635 ","End":"15:06.445","Text":"but this Rho is charged density."},{"Start":"15:06.445 ","End":"15:08.100","Text":"Specifically in this equation,"},{"Start":"15:08.100 ","End":"15:10.390","Text":"we\u0027re speaking about charge density Rho,"},{"Start":"15:10.390 ","End":"15:15.460","Text":"where we\u0027ve already spoken about Rho as resistivity."},{"Start":"15:15.460 ","End":"15:17.020","Text":"These are 2 different Rhos."},{"Start":"15:17.020 ","End":"15:20.020","Text":"They\u0027re just symbolized with the same thing."},{"Start":"15:20.020 ","End":"15:24.295","Text":"But just remember in this equation over here,"},{"Start":"15:24.295 ","End":"15:27.415","Text":"this Rho symbolizes charged density,"},{"Start":"15:27.415 ","End":"15:30.925","Text":"very important, multiplied by v,"},{"Start":"15:30.925 ","End":"15:33.145","Text":"which over here is velocity."},{"Start":"15:33.145 ","End":"15:35.740","Text":"Not to be confused with voltage, which again,"},{"Start":"15:35.740 ","End":"15:38.995","Text":"we\u0027ve been using previously in this chapter."},{"Start":"15:38.995 ","End":"15:40.960","Text":"But here we\u0027re speaking about velocity."},{"Start":"15:40.960 ","End":"15:50.545","Text":"We have that current density is equal to charge density multiplied by velocity."},{"Start":"15:50.545 ","End":"15:53.035","Text":"Don\u0027t get confused."},{"Start":"15:53.035 ","End":"15:58.645","Text":"If we have some charge and we move it at a certain velocity,"},{"Start":"15:58.645 ","End":"16:01.915","Text":"then we\u0027re going to get current density."},{"Start":"16:01.915 ","End":"16:04.780","Text":"When we\u0027re dealing with a conductor,"},{"Start":"16:04.780 ","End":"16:09.700","Text":"this equation is equal to n multiplied by q"},{"Start":"16:09.700 ","End":"16:16.014","Text":"multiplied by v. This is specifically for a conductor."},{"Start":"16:16.014 ","End":"16:19.180","Text":"This equation isn\u0027t very useful for this course,"},{"Start":"16:19.180 ","End":"16:20.530","Text":"but just so that you know it,"},{"Start":"16:20.530 ","End":"16:22.075","Text":"have it at the back of the head,"},{"Start":"16:22.075 ","End":"16:23.650","Text":"be a bit familiar with it."},{"Start":"16:23.650 ","End":"16:27.580","Text":"We have n, which is the number of charges that can move."},{"Start":"16:27.580 ","End":"16:33.340","Text":"If we remember we have a conductor and not all of the charges are dynamics."},{"Start":"16:33.340 ","End":"16:37.090","Text":"Some of them are tied or stuck in a certain positions."},{"Start":"16:37.090 ","End":"16:40.810","Text":"N refers to the number of charges that can move,"},{"Start":"16:40.810 ","End":"16:46.360","Text":"q is of course the charge and V again is the velocity."},{"Start":"16:46.360 ","End":"16:49.195","Text":"You can also call it V drift."},{"Start":"16:49.195 ","End":"16:51.061","Text":"That just means the drift velocity,"},{"Start":"16:51.061 ","End":"16:57.860","Text":"that\u0027s just the velocity with which the charges move through the material."},{"Start":"16:58.440 ","End":"17:01.810","Text":"This is the equation that you really have to write"},{"Start":"17:01.810 ","End":"17:04.914","Text":"down as well as the others that I squared."},{"Start":"17:04.914 ","End":"17:07.690","Text":"This is of course just the answer to"},{"Start":"17:07.690 ","End":"17:10.690","Text":"the question so don\u0027t copy this into your equation sheets."},{"Start":"17:10.690 ","End":"17:14.785","Text":"But the ones that are in proper squares copy them out."},{"Start":"17:14.785 ","End":"17:18.190","Text":"We\u0027re going to look at some examples that use this equation and"},{"Start":"17:18.190 ","End":"17:22.225","Text":"the other equations for current density."},{"Start":"17:22.225 ","End":"17:26.305","Text":"This is just another equation that you can have at the back of your mind."},{"Start":"17:26.305 ","End":"17:29.030","Text":"That\u0027s the end of this lesson."}],"ID":22415},{"Watched":false,"Name":"Exercise 1","Duration":"13m 2s","ChapterTopicVideoID":21517,"CourseChapterTopicPlaylistID":99491,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.800","Text":"Hello. In this lesson,"},{"Start":"00:01.800 ","End":"00:03.810","Text":"we\u0027re going to be answering the following question."},{"Start":"00:03.810 ","End":"00:09.750","Text":"A cylindrical shell of inner radius a and outer radius b is given,"},{"Start":"00:09.750 ","End":"00:12.045","Text":"between 0 and a,"},{"Start":"00:12.045 ","End":"00:16.065","Text":"the space is empty and between a and b,"},{"Start":"00:16.065 ","End":"00:21.370","Text":"the cylinder is filled with a material of given resistivity Rho."},{"Start":"00:21.370 ","End":"00:26.685","Text":"A voltage source V_0 is connected to the cylindrical shell as shown,"},{"Start":"00:26.685 ","End":"00:29.310","Text":"where the positive side is connected to"},{"Start":"00:29.310 ","End":"00:33.390","Text":"the inner wall and the negative side is connected to the outer wall."},{"Start":"00:33.390 ","End":"00:38.800","Text":"Question number 1 is to calculate the total resistance."},{"Start":"00:39.320 ","End":"00:45.799","Text":"We remember that our equation for resistance is the resistivity multiplied"},{"Start":"00:45.799 ","End":"00:51.780","Text":"by the length of the resistor divided by the cross-sectional area."},{"Start":"00:51.780 ","End":"00:55.205","Text":"We said that we can use this equation in its current form,"},{"Start":"00:55.205 ","End":"01:01.850","Text":"if our resistivity is a constant and if our cross-sectional area is a constant."},{"Start":"01:01.850 ","End":"01:06.665","Text":"In the question, we\u0027re being told that our resistivity is a constant."},{"Start":"01:06.665 ","End":"01:10.070","Text":"However, let\u0027s take a look at our cross-sectional area."},{"Start":"01:10.070 ","End":"01:18.830","Text":"We can see that we\u0027re going through the cross-sectional area of a cylinder,"},{"Start":"01:18.830 ","End":"01:24.785","Text":"so we go through this inner shell over here,"},{"Start":"01:24.785 ","End":"01:29.800","Text":"and then we keep on traveling through until we reach the outer shell."},{"Start":"01:29.800 ","End":"01:33.470","Text":"We can see that the area of the inner shell and the outer shell,"},{"Start":"01:33.470 ","End":"01:36.650","Text":"are the surface areas of a cylinder."},{"Start":"01:36.650 ","End":"01:39.860","Text":"What is the surface area of a cylinder?"},{"Start":"01:39.860 ","End":"01:43.780","Text":"We know that it is equal to,"},{"Start":"01:47.450 ","End":"01:51.075","Text":"just the circumference of a circle,"},{"Start":"01:51.075 ","End":"01:55.265","Text":"rather 2Pir, because we\u0027re dealing with a shell,"},{"Start":"01:55.265 ","End":"01:57.755","Text":"so we\u0027re looking at, let\u0027s say,"},{"Start":"01:57.755 ","End":"02:00.826","Text":"this over here,"},{"Start":"02:00.826 ","End":"02:04.715","Text":"what I\u0027ve drawn over here in black."},{"Start":"02:04.715 ","End":"02:09.260","Text":"We can see that it\u0027s the surface area around the cylinder."},{"Start":"02:09.260 ","End":"02:13.605","Text":"It\u0027s 2Pir, the circumference of a circle, so that\u0027s that."},{"Start":"02:13.605 ","End":"02:17.605","Text":"Then we submit all along the height of the cylinder."},{"Start":"02:17.605 ","End":"02:23.700","Text":"Let\u0027s imagine that the cylinder has a height of h,"},{"Start":"02:24.350 ","End":"02:27.990","Text":"so this is the surface area."},{"Start":"02:27.990 ","End":"02:31.565","Text":"We can see that if we\u0027re specifically looking at"},{"Start":"02:31.565 ","End":"02:35.805","Text":"the surface area of what I\u0027ve colored over here in black,"},{"Start":"02:35.805 ","End":"02:38.160","Text":"our radius at this point is b,"},{"Start":"02:38.160 ","End":"02:39.880","Text":"so the surface area,"},{"Start":"02:39.880 ","End":"02:43.505","Text":"if we wrap something around the outer cylinder,"},{"Start":"02:43.505 ","End":"02:46.115","Text":"is going to be 2Pibh."},{"Start":"02:46.115 ","End":"02:53.255","Text":"However, if we wrap something around the inner cylindrical wall over here,"},{"Start":"02:53.255 ","End":"02:57.685","Text":"we take the surface area of this little cylinder inside,"},{"Start":"02:57.685 ","End":"03:03.680","Text":"our surface area will be 2Piah because the radius over here is a."},{"Start":"03:03.680 ","End":"03:09.810","Text":"What we can see is that our radius is constantly changing."},{"Start":"03:10.750 ","End":"03:14.060","Text":"This is also equal to a over here."},{"Start":"03:14.060 ","End":"03:16.790","Text":"If our radius is constantly changing,"},{"Start":"03:16.790 ","End":"03:22.470","Text":"we can see that our surface area is constantly changing as well."},{"Start":"03:23.360 ","End":"03:28.065","Text":"R is changing. Therefore,"},{"Start":"03:28.065 ","End":"03:32.850","Text":"A isn\u0027t equal to a constant."},{"Start":"03:32.850 ","End":"03:38.440","Text":"That means that we have to do something in order to be able to use this equation,"},{"Start":"03:38.440 ","End":"03:43.510","Text":"because this equation can only be used if Rho and A are constants."},{"Start":"03:43.510 ","End":"03:48.040","Text":"What we\u0027re going to do is the trick that we\u0027ve seen throughout this course,"},{"Start":"03:48.040 ","End":"03:50.840","Text":"is we\u0027re going to split"},{"Start":"03:51.020 ","End":"03:56.960","Text":"this thick cylindrical shell into many different cylindrical shells."},{"Start":"03:57.690 ","End":"04:00.430","Text":"Let\u0027s do this in green."},{"Start":"04:00.430 ","End":"04:04.560","Text":"We\u0027re going to take some random inner shell at"},{"Start":"04:04.560 ","End":"04:09.605","Text":"radius r. Give it some thickness over here,"},{"Start":"04:09.605 ","End":"04:17.380","Text":"where this thickness has to be extremely small and we\u0027ll call this thickness dr."},{"Start":"04:17.380 ","End":"04:21.710","Text":"It\u0027s a thickness which is approaching 0 and it\u0027s so"},{"Start":"04:21.710 ","End":"04:26.090","Text":"small that we can consider the radius between these 2 points,"},{"Start":"04:26.090 ","End":"04:27.560","Text":"so between the inner wall,"},{"Start":"04:27.560 ","End":"04:33.355","Text":"and between the outer wall as being the same because dr is so tiny."},{"Start":"04:33.355 ","End":"04:40.840","Text":"Of course, this is at radius r, which is changing."},{"Start":"04:40.880 ","End":"04:45.245","Text":"Now the next thing that we\u0027re going to look at is our voltage source."},{"Start":"04:45.245 ","End":"04:48.290","Text":"Given that the positive side is over here,"},{"Start":"04:48.290 ","End":"04:53.765","Text":"we know that our current is going to be traveling in this direction."},{"Start":"04:53.765 ","End":"04:57.590","Text":"Then what we\u0027re going to see is that the current is going to"},{"Start":"04:57.590 ","End":"05:01.673","Text":"go through like so in this direction,"},{"Start":"05:01.673 ","End":"05:03.658","Text":"and then out to this end,"},{"Start":"05:03.658 ","End":"05:07.060","Text":"and back to our voltage source."},{"Start":"05:07.060 ","End":"05:09.645","Text":"It\u0027s traveling like so."},{"Start":"05:09.645 ","End":"05:15.815","Text":"We can see that the current is traveling in a radial direction. We can draw it."},{"Start":"05:15.815 ","End":"05:17.270","Text":"Let\u0027s say over here,"},{"Start":"05:17.270 ","End":"05:21.580","Text":"it\u0027s going to be traveling like so,"},{"Start":"05:21.580 ","End":"05:27.440","Text":"and all the way until it hits the outer wall at radius b,"},{"Start":"05:27.440 ","End":"05:30.665","Text":"and then it travels back to the voltage source."},{"Start":"05:30.665 ","End":"05:34.745","Text":"Again, all of this is radial."},{"Start":"05:34.745 ","End":"05:41.170","Text":"It\u0027s happening all around 360 degrees around the cylinder."},{"Start":"05:41.170 ","End":"05:44.165","Text":"What we\u0027re going to do, the first thing,"},{"Start":"05:44.165 ","End":"05:49.130","Text":"is we\u0027re going to find the resistance of this inner cylinder."},{"Start":"05:49.130 ","End":"05:56.060","Text":"Of course, it continues down like so because it\u0027s a cylindrical shell."},{"Start":"05:56.060 ","End":"05:59.510","Text":"Like so."},{"Start":"05:59.510 ","End":"06:06.740","Text":"We\u0027re going to calculate the resistance of this."},{"Start":"06:06.740 ","End":"06:09.030","Text":"First all we\u0027ll call it dR,"},{"Start":"06:09.030 ","End":"06:13.005","Text":"because it\u0027s of the small, tiny cylinder."},{"Start":"06:13.005 ","End":"06:19.055","Text":"This is equal to the resistivity multiplied by the length. What is the length?"},{"Start":"06:19.055 ","End":"06:20.345","Text":"We spoke about it,"},{"Start":"06:20.345 ","End":"06:26.960","Text":"that the length of the path taken by the current through."},{"Start":"06:26.960 ","End":"06:32.365","Text":"We can see that the current is traveling through over here like so,"},{"Start":"06:32.365 ","End":"06:36.309","Text":"right here through this thin shell."},{"Start":"06:36.309 ","End":"06:39.290","Text":"Distance is the thickness of this thin shell,"},{"Start":"06:39.290 ","End":"06:41.310","Text":"which is dR,"},{"Start":"06:41.660 ","End":"06:46.560","Text":"and divided by the cross-sectional area."},{"Start":"06:46.560 ","End":"06:48.110","Text":"What is the area?"},{"Start":"06:48.110 ","End":"06:52.120","Text":"The area is just what we need to wrap,"},{"Start":"06:52.120 ","End":"06:57.380","Text":"or the surface area that we need to wrap this green cylinder with."},{"Start":"06:57.380 ","End":"06:59.885","Text":"That is going to be what we said over here,"},{"Start":"06:59.885 ","End":"07:01.805","Text":"it\u0027s equal to 2Pi,"},{"Start":"07:01.805 ","End":"07:06.305","Text":"where the radius of r multiplied by h,"},{"Start":"07:06.305 ","End":"07:09.720","Text":"the height of the cylinder."},{"Start":"07:10.940 ","End":"07:14.795","Text":"Now what do we want to do is we want to integrate"},{"Start":"07:14.795 ","End":"07:18.710","Text":"along this to sum up to find the total resistance."},{"Start":"07:18.710 ","End":"07:22.909","Text":"The first thing we need to do is we need to see how all of these cylinders,"},{"Start":"07:22.909 ","End":"07:26.660","Text":"if I sum up from inner radius a until outer radius b,"},{"Start":"07:26.660 ","End":"07:29.405","Text":"all of these cylindrical shells,"},{"Start":"07:29.405 ","End":"07:32.434","Text":"how they are connected to one another as resistors,"},{"Start":"07:32.434 ","End":"07:35.390","Text":"are they connected in parallel or in series?"},{"Start":"07:35.390 ","End":"07:38.620","Text":"Then I\u0027ll know how to do the integration."},{"Start":"07:38.620 ","End":"07:41.865","Text":"As we can see, the current will pass through,"},{"Start":"07:41.865 ","End":"07:44.715","Text":"let\u0027s draw just sections of the shells."},{"Start":"07:44.715 ","End":"07:46.715","Text":"The current will pass through here."},{"Start":"07:46.715 ","End":"07:48.050","Text":"This is the first resistor,"},{"Start":"07:48.050 ","End":"07:49.775","Text":"this is the second."},{"Start":"07:49.775 ","End":"07:51.540","Text":"Then the third,"},{"Start":"07:51.540 ","End":"07:54.475","Text":"and it\u0027s traveling just straight through all of them,"},{"Start":"07:54.475 ","End":"07:57.125","Text":"the voltage and the current doesn\u0027t split,"},{"Start":"07:57.125 ","End":"08:00.035","Text":"which means that they\u0027re connected in series."},{"Start":"08:00.035 ","End":"08:05.840","Text":"It has to be the same current that is flowing through each one of these tiny resistors."},{"Start":"08:05.840 ","End":"08:11.225","Text":"Which means that in order to calculate the total resistance,"},{"Start":"08:11.225 ","End":"08:14.150","Text":"we just have to sum this up."},{"Start":"08:14.150 ","End":"08:17.315","Text":"We add in our integration signs and, of course,"},{"Start":"08:17.315 ","End":"08:21.765","Text":"we\u0027re summing from where the resistivity material is."},{"Start":"08:21.765 ","End":"08:24.810","Text":"That is from a radius of a,"},{"Start":"08:24.810 ","End":"08:26.670","Text":"and we\u0027re integrating along the radius,"},{"Start":"08:26.670 ","End":"08:27.930","Text":"or from a radius of a,"},{"Start":"08:27.930 ","End":"08:30.550","Text":"until a radius of b."},{"Start":"08:30.560 ","End":"08:36.500","Text":"We\u0027re integrating all of this as constants aside from dr divided by r,"},{"Start":"08:36.500 ","End":"08:38.765","Text":"which as we know,"},{"Start":"08:38.765 ","End":"08:43.709","Text":"is in the shape we get when we integrate dr divided by r, we get ln."},{"Start":"08:43.709 ","End":"08:52.725","Text":"What we\u0027re going to have is Rho divided by 2Pih of ln,"},{"Start":"08:52.725 ","End":"08:57.410","Text":"and then we have ln of b divided by a."},{"Start":"08:57.410 ","End":"09:03.255","Text":"We\u0027ve seen how the rule of lns work in this type of situation,"},{"Start":"09:03.255 ","End":"09:10.285","Text":"so this is the total resistance of this resistor."},{"Start":"09:10.285 ","End":"09:13.730","Text":"This is the answer to question number 1."},{"Start":"09:13.730 ","End":"09:16.805","Text":"Now let\u0027s answer question number 2."},{"Start":"09:16.805 ","End":"09:19.700","Text":"Calculate the current density."},{"Start":"09:19.700 ","End":"09:22.130","Text":"We\u0027re trying to calculate J."},{"Start":"09:22.130 ","End":"09:26.810","Text":"First of all we can assume that this resistor abides by Ohm\u0027s law,"},{"Start":"09:26.810 ","End":"09:36.105","Text":"which means that the voltage source is equal to I multiplied by the resistance."},{"Start":"09:36.105 ","End":"09:46.685","Text":"Therefore, we can say that our current is equal to V naught divided by our resistance,"},{"Start":"09:46.685 ","End":"09:49.130","Text":"which we just calculated."},{"Start":"09:49.130 ","End":"09:53.350","Text":"Then in order to find the current density J,"},{"Start":"09:53.350 ","End":"09:55.940","Text":"first of all, it\u0027s a vector."},{"Start":"09:55.940 ","End":"10:04.240","Text":"This is equal to the current divided by the cross-sectional area,"},{"Start":"10:04.240 ","End":"10:06.130","Text":"which is what we saw over here,"},{"Start":"10:06.130 ","End":"10:09.115","Text":"it\u0027s the cross-sectional area of a cylindrical shell."},{"Start":"10:09.115 ","End":"10:11.695","Text":"Our I is given by this."},{"Start":"10:11.695 ","End":"10:17.865","Text":"Our I is equal to V naught divided by the resistance."},{"Start":"10:17.865 ","End":"10:22.355","Text":"What we\u0027re going to have is divided by"},{"Start":"10:22.355 ","End":"10:30.200","Text":"Rho ln of b over a divided by 2Pih,"},{"Start":"10:30.270 ","End":"10:35.750","Text":"and all of this is divided by 2Pirh."},{"Start":"10:37.970 ","End":"10:41.240","Text":"Of course, this is in the radial direction because"},{"Start":"10:41.240 ","End":"10:44.735","Text":"our current is in the radial direction."},{"Start":"10:44.735 ","End":"10:47.270","Text":"Then we can just rearrange this."},{"Start":"10:47.270 ","End":"10:53.100","Text":"What we get is V_0 multiplied by"},{"Start":"10:53.230 ","End":"11:05.050","Text":"2Pih divided by Rho ln b divided by a multiplied by 2Pirh,"},{"Start":"11:05.350 ","End":"11:10.025","Text":"so 2Pih, 2Pih can cancel out."},{"Start":"11:10.025 ","End":"11:15.325","Text":"What we\u0027re left with is that J is equal to V_0 divided by"},{"Start":"11:15.325 ","End":"11:22.455","Text":"Rho r ln of b divided by a."},{"Start":"11:22.455 ","End":"11:25.410","Text":"This is the answer to question number 2."},{"Start":"11:25.410 ","End":"11:32.070","Text":"This is the current density in the radial direction."},{"Start":"11:32.070 ","End":"11:33.520","Text":"Question number 3,"},{"Start":"11:33.520 ","End":"11:37.835","Text":"is to calculate the electric field inside of the conductor."},{"Start":"11:37.835 ","End":"11:41.885","Text":"We know that the equation for the electric field inside of the conductor"},{"Start":"11:41.885 ","End":"11:49.950","Text":"given resistivity is equal to Rho multiplied by the current density."},{"Start":"11:49.950 ","End":"11:54.195","Text":"That is simply going to be Rho multiplied by this."},{"Start":"11:54.195 ","End":"12:01.130","Text":"We have Rho V_0 divided by Rho r ln of b divided by a."},{"Start":"12:01.130 ","End":"12:02.930","Text":"We can cancel out the Rhos."},{"Start":"12:02.930 ","End":"12:07.090","Text":"We get that this is equal to V_0 divided by r,"},{"Start":"12:07.090 ","End":"12:10.300","Text":"ln b over a."},{"Start":"12:11.120 ","End":"12:16.715","Text":"Of course, the E field is also a vector field in the radial direction."},{"Start":"12:16.715 ","End":"12:19.325","Text":"That is the end of this lesson."},{"Start":"12:19.325 ","End":"12:24.830","Text":"What you just need to remember when calculating the resistance of this is to"},{"Start":"12:24.830 ","End":"12:30.307","Text":"notice that a is not uniform because r is constantly changing,"},{"Start":"12:30.307 ","End":"12:33.440","Text":"and therefore, you have to split this up into lots of"},{"Start":"12:33.440 ","End":"12:38.720","Text":"different cylindrical shells where we can consider the radius as uniform,"},{"Start":"12:38.720 ","End":"12:40.685","Text":"or as a constant throughout,"},{"Start":"12:40.685 ","End":"12:43.520","Text":"meaning that the cross-sectional area would be constant,"},{"Start":"12:43.520 ","End":"12:45.500","Text":"and then sum up,"},{"Start":"12:45.500 ","End":"12:51.050","Text":"not forgetting that when we sum up their resistances on each cylindrical shell,"},{"Start":"12:51.050 ","End":"12:56.135","Text":"we have to note how the cylindrical shells are connected to one another."},{"Start":"12:56.135 ","End":"12:57.620","Text":"In this example, specifically,"},{"Start":"12:57.620 ","End":"12:59.915","Text":"it was in series."},{"Start":"12:59.915 ","End":"13:02.640","Text":"That\u0027s the end of this lesson."}],"ID":22416},{"Watched":false,"Name":"Exercise 2","Duration":"14m 27s","ChapterTopicVideoID":21518,"CourseChapterTopicPlaylistID":99491,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.080","Text":"Hello. In this lesson we\u0027re going to be answering the following question."},{"Start":"00:04.080 ","End":"00:06.900","Text":"A, cylindrical shell of height h,"},{"Start":"00:06.900 ","End":"00:08.385","Text":"inner radius a,"},{"Start":"00:08.385 ","End":"00:11.115","Text":"and outer radius b is given."},{"Start":"00:11.115 ","End":"00:14.115","Text":"Between 0 and a,"},{"Start":"00:14.115 ","End":"00:17.925","Text":"the space is empty and between a and b,"},{"Start":"00:17.925 ","End":"00:22.530","Text":"the cylinder is filled by a material of given resistivity Rho."},{"Start":"00:22.530 ","End":"00:26.925","Text":"The current flows in the Theta direction."},{"Start":"00:26.925 ","End":"00:30.548","Text":"This question is very similar to the previous question,"},{"Start":"00:30.548 ","End":"00:34.660","Text":"but in the previous question the current was flowing in the radial direction,"},{"Start":"00:34.660 ","End":"00:39.005","Text":"and here the current is flowing in the Theta direction."},{"Start":"00:39.005 ","End":"00:45.185","Text":"Question Number 1 is to calculate the total resistance."},{"Start":"00:45.185 ","End":"00:48.739","Text":"We have our equation for resistance,"},{"Start":"00:48.739 ","End":"00:53.995","Text":"which is equal to Rho multiplied by L divided by A."},{"Start":"00:53.995 ","End":"00:57.815","Text":"Now over here, what we can see is that"},{"Start":"00:57.815 ","End":"01:04.714","Text":"the path length that the current takes is what is changing,"},{"Start":"01:04.714 ","End":"01:07.955","Text":"isn\u0027t uniform and it isn\u0027t constant."},{"Start":"01:07.955 ","End":"01:15.215","Text":"We can see that around the inner cylinder over here,"},{"Start":"01:15.215 ","End":"01:22.220","Text":"the path taken L is the path of the circle around the inner cylinder,"},{"Start":"01:22.220 ","End":"01:25.100","Text":"which is equal to 2Pia."},{"Start":"01:25.100 ","End":"01:34.200","Text":"However, the path taken by the current around the outer cylinder will be 2Pib."},{"Start":"01:35.270 ","End":"01:40.350","Text":"What we can see is that our radius is changing."},{"Start":"01:40.690 ","End":"01:45.040","Text":"Our path length is equal to 2Pir,"},{"Start":"01:45.040 ","End":"01:48.875","Text":"where the radius is constantly changing"},{"Start":"01:48.875 ","End":"01:54.150","Text":"depending on how deep we are inside the cylindrical shell."},{"Start":"01:54.320 ","End":"01:57.815","Text":"Anytime we have something changing,"},{"Start":"01:57.815 ","End":"02:00.230","Text":"where we have some variable,"},{"Start":"02:00.230 ","End":"02:05.975","Text":"that means that we have to split up our resistor into lots of tiny pieces."},{"Start":"02:05.975 ","End":"02:09.815","Text":"What we\u0027re going to do is just like in the previous question,"},{"Start":"02:09.815 ","End":"02:19.850","Text":"we\u0027re going to split this up into lots of very thin cylindrical shells like so,"},{"Start":"02:19.850 ","End":"02:25.145","Text":"where the width over here is dr."},{"Start":"02:25.145 ","End":"02:29.040","Text":"The width is very small, it\u0027s approaching zero."