A solid conducting sphere of radius a is concentric with a...

Name: A solid conducting sphere of radius a is concentric with a spherical insulating shell of inner radius b = 2a and outer radius c = 3a. The sphere has a net charge +q; the shell has a net uniform charge -q. What is the magnitude of the electric field at radial distances r when (a) r < a, (b) a < r < b, (c) b < r < c, (d) c < r?
Uploaded: 2023-09-15T15:33:01-08:00
Duration: 5 min 26 s
Channel: Justin Swantek
Description: A solid conducting sphere of radius a is concentric with a spherical insulating shell of inner radius b = 2a and outer radius c = 3a. The sphere has a net charge +q; the shell has a net uniform charge -q. What is the magnitude of the electric field at radial distances r when (a) r < a, (b) a < r < b, (c) b < r < c, (d) c < r?

A solid conducting sphere of radius a is concentric with a spherical insulating shell of inner radius b = 2a and outer radius c = 3a. The sphere has a net charge +q; the shell has a net uniform charge -q. What is the magnitude of the electric field at radial distances r when (a) r < a, (b) a < r < b, (c) b < r < c, (d) c < r?

Transcript

0:00 All right, hello.

00:01 In this question, we're given this setup.

00:03 We have a conducting sphere, which has a positive charge q distributed on it or in it.

00:11 And then we have a conducting shell, spherical shell, that has a charge of negative q evenly distributed through it.

00:20 And this is an insulating, not conducting, shell.

00:23 And we're told what the radii are.

00:25 And we want to find the electric field at radial distances inside here in this region, in this region, and all the way outside.

00:34 So let's go ahead and start with r is less than a.

00:39 In order to do all of this, we're going to use gauss's law, which is that our electric flux is equal to the surface integral of e dotted with da.

00:50 And that's equal to our enclosed charge divided by epsilon 0.

00:53 So i'm going to go ahead and draw my first gaussian surface.

00:56 It's going to be somewhere inside our conducting sphere here.

01:00 Now, because this is a conducting sphere and has a positive charge, all of the charge is going to repel itself and push itself away from one another as much as possible.

01:10 So all of the actual charge on this conducting sphere is going to be distributed right on the edge.

01:18 So inside the sphere, our q enclosed is going to be 0.

01:22 So our electric field for r is less than a is going to be equal to 0 because we're not enclosing any charge.

01:31 All of the charge is on the outside of the shell, or the outside of the sphere, and we're not enclosing anything there.

01:38 So let's go ahead and look between r equals a and r equals 2a.

01:43 So that's in this region between the two right here.

01:49 And if i go ahead and write gauss's law, i'm going to have the integral of e dotted with da equals my enclosed charge over epsilon 0.

01:57 Well, this is going to be e.

01:58 The area i'm enclosing, since it is three -dimensional, is going to be 4 pi r squared.

02:04 And then how much charge am i enclosing? well, i'm enclosing the entirety of this sphere, which has a charge of q.

02:10 And i'm going to divide that by epsilon 0.

02:12 So my electric field for this region is going to be equal to q over 4 pi epsilon 0 r squared.

02:22 Let's go ahead and look at the next region, which is between the plates of this insulating sphere.

02:31 And this one will be the most tricky.

02:33 On the left -hand side, we have e.

02:35 And then the area that we're enclosing is, again, just 4 pi r squared.

02:40 But what is our charge that we're enclosing now? because this is an insulating sphere, its charge q is evenly distributed all the way throughout its area.

02:49 So it has some charge density, sigma.

02:51 That's the charge divided by its total area...

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