Introduction to Electrostatic Pressure
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[{"Name":"Introduction to Electrostatic Pressure","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Exercise 1","Duration":"7m 36s","ChapterTopicVideoID":21292,"CourseChapterTopicPlaylistID":99479,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/21292.jpeg","UploadDate":"2020-04-06T13:54:29.6300000","DurationForVideoObject":"PT7M36S","Description":null,"MetaTitle":"Exercise 1 - Introduction to Electrostatic Pressure: Practice Makes Perfect | Proprep","MetaDescription":"Studied the topic name and want to practice? Here are some exercises on Introduction to Electrostatic Pressure practice questions for you to maximize your understanding.","Canonical":"https://www.proprep.uk/general-modules/all/physics-2-electricity-and-magnetism/electrostatic-pressure/introduction-to-electrostatic-pressure/vid21372","VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.085","Text":"Hello. In this lesson,"},{"Start":"00:02.085 ","End":"00:04.785","Text":"we\u0027re going to be answering the following question."},{"Start":"00:04.785 ","End":"00:07.980","Text":"A point charge q is at the center of"},{"Start":"00:07.980 ","End":"00:15.225","Text":"a charged and flexible spherical shell of a uniform charge density Sigma."},{"Start":"00:15.225 ","End":"00:19.170","Text":"What size must the charge q be in order for"},{"Start":"00:19.170 ","End":"00:23.730","Text":"the spherical shell to stay in its original form?"},{"Start":"00:23.730 ","End":"00:27.030","Text":"As we\u0027ve seen, there\u0027s going to be"},{"Start":"00:27.030 ","End":"00:32.310","Text":"electrostatic pressure being applied to each point on the shell,"},{"Start":"00:32.310 ","End":"00:37.140","Text":"due to all the other points on the shell and"},{"Start":"00:37.140 ","End":"00:41.305","Text":"this charge q applying pressure to every single section."},{"Start":"00:41.305 ","End":"00:46.570","Text":"Let\u0027s say, if we take this area over here,"},{"Start":"00:46.570 ","End":"00:50.390","Text":"dS, so there\u0027s going to be"},{"Start":"00:50.390 ","End":"00:55.520","Text":"electrostatic pressure on this point due to all the other charges over here."},{"Start":"00:55.520 ","End":"00:57.890","Text":"Due to this pressure,"},{"Start":"00:57.890 ","End":"01:01.770","Text":"because a spherical shell is flexible,"},{"Start":"01:01.770 ","End":"01:07.045","Text":"what can happen is this spherical shell will start stretching outwards."},{"Start":"01:07.045 ","End":"01:09.860","Text":"What we\u0027re trying to do is we\u0027re trying to find"},{"Start":"01:09.860 ","End":"01:13.925","Text":"some condition for this point charge over here q,"},{"Start":"01:13.925 ","End":"01:21.060","Text":"such that the spherical shell will remain in its original form, like so."},{"Start":"01:22.760 ","End":"01:28.655","Text":"The equation for electrostatic pressure is,"},{"Start":"01:28.655 ","End":"01:34.980","Text":"dF divided by dS, force over area."},{"Start":"01:34.980 ","End":"01:42.545","Text":"We\u0027re going to have over here acting in this direction,"},{"Start":"01:42.545 ","End":"01:44.650","Text":"this force over here,"},{"Start":"01:44.650 ","End":"01:49.805","Text":"F_1, and inwards from this point,"},{"Start":"01:49.805 ","End":"01:55.270","Text":"we\u0027re going to have this force over here, F_2."},{"Start":"01:55.270 ","End":"01:57.990","Text":"Let\u0027s deal with F_1,"},{"Start":"01:57.990 ","End":"02:06.440","Text":"that is the electrostatic force acting on a point just above the surface,"},{"Start":"02:06.440 ","End":"02:08.980","Text":"and we\u0027re trying to find the force over here,"},{"Start":"02:08.980 ","End":"02:15.050","Text":"it\u0027s similar or it\u0027s exactly the same as what we did in the previous lesson."