},{"Start":"02:29.040 ","End":"02:32.270","Text":"We can say that the difference in the radius between"},{"Start":"02:32.270 ","End":"02:35.810","Text":"this point on the inner shell and this point on"},{"Start":"02:35.810 ","End":"02:39.200","Text":"the outer shell is so small that we can consider"},{"Start":"02:39.200 ","End":"02:43.770","Text":"the radius as being the same at these 2 points."},{"Start":"02:45.830 ","End":"02:54.740","Text":"Here we can see that the path length is going to be around here, inside this circle."},{"Start":"02:54.740 ","End":"02:59.655","Text":"This is our path length, like so."},{"Start":"02:59.655 ","End":"03:01.935","Text":"This is the path that our current takes."},{"Start":"03:01.935 ","End":"03:08.750","Text":"It is equal to the path length of the circle surrounding this green cylinder,"},{"Start":"03:08.750 ","End":"03:12.095","Text":"which is 2Pi multiplied by the radius of the cylinder,"},{"Start":"03:12.095 ","End":"03:14.510","Text":"r, which is a variable."},{"Start":"03:14.510 ","End":"03:20.015","Text":"Now let\u0027s take a look at what our a is, our cross-sectional area."},{"Start":"03:20.015 ","End":"03:26.345","Text":"Our cross-sectional area is going to be perpendicular to this path length,"},{"Start":"03:26.345 ","End":"03:32.940","Text":"which means that it\u0027s going to be just something like so."},{"Start":"03:33.740 ","End":"03:40.330","Text":"It\u0027s just going to be this pink rectangle that connects"},{"Start":"03:40.330 ","End":"03:50.700","Text":"the inner cylinder wall to the outer cylinder wall and the bottom side to the top side."},{"Start":"03:50.700 ","End":"03:54.000","Text":"Now, this is our cross-sectional area."},{"Start":"03:54.000 ","End":"04:01.235","Text":"We can see that the current is going to travel through this over here."},{"Start":"04:01.235 ","End":"04:04.570","Text":"This is the cross-sectional area and it\u0027s perpendicular"},{"Start":"04:04.570 ","End":"04:09.340","Text":"to the direction of travel of the current. That is great."},{"Start":"04:09.340 ","End":"04:13.255","Text":"We can see that we just have to work out the cross sectional area"},{"Start":"04:13.255 ","End":"04:17.220","Text":"which is that of a rectangle."},{"Start":"04:17.220 ","End":"04:20.660","Text":"The dimensions of the rectangle we can see that this length"},{"Start":"04:20.660 ","End":"04:23.750","Text":"over here is h and the width of"},{"Start":"04:23.750 ","End":"04:30.200","Text":"the rectangle is the width of this tiny cylinder which is dr."},{"Start":"04:30.200 ","End":"04:32.605","Text":"It\u0027s equal to hdr."},{"Start":"04:32.605 ","End":"04:36.545","Text":"Now, what we\u0027re going to do is we can calculate"},{"Start":"04:36.545 ","End":"04:41.595","Text":"the resistance of this tiny cylindrical shell."},{"Start":"04:41.595 ","End":"04:48.020","Text":"We have that dR is equal to the resistivity multiplied by our path length,"},{"Start":"04:48.020 ","End":"04:56.310","Text":"which is 2Pir divided by our cross-sectional area, which is hdr."},{"Start":"04:57.890 ","End":"05:03.129","Text":"Now, in order to find the total resistance of this resistor,"},{"Start":"05:03.129 ","End":"05:10.290","Text":"we just have to sum up the resistances on each of these sub cylinders."},{"Start":"05:10.290 ","End":"05:12.085","Text":"In order to sum them up,"},{"Start":"05:12.085 ","End":"05:13.675","Text":"first of all we have to see if"},{"Start":"05:13.675 ","End":"05:20.176","Text":"these different resistors are connected in series or in parallel,"},{"Start":"05:20.176 ","End":"05:23.520","Text":"because we have this resistor over here,"},{"Start":"05:23.520 ","End":"05:26.103","Text":"and then we\u0027ll have a resistor over here,"},{"Start":"05:26.103 ","End":"05:28.665","Text":"and another one over here."},{"Start":"05:28.665 ","End":"05:36.905","Text":"What we can see if I draw this in pink is that if we have some charge over here."},{"Start":"05:36.905 ","End":"05:39.940","Text":"We have a charge over here in a circle,"},{"Start":"05:39.940 ","End":"05:42.550","Text":"and it\u0027s traveling around over here."},{"Start":"05:42.550 ","End":"05:45.907","Text":"Just like the current, it\u0027s traveling in this Theta direction,"},{"Start":"05:45.907 ","End":"05:47.810","Text":"and it\u0027s just going to the same charge,"},{"Start":"05:47.810 ","End":"05:53.750","Text":"is going to carry on traveling around inside of this resistor."},{"Start":"05:53.750 ","End":"05:57.170","Text":"Similarly in this next resistor,"},{"Start":"05:57.170 ","End":"05:58.490","Text":"if we have a charge,"},{"Start":"05:58.490 ","End":"06:06.810","Text":"it will travel around like so in a circle around this resistor and similarly over here."},{"Start":"06:06.810 ","End":"06:13.055","Text":"What we can see is that this charge or the current circle, the current loop,"},{"Start":"06:13.055 ","End":"06:17.801","Text":"doesn\u0027t have to travel through each and every single resistor,"},{"Start":"06:17.801 ","End":"06:22.100","Text":"it just travels through its own resistor in a circle like so."},{"Start":"06:22.100 ","End":"06:27.665","Text":"The charges aren\u0027t moving between these green cylinders,"},{"Start":"06:27.665 ","End":"06:32.175","Text":"they stay within each separate cylinder."},{"Start":"06:32.175 ","End":"06:37.970","Text":"What does that mean? That means that our voltage source that is connected to this"},{"Start":"06:37.970 ","End":"06:44.465","Text":"causing the current move or causing a current and causing the charges to move rather."},{"Start":"06:44.465 ","End":"06:48.980","Text":"That means that we have a voltage split between each"},{"Start":"06:48.980 ","End":"06:54.105","Text":"one of these sub resistors that we\u0027ve formed over here."},{"Start":"06:54.105 ","End":"06:56.810","Text":"In that case, if we have a voltage split,"},{"Start":"06:56.810 ","End":"07:03.570","Text":"that means that each of these sub resistors is connected in parallel."},{"Start":"07:04.560 ","End":"07:06.850","Text":"In a parallel connection,"},{"Start":"07:06.850 ","End":"07:10.135","Text":"how do we add up to find the total resistance?"},{"Start":"07:10.135 ","End":"07:15.010","Text":"We know that 1 divided by the total resistance is equal"},{"Start":"07:15.010 ","End":"07:20.515","Text":"to the sum of the reciprocals of the total resistances."},{"Start":"07:20.515 ","End":"07:24.730","Text":"But here, because we\u0027re dealing with such small values,"},{"Start":"07:24.730 ","End":"07:30.710","Text":"we\u0027re dealing with dr, so this summation sign becomes an integral sign."},{"Start":"07:30.840 ","End":"07:36.940","Text":"Therefore, we can say that 1 divided by the total resistance"},{"Start":"07:36.940 ","End":"07:42.690","Text":"is equal to the integral of 1 divided by the total resistance,"},{"Start":"07:42.690 ","End":"07:44.880","Text":"which in our case is,"},{"Start":"07:44.880 ","End":"07:47.670","Text":"we just take the reciprocal of this,"},{"Start":"07:47.670 ","End":"07:55.030","Text":"like so, of course over here it\u0027s also dr because we changed this to a summation."},{"Start":"07:55.030 ","End":"08:05.120","Text":"What we\u0027re going to have is an integral of hdr divided by 2rhopir,"},{"Start":"08:06.270 ","End":"08:13.690","Text":"and the bounds are going to be from the inner radius,"},{"Start":"08:13.690 ","End":"08:20.155","Text":"which is a, and up until the outer radius, which is b."},{"Start":"08:20.155 ","End":"08:24.205","Text":"Of course, here everything is a constant aside from"},{"Start":"08:24.205 ","End":"08:28.420","Text":"r integrating of dr divided by r, so we get LAN,"},{"Start":"08:28.420 ","End":"08:32.170","Text":"so we\u0027re simply going to have that this is equal to h divided by"},{"Start":"08:32.170 ","End":"08:40.280","Text":"2rhopi multiplied by LAN of b divided by a."},{"Start":"08:40.800 ","End":"08:45.715","Text":"This is of course, 1 divided by r, so therefore,"},{"Start":"08:45.715 ","End":"08:50.515","Text":"the total resistance is simply the reciprocal of this,"},{"Start":"08:50.515 ","End":"08:53.140","Text":"which is just this."},{"Start":"08:53.140 ","End":"08:55.765","Text":"That\u0027s the answer to Question 1,"},{"Start":"08:55.765 ","End":"09:02.660","Text":"and now let\u0027s answer Question 2 to calculate the current density J."},{"Start":"09:03.510 ","End":"09:09.895","Text":"This time, this question is going to be slightly more complicated, not by much,"},{"Start":"09:09.895 ","End":"09:14.290","Text":"but because all of our tiny resistors are connected in parallel,"},{"Start":"09:14.290 ","End":"09:18.025","Text":"we have to calculate the current in a slightly different way,"},{"Start":"09:18.025 ","End":"09:22.675","Text":"and therefore, the current density is calculated in a different way."},{"Start":"09:22.675 ","End":"09:28.885","Text":"We\u0027ve already gathered that all of these sub cylinders are connected in parallel,"},{"Start":"09:28.885 ","End":"09:35.095","Text":"which means that they have the same voltage across each one."},{"Start":"09:35.095 ","End":"09:36.715","Text":"That\u0027s a definition."},{"Start":"09:36.715 ","End":"09:39.205","Text":"If we have resistors connected in parallel,"},{"Start":"09:39.205 ","End":"09:43.960","Text":"the voltage across each resistor is going to be the same."},{"Start":"09:43.960 ","End":"09:50.545","Text":"Therefore, we can say that the total voltage across this entire cylinder is also"},{"Start":"09:50.545 ","End":"09:58.135","Text":"equal to the voltage across each individual sub resistor that we formed over here."},{"Start":"09:58.135 ","End":"10:03.505","Text":"Then, assuming that each sub resistor abides by Ohm\u0027s law,"},{"Start":"10:03.505 ","End":"10:09.190","Text":"we can therefore say that the total voltage is equal"},{"Start":"10:09.190 ","End":"10:15.520","Text":"to the current multiplied by the total resistance,"},{"Start":"10:15.520 ","End":"10:18.590","Text":"which we just called r,"},{"Start":"10:20.220 ","End":"10:22.375","Text":"and this is of course,"},{"Start":"10:22.375 ","End":"10:28.960","Text":"also equal to the voltage across each sub resistor."},{"Start":"10:28.960 ","End":"10:31.975","Text":"Now, we want to calculate the"},{"Start":"10:31.975 ","End":"10:37.045","Text":"current that flows through each sub resistor because as we saw,"},{"Start":"10:37.045 ","End":"10:40.555","Text":"it\u0027s changing because it\u0027s in the Theta direction."},{"Start":"10:40.555 ","End":"10:44.680","Text":"We saw that the current in this resistor is going to be different"},{"Start":"10:44.680 ","End":"10:48.880","Text":"to the current in this resistor and in this resistor,"},{"Start":"10:48.880 ","End":"10:52.450","Text":"because again, these are connected in parallel."},{"Start":"10:52.450 ","End":"10:53.920","Text":"If they were connected in series,"},{"Start":"10:53.920 ","End":"10:57.505","Text":"then obviously it would be the same current throughout."},{"Start":"10:57.505 ","End":"11:06.370","Text":"Therefore, we can say that the current in each resistor is equal to the voltage,"},{"Start":"11:06.370 ","End":"11:08.560","Text":"so the total voltage,"},{"Start":"11:08.560 ","End":"11:11.575","Text":"or the voltage across each resistor, it makes no difference,"},{"Start":"11:11.575 ","End":"11:19.570","Text":"divided by the resistance of each resistor, which we saw,"},{"Start":"11:19.570 ","End":"11:24.290","Text":"changes depending on our radius,"},{"Start":"11:26.130 ","End":"11:29.275","Text":"so divided by dr."},{"Start":"11:29.275 ","End":"11:33.470","Text":"It\u0027s changing depending on the radius, the resistance."},{"Start":"11:34.230 ","End":"11:41.980","Text":"This is simply going to be equal to V total divided by our dr,"},{"Start":"11:41.980 ","End":"11:45.190","Text":"which is equal 2 once we sorted out,"},{"Start":"11:45.190 ","End":"11:47.630","Text":"we have 2rhopir,"},{"Start":"11:48.870 ","End":"11:54.105","Text":"multiply it by over here hdr."},{"Start":"11:54.105 ","End":"11:56.800","Text":"I just rearranged this over here because"},{"Start":"11:56.800 ","End":"12:00.580","Text":"the denominator in the denominator moves up to the numerator."},{"Start":"12:00.580 ","End":"12:02.155","Text":"This is of course,"},{"Start":"12:02.155 ","End":"12:06.100","Text":"dI, the current in each resistor."},{"Start":"12:06.100 ","End":"12:08.455","Text":"This also makes sense that,"},{"Start":"12:08.455 ","End":"12:14.155","Text":"if we have lots and lots of tiny resistors inside this entire cylinder,"},{"Start":"12:14.155 ","End":"12:17.710","Text":"it makes sense that the current flowing through each resistor"},{"Start":"12:17.710 ","End":"12:22.090","Text":"is going to also be very small."},{"Start":"12:22.090 ","End":"12:24.475","Text":"This is the current in each resistor,"},{"Start":"12:24.475 ","End":"12:30.350","Text":"and now what we want to do is we want to calculate the current density J."},{"Start":"12:30.750 ","End":"12:38.395","Text":"So, the current density in each resistor is simply equal to the"},{"Start":"12:38.395 ","End":"12:45.745","Text":"current in each resistor divided by the cross-sectional area of each resistor,"},{"Start":"12:45.745 ","End":"12:48.700","Text":"where dA, I\u0027m reminding you,"},{"Start":"12:48.700 ","End":"12:51.925","Text":"is simply equal to our hdr,"},{"Start":"12:51.925 ","End":"12:56.500","Text":"remember this pink rectangle over here."},{"Start":"12:56.500 ","End":"13:02.365","Text":"Our dI is equal to our V total,"},{"Start":"13:02.365 ","End":"13:05.650","Text":"which you can just plug in from this over here,"},{"Start":"13:05.650 ","End":"13:08.155","Text":"multiplied by hdr,"},{"Start":"13:08.155 ","End":"13:11.120","Text":"divided by 2rhopir,"},{"Start":"13:11.160 ","End":"13:16.585","Text":"and then our dA, will give us an hdr also over here."},{"Start":"13:16.585 ","End":"13:25.100","Text":"So, these two cancel out and what we are left with is V total divided by 2rhopir,"},{"Start":"13:25.980 ","End":"13:28.240","Text":"and of course this is in"},{"Start":"13:28.240 ","End":"13:37.030","Text":"the Theta direction because"},{"Start":"13:37.030 ","End":"13:41.840","Text":"our current is flowing in the Theta direction."},{"Start":"13:42.360 ","End":"13:45.130","Text":"This is the answer to Question 2,"},{"Start":"13:45.130 ","End":"13:47.410","Text":"and of course you can sub in V total,"},{"Start":"13:47.410 ","End":"13:51.145","Text":"which is just equal to this over here."},{"Start":"13:51.145 ","End":"13:53.170","Text":"Then for Question 3,"},{"Start":"13:53.170 ","End":"13:56.320","Text":"calculate the electric field inside of the conductor,"},{"Start":"13:56.320 ","End":"14:03.340","Text":"so the E field is equal to the resistivity multiplied by J."},{"Start":"14:03.340 ","End":"14:07.915","Text":"Then, what we\u0027ll get is resistivity multiplied by this,"},{"Start":"14:07.915 ","End":"14:12.800","Text":"so we have rho in the numerator and rho in the denominator, so they\u0027ll cancel out."},{"Start":"14:12.800 ","End":"14:21.760","Text":"So, our resistivity is simply equal to V total divided by 2pir in the Theta direction."},{"Start":"14:21.760 ","End":"14:24.925","Text":"That\u0027s the answer to Question 3,"},{"Start":"14:24.925 ","End":"14:27.980","Text":"and that is the end of this lesson."}],"ID":22417},{"Watched":false,"Name":"Surface Current Density","Duration":"10m ","ChapterTopicVideoID":21312,"CourseChapterTopicPlaylistID":99491,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.845","Text":"Hello. In this lesson,"},{"Start":"00:01.845 ","End":"00:05.699","Text":"we\u0027re going to be speaking about surface current density."},{"Start":"00:05.699 ","End":"00:08.595","Text":"Before we speak about surface current density,"},{"Start":"00:08.595 ","End":"00:12.300","Text":"let\u0027s speak a little bit more about current density."},{"Start":"00:12.300 ","End":"00:17.340","Text":"We\u0027ve already seen current density denoted by the letter J,"},{"Start":"00:17.340 ","End":"00:22.010","Text":"where if we have current traveling in this direction,"},{"Start":"00:22.010 ","End":"00:24.900","Text":"we know that it\u0027s going to travel all the way through"},{"Start":"00:24.900 ","End":"00:28.800","Text":"this conductor and exit over here on the other side,"},{"Start":"00:28.800 ","End":"00:31.680","Text":"where the current is going to spread out across"},{"Start":"00:31.680 ","End":"00:35.640","Text":"this entire cross-sectional surface area and, of course,"},{"Start":"00:35.640 ","End":"00:38.670","Text":"the current density through here is going to"},{"Start":"00:38.670 ","End":"00:43.010","Text":"be the same as the current density through here,"},{"Start":"00:43.010 ","End":"00:47.180","Text":"because this is the same cross-sectional surface area,"},{"Start":"00:47.180 ","End":"00:54.600","Text":"so we\u0027re just going to have different charges going through like so."},{"Start":"00:54.950 ","End":"01:02.480","Text":"If we take a closer look at this cross-sectional area,"},{"Start":"01:02.480 ","End":"01:05.390","Text":"what we could see, what J was,"},{"Start":"01:05.390 ","End":"01:10.474","Text":"was if we looked at tiny unit areas,"},{"Start":"01:10.474 ","End":"01:18.430","Text":"we would count how many charged particles were passing through this unit area of area"},{"Start":"01:18.430 ","End":"01:22.490","Text":"ds and we would look at it for all of"},{"Start":"01:22.490 ","End":"01:27.785","Text":"the unit areas and the surface area and then this would be J,"},{"Start":"01:27.785 ","End":"01:31.804","Text":"how many charges are passing through this unit area and therefore,"},{"Start":"01:31.804 ","End":"01:36.214","Text":"we would get that the current I is equal to the summation"},{"Start":"01:36.214 ","End":"01:40.700","Text":"of all of the charged particles traveling through all of these unit areas,"},{"Start":"01:40.700 ","End":"01:44.285","Text":"which make up this total cross-sectional area."},{"Start":"01:44.285 ","End":"01:48.960","Text":"In other words, the integral on J.ds."},{"Start":"01:49.490 ","End":"01:52.235","Text":"This we\u0027ve already seen."},{"Start":"01:52.235 ","End":"01:57.664","Text":"Now what I want to do is I want to speak about surface current density."},{"Start":"01:57.664 ","End":"02:02.820","Text":"Surface current density is denoted by the letter k."},{"Start":"02:03.020 ","End":"02:10.760","Text":"Surface current density is also a vector quantity and it\u0027s very similar to J,"},{"Start":"02:10.760 ","End":"02:15.155","Text":"where J deals with area."},{"Start":"02:15.155 ","End":"02:17.194","Text":"We look at unit areas,"},{"Start":"02:17.194 ","End":"02:21.590","Text":"and k deals with length."},{"Start":"02:21.590 ","End":"02:26.324","Text":"Here we\u0027re dealing with unit lengths."},{"Start":"02:26.324 ","End":"02:29.780","Text":"Let\u0027s again look at this cross-sectional area."},{"Start":"02:29.780 ","End":"02:35.170","Text":"Let\u0027s imagine that this whole thing is empty."},{"Start":"02:35.170 ","End":"02:40.950","Text":"We just have this outline of the square,"},{"Start":"02:40.950 ","End":"02:43.455","Text":"but there\u0027s nothing in the middle."},{"Start":"02:43.455 ","End":"02:45.450","Text":"There was just air in the middle."},{"Start":"02:45.450 ","End":"02:48.635","Text":"Of course, if there\u0027s current going through,"},{"Start":"02:48.635 ","End":"02:52.055","Text":"the current is still going to flow through this."},{"Start":"02:52.055 ","End":"02:59.399","Text":"However, it\u0027s going to flow through this surface over here, so the outline."},{"Start":"02:59.399 ","End":"03:06.315","Text":"You can imagine this being some kind of hollow rectangular-shaped tube."},{"Start":"03:06.315 ","End":"03:09.110","Text":"The current is just going to flow along here,"},{"Start":"03:09.110 ","End":"03:11.045","Text":"and if I draw this in blue,"},{"Start":"03:11.045 ","End":"03:13.865","Text":"we\u0027re just going to get charged particles"},{"Start":"03:13.865 ","End":"03:19.100","Text":"instead of at unit areas in the middle over here,"},{"Start":"03:19.100 ","End":"03:24.090","Text":"the current can only travel along this black outline like so."},{"Start":"03:25.090 ","End":"03:34.345","Text":"What we\u0027re going to have is a situation like so where we have this frame,"},{"Start":"03:34.345 ","End":"03:37.580","Text":"where we\u0027re just looking at the cross section and this is hollow."},{"Start":"03:37.580 ","End":"03:47.260","Text":"We have this frame where we only have charges traveling through the outline over here."},{"Start":"03:47.960 ","End":"03:56.510","Text":"This is k. What we do is if we look at certain regions,"},{"Start":"03:56.510 ","End":"03:59.585","Text":"if we look at this area over here,"},{"Start":"03:59.585 ","End":"04:01.640","Text":"or this length rather over here,"},{"Start":"04:01.640 ","End":"04:03.800","Text":"and we call this length dL,"},{"Start":"04:03.800 ","End":"04:05.570","Text":"and it\u0027s a very small length."},{"Start":"04:05.570 ","End":"04:14.585","Text":"We count how many charges are passing through this length and similarly over here,"},{"Start":"04:14.585 ","End":"04:17.900","Text":"this is also of length dL and we count"},{"Start":"04:17.900 ","End":"04:22.765","Text":"how many charges are passing through this length over here, dL."},{"Start":"04:22.765 ","End":"04:29.140","Text":"Then what we\u0027ll get is that the current that is flowing through this over here,"},{"Start":"04:29.140 ","End":"04:32.965","Text":"this hollow rectangular tube,"},{"Start":"04:32.965 ","End":"04:42.650","Text":"is simply going to be equal to the integral along kdl."},{"Start":"04:43.860 ","End":"04:48.700","Text":"J is the current density per unit area,"},{"Start":"04:48.700 ","End":"04:51.850","Text":"and k is the surface current density,"},{"Start":"04:51.850 ","End":"04:55.060","Text":"which just means the current density per unit length."},{"Start":"04:55.060 ","End":"04:59.425","Text":"We have 1 dimensions and we\u0027re counting"},{"Start":"04:59.425 ","End":"05:05.260","Text":"how many currents we have passing over this 1 dimension."},{"Start":"05:05.260 ","End":"05:11.395","Text":"This over here represents k and alternatively,"},{"Start":"05:11.395 ","End":"05:17.085","Text":"this is over here, dL."},{"Start":"05:17.085 ","End":"05:20.910","Text":"If we have an area where this is ds,"},{"Start":"05:20.910 ","End":"05:23.225","Text":"J on the other hand,"},{"Start":"05:23.225 ","End":"05:28.715","Text":"is counting how many currents are in this area,"},{"Start":"05:28.715 ","End":"05:31.615","Text":"and this is J."},{"Start":"05:31.615 ","End":"05:41.161","Text":"Now, of course, we could have a situation where we have k and J. Let\u0027s draw."},{"Start":"05:41.161 ","End":"05:45.500","Text":"Here, we also have charges passing through in the middle,"},{"Start":"05:45.500 ","End":"05:48.650","Text":"now this isn\u0027t hollow, it\u0027s filled."},{"Start":"05:48.650 ","End":"05:50.390","Text":"Therefore in that case,"},{"Start":"05:50.390 ","End":"05:54.475","Text":"if we have both k and J,"},{"Start":"05:54.475 ","End":"06:00.605","Text":"our total current therefore is going to be equal to the integral"},{"Start":"06:00.605 ","End":"06:09.495","Text":"on k.dL plus the integral on J.ds."},{"Start":"06:09.495 ","End":"06:14.675","Text":"We sum up the charges passing through the length over here,"},{"Start":"06:14.675 ","End":"06:18.260","Text":"this 1 dimensional loop over here and the charges"},{"Start":"06:18.260 ","End":"06:23.100","Text":"passing through this area over here in the middle."},{"Start":"06:23.540 ","End":"06:26.234","Text":"Let\u0027s look at an example."},{"Start":"06:26.234 ","End":"06:33.400","Text":"Here we have a hollow tube where we\u0027re being told that we have charges over"},{"Start":"06:33.400 ","End":"06:40.492","Text":"here that are traveling just along this outline over here like so,"},{"Start":"06:40.492 ","End":"06:44.350","Text":"and that this is a uniform throughout and then these charges just carry on"},{"Start":"06:44.350 ","End":"06:50.090","Text":"traveling down along the surface area of this hollow tube."},{"Start":"06:50.090 ","End":"07:01.050","Text":"What we want to do in this question is to calculate k. This is an example. What is k?"},{"Start":"07:01.050 ","End":"07:07.715","Text":"We know from here that I is equal to the integral and k.dL,"},{"Start":"07:07.715 ","End":"07:12.290","Text":"but here we were told that this is uniform over here."},{"Start":"07:12.290 ","End":"07:14.015","Text":"Instead of doing an integral,"},{"Start":"07:14.015 ","End":"07:20.585","Text":"we can say that I is equal to k multiplied by the length. What is the length?"},{"Start":"07:20.585 ","End":"07:26.305","Text":"The length is this over here, this entire loop."},{"Start":"07:26.305 ","End":"07:27.830","Text":"Just like in this example,"},{"Start":"07:27.830 ","End":"07:32.075","Text":"we saw that dL are lengths along this outline."},{"Start":"07:32.075 ","End":"07:38.300","Text":"The total length is going to be the total outline of this rectangle over here."},{"Start":"07:38.300 ","End":"07:43.780","Text":"Here our dL is like so."},{"Start":"07:43.780 ","End":"07:47.605","Text":"This is dL,"},{"Start":"07:47.605 ","End":"07:52.910","Text":"so the total length is just the circumference of this circle."