A small conducting spherical shell with inner radius a and...

A small conducting spherical shell with inner radius $a$ and outer radius $b$ is concentric with a larger conducting spherical shell with inner radius $c$ and outer radius $d$ ($\textbf{Fig. P22.45}$). The inner shell has total charge +2$q$, and the outer shell has charge +4$q$. (a) Calculate the electric field $\overrightarrow{E}$ (magnitude and direction) in terms of $q$ and the distance $r$ from the common center of the two shells for (i)$ r < a; \mathrm{(ii)} a < r < b; \mathrm{(iii)} b < r < c; \mathrm{(iv)} c < r < d; \mathrm{(v)} r > d$. Graph the radial component of $\overrightarrow{E}$ as a function of $r$. (b) What is the total charge on the (i) inner surface of the small shell; (ii) outer surface of the small shell; (iii) inner surface of the large shell; (iv) outer surface of the large shell?

Transcript

00:01 We're going to review what is true about how conductors behave in electrostatics.

00:07 And in particular, we're going to look in an example with two charged conducting spheres, one nested in the other, and actually these are shells.

00:18 So they have hollows in them.

00:21 So the first thing to realize is that conductors, of course, have charges that can readily move around in them fairly rapidly.

00:31 And what this makes happen is that the electric field inside the volume of a metal is always equal to zero.

00:45 The electric field is zero.

00:47 And this is because the charges can move around.

00:50 So what they try to do is get away from each other and if they can get as far as possible from each other on the surface.

01:01 So a corollary to that is that the charges reside solely on the surface or surfaces.

01:13 So it could be an inner or an outer surface.

01:17 And one of the things that tends to happen because of this is that electric field lines terminate on conducting surfaces, meaning they begin and end on those surfaces.

01:39 So let's kind of take a look at an example of how this works, and we'll probably have to pull in gauss's law for conductors.

01:50 Let's pretend that we started off with two spheres that were separately charged.

01:57 So the outer sphere, separate.

02:00 Got plus 6q on it.

02:05 And the inner separately got zapped with positive 2q.

02:10 Sorry, let's make the outer 4q just for fun.

02:21 And then we somehow managed to hold the inner sphere steady with an insulator and manage to place it in the hollow of the first sphere.

02:35 Will anything happen to either sphere? well, we're assuming that all the charge stays on each sphere.

02:42 But what will happen to the outer sphere is that it will have negative charges come whooshing to its inner surface in the exact amount of the two q, but negative.

03:04 And where those negative charges came from, they came from somewhere on the interior, but what is true is that the total charge of that outer sphere still has to remain plus 4q, which means the outer surface then has been drained a little bit.

03:26 This is kind of how grounding works.

03:29 Has been drained of negative to q, which means it will get up to plus 6 q on the surface.

03:42 So that the total charge on that sphere remains positive for a q, and all the charge is on its surface.

03:51 But why that must happen? we can think a little bit about why that happens by looking at gauss's law.

03:58 Galses law guarantees that the electric flux through a surface, which is an electric field dotted into a surface area.

04:10 So it's the flow of electric lines straight through a surface, that that is proportional to the enclosed charge over the electrical constant epsilon zero.

04:25 If i draw a surface right inside, so gaussian surfaces are imaginary, but if i draw one right inside the inner surface of the outer sphere, that surface must enclose zero charge to have the electric field equal to zero.

04:59 And that means that, yes, the negative 2q had to scoot to that surface and cancel the positive 2q.

05:11 In other words, electric field lines had to bust out of that smaller inner sphere and turn, on the conducting outer sphere.

05:21 And i'm going to show a quick schematic of the electric field lines inside this double sphere as well as outside of it, and then we can do some quick calculations with galses ' law that aren't too hard to understand this.

05:38 But if you go into the inner sphere, there is nothing there, and the electric field is zero.

05:46 So the first place that you have an electric field is from the positive 2q sitting on that inner sphere.

05:59 And that electric field is going to basically stop, terminate on the inner surface of the outer sphere.

06:10 That electric field will disappear for a while, and it will come back on the outside.

06:17 So there are two places where there is electric field that's not zero.

06:28 So it's zero inside the metal in both instances.

06:33 It's zero in the inner hollow because there's no charge place there.

06:38 If you did put a charge there, you'd have a triply dested situation.

06:45 But finding the electric field in the region are greater than b, less than that's c, so we're just following the radii, a, b, c, d.

06:58 What we have is the electric field times 4 pi r squared, r is the radius of an arbitrary surface in there, the green surface, and that has got to equal the enclosed charge, which is plus 2q, and then over epsilon knot...

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