},{"Start":"02:15.050 ","End":"02:18.230","Text":"We saw that the force is equal to"},{"Start":"02:18.230 ","End":"02:22.610","Text":"dq multiplied by the electric field at this point over here,"},{"Start":"02:22.610 ","End":"02:24.409","Text":"just above the surface."},{"Start":"02:24.409 ","End":"02:31.020","Text":"Then we said that this was equal to Sigma dS multiplied by the electric field."},{"Start":"02:31.020 ","End":"02:35.180","Text":"We found later that the electric field due to"},{"Start":"02:35.180 ","End":"02:43.245","Text":"this tiny portion or unit area of dS is equal to Sigma divided by 2 epsilon naught,"},{"Start":"02:43.245 ","End":"02:47.850","Text":"so we got that this was equal to Sigma dS"},{"Start":"02:47.850 ","End":"02:53.370","Text":"multiplied by Sigma divided by 2 Epsilon naught."},{"Start":"02:53.370 ","End":"02:59.010","Text":"Up until here, this is exactly what we did in the previous lesson."},{"Start":"03:00.560 ","End":"03:04.475","Text":"F_1 over here to just explain,"},{"Start":"03:04.475 ","End":"03:10.280","Text":"is the force being applied by every single one of the points on the spherical shell,"},{"Start":"03:10.280 ","End":"03:13.100","Text":"on this point dS over here."},{"Start":"03:13.100 ","End":"03:17.855","Text":"F_1 is the force applied by"},{"Start":"03:17.855 ","End":"03:24.735","Text":"all the points on the spherical shell onto this unit area dS."},{"Start":"03:24.735 ","End":"03:28.890","Text":"Now let\u0027s deal with F_2."},{"Start":"03:28.890 ","End":"03:32.984","Text":"F_2 is the force being applied on dS"},{"Start":"03:32.984 ","End":"03:41.370","Text":"by this charge q,"},{"Start":"03:41.370 ","End":"03:48.525","Text":"and what we want is for our force F_2 to balance out our force F_1."},{"Start":"03:48.525 ","End":"03:52.680","Text":"Because if the forces on dS are balanced,"},{"Start":"03:52.680 ","End":"03:54.415","Text":"F_2 is equal to F_1,"},{"Start":"03:54.415 ","End":"04:00.290","Text":"then that means that this point over here won\u0027t move outwards and won\u0027t move"},{"Start":"04:00.290 ","End":"04:07.995","Text":"inwards.We can make that as a rule on every single point on the spherical shell."},{"Start":"04:07.995 ","End":"04:12.905","Text":"F_2, we\u0027re just dealing with Coulomb\u0027s law."},{"Start":"04:12.905 ","End":"04:16.535","Text":"We have k, q1, q2,"},{"Start":"04:16.535 ","End":"04:23.540","Text":"Q1 is just q and q2 is this dq,"},{"Start":"04:23.540 ","End":"04:29.465","Text":"the charge on this unit area and divided by the radius squared,"},{"Start":"04:29.465 ","End":"04:32.855","Text":"so that\u0027s capital R^2."},{"Start":"04:32.855 ","End":"04:36.980","Text":"Then dq, we know is equal to Sigma dS,"},{"Start":"04:36.980 ","End":"04:39.045","Text":"so we have kq"},{"Start":"04:39.045 ","End":"04:47.180","Text":"Sigma dS divided by R^2."},{"Start":"04:47.180 ","End":"04:52.130","Text":"These are all in the radial directions."},{"Start":"04:52.130 ","End":"04:57.180","Text":"But what we can do is we can just work out their size."},{"Start":"05:00.220 ","End":"05:05.690","Text":"We want these 2 to be equal to 1 another,"},{"Start":"05:05.690 ","End":"05:09.335","Text":"so that when we do F_2 minus F_1,"},{"Start":"05:09.335 ","End":"05:14.519","Text":"or F_1 minus F_2 we\u0027ll get 0."},{"Start":"05:16.760 ","End":"05:27.360","Text":"Therefore, we\u0027re going to say that F_1 has to be equal to negative F_2."