},{"Start":"07:52.910 ","End":"07:59.720","Text":"That is simply just going to be equal to k multiplied by 2 Pi R,"},{"Start":"07:59.720 ","End":"08:04.895","Text":"where the radius is R. If we want to calculate Rk,"},{"Start":"08:04.895 ","End":"08:09.275","Text":"k is simply equal to I divided by"},{"Start":"08:09.275 ","End":"08:16.050","Text":"2 Pi R. This is what our k would be equal"},{"Start":"08:16.050 ","End":"08:23.570","Text":"to and then what would happen if we actually wanted to know what I was equal to."},{"Start":"08:23.570 ","End":"08:26.645","Text":"In this case we\u0027re going to integrate,"},{"Start":"08:26.645 ","End":"08:29.915","Text":"we have the integration of k,"},{"Start":"08:29.915 ","End":"08:36.230","Text":"which here we got was equal to I divided by 2 Pi R. We\u0027re assuming that in this question"},{"Start":"08:36.230 ","End":"08:42.842","Text":"we\u0027re given k and then we have to multiply this by dL,"},{"Start":"08:42.842 ","End":"08:44.930","Text":"so dL is this arc length."},{"Start":"08:44.930 ","End":"08:51.430","Text":"Each arc length here we\u0027re dealing with polar coordinates is equal to rd Theta,"},{"Start":"08:51.430 ","End":"08:56.030","Text":"that is dL when we\u0027re dealing with polar coordinates and of course we\u0027re"},{"Start":"08:56.030 ","End":"09:03.465","Text":"going from radians 0 until 2 Pi radians."},{"Start":"09:03.465 ","End":"09:05.519","Text":"That means a full circle,"},{"Start":"09:05.519 ","End":"09:11.540","Text":"and then what you\u0027ll see is that you get back I over here."},{"Start":"09:11.540 ","End":"09:14.515","Text":"That is what you do."},{"Start":"09:14.515 ","End":"09:16.790","Text":"That is the end of the lesson."},{"Start":"09:16.790 ","End":"09:22.280","Text":"Don\u0027t forget to write out this equation in your equation sheets,"},{"Start":"09:22.280 ","End":"09:27.560","Text":"remembering that k is your surface current density.dL,"},{"Start":"09:27.560 ","End":"09:31.305","Text":"where dL is the length of the outline."},{"Start":"09:31.305 ","End":"09:33.635","Text":"Instead of taking the cross sectional area,"},{"Start":"09:33.635 ","End":"09:39.050","Text":"you take the parameter where you would have had cross-sectional area and"},{"Start":"09:39.050 ","End":"09:41.360","Text":"if you\u0027re told that you also have"},{"Start":"09:41.360 ","End":"09:45.140","Text":"current density that goes through the cross-sectional area itself,"},{"Start":"09:45.140 ","End":"09:49.825","Text":"then you have to add that on just like we previously learned."},{"Start":"09:49.825 ","End":"09:54.890","Text":"k is speaking about all of the charges going across the parameter of what"},{"Start":"09:54.890 ","End":"10:00.320","Text":"would have been the cross-sectional area and that is the end of this lesson."}],"ID":21392},{"Watched":false,"Name":"Calculating Current Density from Charge Density and Velocity","Duration":"21m 32s","ChapterTopicVideoID":21519,"CourseChapterTopicPlaylistID":99491,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.785","Text":"Hello. In this lesson,"},{"Start":"00:01.785 ","End":"00:05.790","Text":"we\u0027re going to learn different equations for how to"},{"Start":"00:05.790 ","End":"00:10.664","Text":"find the current density from a moving charge."},{"Start":"00:10.664 ","End":"00:13.215","Text":"A few lessons ago,"},{"Start":"00:13.215 ","End":"00:16.890","Text":"we learned or saw an equation for J,"},{"Start":"00:16.890 ","End":"00:20.490","Text":"the current density, which was given as Rho,"},{"Start":"00:20.490 ","End":"00:26.010","Text":"where here, Rho was charge density per unit volume,"},{"Start":"00:26.010 ","End":"00:32.745","Text":"multiplied by the velocity at which it is traveling."},{"Start":"00:32.745 ","End":"00:36.255","Text":"In the previous lesson, we learned about K,"},{"Start":"00:36.255 ","End":"00:41.615","Text":"which is the surface charge density and similarly to J,"},{"Start":"00:41.615 ","End":"00:43.835","Text":"this is equal to Sigma,"},{"Start":"00:43.835 ","End":"00:52.350","Text":"which is the charge density per unit area multiplied by the velocity."},{"Start":"00:52.730 ","End":"00:55.865","Text":"Let\u0027s take this example over here,"},{"Start":"00:55.865 ","End":"00:58.340","Text":"where we have some box."},{"Start":"00:58.340 ","End":"01:02.525","Text":"It\u0027s 3 dimensional and it\u0027s made out of charges."},{"Start":"01:02.525 ","End":"01:07.640","Text":"It\u0027s filled with a material with charge density per unit volume, Rho."},{"Start":"01:07.640 ","End":"01:11.450","Text":"Let\u0027s say that here we have some opening,"},{"Start":"01:11.450 ","End":"01:17.195","Text":"some hole and this material is traveling through this hole with"},{"Start":"01:17.195 ","End":"01:23.860","Text":"a velocity of v. What we can see is that we will have a current."},{"Start":"01:23.860 ","End":"01:28.160","Text":"Then, if we look at this hole over here."},{"Start":"01:28.160 ","End":"01:30.440","Text":"Let\u0027s just draw it in blue."},{"Start":"01:30.440 ","End":"01:35.915","Text":"We\u0027re looking at this hole and we have this box over here,"},{"Start":"01:35.915 ","End":"01:37.625","Text":"which is traveling through."},{"Start":"01:37.625 ","End":"01:44.760","Text":"We could calculate the current density traveling through this blue hole."},{"Start":"01:44.840 ","End":"01:48.170","Text":"That is simply going to be equal to Rho,"},{"Start":"01:48.170 ","End":"01:53.300","Text":"the number of charges traveling through this slit,"},{"Start":"01:53.300 ","End":"01:57.940","Text":"multiplied by v, the velocity at which they\u0027re moving through."},{"Start":"01:57.940 ","End":"02:02.540","Text":"Now let\u0027s look at the example dealing with K. Here we have"},{"Start":"02:02.540 ","End":"02:07.520","Text":"some plane and it has charge density per unit area Sigma."},{"Start":"02:07.520 ","End":"02:16.490","Text":"It is also traveling through this hole with a velocity of v. As we remember,"},{"Start":"02:16.490 ","End":"02:23.420","Text":"K is the surface current density and it has units."},{"Start":"02:23.420 ","End":"02:26.525","Text":"Remember it\u0027s dL units of length."},{"Start":"02:26.525 ","End":"02:28.580","Text":"Well, it doesn\u0027t have units of length,"},{"Start":"02:28.580 ","End":"02:31.510","Text":"but it deals with length."},{"Start":"02:31.510 ","End":"02:36.215","Text":"What we can see is that what\u0027s going through is just a line over here."},{"Start":"02:36.215 ","End":"02:39.860","Text":"We can see it\u0027s just this line because this is a plane and it\u0027s"},{"Start":"02:39.860 ","End":"02:44.960","Text":"just a line passing through this blue hollow loop over here."},{"Start":"02:44.960 ","End":"02:47.780","Text":"If we look at it again from the side,"},{"Start":"02:47.780 ","End":"02:52.605","Text":"here is our blue hole and we just have this line,"},{"Start":"02:52.605 ","End":"02:56.240","Text":"this is if we\u0027re looking at this plane from the side over here."},{"Start":"02:56.240 ","End":"03:00.925","Text":"It\u0027s just a line coming through or traveling through this hole."},{"Start":"03:00.925 ","End":"03:04.865","Text":"Then if we split up this line into lengths,"},{"Start":"03:04.865 ","End":"03:09.380","Text":"so here is our dL and we count how many charges are in this dL."},{"Start":"03:09.380 ","End":"03:12.790","Text":"That is going to be our K,"},{"Start":"03:12.790 ","End":"03:14.915","Text":"our surface charge density,"},{"Start":"03:14.915 ","End":"03:19.880","Text":"which is also equal to Sigma multiplied by v. Of course,"},{"Start":"03:19.880 ","End":"03:26.660","Text":"if I sum up this K along this entire length of the plane that is going through the hole,"},{"Start":"03:26.660 ","End":"03:29.285","Text":"then I will get my current."},{"Start":"03:29.285 ","End":"03:33.095","Text":"Let\u0027s look at another example."},{"Start":"03:33.095 ","End":"03:36.650","Text":"Here we have a cylinder made out of"},{"Start":"03:36.650 ","End":"03:40.505","Text":"a material where the charge density per unit volume Rho."},{"Start":"03:40.505 ","End":"03:49.690","Text":"This cylinder is rotating around its axis of symmetry with an angular velocity of omega."},{"Start":"03:49.760 ","End":"03:54.365","Text":"If we look at some piece over here,"},{"Start":"03:54.365 ","End":"04:01.580","Text":"so we know that we have volume over here and so we\u0027re looking at some small cube."},{"Start":"04:01.580 ","End":"04:06.800","Text":"The distance, the small cube of Rho is from the axis"},{"Start":"04:06.800 ","End":"04:11.880","Text":"of symmetry is a radius of r. Our J,"},{"Start":"04:11.880 ","End":"04:14.630","Text":"our current density is equal to,"},{"Start":"04:14.630 ","End":"04:16.670","Text":"as we know, Rho v,"},{"Start":"04:16.670 ","End":"04:20.255","Text":"where in this question that we\u0027re dealing with,"},{"Start":"04:20.255 ","End":"04:23.105","Text":"our velocity is angular velocity."},{"Start":"04:23.105 ","End":"04:31.170","Text":"This is going to become Rho and then v is equal to omega r,"},{"Start":"04:31.170 ","End":"04:32.600","Text":"and of course,"},{"Start":"04:32.600 ","End":"04:36.160","Text":"this is traveling in the Theta direction."},{"Start":"04:36.160 ","End":"04:45.280","Text":"What we can see is that this unit volume is traveling through some area like so,"},{"Start":"04:45.280 ","End":"04:49.890","Text":"and this area is like a plane,"},{"Start":"04:49.890 ","End":"04:52.965","Text":"in the r, z plane."},{"Start":"04:52.965 ","End":"04:59.810","Text":"That means that the ds for this blue square over here is"},{"Start":"04:59.810 ","End":"05:06.395","Text":"going to be equal to some change in r. This over here some change in r,"},{"Start":"05:06.395 ","End":"05:09.200","Text":"multiplied by this over here,"},{"Start":"05:09.200 ","End":"05:10.640","Text":"which is in the z direction."},{"Start":"05:10.640 ","End":"05:17.885","Text":"Multiplied by some change in z and its vector direction is of course,"},{"Start":"05:17.885 ","End":"05:20.615","Text":"in the direction perpendicular to the plane,"},{"Start":"05:20.615 ","End":"05:22.469","Text":"which is the Theta direction."},{"Start":"05:22.469 ","End":"05:25.430","Text":"This is the ds."},{"Start":"05:26.020 ","End":"05:33.380","Text":"Now let\u0027s look at the example where we deal with K. Here we have a cylindrical shell of"},{"Start":"05:33.380 ","End":"05:38.945","Text":"radius R and the cylindrical shell"},{"Start":"05:38.945 ","End":"05:44.690","Text":"is made out of a material with charge density per unit area, Sigma."},{"Start":"05:44.690 ","End":"05:49.530","Text":"Our K is equal to Sigma, v,"},{"Start":"05:49.530 ","End":"05:53.990","Text":"where v, we are dealing with angular momentum over here."},{"Start":"05:53.990 ","End":"06:01.310","Text":"It\u0027s going to be equal to Sigma multiplied by omega multiplied by r. It\u0027s just"},{"Start":"06:01.310 ","End":"06:03.950","Text":"multiplied by capital R this time and not lowercase"},{"Start":"06:03.950 ","End":"06:09.275","Text":"r. That\u0027s because in our example with J,"},{"Start":"06:09.275 ","End":"06:11.510","Text":"because the whole cylinder was filled,"},{"Start":"06:11.510 ","End":"06:19.325","Text":"our r was changing depending on what cube of unit volume we chose where it was."},{"Start":"06:19.325 ","End":"06:24.390","Text":"It could have been here closer to the axis of symmetry or here further away,"},{"Start":"06:24.390 ","End":"06:27.724","Text":"so the r over here and the J is a variable."},{"Start":"06:27.724 ","End":"06:33.455","Text":"However, in the K, when we only have charges along the shell,"},{"Start":"06:33.455 ","End":"06:38.060","Text":"that means that the charges can only be located at 1 radius,"},{"Start":"06:38.060 ","End":"06:40.600","Text":"and that is the radius of the shell."},{"Start":"06:40.600 ","End":"06:45.920","Text":"Of course, the direction is still in the Theta direction."},{"Start":"06:45.920 ","End":"06:49.490","Text":"We\u0027re looking at a piece of area over here,"},{"Start":"06:49.490 ","End":"06:51.800","Text":"and it\u0027s traveling in this direction,"},{"Start":"06:51.800 ","End":"06:55.020","Text":"which is the Theta direction."},{"Start":"06:55.190 ","End":"07:03.440","Text":"Then if I wanted to see what piece of length this charge is traveling through,"},{"Start":"07:03.440 ","End":"07:08.455","Text":"I could choose this length, which is dL."},{"Start":"07:08.455 ","End":"07:12.905","Text":"Then I can count how many charges go through this length."},{"Start":"07:12.905 ","End":"07:23.470","Text":"In our case, dL would simply be equal to dz because it\u0027s along the z-axis."},{"Start":"07:23.470 ","End":"07:28.175","Text":"Of course, these are also vector quantities."},{"Start":"07:28.175 ","End":"07:32.360","Text":"The last thing that I want to speak about is what if we"},{"Start":"07:32.360 ","End":"07:37.110","Text":"have charge density per unit length?"},{"Start":"07:38.150 ","End":"07:45.080","Text":"We have charge density per unit length Lambda multiplied by the velocity."},{"Start":"07:45.080 ","End":"07:46.865","Text":"Let\u0027s take a look at this."},{"Start":"07:46.865 ","End":"07:54.035","Text":"Here, we will have some wire with charge density per unit length of Lambda."},{"Start":"07:54.035 ","End":"07:56.494","Text":"This wire, just like before,"},{"Start":"07:56.494 ","End":"08:06.885","Text":"is passing through this blue frame or this blue hole over here with a velocity of v."},{"Start":"08:06.885 ","End":"08:12.499","Text":"What we will see from the side is our blue frame"},{"Start":"08:12.499 ","End":"08:19.920","Text":"and then we will see this wire with charges passing through."},{"Start":"08:20.450 ","End":"08:27.200","Text":"Then what we\u0027ll get is simply just a wire with charges traveling through it."},{"Start":"08:27.200 ","End":"08:30.200","Text":"That\u0027s essentially what we have over here."},{"Start":"08:30.200 ","End":"08:32.300","Text":"What we\u0027ll get is simply I,"},{"Start":"08:32.300 ","End":"08:35.695","Text":"which as we know, is current."},{"Start":"08:35.695 ","End":"08:40.855","Text":"This is literally just the current passing through a wire."},{"Start":"08:40.855 ","End":"08:43.105","Text":"As we can see, with the J,"},{"Start":"08:43.105 ","End":"08:47.319","Text":"I have to multiply by an area to get the current,"},{"Start":"08:47.319 ","End":"08:52.290","Text":"with K, I have to multiply by the total length to get the current."},{"Start":"08:52.290 ","End":"08:56.225","Text":"When I\u0027m dealing with charge density per unit length,"},{"Start":"08:56.225 ","End":"08:59.315","Text":"and I multiply it simply by the velocity,"},{"Start":"08:59.315 ","End":"09:03.170","Text":"just this, I automatically get the current."},{"Start":"09:03.170 ","End":"09:07.950","Text":"I don\u0027t have to then multiply this expression by something else to get this."},{"Start":"09:08.540 ","End":"09:14.720","Text":"Now we\u0027re going to look at another way to calculate the current and it is"},{"Start":"09:14.720 ","End":"09:21.450","Text":"based on the fact that current is equal to the amount of charge in a space of time."},{"Start":"09:21.800 ","End":"09:28.695","Text":"As we know, I is equal to dq by dt."},{"Start":"09:28.695 ","End":"09:34.735","Text":"We can calculate dq exactly how we did in statics."},{"Start":"09:34.735 ","End":"09:38.765","Text":"We have 3 different options."},{"Start":"09:38.765 ","End":"09:42.020","Text":"We either have Rho dv,"},{"Start":"09:42.020 ","End":"09:48.010","Text":"Sigma ds, or Lambda dL."},{"Start":"09:48.010 ","End":"09:51.290","Text":"Where dl, for instance,"},{"Start":"09:51.290 ","End":"09:57.260","Text":"can be equal to many different things including dx or dy,"},{"Start":"09:57.260 ","End":"09:59.540","Text":"or depending on the coordinates,"},{"Start":"09:59.540 ","End":"10:03.035","Text":"it can be equal to rd Theta."},{"Start":"10:03.035 ","End":"10:05.945","Text":"Ds can be equal to dx dy,"},{"Start":"10:05.945 ","End":"10:11.500","Text":"it can be equal to rd rd Theta, and so on."},{"Start":"10:11.500 ","End":"10:14.535","Text":"and similarly for dv."},{"Start":"10:14.535 ","End":"10:19.790","Text":"What we can do is we can substitute in 1 of these for dq and then,"},{"Start":"10:19.790 ","End":"10:23.570","Text":"we do something that isn\u0027t exactly taking the derivative."},{"Start":"10:23.570 ","End":"10:31.850","Text":"We look at dq as some variable like a regular variable and we\u0027re just dividing it by dt."},{"Start":"10:31.850 ","End":"10:35.877","Text":"It\u0027s not exactly taking the derivative."},{"Start":"10:35.877 ","End":"10:39.669","Text":"Here are some of the examples that we looked at previously."},{"Start":"10:39.669 ","End":"10:41.845","Text":"Here we have our J."},{"Start":"10:41.845 ","End":"10:46.470","Text":"Our J in this example is as we know,"},{"Start":"10:46.470 ","End":"10:52.710","Text":"equal to Rho multiplied by v. Where if we say that this is the x-direction,"},{"Start":"10:52.710 ","End":"10:58.860","Text":"will get that Rho is equal to v in the x-direction."},{"Start":"10:58.860 ","End":"11:01.165","Text":"With this second equation,"},{"Start":"11:01.165 ","End":"11:06.505","Text":"I is equal to dq by dt."},{"Start":"11:06.505 ","End":"11:08.530","Text":"Let\u0027s take a look at what dq is."},{"Start":"11:08.530 ","End":"11:13.615","Text":"dq over here is equal to Rho dv,"},{"Start":"11:13.615 ","End":"11:17.230","Text":"which because we\u0027re using Cartesian coordinates,"},{"Start":"11:17.230 ","End":"11:22.465","Text":"is equal to Rho dx dy dz."},{"Start":"11:22.465 ","End":"11:24.400","Text":"Now let\u0027s plug this in."},{"Start":"11:24.400 ","End":"11:25.870","Text":"We have dq,"},{"Start":"11:25.870 ","End":"11:33.020","Text":"so Rho dx dy dz divided by dt."},{"Start":"11:34.170 ","End":"11:41.995","Text":"Velocity is equal to distance over time."},{"Start":"11:41.995 ","End":"11:45.760","Text":"Over here, we\u0027re covering a distance in the x-direction."},{"Start":"11:45.760 ","End":"11:48.100","Text":"We define this as the x-direction,"},{"Start":"11:48.100 ","End":"11:52.150","Text":"the direction that our block over here is traveling in."},{"Start":"11:52.150 ","End":"11:56.710","Text":"We have that, we can say that our velocity in"},{"Start":"11:56.710 ","End":"12:03.115","Text":"the x-direction is the distance traveled in the x-direction and the time taken."},{"Start":"12:03.115 ","End":"12:06.370","Text":"That is exactly this over here,"},{"Start":"12:06.370 ","End":"12:09.260","Text":"dx divided by dt."},{"Start":"12:09.750 ","End":"12:14.470","Text":"This is of course the x-component of the velocity,"},{"Start":"12:14.470 ","End":"12:15.895","Text":"which is what we have here."},{"Start":"12:15.895 ","End":"12:23.440","Text":"What we in fact have is that this is equal to Rho multiplied by vx,"},{"Start":"12:23.440 ","End":"12:25.555","Text":"the x component of the velocity,"},{"Start":"12:25.555 ","End":"12:27.775","Text":"multiplied by the area,"},{"Start":"12:27.775 ","End":"12:31.610","Text":"which is dy dz."},{"Start":"12:33.000 ","End":"12:36.535","Text":"I can see that I have differentials over here."},{"Start":"12:36.535 ","End":"12:38.305","Text":"that means that over here,"},{"Start":"12:38.305 ","End":"12:40.705","Text":"I\u0027m left with dI."},{"Start":"12:40.705 ","End":"12:43.315","Text":"That\u0027s just something we write."},{"Start":"12:43.315 ","End":"12:48.370","Text":"Now, what we can see is that this over here,"},{"Start":"12:48.370 ","End":"12:57.400","Text":"Rho vx is just my charge density per unit volume multiplied by my velocity,"},{"Start":"12:57.400 ","End":"13:00.340","Text":"so this is just J."},{"Start":"13:00.340 ","End":"13:06.070","Text":"This, dy dz, is my ds."},{"Start":"13:06.070 ","End":"13:09.460","Text":"This is the area that I\u0027m traveling through."},{"Start":"13:09.460 ","End":"13:13.705","Text":"It would be the area of this blue thing over here."},{"Start":"13:13.705 ","End":"13:16.975","Text":"We\u0027ve seen that before. It\u0027s also perpendicular."},{"Start":"13:16.975 ","End":"13:21.595","Text":"The area always has to be perpendicular to the direction of travel of the charges."},{"Start":"13:21.595 ","End":"13:24.205","Text":"If the charges are traveling in the x-direction,"},{"Start":"13:24.205 ","End":"13:30.025","Text":"the yz-plane is perpendicular to the x-direction so that is great."},{"Start":"13:30.025 ","End":"13:37.070","Text":"Then what we have is J multiplied by ds."},{"Start":"13:37.140 ","End":"13:41.815","Text":"if we remember, we know that I is,"},{"Start":"13:41.815 ","End":"13:48.070","Text":"the current is also equal to the integral of J.ds,"},{"Start":"13:48.070 ","End":"13:51.670","Text":"which is exactly what we have over here,"},{"Start":"13:51.670 ","End":"13:54.040","Text":"which we said is equal to dI."},{"Start":"13:54.040 ","End":"13:55.960","Text":"What we have over here,"},{"Start":"13:55.960 ","End":"13:57.760","Text":"J.ds is equal to dI."},{"Start":"13:57.760 ","End":"14:02.200","Text":"We can see that I is the integral on dI."},{"Start":"14:02.200 ","End":"14:08.470","Text":"We just sum up the currents along these areas passing through here,"},{"Start":"14:08.470 ","End":"14:11.785","Text":"and then we get the total current passing through."},{"Start":"14:11.785 ","End":"14:21.610","Text":"Similarly, when dealing with Sigma and K. Let\u0027s say that this is still the x-direction."},{"Start":"14:21.610 ","End":"14:27.370","Text":"Let\u0027s say that this"},{"Start":"14:27.370 ","End":"14:33.080","Text":"is the y-direction and that this is the z-direction."},{"Start":"14:35.370 ","End":"14:41.530","Text":"This pointing off is the z and here going to the side is the y-direction."},{"Start":"14:41.530 ","End":"14:45.625","Text":"What we\u0027ll have is that,"},{"Start":"14:45.625 ","End":"14:51.475","Text":"we know that K is equal to Sigma"},{"Start":"14:51.475 ","End":"15:01.540","Text":"v. Here that is equal to Sigma multiplied by v in the x-direction still."},{"Start":"15:01.540 ","End":"15:08.220","Text":"Then if we look at I is equal to dq by dt."},{"Start":"15:08.220 ","End":"15:14.385","Text":"Here, dq is going to be equal to Sigma multiplied by the surface area,"},{"Start":"15:14.385 ","End":"15:18.760","Text":"which is in the xy-plane."},{"Start":"15:18.760 ","End":"15:25.090","Text":"This is in the x and this is in the y. Sigma dx dy."},{"Start":"15:25.090 ","End":"15:27.175","Text":"Then we can plug this in here."},{"Start":"15:27.175 ","End":"15:33.475","Text":"We have Sigma dx dy divided by dt,"},{"Start":"15:33.475 ","End":"15:37.315","Text":"where of course were traveling in the x-direction."},{"Start":"15:37.315 ","End":"15:42.055","Text":"This over here is going to be our velocity component."},{"Start":"15:42.055 ","End":"15:46.810","Text":"This is equal to Sigma vx,"},{"Start":"15:46.810 ","End":"15:52.190","Text":"the x component of the velocity multiplied by dy."},{"Start":"15:53.750 ","End":"15:59.535","Text":"Of course, because we have differentials over here, this is dI."},{"Start":"15:59.535 ","End":"16:04.230","Text":"Similarly, this over here is our K,"},{"Start":"16:04.230 ","End":"16:08.890","Text":"and this over here is our dl."},{"Start":"16:11.610 ","End":"16:14.830","Text":"That\u0027s exactly what we expected."},{"Start":"16:14.830 ","End":"16:17.665","Text":"Then of course, to get our I,"},{"Start":"16:17.665 ","End":"16:23.110","Text":"we know that this is the integral on K.dl,"},{"Start":"16:23.110 ","End":"16:31.940","Text":"which in this case is simply equal to the integral along this d over here."},{"Start":"16:32.670 ","End":"16:38.695","Text":"Then of course, we can look at this example where we had our solid cylinder"},{"Start":"16:38.695 ","End":"16:45.355","Text":"rotating around its axis of symmetry with an angular velocity of Omega."},{"Start":"16:45.355 ","End":"16:47.335","Text":"Then here we took,"},{"Start":"16:47.335 ","End":"16:51.920","Text":"remember some cube over here,"},{"Start":"16:51.990 ","End":"16:58.465","Text":"which was traveling in the Theta direction through this blue hole."},{"Start":"16:58.465 ","End":"17:02.140","Text":"In this example over here,"},{"Start":"17:02.140 ","End":"17:07.510","Text":"our dq is festival its volume,"},{"Start":"17:07.510 ","End":"17:12.265","Text":"but we\u0027re also using cylindrical coordinates."},{"Start":"17:12.265 ","End":"17:16.420","Text":"What we have is that our dq is equal to Rho dv,"},{"Start":"17:16.420 ","End":"17:25.760","Text":"which in cylindrical coordinates is equal to Rho r dr, d Theta dz."},{"Start":"17:27.030 ","End":"17:33.655","Text":"Then we know that I is equal to dq by dt,"},{"Start":"17:33.655 ","End":"17:37.675","Text":"which is equal to Rho r, dr,"},{"Start":"17:37.675 ","End":"17:43.550","Text":"d Theta dz divided by dt."},{"Start":"17:44.640 ","End":"17:47.379","Text":"Then if we look closely,"},{"Start":"17:47.379 ","End":"17:51.475","Text":"we have this d Theta divided by dt over here,"},{"Start":"17:51.475 ","End":"17:52.750","Text":"which as we know,"},{"Start":"17:52.750 ","End":"17:54.265","Text":"is equal to Omega."},{"Start":"17:54.265 ","End":"17:57.160","Text":"Omega is equal to d Theta by dt."