},{"Start":"05:27.880 ","End":"05:36.905","Text":"They\u0027re the same but with the opposite sign when we add them up,"},{"Start":"05:36.905 ","End":"05:40.230","Text":"the forces over here will cancel out."},{"Start":"05:41.000 ","End":"05:50.315","Text":"F_1 is Sigma squared dS divided by 2 Epsilon naught,"},{"Start":"05:50.315 ","End":"05:53.000","Text":"and this is equal to negative F_2,"},{"Start":"05:53.000 ","End":"06:01.110","Text":"so negative kq Sigma dS divided by R^2,"},{"Start":"06:01.110 ","End":"06:05.640","Text":"so we can cancel off 1 of the Sigmas and the dS."},{"Start":"06:05.640 ","End":"06:11.985","Text":"We can remember that Epsilon naught is equal to 1"},{"Start":"06:11.985 ","End":"06:18.570","Text":"over 4 Pi k. Therefore,"},{"Start":"06:18.570 ","End":"06:23.775","Text":"we can say that Sigma divided by"},{"Start":"06:23.775 ","End":"06:30.330","Text":"2 multiplied by 4 Pi k,"},{"Start":"06:30.330 ","End":"06:32.805","Text":"because the Epsilon is in the denominator,"},{"Start":"06:32.805 ","End":"06:41.490","Text":"this is equal to negative kq divided by R^2."},{"Start":"06:41.490 ","End":"06:44.925","Text":"We can cancel that,"},{"Start":"06:44.925 ","End":"06:52.770","Text":"we can divide both sides by k. What we\u0027ll get is that 2 of"},{"Start":"06:52.770 ","End":"07:02.130","Text":"Sigma Pi is equal to negative q divided by R^2."},{"Start":"07:02.130 ","End":"07:07.440","Text":"We\u0027re trying to find the value of q,"},{"Start":"07:07.440 ","End":"07:10.035","Text":"so that the forces here balance out."},{"Start":"07:10.035 ","End":"07:12.730","Text":"Once we isolate out our q,"},{"Start":"07:12.730 ","End":"07:22.720","Text":"will get that this is equal to negative 2 Pi Sigma R^2."},{"Start":"07:22.720 ","End":"07:27.725","Text":"This is the answer to the question on what size must charge q be"},{"Start":"07:27.725 ","End":"07:33.335","Text":"in order to keep this flexible spherical shell in its original form."},{"Start":"07:33.335 ","End":"07:36.570","Text":"That\u0027s the end of this lesson."}],"ID":21372},{"Watched":false,"Name":"Exercise 2","Duration":"10m 5s","ChapterTopicVideoID":21403,"CourseChapterTopicPlaylistID":99479,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.115","Text":"Hello. In this lesson,"},{"Start":"00:02.115 ","End":"00:04.770","Text":"we\u0027re going to be answering the following question."},{"Start":"00:04.770 ","End":"00:07.950","Text":"Calculate the electrostatic pressure on"},{"Start":"00:07.950 ","End":"00:13.350","Text":"a spherical shell of radius R with a charge density Sigma."},{"Start":"00:13.350 ","End":"00:18.885","Text":"First of all, let\u0027s define what electrostatic pressure is."},{"Start":"00:18.885 ","End":"00:23.850","Text":"Pressure, and it\u0027s defined as the amount of"},{"Start":"00:23.850 ","End":"00:31.030","Text":"force that is acting on the amount of area."},{"Start":"00:31.550 ","End":"00:34.980","Text":"This is the equation that we\u0027re going to use."},{"Start":"00:34.980 ","End":"00:40.020","Text":"Now, what we\u0027re going to do is we\u0027re going to take a tiny little piece of"},{"Start":"00:40.020 ","End":"00:48.300","Text":"area from the spherical shell and this has an area of dS,"},{"Start":"00:48.300 ","End":"00:50.523","Text":"it\u0027s a very small area,"},{"Start":"00:50.523 ","End":"00:53.615","Text":"and what we\u0027re going to do is we\u0027re going to find"},{"Start":"00:53.615 ","End":"00:57.575","Text":"out how much force is acting on this area"},{"Start":"00:57.575 ","End":"01:07.830","Text":"dS due to all the other pieces of area or unit areas in the spherical shell."