},{"Start":"17:57.160 ","End":"18:04.150","Text":"Therefore, I is equal to Rho r Omega,"},{"Start":"18:04.150 ","End":"18:07.760","Text":"multiplied by dr dz."},{"Start":"18:07.760 ","End":"18:11.950","Text":"Because again, those differentials on this side,"},{"Start":"18:11.950 ","End":"18:14.660","Text":"so this is our dI."},{"Start":"18:15.000 ","End":"18:17.950","Text":"Then, as we saw earlier,"},{"Start":"18:17.950 ","End":"18:23.620","Text":"our J is Rho r Omega,"},{"Start":"18:23.620 ","End":"18:25.405","Text":"in the Theta direction."},{"Start":"18:25.405 ","End":"18:28.075","Text":"This is our J."},{"Start":"18:28.075 ","End":"18:32.035","Text":"Of course, our dr dz is our ds,"},{"Start":"18:32.035 ","End":"18:36.800","Text":"our area, which is exactly what we wanted to see."},{"Start":"18:37.230 ","End":"18:41.680","Text":"If we look at our Sigma example that we also looked at where we"},{"Start":"18:41.680 ","End":"18:45.565","Text":"have a radius of r and we\u0027re dealing with Sigma instead."},{"Start":"18:45.565 ","End":"18:48.445","Text":"Therefore, we\u0027re just looking at some kind of"},{"Start":"18:48.445 ","End":"18:54.385","Text":"square area going through this blue hole over here."},{"Start":"18:54.385 ","End":"19:01.120","Text":"What we\u0027ll have in this case is that dq is equal to Sigma ds,"},{"Start":"19:01.120 ","End":"19:10.040","Text":"where here ds is equal to R d Theta dz."},{"Start":"19:10.830 ","End":"19:20.605","Text":"Then what we have is that already write it in dI is equal to R d Theta,"},{"Start":"19:20.605 ","End":"19:24.730","Text":"dz divided by dt."},{"Start":"19:24.730 ","End":"19:28.000","Text":"Again, here we have our Omega."},{"Start":"19:28.000 ","End":"19:32.780","Text":"What we have is R Omega dz."},{"Start":"19:33.780 ","End":"19:36.550","Text":"Of course, here, we have Sigma as well."},{"Start":"19:36.550 ","End":"19:38.630","Text":"I forgot to write it."},{"Start":"19:39.930 ","End":"19:45.115","Text":"Remember the Sigma and if you remember,"},{"Start":"19:45.115 ","End":"19:52.165","Text":"this is also what we got in the earlier part of this lesson,"},{"Start":"19:52.165 ","End":"19:57.860","Text":"where this is our K and this is our dL."},{"Start":"19:58.800 ","End":"20:02.515","Text":"Then to get the total current flowing through,"},{"Start":"20:02.515 ","End":"20:06.460","Text":"we just sum up, we integrate all of this."},{"Start":"20:06.460 ","End":"20:08.275","Text":"Let\u0027s scroll back up."},{"Start":"20:08.275 ","End":"20:13.300","Text":"What we can see is that we have 2 methods for solving these types of questions."},{"Start":"20:13.300 ","End":"20:18.700","Text":"We can either find the current density by multiplying"},{"Start":"20:18.700 ","End":"20:25.075","Text":"the charge density per unit volume area or length multiplied by the velocity."},{"Start":"20:25.075 ","End":"20:32.574","Text":"Or we could use our equation for I is equal to dq by dt and then use this whole story,"},{"Start":"20:32.574 ","End":"20:36.220","Text":"substitute that in for dq and solve."},{"Start":"20:36.220 ","End":"20:39.470","Text":"Which method should I choose?"},{"Start":"20:40.260 ","End":"20:46.810","Text":"To me honest, both methods are pretty easy once you get the hang of each."},{"Start":"20:46.810 ","End":"20:49.075","Text":"However, if you\u0027re still struggling,"},{"Start":"20:49.075 ","End":"20:53.994","Text":"I would say that if you\u0027re given Rho or if you\u0027re given Lambda,"},{"Start":"20:53.994 ","End":"20:56.245","Text":"I would use these equations."},{"Start":"20:56.245 ","End":"20:58.375","Text":"But if you\u0027re given Sigma,"},{"Start":"20:58.375 ","End":"21:03.990","Text":"I would use this version of this equation over here."},{"Start":"21:03.990 ","End":"21:06.360","Text":"Just because sometimes when you\u0027re dealing with K,"},{"Start":"21:06.360 ","End":"21:10.955","Text":"it\u0027s difficult to understand which length you have to sum up along."},{"Start":"21:10.955 ","End":"21:14.135","Text":"Whereas when you use this equation, dq by dt,"},{"Start":"21:14.135 ","End":"21:17.295","Text":"using dq as Sigma ds,"},{"Start":"21:17.295 ","End":"21:19.885","Text":"it just solves it for you."},{"Start":"21:19.885 ","End":"21:22.580","Text":"You don\u0027t have to think about it. That\u0027s it."},{"Start":"21:22.580 ","End":"21:26.590","Text":"But to be honest, all of the equations are great and easy to use."},{"Start":"21:26.590 ","End":"21:30.695","Text":"Don\u0027t forget to write everything in your equation sheets if you haven\u0027t already."},{"Start":"21:30.695 ","End":"21:33.750","Text":"That is the end of this lesson."}],"ID":22418},{"Watched":false,"Name":"Deriving Equation for Current Density","Duration":"6m 48s","ChapterTopicVideoID":21313,"CourseChapterTopicPlaylistID":99491,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.950","Text":"Hello. In this lesson,"},{"Start":"00:01.950 ","End":"00:06.780","Text":"we\u0027re going to be deriving the equation for current density."},{"Start":"00:06.780 ","End":"00:11.760","Text":"Here, we have our equation for current density J, and as we know,"},{"Start":"00:11.760 ","End":"00:13.575","Text":"it\u0027s equal to Rho,"},{"Start":"00:13.575 ","End":"00:15.840","Text":"the charge density per unit volume,"},{"Start":"00:15.840 ","End":"00:20.265","Text":"multiplied by v, the velocity."},{"Start":"00:20.265 ","End":"00:29.985","Text":"Here, we have a cylinder and we can imagine that we have a current flowing like so,"},{"Start":"00:29.985 ","End":"00:33.023","Text":"and that this cylinder is some kind of resistor,"},{"Start":"00:33.023 ","End":"00:37.290","Text":"and we can imagine that the cylinder carries on in this direction and"},{"Start":"00:37.290 ","End":"00:42.630","Text":"we\u0027re just looking at one of its cross sections."},{"Start":"00:42.770 ","End":"00:46.820","Text":"As we know, if the current is flowing in this direction,"},{"Start":"00:46.820 ","End":"00:52.910","Text":"that means that we have lots of little charges that are traveling through like"},{"Start":"00:52.910 ","End":"01:02.590","Text":"so throughout the area of this circle, cross section."},{"Start":"01:02.590 ","End":"01:07.850","Text":"What do we want to do is we want to count how many charges are passing"},{"Start":"01:07.850 ","End":"01:13.495","Text":"through this small area and let\u0027s call this area ds."},{"Start":"01:13.495 ","End":"01:17.990","Text":"Of course, we want to know this as a function of time."},{"Start":"01:17.990 ","End":"01:23.610","Text":"How many charges go through this ds per unit time?"},{"Start":"01:23.930 ","End":"01:28.280","Text":"What I\u0027m going to do is I\u0027m going to take a stopwatch,"},{"Start":"01:28.280 ","End":"01:34.540","Text":"which starts at time t is equal to 0 and I stop the stopwatch at time dt,"},{"Start":"01:34.540 ","End":"01:39.770","Text":"and what I want to do is I want to count how many charges have gone"},{"Start":"01:39.770 ","End":"01:46.520","Text":"through this area ds in that given time frame."},{"Start":"01:46.520 ","End":"01:48.785","Text":"What I want to do is, of course,"},{"Start":"01:48.785 ","End":"01:51.905","Text":"I know that the closer charges will immediately pass through,"},{"Start":"01:51.905 ","End":"01:58.425","Text":"but what I want to do is I want to know what is the last charge that will pass through."},{"Start":"01:58.425 ","End":"02:03.185","Text":"Here, I have a charge going in this direction and"},{"Start":"02:03.185 ","End":"02:09.050","Text":"this distance between this charge and my area ds."},{"Start":"02:09.050 ","End":"02:12.450","Text":"Let\u0027s call this distance dx,"},{"Start":"02:12.530 ","End":"02:17.290","Text":"let\u0027s call this direction the x direction."},{"Start":"02:17.930 ","End":"02:22.785","Text":"In other words, in order for this charged particle,"},{"Start":"02:22.785 ","End":"02:24.450","Text":"let\u0027s draw it in blue,"},{"Start":"02:24.450 ","End":"02:31.205","Text":"for this charged particle to pass through this area ds in this time dt,"},{"Start":"02:31.205 ","End":"02:37.550","Text":"that means that it has to travel a distance dx in time dt."},{"Start":"02:37.550 ","End":"02:38.720","Text":"Or in other words,"},{"Start":"02:38.720 ","End":"02:44.930","Text":"its velocity v has to be equal to its distance traveled dx,"},{"Start":"02:44.930 ","End":"02:48.040","Text":"in the given time dt."},{"Start":"02:48.040 ","End":"02:50.705","Text":"Or in other words,"},{"Start":"02:50.705 ","End":"02:59.225","Text":"I know that if I have a given velocity v and I measure a total time dt,"},{"Start":"02:59.225 ","End":"03:03.380","Text":"so the last particles that will be able to enter this are"},{"Start":"03:03.380 ","End":"03:08.135","Text":"all of the particles up until this line dx,"},{"Start":"03:08.135 ","End":"03:10.100","Text":"because if they\u0027re further than dx,"},{"Start":"03:10.100 ","End":"03:16.350","Text":"they won\u0027t make it to this area in the allotted time dt."},{"Start":"03:16.690 ","End":"03:22.605","Text":"What I can do is I can make some cube shape"},{"Start":"03:22.605 ","End":"03:28.065","Text":"over here that encompasses a certain volume,"},{"Start":"03:28.065 ","End":"03:32.410","Text":"so it looks something like so."},{"Start":"03:32.680 ","End":"03:40.520","Text":"Then I know that all of the particles inside this cube will"},{"Start":"03:40.520 ","End":"03:48.540","Text":"make it in time to reach the surface area ds in my given time dt."},{"Start":"03:49.250 ","End":"03:54.670","Text":"Let\u0027s write out the volume of this cube over here,"},{"Start":"03:54.670 ","End":"04:00.515","Text":"and of course, the q reaches this particle over here because it\u0027s distance dx."},{"Start":"04:00.515 ","End":"04:08.990","Text":"The volume of this cube is going to be equal to this surface area,"},{"Start":"04:08.990 ","End":"04:10.595","Text":"which we said is ds,"},{"Start":"04:10.595 ","End":"04:12.875","Text":"multiplied by this length."},{"Start":"04:12.875 ","End":"04:16.025","Text":"That\u0027s how volume works. What is this length?"},{"Start":"04:16.025 ","End":"04:22.115","Text":"dx. Now, let\u0027s write out our dq."},{"Start":"04:22.115 ","End":"04:28.910","Text":"Our dq is simply going to be equal to our charge density per unit volume,"},{"Start":"04:28.910 ","End":"04:30.470","Text":"which is Rho,"},{"Start":"04:30.470 ","End":"04:34.385","Text":"multiplied by our dv."},{"Start":"04:34.385 ","End":"04:41.990","Text":"That is simply going to be Rho dx.ds."},{"Start":"04:43.400 ","End":"04:50.090","Text":"Another way of writing this is by saying that this is equal to the volume of"},{"Start":"04:50.090 ","End":"04:55.924","Text":"the cube multiplied by the number of charges in the volume,"},{"Start":"04:55.924 ","End":"05:01.180","Text":"multiplied by the charge q."},{"Start":"05:01.580 ","End":"05:07.040","Text":"What we have here is the number of charges in this unit volume multiplied by"},{"Start":"05:07.040 ","End":"05:12.570","Text":"their charge and this is the same thing as Rho."},{"Start":"05:12.770 ","End":"05:18.190","Text":"That\u0027s why it\u0027s the same equation and of course, dv is dx.ds."},{"Start":"05:19.850 ","End":"05:28.060","Text":"In other words, we can say therefore that our J is equal to dq,"},{"Start":"05:28.060 ","End":"05:31.715","Text":"so the charges going through here,"},{"Start":"05:31.715 ","End":"05:37.935","Text":"divided by the surface area and divided by dt."},{"Start":"05:37.935 ","End":"05:40.230","Text":"This is j."},{"Start":"05:40.230 ","End":"05:43.455","Text":"We can see that J is equal to dq by dt,"},{"Start":"05:43.455 ","End":"05:46.680","Text":"which is I, or in other words,"},{"Start":"05:46.680 ","End":"05:51.455","Text":"dI, divided by the surface area ds."},{"Start":"05:51.455 ","End":"05:54.890","Text":"This will give us our current density which is how"},{"Start":"05:54.890 ","End":"05:59.550","Text":"much current is passing per unit of area."},{"Start":"06:00.950 ","End":"06:04.410","Text":"Let\u0027s substitute in our dq."},{"Start":"06:04.410 ","End":"06:11.940","Text":"Our dq is simply equal to Rho dx.ds,"},{"Start":"06:11.940 ","End":"06:16.990","Text":"and then divide it by ds.dt."},{"Start":"06:17.600 ","End":"06:22.195","Text":"The ds is cross out and then we have dx by dt,"},{"Start":"06:22.195 ","End":"06:24.080","Text":"which we look at over here,"},{"Start":"06:24.080 ","End":"06:31.775","Text":"is equal to v. This is simply equal to Rho v. That\u0027s how we get to this equation for J"},{"Start":"06:31.775 ","End":"06:40.210","Text":"and you can do the exact same thing for k to show that k is equal to Sigma v,"},{"Start":"06:40.210 ","End":"06:42.500","Text":"and the I, the current,"},{"Start":"06:42.500 ","End":"06:49.230","Text":"is equal to Lambda multiplied by v. That\u0027s the end of this lesson."}],"ID":21393},{"Watched":false,"Name":"Exercise 3","Duration":"11m 33s","ChapterTopicVideoID":21520,"CourseChapterTopicPlaylistID":99491,"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":"00:04.485","Text":"we\u0027re going to be answer the following question."},{"Start":"00:04.485 ","End":"00:07.710","Text":"A clinical frustum of a length L,"},{"Start":"00:07.710 ","End":"00:10.035","Text":"small base radius a,"},{"Start":"00:10.035 ","End":"00:12.870","Text":"and large base radius b is given."},{"Start":"00:12.870 ","End":"00:14.910","Text":"Its resistivity is Rho,"},{"Start":"00:14.910 ","End":"00:21.675","Text":"and we\u0027re being told that this is the direction of the current."},{"Start":"00:21.675 ","End":"00:27.280","Text":"We\u0027re being asked to calculate the resistance between the 2 bases."},{"Start":"00:27.350 ","End":"00:34.395","Text":"We\u0027ve seen that the equation for resistance is equal to the resistivity"},{"Start":"00:34.395 ","End":"00:41.555","Text":"multiplied by the length of the path taken divided by the cross-sectional area."},{"Start":"00:41.555 ","End":"00:44.900","Text":"We were told that we can only use this equation when each one of"},{"Start":"00:44.900 ","End":"00:48.860","Text":"these is uniform, or constant throughout."},{"Start":"00:48.860 ","End":"00:54.980","Text":"Now, we can see that resistivity is given to us and it appears to be a constant."},{"Start":"00:54.980 ","End":"00:58.115","Text":"We can see that the length is this length L,"},{"Start":"00:58.115 ","End":"01:00.320","Text":"this is also a constant."},{"Start":"01:00.320 ","End":"01:02.585","Text":"The path taken by the current,"},{"Start":"01:02.585 ","End":"01:04.640","Text":"it\u0027s this length L. However,"},{"Start":"01:04.640 ","End":"01:08.240","Text":"the cross-sectional area is changing."},{"Start":"01:08.240 ","End":"01:10.940","Text":"We can see that at the start,"},{"Start":"01:10.940 ","End":"01:15.095","Text":"the cross-sectional area is this circle of radius a,"},{"Start":"01:15.095 ","End":"01:17.000","Text":"which is smaller and at the end,"},{"Start":"01:17.000 ","End":"01:20.240","Text":"the radius keeps on increasing up until we get to"},{"Start":"01:20.240 ","End":"01:24.115","Text":"a cross-sectional area of this circle of radius b."},{"Start":"01:24.115 ","End":"01:29.809","Text":"What we have to do is we have to do our usual trick where we cut"},{"Start":"01:29.809 ","End":"01:36.040","Text":"this conical frustum in 2 little pieces,"},{"Start":"01:36.040 ","End":"01:40.750","Text":"where they\u0027re very thin conical frustums,"},{"Start":"01:40.750 ","End":"01:45.660","Text":"where this change in line with over here,"},{"Start":"01:45.660 ","End":"01:47.565","Text":"let\u0027s call this the x-axis."},{"Start":"01:47.565 ","End":"01:50.630","Text":"This length over here is dx."},{"Start":"01:50.630 ","End":"01:55.550","Text":"If we imagine that this dx is very small and is approaching 0,"},{"Start":"01:55.550 ","End":"02:00.620","Text":"then we can assume that the surface area of this circle"},{"Start":"02:00.620 ","End":"02:06.540","Text":"over here is equal to the surface area of this over here."},{"Start":"02:06.860 ","End":"02:10.809","Text":"The first thing that we\u0027re going to do is we\u0027re going to calculate"},{"Start":"02:10.809 ","End":"02:14.815","Text":"the resistance of this disk over here."},{"Start":"02:14.815 ","End":"02:22.850","Text":"Let\u0027s imagine that it is of radius R. The resistance of this thin disk dR,"},{"Start":"02:22.850 ","End":"02:28.975","Text":"is equal to the resistivity multiplied by the distance of the path taken,"},{"Start":"02:28.975 ","End":"02:30.810","Text":"the current\u0027s path taken through,"},{"Start":"02:30.810 ","End":"02:33.505","Text":"so that is dx;"},{"Start":"02:33.505 ","End":"02:36.445","Text":"divided by the cross-sectional area,"},{"Start":"02:36.445 ","End":"02:40.420","Text":"which is just the cross-sectional area of a circle of radius r,"},{"Start":"02:40.420 ","End":"02:42.025","Text":"so that\u0027s equal to Pi r^2."},{"Start":"02:42.025 ","End":"02:51.075","Text":"If we assume that our origin is over here,"},{"Start":"02:51.075 ","End":"02:59.780","Text":"then what we can do is we can say that this distance over here is a distance x."},{"Start":"02:59.780 ","End":"03:05.225","Text":"Now, what we want to do is we want to see what r is equal to,"},{"Start":"03:05.225 ","End":"03:07.800","Text":"what this radius is."},{"Start":"03:08.990 ","End":"03:11.030","Text":"As we can see,"},{"Start":"03:11.030 ","End":"03:16.610","Text":"I\u0027m going to be integrating with this variable of dx or x,"},{"Start":"03:16.610 ","End":"03:18.620","Text":"and my r,"},{"Start":"03:18.620 ","End":"03:22.160","Text":"I have to try and find the connection between"},{"Start":"03:22.160 ","End":"03:26.640","Text":"it\u0027s an x in order to do at the integration properly."},{"Start":"03:26.680 ","End":"03:31.070","Text":"The first thing that I\u0027m going to do is I\u0027m going to try,"},{"Start":"03:31.070 ","End":"03:34.895","Text":"and connect my x with my r. I\u0027m going to draw this line,"},{"Start":"03:34.895 ","End":"03:41.495","Text":"which is going to reach this point over here at 90 degrees."},{"Start":"03:41.495 ","End":"03:42.830","Text":"Of course, here,"},{"Start":"03:42.830 ","End":"03:47.105","Text":"there\u0027s also a 90 degree angle over here and everything."},{"Start":"03:47.105 ","End":"03:53.250","Text":"I\u0027m going to say that this angle over here is equal to Theta."},{"Start":"03:54.260 ","End":"03:59.105","Text":"Now, what we can see is that"},{"Start":"03:59.105 ","End":"04:05.760","Text":"this distance from this line up until this point is a distance of a."},{"Start":"04:06.100 ","End":"04:10.255","Text":"Then I have an addition of,"},{"Start":"04:10.255 ","End":"04:12.970","Text":"let\u0027s call this Delta r,"},{"Start":"04:12.970 ","End":"04:14.459","Text":"a change in radius,"},{"Start":"04:14.459 ","End":"04:21.690","Text":"and A plus this Delta r is going to make up my radius over here."},{"Start":"04:21.790 ","End":"04:27.250","Text":"If I draw a straight line over here,"},{"Start":"04:27.250 ","End":"04:32.525","Text":"if I move this Delta r to over here at this point, so right now,"},{"Start":"04:32.525 ","End":"04:38.120","Text":"this is my Delta r. I know that this length over here is x,"},{"Start":"04:38.120 ","End":"04:45.085","Text":"which means that this length also over here is x. I\u0027ve just moved it up."},{"Start":"04:45.085 ","End":"04:48.500","Text":"Now, using trigonometry,"},{"Start":"04:48.500 ","End":"04:53.315","Text":"I can say that my Delta r over here,"},{"Start":"04:53.315 ","End":"04:57.750","Text":"divided by this side over here, so divided by x."},{"Start":"04:57.750 ","End":"05:07.190","Text":"What I have is opposite divided by adjacent is equal to tangent of my angle here, Theta."},{"Start":"05:08.930 ","End":"05:15.920","Text":"This is a similar triangle to if I drew another line over here,"},{"Start":"05:15.920 ","End":"05:22.710","Text":"where this length over here is simply b minus a."},{"Start":"05:23.990 ","End":"05:28.250","Text":"Then this entire length over here, as we know,"},{"Start":"05:28.250 ","End":"05:32.480","Text":"is L. This length of the entire green line over here"},{"Start":"05:32.480 ","End":"05:36.980","Text":"is L. Because these are similar triangles,"},{"Start":"05:36.980 ","End":"05:41.900","Text":"I know that Tan of Theta is also equal to b"},{"Start":"05:41.900 ","End":"05:47.670","Text":"minus a divided by L. Now,"},{"Start":"05:47.670 ","End":"05:52.970","Text":"I want to find my Delta r. What I get is that my Delta r is simply"},{"Start":"05:52.970 ","End":"05:59.000","Text":"equal to b minus a divided by L multiplied by x."},{"Start":"05:59.000 ","End":"06:08.855","Text":"Therefore, my r is equal to a plus Delta r. It\u0027s equal to a plus Delta r,"},{"Start":"06:08.855 ","End":"06:17.310","Text":"which is equal to a plus b minus a divided by L multiplied by x."},{"Start":"06:17.310 ","End":"06:19.625","Text":"Now, I can plug all of that in."},{"Start":"06:19.625 ","End":"06:28.400","Text":"I have that my dR is equal to Rho dx divided by Pi multiplied by r^2,"},{"Start":"06:28.400 ","End":"06:37.635","Text":"where all of this is r, so multiplied by a plus b minus a divided by L multiplied by x,"},{"Start":"06:37.635 ","End":"06:40.140","Text":"and this is squared."},{"Start":"06:40.140 ","End":"06:45.515","Text":"Now, I want to add up the resistance of all of these thin discs."},{"Start":"06:45.515 ","End":"06:47.840","Text":"Of course, all I have to do in order to find"},{"Start":"06:47.840 ","End":"06:52.145","Text":"the total resistance is I have to integrate along this."},{"Start":"06:52.145 ","End":"06:56.060","Text":"What we can see is that all of these thin disks are connected in"},{"Start":"06:56.060 ","End":"07:00.350","Text":"series because the current has to go through each and every single one of them."},{"Start":"07:00.350 ","End":"07:02.000","Text":"It doesn\u0027t split."},{"Start":"07:02.000 ","End":"07:04.819","Text":"That means that the disks are connected in series,"},{"Start":"07:04.819 ","End":"07:06.620","Text":"and then to add up the resistances,"},{"Start":"07:06.620 ","End":"07:09.780","Text":"we just sum them up."},{"Start":"07:10.010 ","End":"07:12.345","Text":"What are the borders?"},{"Start":"07:12.345 ","End":"07:15.425","Text":"We\u0027re going from here, which is the origin of 0,"},{"Start":"07:15.425 ","End":"07:18.665","Text":"up until this length over here,"},{"Start":"07:18.665 ","End":"07:25.460","Text":"which is L. We\u0027re summing up the full length of this frustum."},{"Start":"07:25.460 ","End":"07:33.180","Text":"Now, what we\u0027re going to do is we\u0027re going to use substitution to solve this integral."},{"Start":"07:33.180 ","End":"07:41.420","Text":"Let\u0027s call everything inside the brackets t. What we"},{"Start":"07:41.420 ","End":"07:50.285","Text":"have is that t is equal to a plus b minus a divided by L multiplied by x."},{"Start":"07:50.285 ","End":"07:58.140","Text":"Therefore, dt is equal to b minus a divided by L dx."},{"Start":"07:58.730 ","End":"08:05.100","Text":"Then if we take out all of our constants and substitute in t,"},{"Start":"08:05.100 ","End":"08:07.850","Text":"of course, what we have is Rho divided by Pi,"},{"Start":"08:07.850 ","End":"08:10.445","Text":"and then we have an integral."},{"Start":"08:10.445 ","End":"08:15.055","Text":"I won\u0027t put in the bounds because we\u0027re plugging in t over here."},{"Start":"08:15.055 ","End":"08:18.000","Text":"I want to get in terms of dx."},{"Start":"08:18.000 ","End":"08:28.020","Text":"What I\u0027ll have is I\u0027ll have Ldt divided by b minus a,"},{"Start":"08:28.020 ","End":"08:31.560","Text":"and then I have here t squared."},{"Start":"08:31.560 ","End":"08:34.340","Text":"What I did is I just took this equation over here,"},{"Start":"08:34.340 ","End":"08:36.860","Text":"and I isolated out dx."},{"Start":"08:36.860 ","End":"08:41.015","Text":"Now, again, I\u0027m going to take the constants out."},{"Start":"08:41.015 ","End":"08:47.190","Text":"What I have is Rho L divided by Pi b minus a."},{"Start":"08:47.190 ","End":"08:52.445","Text":"Then I\u0027m integrating along t to the power of negative 2 dt."},{"Start":"08:52.445 ","End":"08:58.100","Text":"What I\u0027m going to get is Rho L divided by Pi b"},{"Start":"08:58.100 ","End":"09:04.940","Text":"minus a of t to the power of negative 1,"},{"Start":"09:04.940 ","End":"09:09.515","Text":"and of course, there has to be a minus over here."},{"Start":"09:09.515 ","End":"09:12.290","Text":"I have to plug in the bounds."},{"Start":"09:12.290 ","End":"09:15.980","Text":"Now, what I\u0027m going to do is I\u0027m going to substitute"},{"Start":"09:15.980 ","End":"09:19.280","Text":"back what t is equal to instead of trying to work out"},{"Start":"09:19.280 ","End":"09:26.330","Text":"the bounds with this variable t. I\u0027m just writing out my t. What I"},{"Start":"09:26.330 ","End":"09:34.685","Text":"have is negative Rho L divided by Pi b minus a."},{"Start":"09:34.685 ","End":"09:37.715","Text":"Instead of t,"},{"Start":"09:37.715 ","End":"09:46.545","Text":"I have a plus b minus a divided by L x."},{"Start":"09:46.545 ","End":"09:50.375","Text":"This is my t_negative 1."},{"Start":"09:50.375 ","End":"10:01.230","Text":"My bounds are from 0 to L. Let\u0027s plug in our bounds."},{"Start":"10:01.230 ","End":"10:03.105","Text":"Let\u0027s just scroll over here."},{"Start":"10:03.105 ","End":"10:10.290","Text":"What I\u0027ll have is negative Rho L divided by Pi b minus a multiplied by,"},{"Start":"10:10.290 ","End":"10:13.880","Text":"so let\u0027s substitute an L. I have a plus b minus a"},{"Start":"10:13.880 ","End":"10:18.230","Text":"divided by L multiplied by L. I have a plus b minus a."},{"Start":"10:18.230 ","End":"10:24.495","Text":"What I\u0027ll have is b_negative 1,"},{"Start":"10:24.495 ","End":"10:26.250","Text":"so 1 divided by b."},{"Start":"10:26.250 ","End":"10:29.840","Text":"Then minus if I substitute 0 and over here,"},{"Start":"10:29.840 ","End":"10:31.970","Text":"so I have a plus 0,"},{"Start":"10:31.970 ","End":"10:35.460","Text":"minus 1 divided by a."},{"Start":"10:35.720 ","End":"10:43.630","Text":"Then what I can do is I can put this negative sign inside the brackets over here."},{"Start":"10:43.630 ","End":"10:46.850","Text":"This will become a positive and this will become a negative."},{"Start":"10:46.850 ","End":"10:54.720","Text":"Then what I can do is I can make this into a common denominator."},{"Start":"10:54.720 ","End":"11:03.500","Text":"What I\u0027ll get is Rho L divided by Pi b minus a multiplied by,"},{"Start":"11:03.500 ","End":"11:10.680","Text":"and then here, I\u0027ll have b minus a divided by ab."},{"Start":"11:12.620 ","End":"11:16.230","Text":"This b minus a cancels with this."},{"Start":"11:16.230 ","End":"11:21.500","Text":"My final answer for the total resistance of"},{"Start":"11:21.500 ","End":"11:28.277","Text":"this conical frustum is equal to Rho L divided by Pi ab."},{"Start":"11:28.277 ","End":"11:31.438","Text":"That\u0027s the onset,"},{"Start":"11:31.438 ","End":"11:34.380","Text":"and this is the end of the question."}],"ID":22419},{"Watched":false,"Name":"Exercise 4","Duration":"17m 37s","ChapterTopicVideoID":21521,"CourseChapterTopicPlaylistID":99491,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.890","Text":"Hello. In this lesson,"},{"Start":"00:01.890 ","End":"00:04.245","Text":"we\u0027re going to answering the following question."},{"Start":"00:04.245 ","End":"00:11.925","Text":"What we have here is a resistor in the shape of a spherical shell of inner radius R_1 and"},{"Start":"00:11.925 ","End":"00:16.365","Text":"outer radius R_2 and it is made of"},{"Start":"00:16.365 ","End":"00:23.265","Text":"a material with resistivity Rho_1 in the region between R_1 and R_3."},{"Start":"00:23.265 ","End":"00:27.255","Text":"Where R_3 is somewhere in the middle of between R_1 and R_2,"},{"Start":"00:27.255 ","End":"00:35.130","Text":"and a resistivity of Rho_2 in the region between R_3 and R_2."},{"Start":"00:35.130 ","End":"00:37.150","Text":"Question number 1,"},{"Start":"00:37.150 ","End":"00:42.260","Text":"we\u0027re being told that the current is flowing in the radial direction,"},{"Start":"00:42.260 ","End":"00:45.440","Text":"so coming from the center and moving outwards and we\u0027re"},{"Start":"00:45.440 ","End":"00:49.980","Text":"being asked to calculate the total resistance of the shell."},{"Start":"00:50.300 ","End":"00:53.690","Text":"We\u0027ve done a lot of practice on these types of questions,"},{"Start":"00:53.690 ","End":"00:56.300","Text":"we know that we start off by reminding ourselves"},{"Start":"00:56.300 ","End":"00:59.000","Text":"that the equation for resistance is equal to"},{"Start":"00:59.000 ","End":"01:04.985","Text":"the resistivity multiplied by the length of the path taken by the current,"},{"Start":"01:04.985 ","End":"01:09.770","Text":"divided by the cross sectional area and we remember that we can"},{"Start":"01:09.770 ","End":"01:14.510","Text":"only use this equation when all of these over here are uniform."},{"Start":"01:14.510 ","End":"01:16.580","Text":"What we can see number 1 is that there\u0027s"},{"Start":"01:16.580 ","End":"01:21.590","Text":"2 different regions of material with different resistivities."},{"Start":"01:21.590 ","End":"01:26.015","Text":"However, when we\u0027re located inside each individual region,"},{"Start":"01:26.015 ","End":"01:29.750","Text":"the resistivity is constant."},{"Start":"01:29.750 ","End":"01:33.170","Text":"The second thing that we can notice is that the length of"},{"Start":"01:33.170 ","End":"01:37.430","Text":"the path taken is a constant or uniform throughout however,"},{"Start":"01:37.430 ","End":"01:39.470","Text":"the cross-sectional area is not."},{"Start":"01:39.470 ","End":"01:42.605","Text":"We can see that here,"},{"Start":"01:42.605 ","End":"01:45.330","Text":"if we\u0027re located at radius R_1,"},{"Start":"01:45.330 ","End":"01:49.220","Text":"we\u0027re going to have some cross-sectional area that we\u0027re passing through,"},{"Start":"01:49.220 ","End":"01:52.535","Text":"which is going to be smaller than when we\u0027re passing"},{"Start":"01:52.535 ","End":"01:57.360","Text":"over here at this outer radius of R_2."},{"Start":"01:57.360 ","End":"02:01.280","Text":"We\u0027ve seen in all of these types of questions that what we have to do"},{"Start":"02:01.280 ","End":"02:04.598","Text":"is we have to split up our sphere,"},{"Start":"02:04.598 ","End":"02:06.065","Text":"here we\u0027re dealing with a sphere,"},{"Start":"02:06.065 ","End":"02:09.875","Text":"into lots of different spherical shells."},{"Start":"02:09.875 ","End":"02:13.565","Text":"Calculate the resistance of"},{"Start":"02:13.565 ","End":"02:19.760","Text":"each individual tiny spherical shell and add them up to get the total resistance."},{"Start":"02:20.510 ","End":"02:26.810","Text":"Here we have the spherical shell with some very, very tiny width."},{"Start":"02:26.810 ","End":"02:31.485","Text":"This width is, let\u0027s call it dr."},{"Start":"02:31.485 ","End":"02:33.258","Text":"It\u0027s very, very small."},{"Start":"02:33.258 ","End":"02:36.950","Text":"It\u0027s so small that if we look at the change in radius"},{"Start":"02:36.950 ","End":"02:42.290","Text":"between the inner side of the spherical shell and the outer side,"},{"Start":"02:42.290 ","End":"02:44.690","Text":"it\u0027s such a small difference dr,"},{"Start":"02:44.690 ","End":"02:47.630","Text":"that we can consider the surface area of"},{"Start":"02:47.630 ","End":"02:52.890","Text":"this inner green line and the surface area of the outer green line as being equal."},{"Start":"02:52.890 ","End":"02:55.230","Text":"We\u0027ve seen this trick a few times."},{"Start":"02:55.230 ","End":"02:58.640","Text":"We can say that the surface area of each one of"},{"Start":"02:58.640 ","End":"03:03.605","Text":"these green circles is equal to the surface area of a sphere,"},{"Start":"03:03.605 ","End":"03:08.700","Text":"which is equal to 4Pir^2."},{"Start":"03:08.960 ","End":"03:15.860","Text":"Now therefore we know that the resistance of each one of these tiny,"},{"Start":"03:15.860 ","End":"03:18.335","Text":"tiny spherical shells, so we\u0027ll call it dR,"},{"Start":"03:18.335 ","End":"03:21.305","Text":"is equal to the resistivity,"},{"Start":"03:21.305 ","End":"03:22.550","Text":"either 1 or 2,"},{"Start":"03:22.550 ","End":"03:31.800","Text":"depending on which case multiplied by L and divided by 4Pir^2."},{"Start":"03:31.800 ","End":"03:36.260","Text":"Now we can see that if we draw more of these spherical shells,"},{"Start":"03:36.260 ","End":"03:40.130","Text":"we can see that the current is not splitting and"},{"Start":"03:40.130 ","End":"03:44.195","Text":"it goes through each one of these individual spherical shells,"},{"Start":"03:44.195 ","End":"03:50.462","Text":"which means that we can consider each spherical shell as connected in series."},{"Start":"03:50.462 ","End":"03:54.740","Text":"Each tiny resistor is connected in series to the next 1."},{"Start":"03:54.740 ","End":"03:58.805","Text":"Which means that in order to find the total resistance,"},{"Start":"03:58.805 ","End":"04:02.300","Text":"we just have to integrate along dr,"},{"Start":"04:02.300 ","End":"04:06.845","Text":"we just add up all the resistances because they\u0027re connected in series."},{"Start":"04:06.845 ","End":"04:13.565","Text":"What we\u0027re going to have is in the region between R_1 and R_3,"},{"Start":"04:13.565 ","End":"04:16.800","Text":"we have resistivity Rho_1."},{"Start":"04:21.140 ","End":"04:27.440","Text":"The length of the path taken is going through the width of each shells,"},{"Start":"04:27.440 ","End":"04:30.340","Text":"so it\u0027s not L if we got to substitute it in,"},{"Start":"04:30.340 ","End":"04:33.125","Text":"what\u0027s the width of each individual\u0027s shell?"},{"Start":"04:33.125 ","End":"04:38.765","Text":"It\u0027s dr. This is our first integral and then we add"},{"Start":"04:38.765 ","End":"04:46.055","Text":"in the second region where our bounds are between R_3 and R_2."},{"Start":"04:46.055 ","End":"04:50.150","Text":"Here our resistivity is Rho_2 and again,"},{"Start":"04:50.150 ","End":"04:56.675","Text":"the width of each shell is dr divided by the cross-sectional area, which is 4Pir^2."},{"Start":"04:56.675 ","End":"04:59.735","Text":"What we can notice is that everything is a constant and we\u0027re just"},{"Start":"04:59.735 ","End":"05:05.940","Text":"integrating along dr divided by r^2."},{"Start":"05:05.940 ","End":"05:07.610","Text":"What we\u0027re going to get,"},{"Start":"05:07.610 ","End":"05:10.580","Text":"I\u0027m not going to do the whole integration."},{"Start":"05:10.580 ","End":"05:14.060","Text":"But if you don\u0027t see how I got to this,"},{"Start":"05:14.060 ","End":"05:16.985","Text":"please pause the video and calculate it."},{"Start":"05:16.985 ","End":"05:20.900","Text":"But what will essentially get is Rho_1 divided by"},{"Start":"05:20.900 ","End":"05:27.455","Text":"4Pi multiplied by 1 divided by R_1 minus 1 divided by"},{"Start":"05:27.455 ","End":"05:35.460","Text":"R_3 and then plus Rho_2 divided by 4Pi multiplied by"},{"Start":"05:35.460 ","End":"05:44.620","Text":"1 divided by R_3 minus 1 divided by R_2."},{"Start":"05:45.560 ","End":"05:49.070","Text":"This is the answer to Question number 1."},{"Start":"05:49.070 ","End":"05:52.645","Text":"Now let\u0027s move on to Question number 2."},{"Start":"05:52.645 ","End":"05:58.790","Text":"Question number 2 is to calculate the current density in the resistor if it is connected"},{"Start":"05:58.790 ","End":"06:04.925","Text":"to a voltage source V. Assuming that the resistor abides by Ohm\u0027s law,"},{"Start":"06:04.925 ","End":"06:10.865","Text":"we know that V is equal to IR and we\u0027re given our V,"},{"Start":"06:10.865 ","End":"06:19.200","Text":"we just calculated our R. Therefore our I is equal to V divided by R."},{"Start":"06:19.790 ","End":"06:25.760","Text":"Because we discussed already that each one of these tiny resistors that"},{"Start":"06:25.760 ","End":"06:31.156","Text":"we formed is connected in series to all the others,"},{"Start":"06:31.156 ","End":"06:33.560","Text":"if the resistors are connected in series,"},{"Start":"06:33.560 ","End":"06:37.355","Text":"that means that each mini resistor has the same"},{"Start":"06:37.355 ","End":"06:43.420","Text":"current and we can of course just carry this on until infinity."},{"Start":"06:43.420 ","End":"06:48.485","Text":"In that case, we can therefore say that J,"},{"Start":"06:48.485 ","End":"06:50.840","Text":"which is the current density,"},{"Start":"06:50.840 ","End":"06:57.361","Text":"is equal to the current which we\u0027ve already seen is the same across each one,"},{"Start":"06:57.361 ","End":"07:01.595","Text":"flowing through each one divided by A, the cross-sectional area."},{"Start":"07:01.595 ","End":"07:06.290","Text":"We have I, which is this over here, where this is of course,"},{"Start":"07:06.290 ","End":"07:10.115","Text":"the resistance divided by the cross-sectional area of a sphere,"},{"Start":"07:10.115 ","End":"07:19.385","Text":"which is 4Pir^2 and this is of course traveling in the radial direction."},{"Start":"07:19.385 ","End":"07:26.900","Text":"Of course, if we put our voltage source where the positive end is connected to"},{"Start":"07:26.900 ","End":"07:34.898","Text":"the inner shell and the negative side is connected to the outer shell,"},{"Start":"07:34.898 ","End":"07:40.130","Text":"then we will get this in the positive radial direction and if this"},{"Start":"07:40.130 ","End":"07:42.740","Text":"was the other way round where the positive side was"},{"Start":"07:42.740 ","End":"07:45.815","Text":"connected to the outer shell of the voltage source,"},{"Start":"07:45.815 ","End":"07:48.534","Text":"then this would be in the negative radial direction."},{"Start":"07:48.534 ","End":"07:51.270","Text":"Because we were told that the current is in the radial direction,"},{"Start":"07:51.270 ","End":"07:54.635","Text":"we can assume that that\u0027s in the positive radial direction so"},{"Start":"07:54.635 ","End":"07:59.400","Text":"we can assume that this is what the voltage source looks like."},{"Start":"07:59.600 ","End":"08:03.140","Text":"That\u0027s with regards to the direction."},{"Start":"08:03.140 ","End":"08:06.200","Text":"The next thing that I just wanted to mention is that we can see that"},{"Start":"08:06.200 ","End":"08:11.000","Text":"J is a uniform on the specific shell that it is."},{"Start":"08:11.000 ","End":"08:16.335","Text":"We can see what does that mean that J is with respect"},{"Start":"08:16.335 ","End":"08:22.790","Text":"to r. J is determined by the radius that we are at."},{"Start":"08:22.790 ","End":"08:28.340","Text":"But what we can see is that if we\u0027re at 1 constant radius r,"},{"Start":"08:28.340 ","End":"08:31.260","Text":"J across it is uniform."},{"Start":"08:31.260 ","End":"08:34.670","Text":"There\u0027s not going to be certain areas on the sphere"},{"Start":"08:34.670 ","End":"08:38.525","Text":"that has a higher current density than others."},{"Start":"08:38.525 ","End":"08:44.344","Text":"The current density across the specific sphere of radius r is going to be uniform."},{"Start":"08:44.344 ","End":"08:48.260","Text":"But of course, the current density at a larger radius is going to"},{"Start":"08:48.260 ","End":"08:53.040","Text":"be smaller because of course the surface area is greater."},{"Start":"08:53.040 ","End":"08:58.279","Text":"That\u0027s also something to just keep in mind and remember."},{"Start":"08:58.279 ","End":"09:00.950","Text":"This is the answer to Question number 2."},{"Start":"09:00.950 ","End":"09:02.780","Text":"Now let\u0027s move on to Question number 3."},{"Start":"09:02.780 ","End":"09:05.935","Text":"What is the electric field in the resistor?"},{"Start":"09:05.935 ","End":"09:09.200","Text":"We are dealing with resistivity so that"},{"Start":"09:09.200 ","End":"09:11.870","Text":"means that our equation for the electric field is equal"},{"Start":"09:11.870 ","End":"09:18.525","Text":"to the resistivity multiplied by J, the current density."},{"Start":"09:18.525 ","End":"09:23.585","Text":"Therefore, we can say that our electric field"},{"Start":"09:23.585 ","End":"09:28.990","Text":"is different in 2 different regions depending on the resistivity."},{"Start":"09:28.990 ","End":"09:30.625","Text":"In the first region,"},{"Start":"09:30.625 ","End":"09:35.535","Text":"it\u0027s Rho_1 multiplied by J, which is I,"},{"Start":"09:35.535 ","End":"09:42.785","Text":"divided by 4Pir^2 in the radial direction and this is in the region of"},{"Start":"09:42.785 ","End":"09:50.295","Text":"R_1 and R_3 and in the second region we have Rho_2,"},{"Start":"09:50.295 ","End":"09:54.240","Text":"again multiplied by the same J. I divided by"},{"Start":"09:54.240 ","End":"10:03.800","Text":"4Pir^2 in the radial direction and this is between R_3 and R_2."},{"Start":"10:03.800 ","End":"10:08.050","Text":"This is the answer to Question number 3,"},{"Start":"10:08.050 ","End":"10:11.125","Text":"and now let\u0027s take a look at Question number 4."},{"Start":"10:11.125 ","End":"10:17.545","Text":"Calculate the surface and volumetric charge distribution of the shell."},{"Start":"10:17.545 ","End":"10:20.815","Text":"We\u0027re finding Rho now,"},{"Start":"10:20.815 ","End":"10:23.740","Text":"don\u0027t get confused between Rho for"},{"Start":"10:23.740 ","End":"10:30.039","Text":"volumetric charge distribution and the Rho of this chapter,"},{"Start":"10:30.039 ","End":"10:32.350","Text":"which is dealing with resistivity."},{"Start":"10:32.350 ","End":"10:35.890","Text":"What we\u0027re going to do is we\u0027re going to mark this with a tilde."},{"Start":"10:35.890 ","End":"10:40.370","Text":"This is volumetric charge distribution."},{"Start":"10:40.430 ","End":"10:47.095","Text":"Rho tilde volumetric charge distribution is equal to,"},{"Start":"10:47.095 ","End":"10:52.090","Text":"so we had a whole chapter speaking about how to find the charge distribution."},{"Start":"10:52.090 ","End":"10:53.380","Text":"But if you don\u0027t remember,"},{"Start":"10:53.380 ","End":"10:58.510","Text":"volumetric charge distribution is calculated by this equation,"},{"Start":"10:58.510 ","End":"11:06.500","Text":"Epsilon_naught multiplied by the divergence of the electric field."},{"Start":"11:07.200 ","End":"11:10.390","Text":"The divergence of E,"},{"Start":"11:10.390 ","End":"11:16.465","Text":"so we\u0027re dealing with spherical coordinates because we have a spherical shell."},{"Start":"11:16.465 ","End":"11:19.840","Text":"The divergence, and spherical coordinates is equal to,"},{"Start":"11:19.840 ","End":"11:27.250","Text":"so we have Epsilon_naught multiplied by 1 divided by r^2 d by"},{"Start":"11:27.250 ","End":"11:35.980","Text":"dr of r^2 multiplied by the radial component of the E field,"},{"Start":"11:35.980 ","End":"11:45.445","Text":"so luckily enough, we have in our equation for the E field just a radial component,"},{"Start":"11:45.445 ","End":"11:47.740","Text":"and of course you do this,"},{"Start":"11:47.740 ","End":"11:52.165","Text":"minus the same because we\u0027re looking at the change in the E field,"},{"Start":"11:52.165 ","End":"11:58.285","Text":"so you do this minus this one for the E field in the second region."},{"Start":"11:58.285 ","End":"12:00.130","Text":"But in both cases,"},{"Start":"12:00.130 ","End":"12:01.900","Text":"we have the same equation."},{"Start":"12:01.900 ","End":"12:04.375","Text":"What we\u0027ll see is that this is equal to 0,"},{"Start":"12:04.375 ","End":"12:10.615","Text":"so we have Epsilon_naught multiplied by 1 divided by r-squared of d by dr,"},{"Start":"12:10.615 ","End":"12:15.265","Text":"and then we have the radial component,"},{"Start":"12:15.265 ","End":"12:17.020","Text":"which is loads of constants,"},{"Start":"12:17.020 ","End":"12:19.630","Text":"a bunch of constants divided by r^2."},{"Start":"12:19.630 ","End":"12:22.225","Text":"But once they\u0027re multiplied by this r^2,"},{"Start":"12:22.225 ","End":"12:23.800","Text":"the r^2 cancel out."},{"Start":"12:23.800 ","End":"12:25.160","Text":"Then we have row 1,"},{"Start":"12:25.160 ","End":"12:29.515","Text":"I divided by 4 Phi and of course,"},{"Start":"12:29.515 ","End":"12:32.905","Text":"d by dr of this constant is going to be equal to 0,"},{"Start":"12:32.905 ","End":"12:35.590","Text":"which is the same that we would have gotten for"},{"Start":"12:35.590 ","End":"12:42.505","Text":"the volumetric charge distribution or the divergence of the E field in this region."},{"Start":"12:42.505 ","End":"12:49.700","Text":"Therefore, we can see that the volumetric charge distribution is equal to 0."},{"Start":"12:50.490 ","End":"12:54.400","Text":"Now we\u0027re dealing with Sigma."},{"Start":"12:54.400 ","End":"12:59.346","Text":"Of course, Sigma in this chapter refers to conductivity,"},{"Start":"12:59.346 ","End":"13:02.530","Text":"so we\u0027re going to put another tilde over here,"},{"Start":"13:02.530 ","End":"13:09.385","Text":"and this is just going to be surface charge distribution."},{"Start":"13:09.385 ","End":"13:12.535","Text":"Just remember this surface charge distribution"},{"Start":"13:12.535 ","End":"13:15.340","Text":"and in our chapter dealing with charges distribution,"},{"Start":"13:15.340 ","End":"13:19.060","Text":"we saw that this is equal to Epsilon_naught"},{"Start":"13:19.060 ","End":"13:24.490","Text":"multiplied by the change in the perpendicular electric field."},{"Start":"13:24.490 ","End":"13:31.780","Text":"What does it mean? It means that I\u0027m looking at regions over here where there\u0027s"},{"Start":"13:31.780 ","End":"13:39.430","Text":"going to be some potential jump in the electric field due to changes along the surface."},{"Start":"13:39.430 ","End":"13:45.820","Text":"For instance, I can see that between a radius of 0 up until R_1,"},{"Start":"13:45.820 ","End":"13:48.010","Text":"I have nothing in the middle,"},{"Start":"13:48.010 ","End":"13:53.440","Text":"so that means that my E field in this region is equal to 0."},{"Start":"13:53.440 ","End":"13:58.600","Text":"But then I know that I have an E field and the region between R_1 and R_3,"},{"Start":"13:58.600 ","End":"14:01.150","Text":"which is what I calculated over here."},{"Start":"14:01.150 ","End":"14:06.966","Text":"Similarly, between R_1 and R_3, and R_2,"},{"Start":"14:06.966 ","End":"14:10.180","Text":"I also have a change in the electric field,"},{"Start":"14:10.180 ","End":"14:12.370","Text":"so that means that over here,"},{"Start":"14:12.370 ","End":"14:14.215","Text":"the radius R_3,"},{"Start":"14:14.215 ","End":"14:17.635","Text":"there\u0027s going to be a jump in the electric field."},{"Start":"14:17.635 ","End":"14:21.865","Text":"Similarly, I know that at a radius greater than R_2,"},{"Start":"14:21.865 ","End":"14:23.455","Text":"so in this region over here,"},{"Start":"14:23.455 ","End":"14:27.250","Text":"my electric field is also equal to 0 because I have nothing over here."},{"Start":"14:27.250 ","End":"14:32.530","Text":"But I know that I do have an E field in this light gray region over here."},{"Start":"14:32.530 ","End":"14:33.808","Text":"We just calculated it,"},{"Start":"14:33.808 ","End":"14:37.465","Text":"so I\u0027m assuming that on the surface over here,"},{"Start":"14:37.465 ","End":"14:38.995","Text":"there\u0027s also going to be"},{"Start":"14:38.995 ","End":"14:44.330","Text":"some charge distribution causing this jump in the electric field."},{"Start":"14:44.330 ","End":"14:49.805","Text":"The regions that I\u0027m expecting a jump is in this region over here,"},{"Start":"14:49.805 ","End":"14:51.400","Text":"this region over here,"},{"Start":"14:51.400 ","End":"14:54.025","Text":"and this region over here."},{"Start":"14:54.025 ","End":"14:56.140","Text":"Let\u0027s do this."},{"Start":"14:56.140 ","End":"15:00.460","Text":"My surface charge distribution at R_1,"},{"Start":"15:00.460 ","End":"15:02.875","Text":"I\u0027m expecting a jump is going to be equal to"},{"Start":"15:02.875 ","End":"15:07.240","Text":"Epsilon_naught multiplied by the change in the E field."},{"Start":"15:07.240 ","End":"15:13.269","Text":"I\u0027m going to take the E field at radius R_1 plus,"},{"Start":"15:13.269 ","End":"15:16.360","Text":"slightly above this radius R_1,"},{"Start":"15:16.360 ","End":"15:18.685","Text":"so I\u0027m located in this region over here,"},{"Start":"15:18.685 ","End":"15:20.755","Text":"minus the E field,"},{"Start":"15:20.755 ","End":"15:23.350","Text":"just under my radius R_1,"},{"Start":"15:23.350 ","End":"15:24.670","Text":"so at this region over here,"},{"Start":"15:24.670 ","End":"15:27.894","Text":"so let\u0027s call this R_1 minus."},{"Start":"15:27.894 ","End":"15:32.410","Text":"I have Epsilon_naught multiplied by my E field here,"},{"Start":"15:32.410 ","End":"15:39.880","Text":"which is Rho_1 I divided by 4 Phi R_1^2."},{"Start":"15:39.880 ","End":"15:43.300","Text":"Then I\u0027m subtracting the E field just below,"},{"Start":"15:43.300 ","End":"15:44.575","Text":"so in this region here,"},{"Start":"15:44.575 ","End":"15:46.150","Text":"where my E field is 0,"},{"Start":"15:46.150 ","End":"15:47.650","Text":"so I subtract 0,"},{"Start":"15:47.650 ","End":"15:50.725","Text":"or in other words, this is my onset."},{"Start":"15:50.725 ","End":"15:58.600","Text":"Then my surface charge distribution at my next region is at R_3."},{"Start":"15:58.600 ","End":"16:04.030","Text":"I\u0027m going to have Epsilon_naught multiplied by the E field just above"},{"Start":"16:04.030 ","End":"16:11.890","Text":"my R_3 radius minus the E field just below my R_3 radius."},{"Start":"16:11.890 ","End":"16:13.765","Text":"What I\u0027m going to have,"},{"Start":"16:13.765 ","End":"16:15.370","Text":"I\u0027m not going to do the whole calculation,"},{"Start":"16:15.370 ","End":"16:19.930","Text":"but I\u0027m just substituting R_3 into these 2 and subtracting them,"},{"Start":"16:19.930 ","End":"16:28.360","Text":"so what we are going to get is I multiplied by Epsilon_naught divided"},{"Start":"16:28.360 ","End":"16:37.750","Text":"by 4 Phi R_3^2 multiplied by Rho_2 minus Rho_1 where,"},{"Start":"16:37.750 ","End":"16:40.600","Text":"of course these are the resistivity."},{"Start":"16:40.600 ","End":"16:48.460","Text":"I just subtracted this from this."},{"Start":"16:48.460 ","End":"16:58.270","Text":"My next region is that this radius R_2 Sigma surface charge distribution R_2,"},{"Start":"16:58.270 ","End":"17:00.835","Text":"which is going to be Epsilon_naught."