},{"Start":"01:08.020 ","End":"01:13.430","Text":"This unit area over here belonging to the spherical shell"},{"Start":"01:13.430 ","End":"01:18.907","Text":"isn\u0027t influenced by its own force or electric field,"},{"Start":"01:18.907 ","End":"01:24.731","Text":"however, it is influenced by all the other unit areas in the shell."},{"Start":"01:24.731 ","End":"01:29.610","Text":"So we\u0027re trying to find out how much that is."},{"Start":"01:30.710 ","End":"01:33.990","Text":"We\u0027re trying to find this force."},{"Start":"01:33.990 ","End":"01:36.750","Text":"It\u0027s the force on this unit area,"},{"Start":"01:36.750 ","End":"01:38.919","Text":"so it\u0027s going to be dF."},{"Start":"01:38.919 ","End":"01:47.825","Text":"This is going to be equal to as usual force is to do with the charge and the E-field."},{"Start":"01:47.825 ","End":"01:51.409","Text":"It\u0027s going to be multiplied by the charge over here,"},{"Start":"01:51.409 ","End":"01:54.210","Text":"which is dq,"},{"Start":"01:54.280 ","End":"01:59.295","Text":"and then multiplied by the E-field."},{"Start":"01:59.295 ","End":"02:05.970","Text":"We know that there is an E-field acting over here."},{"Start":"02:07.610 ","End":"02:10.730","Text":"We\u0027re not given the charge,"},{"Start":"02:10.730 ","End":"02:14.480","Text":"but we are given the charge density as Sigma."},{"Start":"02:14.480 ","End":"02:20.615","Text":"We know that the charge is equal to the charge density multiplied by the area,"},{"Start":"02:20.615 ","End":"02:24.157","Text":"where area over here is unit area ds,"},{"Start":"02:24.157 ","End":"02:29.040","Text":"and then we\u0027re multiplying it by the E-field."},{"Start":"02:30.010 ","End":"02:35.745","Text":"Let\u0027s go back to our equation here for electrostatic pressure."},{"Start":"02:35.745 ","End":"02:41.430","Text":"That is equal to dF divided by dS."},{"Start":"02:41.430 ","End":"02:48.380","Text":"If we divide both sides by dS we will get that dF divided"},{"Start":"02:48.380 ","End":"02:56.940","Text":"by dS is equal to Sigma multiplied by E. That\u0027s what we plug in here."},{"Start":"03:00.020 ","End":"03:08.180","Text":"This means that the electrostatic pressure acting in this unit of area is equal to Sigma"},{"Start":"03:08.180 ","End":"03:11.870","Text":"multiplied by E. Now the only problem is we don\u0027t"},{"Start":"03:11.870 ","End":"03:16.830","Text":"know what our E-field is so we have to calculate that."},{"Start":"03:17.150 ","End":"03:23.480","Text":"We can\u0027t just calculate the electric field of a spherical shell and that\u0027s because we\u0027re"},{"Start":"03:23.480 ","End":"03:30.120","Text":"trying to find the electric field acting just on this point."},{"Start":"03:30.120 ","End":"03:35.996","Text":"The electric field of this point isn\u0027t acting on itself,"},{"Start":"03:35.996 ","End":"03:42.600","Text":"so we\u0027re trying to find all the rest of the electric field acting on this point."},{"Start":"03:43.310 ","End":"03:49.670","Text":"We\u0027re trying to find the electric field without this area over here,"},{"Start":"03:49.670 ","End":"03:53.250","Text":"so now let\u0027s rub out this unit area."},{"Start":"03:53.840 ","End":"04:02.720","Text":"What I\u0027m trying to find is the electric field at this point due to"},{"Start":"04:02.720 ","End":"04:08.000","Text":"all of the unit areas or all"},{"Start":"04:08.000 ","End":"04:14.370","Text":"of these pieces over here exerting or producing an electric field at this point."},{"Start":"04:14.420 ","End":"04:20.870","Text":"What I\u0027m going to do is I\u0027m going to calculate the electric field right over"},{"Start":"04:20.870 ","End":"04:27.230","Text":"here in green where it\u0027s only slightly above this red point over here."},{"Start":"04:27.230 ","End":"04:30.710","Text":"Then I can say that the E-field at the green point"},{"Start":"04:30.710 ","End":"04:33.725","Text":"and the E-field at the red point is the same"},{"Start":"04:33.725 ","End":"04:37.530","Text":"because these 2 points are very close to"},{"Start":"04:37.530 ","End":"04:42.360","Text":"1 another and the distance between them is approaching 0."},{"Start":"04:43.460 ","End":"04:47.860","Text":"First, let\u0027s calculate the electric field on"},{"Start":"04:47.860 ","End":"04:54.480","Text":"the surface due to the whole shell being charged."},{"Start":"04:54.480 ","End":"04:57.585","Text":"This is our usual E-field,"},{"Start":"04:57.585 ","End":"05:04.230","Text":"so it\u0027s equal to kQ divided by R^2,"},{"Start":"05:04.230 ","End":"05:11.015","Text":"I\u0027m going in the radial direction because we\u0027re imagining that we\u0027re on the surface."},{"Start":"05:11.015 ","End":"05:12.920","Text":"The surface is at a radius R,"},{"Start":"05:12.920 ","End":"05:15.900","Text":"it\u0027s a distance R away from the center."},{"Start":"05:16.280 ","End":"05:23.135","Text":"This is the E-field due to a spherical shell."},{"Start":"05:23.135 ","End":"05:30.459","Text":"Now, my Q, we worked out over here was Sigma dS,"},{"Start":"05:30.459 ","End":"05:32.411","Text":"so let\u0027s write it out,"},{"Start":"05:32.411 ","End":"05:36.093","Text":"so it\u0027s Sigma multiplied by the surface area,"},{"Start":"05:36.093 ","End":"05:41.390","Text":"so the surface area of a sphere is equal to 4 Pi"},{"Start":"05:41.390 ","End":"05:47.743","Text":"multiplied by R^2 and then all of this is divided by R^2,"},{"Start":"05:47.743 ","End":"05:50.300","Text":"so now we can cross out and this is of course still"},{"Start":"05:50.300 ","End":"05:53.345","Text":"in the radial direction, these 2 cancel."},{"Start":"05:53.345 ","End":"05:55.835","Text":"Then if you remember,"},{"Start":"05:55.835 ","End":"06:01.080","Text":"Epsilon Naught is 1/4 Pi K. So 4,"},{"Start":"06:01.080 ","End":"06:02.730","Text":"Pi, and K,"},{"Start":"06:02.730 ","End":"06:06.170","Text":"so therefore we can write this as Sigma"},{"Start":"06:06.170 ","End":"06:11.010","Text":"divided by Epsilon Naught in the radial direction."},{"Start":"06:12.380 ","End":"06:16.916","Text":"This is the E-field of the entire spherical shell,"},{"Start":"06:16.916 ","End":"06:20.105","Text":"and now what we want to do is we want to find"},{"Start":"06:20.105 ","End":"06:24.530","Text":"the E-field just of this little piece over here,"},{"Start":"06:24.530 ","End":"06:28.130","Text":"and then what we\u0027ll do is we\u0027ll subtract the E-field from"},{"Start":"06:28.130 ","End":"06:32.570","Text":"this little piece from the E-field of the entire spherical shell,"},{"Start":"06:32.570 ","End":"06:40.090","Text":"and then we will substitute that in over here and then we\u0027ll get our answer."},{"Start":"06:41.030 ","End":"06:43.745","Text":"The E-field of dS,"},{"Start":"06:43.745 ","End":"06:45.335","Text":"of this little piece over here,"},{"Start":"06:45.335 ","End":"06:48.785","Text":"is the same as the E-field of an infinite plane."},{"Start":"06:48.785 ","End":"06:51.635","Text":"If we look at this little section over here,"},{"Start":"06:51.635 ","End":"06:53.435","Text":"if we zoom in,"},{"Start":"06:53.435 ","End":"06:55.235","Text":"it will look something like this,"},{"Start":"06:55.235 ","End":"06:57.650","Text":"or if I draw it with the angle,"},{"Start":"06:57.650 ","End":"07:01.435","Text":"it will look something like this, zoomed in."