},{"Start":"17:00.835 ","End":"17:05.230","Text":"Then I\u0027m looking at the electric field at this point just above R_2,"},{"Start":"17:05.230 ","End":"17:07.690","Text":"which we know is equal to 0 so I\u0027m just not going to write it"},{"Start":"17:07.690 ","End":"17:10.990","Text":"out minus the electric field just below this point,"},{"Start":"17:10.990 ","End":"17:13.150","Text":"which is over here in this region,"},{"Start":"17:13.150 ","End":"17:15.550","Text":"so I have a negative sign."},{"Start":"17:15.550 ","End":"17:19.908","Text":"First of all negative Epsilon_naught multiplied by I"},{"Start":"17:19.908 ","End":"17:29.305","Text":"Rho_2 divided by 4 Phi R_2^2."},{"Start":"17:29.305 ","End":"17:33.775","Text":"These are the answers to Question number 4,"},{"Start":"17:33.775 ","End":"17:37.370","Text":"and that is the end of this lesson."}],"ID":22420},{"Watched":false,"Name":"Exercise 5","Duration":"20m 35s","ChapterTopicVideoID":21314,"CourseChapterTopicPlaylistID":99491,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.890","Text":"Hello. In this lesson,"},{"Start":"00:01.890 ","End":"00:04.380","Text":"we\u0027re going to be answering the following question."},{"Start":"00:04.380 ","End":"00:09.525","Text":"A parallel plate capacitor of charge Q is connected to"},{"Start":"00:09.525 ","End":"00:15.450","Text":"a resistor of resistance R. The radius of the capacitor plate is a,"},{"Start":"00:15.450 ","End":"00:18.915","Text":"and the distance between the plates is d,"},{"Start":"00:18.915 ","End":"00:20.670","Text":"such that d is much,"},{"Start":"00:20.670 ","End":"00:23.205","Text":"much smaller than a."},{"Start":"00:23.205 ","End":"00:28.270","Text":"Question number 1 is to calculate the current in the circuit."},{"Start":"00:28.880 ","End":"00:35.510","Text":"The first thing we can see is that we have a circuit with a capacitor and a resistor,"},{"Start":"00:35.510 ","End":"00:36.860","Text":"or in other words,"},{"Start":"00:36.860 ","End":"00:40.250","Text":"what we\u0027ve learned from a previous chapter is that we\u0027re dealing"},{"Start":"00:40.250 ","End":"00:44.245","Text":"with a discharging RC circuit."},{"Start":"00:44.245 ","End":"00:50.839","Text":"First of all, we\u0027ll write down what are capacitance is on this capacitor."},{"Start":"00:50.839 ","End":"00:53.720","Text":"The general equation for capacitance of"},{"Start":"00:53.720 ","End":"00:57.679","Text":"a parallel plate capacitor is Epsilon Naught multiplied"},{"Start":"00:57.679 ","End":"01:00.530","Text":"by the cross-sectional area of"},{"Start":"01:00.530 ","End":"01:04.655","Text":"the capacitor divided by the distance between the 2 plates,"},{"Start":"01:04.655 ","End":"01:06.890","Text":"which in our case, we have Epsilon Naught,"},{"Start":"01:06.890 ","End":"01:09.200","Text":"the cross-sectional area of this capacitor,"},{"Start":"01:09.200 ","End":"01:11.075","Text":"as we have circular plates,"},{"Start":"01:11.075 ","End":"01:13.430","Text":"is Pi a^2,"},{"Start":"01:13.430 ","End":"01:18.160","Text":"where a is the radius and divided by the distance between the 2 plates,"},{"Start":"01:18.160 ","End":"01:21.200","Text":"which is d."},{"Start":"01:26.150 ","End":"01:28.690","Text":"Our charge"},{"Start":"01:28.690 ","End":"01:31.150","Text":"at t is equal to 0."},{"Start":"01:31.150 ","End":"01:35.295","Text":"We\u0027re told in the question is equal to q."},{"Start":"01:35.295 ","End":"01:39.970","Text":"Then of course, our capacitor will begin to discharge."},{"Start":"01:39.970 ","End":"01:42.490","Text":"Therefore, we can work out"},{"Start":"01:42.490 ","End":"01:46.510","Text":"the differential equation for Q as a function of t. I\u0027m not going to"},{"Start":"01:46.510 ","End":"01:48.340","Text":"calculate it now because"},{"Start":"01:48.340 ","End":"01:53.305","Text":"the previous chapter deals with this calculation solving the differential equation."},{"Start":"01:53.305 ","End":"01:57.495","Text":"But what we\u0027ll get is that the charge on the capacitor as a function of time,"},{"Start":"01:57.495 ","End":"02:00.605","Text":"is equal to Q, the initial charge,"},{"Start":"02:00.605 ","End":"02:06.515","Text":"multiplied by e to the power of negative t divided by RC,"},{"Start":"02:06.515 ","End":"02:09.575","Text":"where R is given to us in the question and C,"},{"Start":"02:09.575 ","End":"02:12.860","Text":"is this over here that we just calculated."},{"Start":"02:12.860 ","End":"02:16.415","Text":"As we know, I is equal to,"},{"Start":"02:16.415 ","End":"02:17.570","Text":"this is what we\u0027re trying to find,"},{"Start":"02:17.570 ","End":"02:18.875","Text":"the current in the circuit,"},{"Start":"02:18.875 ","End":"02:23.225","Text":"I is equal to in general dq by dt."},{"Start":"02:23.225 ","End":"02:27.155","Text":"But as mentioned in the previous chapter dealing with capacitors,"},{"Start":"02:27.155 ","End":"02:29.930","Text":"when we have a discharging circuit,"},{"Start":"02:29.930 ","End":"02:34.760","Text":"that means that the current is equal to negative dq by"},{"Start":"02:34.760 ","End":"02:39.830","Text":"dt because we are in fact speaking about the charge leaving the capacitor plate."},{"Start":"02:39.830 ","End":"02:41.810","Text":"If it\u0027s leaving, it\u0027s a minus."},{"Start":"02:41.810 ","End":"02:44.000","Text":"If the capacitor was charging,"},{"Start":"02:44.000 ","End":"02:46.370","Text":"then charges are joining the plate,"},{"Start":"02:46.370 ","End":"02:49.355","Text":"which means that I would be a positive."},{"Start":"02:49.355 ","End":"02:53.015","Text":"Here, the current is negative dq by dt,"},{"Start":"02:53.015 ","End":"02:57.945","Text":"just because our capacitor is discharging."},{"Start":"02:57.945 ","End":"03:02.360","Text":"Of course, we just have to take the negative derivative of this."},{"Start":"03:02.360 ","End":"03:07.610","Text":"The negative from this over here in a derivative and this will cancel"},{"Start":"03:07.610 ","End":"03:13.035","Text":"out and then what we\u0027re left with is Q divided by RC,"},{"Start":"03:13.035 ","End":"03:14.840","Text":"where C is this over here,"},{"Start":"03:14.840 ","End":"03:21.660","Text":"multiplied by e to the power of negative t divided by RC."},{"Start":"03:22.130 ","End":"03:25.025","Text":"That\u0027s the answer to question number 1."},{"Start":"03:25.025 ","End":"03:28.210","Text":"Now let\u0027s take a look at question number 2"},{"Start":"03:28.210 ","End":"03:32.185","Text":"calculate the current density on the capacitor plate."},{"Start":"03:32.185 ","End":"03:35.110","Text":"Let\u0027s take a look at our capacitor plate."},{"Start":"03:35.110 ","End":"03:40.960","Text":"We know that on the plate we have lots and lots of charges over here."},{"Start":"03:40.960 ","End":"03:42.535","Text":"This is the positive plates."},{"Start":"03:42.535 ","End":"03:44.830","Text":"We have lots and lots of charges."},{"Start":"03:44.830 ","End":"03:48.630","Text":"How does a capacitor discharge?"},{"Start":"03:48.630 ","End":"03:53.455","Text":"Let\u0027s look at our capacitor plate from a bird\u0027s eye view."},{"Start":"03:53.455 ","End":"03:55.420","Text":"Here it\u0027s a radius a."},{"Start":"03:55.420 ","End":"03:56.845","Text":"What happens is,"},{"Start":"03:56.845 ","End":"04:00.385","Text":"so we have all of these charges around and also in the middle."},{"Start":"04:00.385 ","End":"04:04.650","Text":"But what happens when the capacitor plate discharges."},{"Start":"04:04.650 ","End":"04:09.920","Text":"All the charges move in towards the center,"},{"Start":"04:09.920 ","End":"04:12.260","Text":"towards where our wire is."},{"Start":"04:12.260 ","End":"04:15.380","Text":"Our wire is over here in the center of the plate."},{"Start":"04:15.380 ","End":"04:20.645","Text":"All the charges move towards the center of the plate where there\u0027s a wire"},{"Start":"04:20.645 ","End":"04:26.720","Text":"and then the charges will flow in this direction up the wire,"},{"Start":"04:26.720 ","End":"04:31.515","Text":"through the resistor and over onto the negatively charged plate."},{"Start":"04:31.515 ","End":"04:34.285","Text":"That is how the capacitor discharges."},{"Start":"04:34.285 ","End":"04:38.375","Text":"There\u0027s this flow of charge to the center."},{"Start":"04:38.375 ","End":"04:43.685","Text":"That is what is very important to remember and to understand."},{"Start":"04:43.685 ","End":"04:49.370","Text":"What we\u0027re going to do in order to somehow understand this flow of charges,"},{"Start":"04:49.370 ","End":"04:53.585","Text":"we\u0027re going to choose some arbitrary radius,"},{"Start":"04:53.585 ","End":"04:58.400","Text":"and let\u0027s call this radius, r. Now,"},{"Start":"04:58.400 ","End":"05:00.980","Text":"what we want to do is we want to calculate"},{"Start":"05:00.980 ","End":"05:06.230","Text":"how many charges are in this blue region over here at a radius"},{"Start":"05:06.230 ","End":"05:10.220","Text":"greater than r. What we\u0027re going to"},{"Start":"05:10.220 ","End":"05:15.320","Text":"do is first we\u0027re going to calculate the charge distribution."},{"Start":"05:15.460 ","End":"05:18.710","Text":"Remember this isn\u0027t conductivity,"},{"Start":"05:18.710 ","End":"05:22.730","Text":"this is charge distribution and because we haven\u0027t been told otherwise,"},{"Start":"05:22.730 ","End":"05:28.565","Text":"we can assume that the charge that we calculated is evenly distributed throughout."},{"Start":"05:28.565 ","End":"05:32.435","Text":"The charge distribution is going to be equal to"},{"Start":"05:32.435 ","End":"05:38.090","Text":"our charge as a function of time divided by the total area,"},{"Start":"05:38.090 ","End":"05:40.835","Text":"then we have a uniform charge distribution."},{"Start":"05:40.835 ","End":"05:45.480","Text":"What we have is Qe to the power of"},{"Start":"05:45.480 ","End":"05:51.695","Text":"negative t divided by RC divided by the total area,"},{"Start":"05:51.695 ","End":"05:55.210","Text":"which is Pi a^2."},{"Start":"05:55.210 ","End":"06:00.860","Text":"Now, what do we want to do is we want to see what is the charge as"},{"Start":"06:00.860 ","End":"06:07.080","Text":"a function of time in this blue region given this charge distribution."},{"Start":"06:07.520 ","End":"06:12.895","Text":"What we\u0027re going to have is that Q as a function of r,"},{"Start":"06:12.895 ","End":"06:16.520","Text":"so the Q greater than this radius."},{"Start":"06:17.210 ","End":"06:21.300","Text":"Let\u0027s write at a radius greater than,"},{"Start":"06:21.300 ","End":"06:31.060","Text":"this blue r. This is going to be equal to the charge distribution."},{"Start":"06:31.060 ","End":"06:34.160","Text":"That\u0027s Sigma."},{"Start":"06:34.160 ","End":"06:36.100","Text":"Of course,"},{"Start":"06:36.100 ","End":"06:38.755","Text":"Sigma is as a function of time."},{"Start":"06:38.755 ","End":"06:45.490","Text":"Sigma as a function of time multiplied by this area in blue."},{"Start":"06:45.490 ","End":"06:47.545","Text":"What is this area in blue?"},{"Start":"06:47.545 ","End":"06:53.460","Text":"It\u0027s equal to Pi a^2 minus r^2."},{"Start":"06:53.460 ","End":"06:58.760","Text":"That\u0027s this region over here in blue and the outskirts."},{"Start":"06:58.760 ","End":"07:03.361","Text":"Once we just play around with this,"},{"Start":"07:03.361 ","End":"07:07.790","Text":"of course our sigma as a function of time is equal to this, and this is,"},{"Start":"07:07.790 ","End":"07:14.220","Text":"of course, this over here is Q as a function of time."},{"Start":"07:14.220 ","End":"07:21.500","Text":"What we\u0027re going to have is that this is equal to a^2 minus r^2 divided"},{"Start":"07:21.500 ","End":"07:28.820","Text":"by a^2 multiplied by q as a function of time."},{"Start":"07:28.820 ","End":"07:31.445","Text":"Once you substitute in sigma,"},{"Start":"07:31.445 ","End":"07:32.945","Text":"all of this over here,"},{"Start":"07:32.945 ","End":"07:38.010","Text":"and then the Pi\u0027s cancel out and you\u0027re just left with this."},{"Start":"07:38.930 ","End":"07:44.375","Text":"Remember, we\u0027re trying to calculate the current density."},{"Start":"07:44.375 ","End":"07:49.010","Text":"What we want to do is we want to see therefore the current that"},{"Start":"07:49.010 ","End":"07:54.165","Text":"this blue section is providing."},{"Start":"07:54.165 ","End":"07:57.950","Text":"We\u0027re remembering that this is discharging."},{"Start":"07:57.950 ","End":"08:00.605","Text":"Just like in question number 1,"},{"Start":"08:00.605 ","End":"08:05.690","Text":"the current is equal to the amount of charges leaving the blue region."},{"Start":"08:05.690 ","End":"08:08.200","Text":"The amount of charges leaving"},{"Start":"08:08.200 ","End":"08:12.350","Text":"the blue region and traveling towards the center of the plate."},{"Start":"08:12.350 ","End":"08:17.630","Text":"Our current, which is as a function of the radius and the time."},{"Start":"08:17.630 ","End":"08:19.760","Text":"The radius over here and time,"},{"Start":"08:19.760 ","End":"08:23.030","Text":"because Q is as a function of time,"},{"Start":"08:23.030 ","End":"08:30.720","Text":"is simply going to be equal to negative Q dot."},{"Start":"08:30.720 ","End":"08:34.635","Text":"The negative, as we just explained."},{"Start":"08:34.635 ","End":"08:38.385","Text":"This is simply going to be equal to,"},{"Start":"08:38.385 ","End":"08:40.770","Text":"all of these are constants."},{"Start":"08:40.770 ","End":"08:48.180","Text":"a^2 minus r^2 divided by a^2 and then we have the minus here."},{"Start":"08:48.180 ","End":"08:54.155","Text":"This is multiplied by negative q dot as a function of time."},{"Start":"08:54.155 ","End":"08:59.300","Text":"This over here is our variable and this is simply equal"},{"Start":"08:59.300 ","End":"09:05.125","Text":"to the current that we calculated over here in question number 1."},{"Start":"09:05.125 ","End":"09:08.265","Text":"Negative Q dot is this,"},{"Start":"09:08.265 ","End":"09:10.980","Text":"dq by dt is Q dot."},{"Start":"09:10.980 ","End":"09:21.500","Text":"Therefore, we can write that the current as a function of r and time"},{"Start":"09:21.500 ","End":"09:29.510","Text":"in this region is equal to a^2 minus r^2 divided"},{"Start":"09:29.510 ","End":"09:38.296","Text":"by a^2 multiplied by I as a function of t from question number 1."},{"Start":"09:38.296 ","End":"09:44.515","Text":"Now we have the current and we want to calculate the current density."},{"Start":"09:44.515 ","End":"09:47.170","Text":"The first thing that we want to look at is"},{"Start":"09:47.170 ","End":"09:49.585","Text":"what type of current density are we looking at?"},{"Start":"09:49.585 ","End":"09:58.652","Text":"Are we looking at current density because we have volume or surface current density?"},{"Start":"09:58.652 ","End":"10:01.915","Text":"We can see also from the diagram and also because"},{"Start":"10:01.915 ","End":"10:05.815","Text":"we aren\u0027t told otherwise in the question about different dimensions,"},{"Start":"10:05.815 ","End":"10:09.160","Text":"that this is a parallel plate capacitor and it\u0027s"},{"Start":"10:09.160 ","End":"10:13.360","Text":"just a thin disk and it doesn\u0027t have some volume,"},{"Start":"10:13.360 ","End":"10:16.940","Text":"we just have an area over here."},{"Start":"10:17.070 ","End":"10:21.340","Text":"That means that we just have a surface area,"},{"Start":"10:21.340 ","End":"10:23.830","Text":"which means that our current density that we\u0027re trying to"},{"Start":"10:23.830 ","End":"10:27.700","Text":"calculate is going to be our surface current density,"},{"Start":"10:27.700 ","End":"10:31.840","Text":"or in other words k. What does k mean?"},{"Start":"10:31.840 ","End":"10:35.590","Text":"K-means that we\u0027re trying to look at how much current is"},{"Start":"10:35.590 ","End":"10:40.630","Text":"flowing through some boundary or some hole."},{"Start":"10:40.630 ","End":"10:42.100","Text":"In our case over here,"},{"Start":"10:42.100 ","End":"10:46.420","Text":"we\u0027ve defined this blue circle as our boundary or hole."},{"Start":"10:46.420 ","End":"10:51.835","Text":"We\u0027re trying to find how much current is flowing through this circle."},{"Start":"10:51.835 ","End":"10:57.160","Text":"Where of course we\u0027re looking at the perimeter of the circle and not the area."},{"Start":"10:57.160 ","End":"11:06.830","Text":"We remember that k is simply equal to the total current divided by length."},{"Start":"11:07.050 ","End":"11:10.120","Text":"Usually, we\u0027ll divide by DL,"},{"Start":"11:10.120 ","End":"11:18.520","Text":"where we look at some arc along the circle like so."},{"Start":"11:18.520 ","End":"11:21.550","Text":"We\u0027ll see how much current is flowing through here,"},{"Start":"11:21.550 ","End":"11:23.710","Text":"and then we\u0027ll add it all up."},{"Start":"11:23.710 ","End":"11:29.395","Text":"However here we don\u0027t really have to do that step-by-step because the amount of current"},{"Start":"11:29.395 ","End":"11:31.600","Text":"going through this arc shouldn\u0027t be"},{"Start":"11:31.600 ","End":"11:34.990","Text":"different than the amount of current going through this arc or this arc,"},{"Start":"11:34.990 ","End":"11:36.730","Text":"or this arc, or this arc,"},{"Start":"11:36.730 ","End":"11:38.950","Text":"or this arc, or this arc, or this,"},{"Start":"11:38.950 ","End":"11:43.525","Text":"because we have a uniform charge distribution and a"},{"Start":"11:43.525 ","End":"11:48.625","Text":"uniform current so the same amount of charges are going through every single point."},{"Start":"11:48.625 ","End":"11:52.195","Text":"Which means that we don\u0027t have to sum this up with an integral,"},{"Start":"11:52.195 ","End":"11:57.160","Text":"we can just take the total length of this circle."},{"Start":"11:57.160 ","End":"12:00.685","Text":"The total length of the circle is of course the perimeter of the circle,"},{"Start":"12:00.685 ","End":"12:04.600","Text":"which is equal to 2pi multiplied by the radius,"},{"Start":"12:04.600 ","End":"12:13.780","Text":"where the radius is r. Now we can just plug all of this in so what we have is"},{"Start":"12:13.780 ","End":"12:19.360","Text":"a^2 minus r^2 divided by"},{"Start":"12:19.360 ","End":"12:26.455","Text":"2pi ra^2 multiplied over here by this i,"},{"Start":"12:26.455 ","End":"12:31.990","Text":"so multiplied by Q divided by RC multiplied by e to"},{"Start":"12:31.990 ","End":"12:37.930","Text":"the power of negative t divided by RC and that\u0027s it."},{"Start":"12:37.930 ","End":"12:40.420","Text":"That\u0027s all that there is to it."},{"Start":"12:40.420 ","End":"12:42.460","Text":"This is the surface current density,"},{"Start":"12:42.460 ","End":"12:48.955","Text":"how much current is flowing through this circle over here."},{"Start":"12:48.955 ","End":"12:50.920","Text":"In order to solve this question,"},{"Start":"12:50.920 ","End":"12:52.225","Text":"it was quite difficult."},{"Start":"12:52.225 ","End":"12:56.860","Text":"We had to take a look at the charge distribution in general,"},{"Start":"12:56.860 ","End":"13:02.050","Text":"we had to realize how will the capacitor plate discharges in the first place,"},{"Start":"13:02.050 ","End":"13:04.510","Text":"which means that all of the charges are flowing towards"},{"Start":"13:04.510 ","End":"13:08.020","Text":"the center of the plate and then out via the wire."},{"Start":"13:08.020 ","End":"13:10.810","Text":"Then from the charge distribution,"},{"Start":"13:10.810 ","End":"13:13.960","Text":"we have to calculate the charge,"},{"Start":"13:13.960 ","End":"13:18.535","Text":"specifically in this region that is flowing towards the center,"},{"Start":"13:18.535 ","End":"13:22.645","Text":"from that we got the current and then from that,"},{"Start":"13:22.645 ","End":"13:25.510","Text":"we could calculate the current density."},{"Start":"13:25.510 ","End":"13:28.885","Text":"That\u0027s the end of question number 2."},{"Start":"13:28.885 ","End":"13:32.335","Text":"Now let\u0027s answer question number 3."},{"Start":"13:32.335 ","End":"13:34.090","Text":"In question number 3,"},{"Start":"13:34.090 ","End":"13:38.065","Text":"we\u0027re being told that the resistor is removed and"},{"Start":"13:38.065 ","End":"13:43.220","Text":"a material of resistivity Rho fills the gap."},{"Start":"13:43.410 ","End":"13:48.325","Text":"Over here we have Rho fills the gap between the capacitor plates."},{"Start":"13:48.325 ","End":"13:54.340","Text":"Given this information, they want us to re-answer question number 1 and 2."},{"Start":"13:54.340 ","End":"13:57.925","Text":"Remember that this over here is resistivity,"},{"Start":"13:57.925 ","End":"14:03.800","Text":"don\u0027t get confused with charge distribution per unit volume."},{"Start":"14:04.250 ","End":"14:11.280","Text":"In this question, we\u0027re just being told that we have the same capacitor,"},{"Start":"14:11.280 ","End":"14:14.835","Text":"like so, however, between its plates,"},{"Start":"14:14.835 ","End":"14:18.525","Text":"we have a material with some resistivity."},{"Start":"14:18.525 ","End":"14:21.960","Text":"What does that mean if we have a material with resistivity?"},{"Start":"14:21.960 ","End":"14:28.830","Text":"That means that this whole section over here in blue has some resistance."},{"Start":"14:28.830 ","End":"14:31.260","Text":"If something has some resistance,"},{"Start":"14:31.260 ","End":"14:32.550","Text":"what does that mean?"},{"Start":"14:32.550 ","End":"14:36.200","Text":"That means that it is just a resistor."},{"Start":"14:36.200 ","End":"14:42.490","Text":"Here we have a resistor and here we have a capacitor."},{"Start":"14:42.490 ","End":"14:47.530","Text":"We, in fact, have the exact same setup as we had previously."},{"Start":"14:47.530 ","End":"14:51.775","Text":"If the resistor is attached to the capacitor like so,"},{"Start":"14:51.775 ","End":"14:57.220","Text":"or if it\u0027s attached like so,"},{"Start":"14:57.220 ","End":"15:00.880","Text":"there is absolutely no difference."},{"Start":"15:00.880 ","End":"15:03.370","Text":"We\u0027re dealing with the exact same question,"},{"Start":"15:03.370 ","End":"15:09.700","Text":"the only difference being that this time we don\u0027t know what our resistance is."},{"Start":"15:09.700 ","End":"15:13.915","Text":"Before in the question, we were told that the resistor has a resistance R"},{"Start":"15:13.915 ","End":"15:18.490","Text":"and now we need to find the resistance of this resistor in the middle,"},{"Start":"15:18.490 ","End":"15:22.670","Text":"given this value for resistivity."},{"Start":"15:23.580 ","End":"15:27.115","Text":"What is the resistance?"},{"Start":"15:27.115 ","End":"15:30.730","Text":"We remember that the equation for resistance is equal to"},{"Start":"15:30.730 ","End":"15:36.265","Text":"the resistivity multiplied by the path length divided by the cross-sectional area,"},{"Start":"15:36.265 ","End":"15:40.180","Text":"which in our case we\u0027re given resistivity Rho."},{"Start":"15:40.180 ","End":"15:44.890","Text":"The path length is this distance between the 2 plates,"},{"Start":"15:44.890 ","End":"15:47.920","Text":"which we\u0027re being told in the question is d and"},{"Start":"15:47.920 ","End":"15:52.495","Text":"the cross-sectional area is just the cross-sectional area of the plate,"},{"Start":"15:52.495 ","End":"15:55.420","Text":"which is the area of a circle,"},{"Start":"15:55.420 ","End":"15:58.370","Text":"which is Pi a^2."},{"Start":"15:59.130 ","End":"16:01.990","Text":"Now we know the resistance."},{"Start":"16:01.990 ","End":"16:05.530","Text":"In fact, we have all of the details that we had at"},{"Start":"16:05.530 ","End":"16:09.040","Text":"the beginning of the question in order to answer questions 1 and 2,"},{"Start":"16:09.040 ","End":"16:10.750","Text":"we know the resistance."},{"Start":"16:10.750 ","End":"16:14.560","Text":"The capacitance is the same capacitance over here."},{"Start":"16:14.560 ","End":"16:22.750","Text":"Let\u0027s write it, C is equal to Epsilon Naught Pi a^2 divided by d. Now,"},{"Start":"16:22.750 ","End":"16:25.345","Text":"to answer question number 1."},{"Start":"16:25.345 ","End":"16:30.100","Text":"Given this circuit, we calculate the current in the circuit."},{"Start":"16:30.100 ","End":"16:32.620","Text":"It\u0027s just going to be the exact same thing."},{"Start":"16:32.620 ","End":"16:35.695","Text":"The current is equal to negative q dot,"},{"Start":"16:35.695 ","End":"16:41.230","Text":"this is the same Q and so the current is equal to Q divided by, RC,"},{"Start":"16:41.230 ","End":"16:43.810","Text":"where the resistance is this over here that"},{"Start":"16:43.810 ","End":"16:46.870","Text":"we calculated the capacitance was the exact same thing,"},{"Start":"16:46.870 ","End":"16:52.880","Text":"multiplied by e to the negative t divided by RC."},{"Start":"16:53.190 ","End":"16:58.405","Text":"This is the current in this circuit over here."},{"Start":"16:58.405 ","End":"17:02.905","Text":"Now the second part is we have to answer question number 2"},{"Start":"17:02.905 ","End":"17:07.120","Text":"for this different circuit over here."},{"Start":"17:07.120 ","End":"17:11.740","Text":"Question number 2 is to calculate the current density and what\u0027s important to"},{"Start":"17:11.740 ","End":"17:17.260","Text":"note in this question is that they\u0027re asking on the capacitor plates."