},{"Start":"07:01.435 ","End":"07:03.810","Text":"Because it\u0027s part of a spherical shell, so we can look,"},{"Start":"07:03.810 ","End":"07:07.445","Text":"if we come really close to it,"},{"Start":"07:07.445 ","End":"07:11.220","Text":"it will look like an infinite flat plane. Why is this?"},{"Start":"07:11.220 ","End":"07:14.570","Text":"It\u0027s the same reason why when we\u0027re standing on earth,"},{"Start":"07:14.570 ","End":"07:17.930","Text":"we\u0027ll see Earth as if it\u0027s flat."},{"Start":"07:17.930 ","End":"07:19.715","Text":"If we\u0027re standing in a field,"},{"Start":"07:19.715 ","End":"07:24.383","Text":"it will look flat to us even though we know that Earth is a sphere."},{"Start":"07:24.383 ","End":"07:27.830","Text":"We\u0027ve seen images from space and it\u0027s a sphere."},{"Start":"07:27.830 ","End":"07:33.980","Text":"That is the same reason why we can consider the E-field of this piece over here,"},{"Start":"07:33.980 ","End":"07:37.335","Text":"dS, as an infinite plane."},{"Start":"07:37.335 ","End":"07:41.360","Text":"The E-field of an infinite plane, as we already know,"},{"Start":"07:41.360 ","End":"07:50.399","Text":"is equal to Sigma divided by 2Epsilon_0 in the radial direction."},{"Start":"07:51.080 ","End":"07:56.585","Text":"Now we can calculate the E-field acting on dS."},{"Start":"07:56.585 ","End":"08:02.460","Text":"The E-field acting on dS is simply"},{"Start":"08:02.460 ","End":"08:08.180","Text":"equal to the E-field of the entire spherical shell,"},{"Start":"08:08.180 ","End":"08:12.240","Text":"so let\u0027s write here S for Shell,"},{"Start":"08:13.100 ","End":"08:18.245","Text":"minus the E-field of dS itself,"},{"Start":"08:18.245 ","End":"08:26.245","Text":"because we already said that the electric field of dS isn\u0027t acting on itself,"},{"Start":"08:26.245 ","End":"08:31.710","Text":"so minus E and then let\u0027s write here of dS,"},{"Start":"08:31.710 ","End":"08:38.090","Text":"so that is going to be equal to Sigma divided by"},{"Start":"08:38.090 ","End":"08:47.435","Text":"Epsilon Naught minus Sigma divided by 2Epsilon_0 in the r direction,"},{"Start":"08:47.435 ","End":"08:55.540","Text":"and this is simply equal to Sigma divided by 2Epsilon_0 in the r direction."},{"Start":"08:56.530 ","End":"09:02.762","Text":"Now, we were being asked to calculate the electrostatic pressure,"},{"Start":"09:02.762 ","End":"09:07.595","Text":"so all that\u0027s left is to plug this into the equation over here."},{"Start":"09:07.595 ","End":"09:16.505","Text":"Therefore, we\u0027ll get that the pressure is equal to Sigma multiplied by E,"},{"Start":"09:16.505 ","End":"09:18.650","Text":"which is what we just calculated,"},{"Start":"09:18.650 ","End":"09:24.530","Text":"which is Sigma multiplied by Sigma divided by"},{"Start":"09:24.530 ","End":"09:32.660","Text":"2Epsilon_0 which is simply equal to Sigma^2 divided by 2Epsilon_0."},{"Start":"09:32.660 ","End":"09:38.175","Text":"That is the electrostatic pressure acting on this section here."},{"Start":"09:38.175 ","End":"09:43.130","Text":"It comes from the fact that all of these points over here on"},{"Start":"09:43.130 ","End":"09:49.436","Text":"the spherical shell are producing an electric field on this point,"},{"Start":"09:49.436 ","End":"09:57.570","Text":"but the electric field produced by this point itself is not acting on this point."},{"Start":"09:59.150 ","End":"10:05.040","Text":"This is the answer to the question and this is the end of this lesson."}],"ID":22357}],"Thumbnail":null,"ID":99479}]

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