},{"Start":"17:17.260 ","End":"17:20.635","Text":"What is the current density on the capacitor plate?"},{"Start":"17:20.635 ","End":"17:22.705","Text":"In this question, specifically,"},{"Start":"17:22.705 ","End":"17:30.070","Text":"what we can see is that our resistor is connected over here,"},{"Start":"17:30.070 ","End":"17:32.185","Text":"as we said in the middle."},{"Start":"17:32.185 ","End":"17:35.680","Text":"But of course, it isn\u0027t just a wire with a small resistor."},{"Start":"17:35.680 ","End":"17:43.345","Text":"All of this is this material with resistivity Rho, what does that mean?"},{"Start":"17:43.345 ","End":"17:48.490","Text":"That means that we can see that all of the positive charges are"},{"Start":"17:48.490 ","End":"17:53.470","Text":"just going to drop down straight through this resistor,"},{"Start":"17:53.470 ","End":"17:55.825","Text":"that we colored in blue,"},{"Start":"17:55.825 ","End":"17:58.750","Text":"straight down like so,"},{"Start":"17:58.750 ","End":"18:02.380","Text":"until they reach the negatively charged plate"},{"Start":"18:02.380 ","End":"18:05.395","Text":"and that is how the capacitor is going to discharge."},{"Start":"18:05.395 ","End":"18:09.280","Text":"We can see that unlike in the first question where we saw"},{"Start":"18:09.280 ","End":"18:13.030","Text":"that the charges were moving because the wire was connected in the middle,"},{"Start":"18:13.030 ","End":"18:18.355","Text":"the charges were moving towards the center and then up through the wire."},{"Start":"18:18.355 ","End":"18:21.355","Text":"This time in this version of the question,"},{"Start":"18:21.355 ","End":"18:27.580","Text":"the charges are just simply dropping down along the z-axis,"},{"Start":"18:27.580 ","End":"18:30.145","Text":"so in the negative z direction."},{"Start":"18:30.145 ","End":"18:37.075","Text":"In that case, we don\u0027t have any charges moving on the plate itself."},{"Start":"18:37.075 ","End":"18:39.535","Text":"If I made that circle again,"},{"Start":"18:39.535 ","End":"18:44.454","Text":"remember I did the circle and the circle was very important in finding"},{"Start":"18:44.454 ","End":"18:47.200","Text":"the current density on the capacitor plate because I was measuring"},{"Start":"18:47.200 ","End":"18:49.975","Text":"how many charges pass through the circle,"},{"Start":"18:49.975 ","End":"18:53.155","Text":"because the charges were moving in this radial direction."},{"Start":"18:53.155 ","End":"18:56.620","Text":"However, now the charges are moving in the z-direction,"},{"Start":"18:56.620 ","End":"18:58.180","Text":"not in the radial direction,"},{"Start":"18:58.180 ","End":"19:03.055","Text":"which means that there\u0027s no movement of charges on the actual plate."},{"Start":"19:03.055 ","End":"19:06.505","Text":"No charges are moving in the radial direction,"},{"Start":"19:06.505 ","End":"19:12.460","Text":"which means that k is equal to 0."},{"Start":"19:12.460 ","End":"19:21.750","Text":"The surface current density on the capacitor plates in this case is equal to 0."},{"Start":"19:22.440 ","End":"19:25.540","Text":"We\u0027ve established that in this case,"},{"Start":"19:25.540 ","End":"19:30.475","Text":"there is no movement on the plate itself so the surface current density is equal to 0."},{"Start":"19:30.475 ","End":"19:37.375","Text":"However, now if we put some circle over here,"},{"Start":"19:37.375 ","End":"19:41.365","Text":"we can see that there is a movement of charges through here,"},{"Start":"19:41.365 ","End":"19:47.810","Text":"which means that we have a regular old current density J."},{"Start":"19:47.820 ","End":"19:52.660","Text":"We have a J because we have current moving just in"},{"Start":"19:52.660 ","End":"19:58.660","Text":"a different direction where this way it\u0027s not on a surface but flowing through a volume."},{"Start":"19:58.660 ","End":"20:06.175","Text":"Where the J, we know that it is equal to the current divided by the area."},{"Start":"20:06.175 ","End":"20:09.109","Text":"What we have is our current,"},{"Start":"20:09.109 ","End":"20:17.585","Text":"which is Qe to the negative t divided by RC divided by over here,"},{"Start":"20:17.585 ","End":"20:20.495","Text":"RC, and our surface area,"},{"Start":"20:20.495 ","End":"20:22.760","Text":"which is pi a^2,"},{"Start":"20:22.760 ","End":"20:25.290","Text":"the surface area of a circle."},{"Start":"20:25.290 ","End":"20:29.785","Text":"This is our current density."},{"Start":"20:29.785 ","End":"20:33.350","Text":"That\u0027s the final answer for question number 3,"},{"Start":"20:33.350 ","End":"20:36.330","Text":"and that is the end of this lesson."}],"ID":21394},{"Watched":false,"Name":"Exercise 6","Duration":"18m 35s","ChapterTopicVideoID":21522,"CourseChapterTopicPlaylistID":99491,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.140","Text":"Hello. In this lesson we\u0027re going to be answering the following question."},{"Start":"00:04.140 ","End":"00:07.440","Text":"A conducting spherical shell of radius a"},{"Start":"00:07.440 ","End":"00:11.250","Text":"is placed inside a material of conductivity Sigma."},{"Start":"00:11.250 ","End":"00:13.260","Text":"The charge on the shell at t=0,"},{"Start":"00:13.260 ","End":"00:16.605","Text":"is equal to q."},{"Start":"00:16.605 ","End":"00:22.575","Text":"Question number 1 is to calculate the charge on the shell as a function of time."},{"Start":"00:22.575 ","End":"00:25.710","Text":"What we can see is that we have this spherical shell over"},{"Start":"00:25.710 ","End":"00:29.730","Text":"here with an initial charge of q."},{"Start":"00:29.730 ","End":"00:38.645","Text":"Then we have this shell placed inside this infinite material with conductivity Sigma."},{"Start":"00:38.645 ","End":"00:44.060","Text":"What we can see is that our charge is going to start"},{"Start":"00:44.060 ","End":"00:49.865","Text":"discharging from the charge shell and travel outwards in this direction,"},{"Start":"00:49.865 ","End":"00:51.440","Text":"in the radial direction."},{"Start":"00:51.440 ","End":"00:52.760","Text":"Because we have symmetry,"},{"Start":"00:52.760 ","End":"00:57.230","Text":"because we\u0027re dealing with a sphere and we\u0027re not told that"},{"Start":"00:57.230 ","End":"01:03.830","Text":"the charge is distributed unevenly or that Sigma,"},{"Start":"01:03.830 ","End":"01:06.410","Text":"the conductivity is not uniform,"},{"Start":"01:06.410 ","End":"01:08.315","Text":"we can assume that we have symmetry,"},{"Start":"01:08.315 ","End":"01:13.550","Text":"so the charges are just going to flow out and carry on flowing until"},{"Start":"01:13.550 ","End":"01:19.265","Text":"they get to infinity and this shell is completely discharged."},{"Start":"01:19.265 ","End":"01:24.050","Text":"Right now we\u0027re trying to calculate the charge on the shell as a function of time."},{"Start":"01:24.050 ","End":"01:27.050","Text":"What we can do is we can think of this as a charged shell"},{"Start":"01:27.050 ","End":"01:32.120","Text":"located inside a resistor, or in other words,"},{"Start":"01:32.120 ","End":"01:36.935","Text":"we can think that we have some spherical shell resistor,"},{"Start":"01:36.935 ","End":"01:42.125","Text":"where the inner shell of this resistor is at radius a,"},{"Start":"01:42.125 ","End":"01:50.065","Text":"and the outer shell of the resistor is at radius infinity."},{"Start":"01:50.065 ","End":"01:56.605","Text":"Here R is equal to infinity and then the border,"},{"Start":"01:56.605 ","End":"02:02.210","Text":"the outer shell of this spherical resistor is over here."},{"Start":"02:03.770 ","End":"02:06.610","Text":"Let\u0027s find the resistance of"},{"Start":"02:06.610 ","End":"02:13.570","Text":"this spherical shell resistor of inner radius a and outer radius infinity."},{"Start":"02:13.570 ","End":"02:16.329","Text":"We remember our equation for the resistance,"},{"Start":"02:16.329 ","End":"02:20.755","Text":"which is equal to the resistivity multiplied by the path length"},{"Start":"02:20.755 ","End":"02:26.215","Text":"of the current or the moving charges divided by the cross-sectional area,"},{"Start":"02:26.215 ","End":"02:33.845","Text":"and we remember that we can only use this equation if all of these are constants."},{"Start":"02:33.845 ","End":"02:39.200","Text":"What we can see over here is that A,"},{"Start":"02:39.200 ","End":"02:42.205","Text":"our cross-sectional area is not a constant."},{"Start":"02:42.205 ","End":"02:43.750","Text":"If we look over here,"},{"Start":"02:43.750 ","End":"02:48.910","Text":"the cross-sectional area is going to be that of a spherical shell of radius a."},{"Start":"02:48.910 ","End":"02:50.590","Text":"As we head out,"},{"Start":"02:50.590 ","End":"02:53.605","Text":"because the radius is increasing,"},{"Start":"02:53.605 ","End":"02:56.665","Text":"the surface area, which is dependent on the radius,"},{"Start":"02:56.665 ","End":"02:59.390","Text":"is also going to be increasing."},{"Start":"02:59.390 ","End":"03:05.200","Text":"A trick that we\u0027ve seen and used many times is we cut up"},{"Start":"03:05.200 ","End":"03:11.485","Text":"our resistor into many spherical shell resistors."},{"Start":"03:11.485 ","End":"03:14.030","Text":"Here\u0027s one of them."},{"Start":"03:14.510 ","End":"03:19.650","Text":"Of course, these lines are meant to be perfect circles."},{"Start":"03:19.650 ","End":"03:26.280","Text":"Then we say that the distance between the inner shell and the outer shell is dr,"},{"Start":"03:26.280 ","End":"03:29.430","Text":"some small change in the radius."},{"Start":"03:29.430 ","End":"03:35.870","Text":"Given that, if we measure the surface area of the inner shell of radius,"},{"Start":"03:35.870 ","End":"03:37.040","Text":"let\u0027s say that this is of"},{"Start":"03:37.040 ","End":"03:42.365","Text":"radius r and then we measure the surface area of the outer shell,"},{"Start":"03:42.365 ","End":"03:44.825","Text":"which is of a radius of r plus dr,"},{"Start":"03:44.825 ","End":"03:46.670","Text":"where dr is approaching 0,"},{"Start":"03:46.670 ","End":"03:54.305","Text":"the surface area of the inner and outer shell will be the same because dr is so small,"},{"Start":"03:54.305 ","End":"03:58.820","Text":"which means that in this subresistor over here,"},{"Start":"03:58.820 ","End":"04:06.215","Text":"we can consider the cross-sectional area as uniform or constant,"},{"Start":"04:06.215 ","End":"04:09.900","Text":"which means that we can therefore use this equation."},{"Start":"04:10.040 ","End":"04:15.705","Text":"That\u0027s number 1. That\u0027s going to be our surface area."},{"Start":"04:15.705 ","End":"04:20.990","Text":"Let\u0027s write that our surface area is the surface area of a spherical shell,"},{"Start":"04:20.990 ","End":"04:24.905","Text":"which is 4 Pi r^2."},{"Start":"04:24.905 ","End":"04:29.165","Text":"Then we know from previous lessons that the resistivity is equal to"},{"Start":"04:29.165 ","End":"04:33.335","Text":"the reciprocal of the conductivity,"},{"Start":"04:33.335 ","End":"04:39.095","Text":"which is what we\u0027re given in this question and specifically in this example over here,"},{"Start":"04:39.095 ","End":"04:43.550","Text":"our path length is going to be this dr."},{"Start":"04:43.760 ","End":"04:47.465","Text":"In our tiny blue resistor over here,"},{"Start":"04:47.465 ","End":"04:50.915","Text":"the current is just going to travel through here,"},{"Start":"04:50.915 ","End":"04:55.020","Text":"which is the length of this path is dr."},{"Start":"04:55.220 ","End":"05:01.070","Text":"Therefore, we can write that the resistance of this blue spherical shell,"},{"Start":"05:01.070 ","End":"05:04.415","Text":"dr is equal to,"},{"Start":"05:04.415 ","End":"05:13.115","Text":"we have a resistivity which is 1 divided by Sigma multiplied by the path length,"},{"Start":"05:13.115 ","End":"05:18.140","Text":"which is dr and divided by the cross-sectional area,"},{"Start":"05:18.140 ","End":"05:22.375","Text":"which is 4 Pi r^2."},{"Start":"05:22.375 ","End":"05:28.040","Text":"Now what we want to do is we want to add up all of these subresistors."},{"Start":"05:28.040 ","End":"05:32.390","Text":"That means from a radius of a up until a radius of infinity,"},{"Start":"05:32.390 ","End":"05:39.090","Text":"and then we will get the total resistance of this resistor."},{"Start":"05:39.740 ","End":"05:43.340","Text":"In order to add up all of these resistances,"},{"Start":"05:43.340 ","End":"05:48.665","Text":"we first have to see how each of these blue resistors is connected."},{"Start":"05:48.665 ","End":"05:54.050","Text":"What we can see is that if we have a charge that leaves this spherical shell,"},{"Start":"05:54.050 ","End":"05:57.680","Text":"in order to travel outwards,"},{"Start":"05:57.680 ","End":"06:02.420","Text":"it has to travel through each and every single resistor."},{"Start":"06:02.420 ","End":"06:08.120","Text":"If we draw part of one of the other subresistors over here,"},{"Start":"06:08.120 ","End":"06:10.490","Text":"it has to travel through this resistor and"},{"Start":"06:10.490 ","End":"06:13.625","Text":"then the next one and the next one and the next one and so on."},{"Start":"06:13.625 ","End":"06:19.610","Text":"We can see that our current or I charges don\u0027t have to split,"},{"Start":"06:19.610 ","End":"06:22.610","Text":"each one will travel through all of the resistors,"},{"Start":"06:22.610 ","End":"06:28.055","Text":"which means that all the resistors are connected in series."},{"Start":"06:28.055 ","End":"06:30.259","Text":"In order to find the total resistance,"},{"Start":"06:30.259 ","End":"06:32.690","Text":"we just have to sum up all of"},{"Start":"06:32.690 ","End":"06:36.735","Text":"the resistances and because they\u0027re infinitesimal resistances,"},{"Start":"06:36.735 ","End":"06:39.140","Text":"that means that we\u0027re just integrating."},{"Start":"06:39.140 ","End":"06:40.670","Text":"As we said before,"},{"Start":"06:40.670 ","End":"06:42.470","Text":"the bounds are from a,"},{"Start":"06:42.470 ","End":"06:47.850","Text":"from this radius over here until the maximum radius of infinity."},{"Start":"06:48.320 ","End":"06:50.670","Text":"When we do that,"},{"Start":"06:50.670 ","End":"06:55.395","Text":"we can take Sigma and 4 Pi out because they are constants."},{"Start":"06:55.395 ","End":"07:00.540","Text":"We have 1 divided by 4 Pi Sigma."},{"Start":"07:00.540 ","End":"07:05.760","Text":"Then we\u0027re just integrating on dr divided by r^2,"},{"Start":"07:05.760 ","End":"07:11.134","Text":"which will give us negative 1 divided by r. Then when we substitute in our bounds,"},{"Start":"07:11.134 ","End":"07:14.420","Text":"we\u0027re just going to be left with 1 divided by a,"},{"Start":"07:14.420 ","End":"07:17.780","Text":"because 1 divided by infinity is equal to 0,"},{"Start":"07:17.780 ","End":"07:18.950","Text":"so we can just cross that out."},{"Start":"07:18.950 ","End":"07:22.950","Text":"Then the minus a multiplied by"},{"Start":"07:22.950 ","End":"07:26.600","Text":"the minus coefficient over here from"},{"Start":"07:26.600 ","End":"07:31.445","Text":"the integration before we substitute it in the bounds will just disappear."},{"Start":"07:31.445 ","End":"07:37.410","Text":"We\u0027ll be left with this as the resistance of this material."},{"Start":"07:37.880 ","End":"07:45.550","Text":"Here\u0027s a reminder that we\u0027re trying to calculate the charge as a function of time."},{"Start":"07:45.550 ","End":"07:53.390","Text":"What we want to do is we have the resistance so what equation can we put together?"},{"Start":"07:53.390 ","End":"07:55.445","Text":"Charge and resistance."},{"Start":"07:55.445 ","End":"07:59.640","Text":"Charge has something to do with the current and"},{"Start":"07:59.640 ","End":"08:04.490","Text":"we know that an equation is V is equal to IR."},{"Start":"08:04.490 ","End":"08:08.975","Text":"The voltage is equal to the current which is linked to charge."},{"Start":"08:08.975 ","End":"08:12.640","Text":"What we want to do is we\u0027re going to try and isolate out this I."},{"Start":"08:12.640 ","End":"08:16.205","Text":"We have this resistance which we just calculated."},{"Start":"08:16.205 ","End":"08:20.300","Text":"This is great. The only thing is we don\u0027t know what V is equal to."},{"Start":"08:20.300 ","End":"08:22.655","Text":"Just a reminder, V,"},{"Start":"08:22.655 ","End":"08:30.965","Text":"the voltage is equal to the change in the potential or the potential difference."},{"Start":"08:30.965 ","End":"08:36.275","Text":"What we have is the potential at this point over here,"},{"Start":"08:36.275 ","End":"08:39.755","Text":"and the potential at this point over here at infinity."},{"Start":"08:39.755 ","End":"08:44.930","Text":"What we\u0027re trying to find is this potential difference between these 2 points."},{"Start":"08:44.930 ","End":"08:53.030","Text":"What we have is our potential at point a minus our potential at infinity."},{"Start":"08:53.030 ","End":"08:56.930","Text":"Our potential at point a is just the potential of"},{"Start":"08:56.930 ","End":"09:02.000","Text":"a charge located at radius a away from the origin."},{"Start":"09:02.000 ","End":"09:05.870","Text":"We already saw that the potential for a charge or"},{"Start":"09:05.870 ","End":"09:09.650","Text":"for the sphere is kq divided by its radius,"},{"Start":"09:09.650 ","End":"09:12.050","Text":"which here it\u0027s at a radius of a."},{"Start":"09:12.050 ","End":"09:18.430","Text":"But notice I\u0027m writing kq as a function of time divided by a. I\u0027m not"},{"Start":"09:18.430 ","End":"09:25.610","Text":"writing the capital Q because that will just give me the potential at this point at t=0."},{"Start":"09:25.610 ","End":"09:29.480","Text":"But then as this spherical shell discharges,"},{"Start":"09:29.480 ","End":"09:33.095","Text":"my q is changing and then this won\u0027t be irrelevant."},{"Start":"09:33.095 ","End":"09:36.635","Text":"This is an answer which is relevant for all time."},{"Start":"09:36.635 ","End":"09:40.400","Text":"Then the potential at infinity is defined as 0."},{"Start":"09:40.400 ","End":"09:43.225","Text":"When we subtract 0,"},{"Start":"09:43.225 ","End":"09:49.530","Text":"we\u0027re just left with this and this is the potential difference,"},{"Start":"09:49.530 ","End":"09:51.045","Text":"which is also the voltage."},{"Start":"09:51.045 ","End":"09:54.950","Text":"Now we have the voltage and the resistance so we can get the current."},{"Start":"09:54.950 ","End":"10:02.782","Text":"Current is equal to the voltage divided by the resistance,"},{"Start":"10:02.782 ","End":"10:10.670","Text":"and this is also equal to the time derivative of our charge, so q dot."},{"Start":"10:10.670 ","End":"10:16.000","Text":"But remember, because the spherical shell is discharging,"},{"Start":"10:16.000 ","End":"10:21.335","Text":"as in the current is formed from a loss of charge on the spherical shell,"},{"Start":"10:21.335 ","End":"10:24.525","Text":"we have to add in a minus."},{"Start":"10:24.525 ","End":"10:30.065","Text":"If the current is formed from charges being added to the spherical shell,"},{"Start":"10:30.065 ","End":"10:32.690","Text":"then we would have a plus over here but because"},{"Start":"10:32.690 ","End":"10:36.320","Text":"the current is formed from loss of charge over here,"},{"Start":"10:36.320 ","End":"10:40.230","Text":"so that equals a minus."},{"Start":"10:41.200 ","End":"10:45.095","Text":"What we have is negative q dot,"},{"Start":"10:45.095 ","End":"10:48.845","Text":"which is negative dq by dt,"},{"Start":"10:48.845 ","End":"10:52.460","Text":"which is equal to our voltage,"},{"Start":"10:52.460 ","End":"10:55.040","Text":"which is kq divided by a,"},{"Start":"10:55.040 ","End":"10:58.340","Text":"divided by the resistance."},{"Start":"10:58.340 ","End":"11:02.759","Text":"Because the resistance, everything over here is on the denominator,"},{"Start":"11:02.759 ","End":"11:06.410","Text":"it will flip up to the top and the a\u0027s will cancel out."},{"Start":"11:06.410 ","End":"11:16.450","Text":"What we\u0027ll be left with is kq multiplied by 4 Pi Sigma."},{"Start":"11:17.190 ","End":"11:22.945","Text":"Now, and of course q is as a function of t. I just didn\u0027t write that out."},{"Start":"11:22.945 ","End":"11:24.325","Text":"This is what we\u0027ll be left with,"},{"Start":"11:24.325 ","End":"11:29.365","Text":"and now what we want to do is we want to get our q on"},{"Start":"11:29.365 ","End":"11:32.665","Text":"the same side as our function with dq and"},{"Start":"11:32.665 ","End":"11:36.850","Text":"all these constants that we have on the same side as our dt."},{"Start":"11:36.850 ","End":"11:39.070","Text":"We\u0027ll also get rid of our minus,"},{"Start":"11:39.070 ","End":"11:40.990","Text":"we\u0027ll put our minus on this side as well."},{"Start":"11:40.990 ","End":"11:47.470","Text":"What we\u0027ll be left with is dq divided by q is equal to"},{"Start":"11:47.470 ","End":"11:56.800","Text":"negative and then what we\u0027ll have is 4Pik Sigma dt."},{"Start":"11:56.800 ","End":"12:05.605","Text":"Now, what is k. K is equal to 1 divided by 4Pi epsilon naught."},{"Start":"12:05.605 ","End":"12:07.645","Text":"If we substitute that in,"},{"Start":"12:07.645 ","End":"12:11.830","Text":"the 4Pi is going to cancel out."},{"Start":"12:11.830 ","End":"12:13.705","Text":"We can write this out again,"},{"Start":"12:13.705 ","End":"12:17.259","Text":"dq divided by q is equal to negative."},{"Start":"12:17.259 ","End":"12:20.710","Text":"The 4Pis cancel out and so does the k and we\u0027re just left"},{"Start":"12:20.710 ","End":"12:25.400","Text":"with Sigma divided by epsilon naught dt."},{"Start":"12:26.970 ","End":"12:29.860","Text":"Now we just want to integrate,"},{"Start":"12:29.860 ","End":"12:34.210","Text":"so we\u0027ll put in our integration signs and now let\u0027s look at our bounds."},{"Start":"12:34.210 ","End":"12:39.040","Text":"From t we\u0027re integrating from a time is equal to 0 up until"},{"Start":"12:39.040 ","End":"12:43.825","Text":"some random time t. Then when we\u0027re integrating with the charges,"},{"Start":"12:43.825 ","End":"12:46.600","Text":"so we\u0027re integrating from our initial charge,"},{"Start":"12:46.600 ","End":"12:48.340","Text":"which was at t is equal to 0,"},{"Start":"12:48.340 ","End":"12:49.600","Text":"which was equal to,"},{"Start":"12:49.600 ","End":"12:52.390","Text":"as we\u0027re told in the question, capital q,"},{"Start":"12:52.390 ","End":"12:59.275","Text":"and then up until some charge that corresponds to time t. When we integrate this,"},{"Start":"12:59.275 ","End":"13:03.325","Text":"so over here we\u0027re going to have Lan,"},{"Start":"13:03.325 ","End":"13:06.175","Text":"so we\u0027re integrating of 1 divided by q."},{"Start":"13:06.175 ","End":"13:10.090","Text":"We\u0027re going to have Lan of q as a function of t"},{"Start":"13:10.090 ","End":"13:14.455","Text":"divided by capital Q and this is going to be equal to,"},{"Start":"13:14.455 ","End":"13:19.600","Text":"so what we\u0027ll have is negative Sigma divided by epsilon naught,"},{"Start":"13:19.600 ","End":"13:25.010","Text":"and then we\u0027ll have t minus 0."},{"Start":"13:25.410 ","End":"13:32.170","Text":"Then the minus and the minus here we can have a plus and then we\u0027ll have this."},{"Start":"13:32.170 ","End":"13:39.565","Text":"What we will be left with is that this is equal to negative Sigma divided by"},{"Start":"13:39.565 ","End":"13:44.190","Text":"epsilon naught t. Then what we want to do is we want to get rid of"},{"Start":"13:44.190 ","End":"13:49.530","Text":"this Lan so we\u0027re going to raise both sides by e the exponential function."},{"Start":"13:49.530 ","End":"13:54.330","Text":"What we\u0027ll be left with is q as a function of t divided by Q is"},{"Start":"13:54.330 ","End":"13:59.650","Text":"equal to e to the power of negative Sigma divided by epsilon naught t,"},{"Start":"13:59.650 ","End":"14:02.455","Text":"and then we\u0027ll multiply both sides by q,"},{"Start":"14:02.455 ","End":"14:06.610","Text":"and therefore what we\u0027ll get is that our charge as a function of"},{"Start":"14:06.610 ","End":"14:11.170","Text":"time is equal to Q multiplied by e to"},{"Start":"14:11.170 ","End":"14:13.960","Text":"the power of negative Sigma divided by"},{"Start":"14:13.960 ","End":"14:20.725","Text":"epsilon naught t. This is our answer to question Number 1,"},{"Start":"14:20.725 ","End":"14:23.410","Text":"and we can see that we get the exact same answer"},{"Start":"14:23.410 ","End":"14:26.410","Text":"that we would get from a discharging capacitor,"},{"Start":"14:26.410 ","End":"14:30.070","Text":"where our value for RC or the RC time"},{"Start":"14:30.070 ","End":"14:34.960","Text":"constant is just the reciprocal of the coefficient for t,"},{"Start":"14:34.960 ","End":"14:40.870","Text":"so it\u0027s just equal to epsilon naught divided by Sigma."},{"Start":"14:40.870 ","End":"14:43.270","Text":"That\u0027s answer for question Number 1."},{"Start":"14:43.270 ","End":"14:46.465","Text":"Now let\u0027s take a look at question Number 2."},{"Start":"14:46.465 ","End":"14:52.100","Text":"Calculate the current density and the electric field inside the resistor."},{"Start":"14:52.590 ","End":"14:56.365","Text":"Let\u0027s first deal with the current density."},{"Start":"14:56.365 ","End":"14:59.320","Text":"Let\u0027s write out first of all our current."},{"Start":"14:59.320 ","End":"15:04.840","Text":"Our current as we know is equal to negative q dot."},{"Start":"15:04.840 ","End":"15:09.460","Text":"We can just take the negative derivative of this."},{"Start":"15:09.460 ","End":"15:19.419","Text":"What we will be left with is simply Sigma divided by epsilon naught multiplied by"},{"Start":"15:19.419 ","End":"15:24.910","Text":"qe to the negative Sigma divided by"},{"Start":"15:24.910 ","End":"15:33.850","Text":"epsilon naught t. That\u0027s just by taking the negative derivative of this."},{"Start":"15:33.850 ","End":"15:38.470","Text":"Of course we saw earlier that if we just rearrange this,"},{"Start":"15:38.470 ","End":"15:42.190","Text":"so we move the negative over to here and the dt over to here,"},{"Start":"15:42.190 ","End":"15:44.270","Text":"so then we have negative dq by dt,"},{"Start":"15:44.270 ","End":"15:46.855","Text":"which is this negative q dot,"},{"Start":"15:46.855 ","End":"15:49.810","Text":"and then what we\u0027re left with is Sigma divided by"},{"Start":"15:49.810 ","End":"15:54.535","Text":"epsilon naught multiplied by q as a function of time,"},{"Start":"15:54.535 ","End":"15:57.385","Text":"which is also what we see here."},{"Start":"15:57.385 ","End":"16:00.070","Text":"This is q as a function of time."},{"Start":"16:00.070 ","End":"16:02.545","Text":"This is the same expression that we have over here."},{"Start":"16:02.545 ","End":"16:09.370","Text":"We can just write this as Sigma divided by epsilon naught q as a function of time."},{"Start":"16:09.370 ","End":"16:14.410","Text":"Therefore our current density, J,"},{"Start":"16:14.410 ","End":"16:20.515","Text":"which is a vector, is equal to our current divided by the cross-sectional area."},{"Start":"16:20.515 ","End":"16:25.240","Text":"Because we can see that our conductivity is constant throughout,"},{"Start":"16:25.240 ","End":"16:32.885","Text":"there is no reason why over here they\u0027ll be more charges passing through than over here."},{"Start":"16:32.885 ","End":"16:37.365","Text":"We can just say that this is equal to Sigma"},{"Start":"16:37.365 ","End":"16:41.730","Text":"multiplied by our charge as a function of time divided by"},{"Start":"16:41.730 ","End":"16:46.660","Text":"epsilon naught and divided by the cross-sectional area which we saw"},{"Start":"16:46.660 ","End":"16:52.360","Text":"before was equal to 4 Pi r squared."},{"Start":"16:52.360 ","End":"16:55.600","Text":"Of course this is a vector and we can see that the charges are"},{"Start":"16:55.600 ","End":"16:59.570","Text":"flowing out in the radial direction."},{"Start":"16:59.640 ","End":"17:04.120","Text":"Now let\u0027s calculate the electric field inside the resistor."},{"Start":"17:04.120 ","End":"17:07.540","Text":"We know that the electric field is equal to"},{"Start":"17:07.540 ","End":"17:12.175","Text":"the resistivity multiplied by the current density."},{"Start":"17:12.175 ","End":"17:16.960","Text":"The resistivity as we know is equal to 1 divided by the conductivity,"},{"Start":"17:16.960 ","End":"17:23.020","Text":"so 1 divided by Sigma and then our j is sigma q as a function of"},{"Start":"17:23.020 ","End":"17:30.805","Text":"time divided by 4 Pi epsilon naught r squared in the radial direction."},{"Start":"17:30.805 ","End":"17:32.620","Text":"The sigmas cancel out and"},{"Start":"17:32.620 ","End":"17:39.220","Text":"4Pi epsilon naught in the denominator is of course equal to k. We can just rewrite"},{"Start":"17:39.220 ","End":"17:48.800","Text":"this as kq as a function of time divided by r squared in the radial direction."},{"Start":"17:49.530 ","End":"17:52.345","Text":"This is the electric field,"},{"Start":"17:52.345 ","End":"17:54.385","Text":"and this is the answer to Question 2."},{"Start":"17:54.385 ","End":"17:56.260","Text":"Now, what we can see is that"},{"Start":"17:56.260 ","End":"17:59.290","Text":"the electric field is the same that we would have gotten from"},{"Start":"17:59.290 ","End":"18:06.730","Text":"this charged sphere or a charged particle if we would have just used Gauss\u0027s law."},{"Start":"18:06.730 ","End":"18:10.915","Text":"We got the exact same answer so we could have just gotten this straight away."},{"Start":"18:10.915 ","End":"18:13.075","Text":"Then from working out the electric field,"},{"Start":"18:13.075 ","End":"18:14.260","Text":"we could have of course,"},{"Start":"18:14.260 ","End":"18:17.095","Text":"very easily calculated j from that."},{"Start":"18:17.095 ","End":"18:19.570","Text":"We would have substituted it in our E field,"},{"Start":"18:19.570 ","End":"18:23.140","Text":"our row as we know is 1 divided by sigma and then we"},{"Start":"18:23.140 ","End":"18:26.830","Text":"would have just isolate out our j and gotten this exact answer."},{"Start":"18:26.830 ","End":"18:29.245","Text":"We could have worked this out either way."},{"Start":"18:29.245 ","End":"18:32.800","Text":"But this is an alternative method."},{"Start":"18:32.800 ","End":"18:36.380","Text":"That\u0027s it. That\u0027s the end of this lesson."}],"ID":22421},{"Watched":false,"Name":"Exercise 7","Duration":"15m 25s","ChapterTopicVideoID":21523,"CourseChapterTopicPlaylistID":99491,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.630","Text":"Hello. In this lesson we\u0027re going to be answering the following question."},{"Start":"00:03.630 ","End":"00:06.795","Text":"2 parallel plates of dimensions,"},{"Start":"00:06.795 ","End":"00:15.075","Text":"L by L are placed a distance d from one another such that L is much greater than d,"},{"Start":"00:15.075 ","End":"00:17.940","Text":"between the 2 plates are conducting material of"},{"Start":"00:17.940 ","End":"00:22.110","Text":"resistivity Rho_as a function of x and y is inserted and"},{"Start":"00:22.110 ","End":"00:24.600","Text":"we\u0027re being asked to calculate the resistance along"},{"Start":"00:24.600 ","End":"00:30.015","Text":"the y-axis when we have the following 2 values for resistivity."},{"Start":"00:30.015 ","End":"00:34.545","Text":"In Question number 1, we can see that the resistivity is just dependent on y,"},{"Start":"00:34.545 ","End":"00:36.480","Text":"meaning that as y changes,"},{"Start":"00:36.480 ","End":"00:41.910","Text":"our resistivity changes, which means that our resistivity isn\u0027t a constant."},{"Start":"00:41.910 ","End":"00:46.610","Text":"What we do, we\u0027ve seen many examples of this,"},{"Start":"00:46.610 ","End":"00:52.290","Text":"is that we take a thin layer over here,"},{"Start":"00:53.440 ","End":"00:58.025","Text":"we take this thin layer where this width over here"},{"Start":"00:58.025 ","End":"01:02.120","Text":"is dy and dy is of course approaching 0."},{"Start":"01:02.120 ","End":"01:05.990","Text":"When I take this y value and plug it into this equation for"},{"Start":"01:05.990 ","End":"01:11.570","Text":"resistivity and when I take this y value and plug it into the equation for resistivity,"},{"Start":"01:11.570 ","End":"01:14.420","Text":"the difference in the resistivity is so small,"},{"Start":"01:14.420 ","End":"01:15.725","Text":"it\u0027s approaching 0,"},{"Start":"01:15.725 ","End":"01:24.140","Text":"that we can consider the resistivity on this blue sub resistor as being uniform."},{"Start":"01:24.140 ","End":"01:32.615","Text":"In that case, we can calculate the resistance of this sub resistor as being equal to dR,"},{"Start":"01:32.615 ","End":"01:36.080","Text":"and that is equal to the resistivity,"},{"Start":"01:36.080 ","End":"01:40.760","Text":"so as a function of x and y soon we\u0027ll plug this in,"},{"Start":"01:40.760 ","End":"01:45.890","Text":"multiplied by the path taken by the current."},{"Start":"01:45.890 ","End":"01:49.050","Text":"I\u0027m telling you, I forgot to write it in the question,"},{"Start":"01:49.050 ","End":"01:51.290","Text":"this is the path taken by the current."},{"Start":"01:51.290 ","End":"02:00.620","Text":"It goes through the resistor like so the path taken is this length dy and"},{"Start":"02:00.620 ","End":"02:10.210","Text":"divided by the cross-sectional area which is just dimension L by L, which is L^2."},{"Start":"02:10.210 ","End":"02:15.590","Text":"Now what we want to do is we want to find the total resistance along the y-axis,"},{"Start":"02:15.590 ","End":"02:18.140","Text":"so what we\u0027re going to do is we\u0027re going to sum up all of"},{"Start":"02:18.140 ","End":"02:22.025","Text":"these sub resistors along the y-axis."},{"Start":"02:22.025 ","End":"02:24.320","Text":"The first thing that we have to do is we have to see"},{"Start":"02:24.320 ","End":"02:26.600","Text":"how each one of these blue resistors is"},{"Start":"02:26.600 ","End":"02:32.970","Text":"connected to the other ones in order to know how to add up the resistances."},{"Start":"02:33.280 ","End":"02:36.455","Text":"If our current is flowing like so,"},{"Start":"02:36.455 ","End":"02:39.620","Text":"then it\u0027s going to have to flow straight through all of"},{"Start":"02:39.620 ","End":"02:44.300","Text":"the resistors until it reaches the top plate,"},{"Start":"02:44.300 ","End":"02:48.830","Text":"which means that the current doesn\u0027t split and it flows straight through each resistor"},{"Start":"02:48.830 ","End":"02:53.975","Text":"which means that all of these blue resistors connected in series."},{"Start":"02:53.975 ","End":"02:55.850","Text":"If they\u0027re connected in series,"},{"Start":"02:55.850 ","End":"03:00.715","Text":"the way to calculate the total resistance is to just integrate."},{"Start":"03:00.715 ","End":"03:03.466","Text":"We\u0027re just going to integrate along this,"},{"Start":"03:03.466 ","End":"03:06.530","Text":"and our bounds are from y is equal to"},{"Start":"03:06.530 ","End":"03:11.210","Text":"0 until this height over here where the second plate is,"},{"Start":"03:11.210 ","End":"03:17.899","Text":"which as it y is equal to d. Now we can plug in our values."},{"Start":"03:17.899 ","End":"03:23.540","Text":"We\u0027ll get that the resistance is equal to 0 to d of Rho_is a function of x y,"},{"Start":"03:23.540 ","End":"03:26.983","Text":"which is Rho_naught multiplied by sin(Piy divided by"},{"Start":"03:26.983 ","End":"03:36.075","Text":"d) and then we have dy divided by L^2."},{"Start":"03:36.075 ","End":"03:39.240","Text":"Of course, Rho_0 and L^2 are constants,"},{"Start":"03:39.240 ","End":"03:43.520","Text":"so we just have to integrate along a sine of Piy as a function of"},{"Start":"03:43.520 ","End":"03:48.560","Text":"d. We have Rho_naught divided by L^2."},{"Start":"03:48.560 ","End":"03:59.015","Text":"Then here we\u0027re going to have that this is equal to negative cos(Piy divided by d)"},{"Start":"03:59.015 ","End":"04:06.215","Text":"multiplied by the anti-derivative which is simply going to"},{"Start":"04:06.215 ","End":"04:13.655","Text":"be equal to 1 divided by Pi over D,"},{"Start":"04:13.655 ","End":"04:16.805","Text":"which is just D over Pi."},{"Start":"04:16.805 ","End":"04:21.950","Text":"We take the reciprocal because we have to divide by the anti-derivative and then"},{"Start":"04:21.950 ","End":"04:27.270","Text":"of course we have to plug in our bounds from 0 to d. What we\u0027re going to have,"},{"Start":"04:27.270 ","End":"04:30.920","Text":"a move-out, the switch d divided by Pi, which is a constant,"},{"Start":"04:30.920 ","End":"04:36.650","Text":"so we have Rho_0 D divided by Pi L^2,"},{"Start":"04:36.650 ","End":"04:45.245","Text":"and then what we have is negative cosine of Pi multiplied by d divided by d,"},{"Start":"04:45.245 ","End":"04:48.335","Text":"which is Pi minus,"},{"Start":"04:48.335 ","End":"04:53.900","Text":"and then negative cosine with a lower bound where we substitute in 0 into here,"},{"Start":"04:53.900 ","End":"04:58.470","Text":"we have cosine of 0."},{"Start":"05:00.970 ","End":"05:08.030","Text":"Cosine of Pi is equal to negative 1,"},{"Start":"05:08.030 ","End":"05:10.790","Text":"and then the negative over here becomes 1,"},{"Start":"05:10.790 ","End":"05:16.160","Text":"so we have Rho_0 D divided by Pi L^2,"},{"Start":"05:16.160 ","End":"05:20.975","Text":"so all of this over here is equal to 1."},{"Start":"05:20.975 ","End":"05:27.380","Text":"Negative cosine of 0 is negative 1 and then plus the negative over here."},{"Start":"05:27.380 ","End":"05:32.300","Text":"We have 1 plus 1 is 2,"},{"Start":"05:32.300 ","End":"05:35.310","Text":"so we just multiply this by 2,"},{"Start":"05:35.450 ","End":"05:38.880","Text":"so that\u0027s the answer to Question number 1."},{"Start":"05:38.880 ","End":"05:41.880","Text":"Now let\u0027s answer Question number 2."},{"Start":"05:41.880 ","End":"05:48.860","Text":"Here we have the resistivity as a function of both y and x."},{"Start":"05:48.860 ","End":"05:50.405","Text":"What does that mean?"},{"Start":"05:50.405 ","End":"05:54.259","Text":"That means that aside from the resistivity changing,"},{"Start":"05:54.259 ","End":"06:00.110","Text":"as Y changes, also as x changes, the resistivity changes."},{"Start":"06:00.110 ","End":"06:03.830","Text":"If we take some length over here,"},{"Start":"06:03.830 ","End":"06:05.465","Text":"so we can call this length,"},{"Start":"06:05.465 ","End":"06:07.805","Text":"which is also approaching 0,"},{"Start":"06:07.805 ","End":"06:11.960","Text":"this length is equal to dx."},{"Start":"06:11.960 ","End":"06:15.604","Text":"Now we\u0027re just cutting this into strips."},{"Start":"06:15.604 ","End":"06:19.090","Text":"The length of each strip is still L,"},{"Start":"06:19.090 ","End":"06:22.480","Text":"however, the width is dx."},{"Start":"06:22.820 ","End":"06:31.400","Text":"Now what we want to do is we want to find the resistance of each one of these strips."},{"Start":"06:31.400 ","End":"06:34.010","Text":"The first thing that we\u0027re going to do is we\u0027re going to find"},{"Start":"06:34.010 ","End":"06:36.662","Text":"the resistance of one of these strips,"},{"Start":"06:36.662 ","End":"06:40.535","Text":"then we\u0027re going to sum up all of the strips that make up"},{"Start":"06:40.535 ","End":"06:47.300","Text":"this entire length L and then we\u0027re going to sum up all of that,"},{"Start":"06:47.300 ","End":"06:55.310","Text":"all of those resistors to make up this height D. First we\u0027re"},{"Start":"06:55.310 ","End":"06:58.760","Text":"summing all of the resistors that are attached like"},{"Start":"06:58.760 ","End":"07:03.545","Text":"so and then we\u0027re summing along here,"},{"Start":"07:03.545 ","End":"07:06.540","Text":"so all of these."},{"Start":"07:08.660 ","End":"07:14.930","Text":"This isn\u0027t a registered way of writing out this equation,"},{"Start":"07:14.930 ","End":"07:17.465","Text":"but this is how I\u0027m going to define it."},{"Start":"07:17.465 ","End":"07:26.510","Text":"What we have is this in a dR refers to our variable y,"},{"Start":"07:26.510 ","End":"07:29.015","Text":"and then when I have the d of this,"},{"Start":"07:29.015 ","End":"07:30.545","Text":"we\u0027re going to be summing up"},{"Start":"07:30.545 ","End":"07:35.615","Text":"this whole equation along x and then this inner equation along y."},{"Start":"07:35.615 ","End":"07:37.864","Text":"What we have is our resistivity,"},{"Start":"07:37.864 ","End":"07:39.275","Text":"so first let\u0027s write that."},{"Start":"07:39.275 ","End":"07:43.051","Text":"Resistivity is Rho_naught multiplied by sin(Piy divided by"},{"Start":"07:43.051 ","End":"07:47.630","Text":"d) divided"},{"Start":"07:47.630 ","End":"07:53.713","Text":"by sin(Pix divided by L)."},{"Start":"07:53.713 ","End":"07:59.900","Text":"Then all of this is multiplied by the path taken."},{"Start":"07:59.900 ","End":"08:03.360","Text":"Our current is still flowing in this y direction."},{"Start":"08:03.360 ","End":"08:05.650","Text":"The resistance along the y-axis,"},{"Start":"08:05.650 ","End":"08:07.650","Text":"so this is the direction of current."},{"Start":"08:07.650 ","End":"08:14.515","Text":"Our path length is still dy and then divided by our cross-sectional area."},{"Start":"08:14.515 ","End":"08:16.845","Text":"What is our cross-sectional area?"},{"Start":"08:16.845 ","End":"08:24.285","Text":"We\u0027re dealing with dx multiplied by L. The length is L and the width is dx."},{"Start":"08:24.285 ","End":"08:28.480","Text":"We have L.dx."},{"Start":"08:28.600 ","End":"08:33.530","Text":"This equation came from the fact that resistance is equal to"},{"Start":"08:33.530 ","End":"08:38.075","Text":"resistivity multiplied by path length divided by cross-sectional area."},{"Start":"08:38.075 ","End":"08:40.075","Text":"Now what we\u0027re going to do,"},{"Start":"08:40.075 ","End":"08:46.359","Text":"is we\u0027re going to sum up for all of the resistors along this plateau."},{"Start":"08:46.359 ","End":"08:50.400","Text":"We\u0027re going, to sum up also with this resistor over here,"},{"Start":"08:50.400 ","End":"08:54.750","Text":"where this is of course still length dx and this is still length L,"},{"Start":"08:54.750 ","End":"08:56.930","Text":"and we\u0027re going to sum up all of these."},{"Start":"08:56.930 ","End":"09:01.640","Text":"What we can see is that if the current is flowing over here,"},{"Start":"09:01.640 ","End":"09:06.015","Text":"let\u0027s say, the current flows all the way like so."},{"Start":"09:06.015 ","End":"09:07.635","Text":"This is the current."},{"Start":"09:07.635 ","End":"09:11.520","Text":"In order for the current to reach these 2 different green resistors,"},{"Start":"09:11.520 ","End":"09:14.465","Text":"the current has to split,"},{"Start":"09:14.465 ","End":"09:15.910","Text":"as we can see."},{"Start":"09:15.910 ","End":"09:18.060","Text":"If the current has to split,"},{"Start":"09:18.060 ","End":"09:25.270","Text":"that means that all of the green resistors on this plateau are connected in parallel."},{"Start":"09:26.060 ","End":"09:29.585","Text":"In order to find the total resistance,"},{"Start":"09:29.585 ","End":"09:32.585","Text":"so that means what resistance are we looking at now?"},{"Start":"09:32.585 ","End":"09:37.529","Text":"This is the total resistance of this blue resistor."},{"Start":"09:37.529 ","End":"09:43.040","Text":"That\u0027s 1 divided by dR, this over here."},{"Start":"09:43.130 ","End":"09:54.710","Text":"That is going to be equal to the integral of 1 divided by d(dR) of this."},{"Start":"09:55.040 ","End":"09:58.714","Text":"When we connect resistors in parallel,"},{"Start":"09:58.714 ","End":"10:02.970","Text":"this is what the addition of the resistors is going to look like."},{"Start":"10:02.970 ","End":"10:05.845","Text":"I\u0027m just reminding you dR is"},{"Start":"10:05.845 ","End":"10:10.360","Text":"the resistance of each of these blue resistors, these strips,"},{"Start":"10:10.360 ","End":"10:18.660","Text":"these horizontal strips so d(dR) is the resistance of each green resistor,"},{"Start":"10:18.660 ","End":"10:24.370","Text":"and dR is the resistance of each blue resistor."},{"Start":"10:25.080 ","End":"10:27.295","Text":"Let\u0027s write this out."},{"Start":"10:27.295 ","End":"10:33.550","Text":"What we have is the integral from x=0 over here,"},{"Start":"10:33.550 ","End":"10:36.165","Text":"up until the maximum length,"},{"Start":"10:36.165 ","End":"10:37.865","Text":"which is the maximum length of the plate,"},{"Start":"10:37.865 ","End":"10:42.700","Text":"which is L. Then we\u0027re taking the reciprocal of d(dR),"},{"Start":"10:42.700 ","End":"10:44.475","Text":"so the reciprocal of this."},{"Start":"10:44.475 ","End":"10:54.395","Text":"What we\u0027re going to have is L multiplied by sin(Pix divided by"},{"Start":"10:54.395 ","End":"11:04.385","Text":"L)dx divided by Rho_0"},{"Start":"11:04.385 ","End":"11:10.590","Text":"multiplied by sin(Piy divided by"},{"Start":"11:10.590 ","End":"11:21.040","Text":"d)dy."},{"Start":"11:21.040 ","End":"11:25.624","Text":"Let\u0027s scroll down to give us a little bit more space."},{"Start":"11:25.624 ","End":"11:28.800","Text":"Now what we\u0027re going to have,"},{"Start":"11:28.800 ","End":"11:31.445","Text":"therefore, let\u0027s carry it on over here."},{"Start":"11:31.445 ","End":"11:38.650","Text":"We have 1 divided by dR is equal to, of course,"},{"Start":"11:38.650 ","End":"11:41.013","Text":"in this integral we\u0027re integrating with respect to x,"},{"Start":"11:41.013 ","End":"11:44.225","Text":"so everything with a y is constant,"},{"Start":"11:44.225 ","End":"11:47.745","Text":"and of course, L is also constant."},{"Start":"11:47.745 ","End":"11:53.370","Text":"What we have is L divided by"},{"Start":"11:53.370 ","End":"12:03.100","Text":"Rho_0 sin(Piy divided by d)dy."},{"Start":"12:03.100 ","End":"12:06.030","Text":"Then we\u0027re integrating from 0 to"},{"Start":"12:06.030 ","End":"12:13.550","Text":"L sin(Pix divided by L)dx."},{"Start":"12:13.550 ","End":"12:15.500","Text":"All of this integral,"},{"Start":"12:15.500 ","End":"12:21.460","Text":"remember this is going to be equal to cos(Pix divided by L),"},{"Start":"12:21.460 ","End":"12:27.330","Text":"divided by Pi divided by L. Then when you substitute in the bounds,"},{"Start":"12:27.330 ","End":"12:32.405","Text":"this whole integral is simply going to be equal to 2L divided by Pi."},{"Start":"12:32.405 ","End":"12:34.595","Text":"If you don\u0027t see that immediately,"},{"Start":"12:34.595 ","End":"12:37.990","Text":"then please pause the video now and just do this integral."},{"Start":"12:37.990 ","End":"12:40.285","Text":"You will see that this is what we get."},{"Start":"12:40.285 ","End":"12:46.640","Text":"Then what we\u0027re going to have is that 1 divided by dR, remember,"},{"Start":"12:46.640 ","End":"12:52.185","Text":"is equal to 2L multiplied by this L,"},{"Start":"12:52.185 ","End":"12:58.410","Text":"2L^2 divided by Rho_0 Pi,"},{"Start":"12:58.410 ","End":"13:01.160","Text":"multiplied by"},{"Start":"13:01.160 ","End":"13:09.320","Text":"sin(Piy divided by d)dy."},{"Start":"13:09.320 ","End":"13:10.620","Text":"Now what I want to do is,"},{"Start":"13:10.620 ","End":"13:15.070","Text":"I want to find out what my total resistance is."},{"Start":"13:15.070 ","End":"13:21.760","Text":"My total resistance is going to be my integral on dR,"},{"Start":"13:21.760 ","End":"13:24.860","Text":"so here I have the reciprocal,"},{"Start":"13:24.860 ","End":"13:27.185","Text":"I just have to take the reciprocal."},{"Start":"13:27.185 ","End":"13:29.945","Text":"That\u0027s going to be the integral on"},{"Start":"13:29.945 ","End":"13:36.210","Text":"Rho_0 Pi sin(Piy divided"},{"Start":"13:36.210 ","End":"13:42.705","Text":"by d)dy divided by 2L^2."},{"Start":"13:42.705 ","End":"13:47.075","Text":"Of course, Rho_0 Pi divided by 2L^2 are constants, and of course,"},{"Start":"13:47.075 ","End":"13:51.440","Text":"our bounds are again from 0 to d. Let\u0027s just go back to the diagram."},{"Start":"13:51.440 ","End":"13:54.440","Text":"Now we\u0027re integrating along all of these blue plates."},{"Start":"13:54.440 ","End":"13:56.250","Text":"Of course, just like in Question number 1,"},{"Start":"13:56.250 ","End":"13:58.115","Text":"we saw, the current doesn\u0027t have to split,"},{"Start":"13:58.115 ","End":"14:00.630","Text":"the current flows through all the blue plates,"},{"Start":"14:00.630 ","End":"14:04.325","Text":"which means that they\u0027re all connected in series."},{"Start":"14:04.325 ","End":"14:12.705","Text":"This is simply going to be equal to Rho_0 Pi divided by 2L^2,"},{"Start":"14:12.705 ","End":"14:20.630","Text":"and then we saw that the integral in Question number 1 of sin(Piy divided by d)dy."},{"Start":"14:20.630 ","End":"14:26.645","Text":"That\u0027s exactly what we had over here I\u0027ll just do this in blue."},{"Start":"14:26.645 ","End":"14:32.480","Text":"That\u0027s exactly this integral with the exact same bounds that we have over here."},{"Start":"14:32.480 ","End":"14:35.760","Text":"We can just copy out the answer."},{"Start":"14:35.760 ","End":"14:39.190","Text":"What we got, we took out all of the constants and we got"},{"Start":"14:39.190 ","End":"14:42.860","Text":"that this was equal to just 2 inside the brackets,"},{"Start":"14:42.860 ","End":"14:46.265","Text":"so all we have to do is multiply all of this by 2."},{"Start":"14:46.265 ","End":"14:53.710","Text":"The 2\u0027s cancel out and we\u0027re just left with this Rho_0 Pi divided by L^2."},{"Start":"14:54.060 ","End":"14:57.905","Text":"Of course, here you have to multiply."},{"Start":"14:57.905 ","End":"14:59.680","Text":"Here we have d divided by Pi,"},{"Start":"14:59.680 ","End":"15:02.700","Text":"so here we have to add a d divided by Pi."},{"Start":"15:02.700 ","End":"15:08.190","Text":"Remember that comes from the inner derivative of this over here."},{"Start":"15:08.190 ","End":"15:11.180","Text":"Just remember to multiply by that."},{"Start":"15:12.750 ","End":"15:19.540","Text":"This is d and that\u0027s it."},{"Start":"15:19.970 ","End":"15:23.760","Text":"This is the answer to Question number 2,"},{"Start":"15:23.760 ","End":"15:26.540","Text":"and that is the end of this lesson."}],"ID":22422}],"Thumbnail":null,"ID":99491}]

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