# Nonconducting shells

1. Feb 22, 2010

### reising1

The electric field inside a conducting shell is 0 because the charge moves to the surface, and thus there is no charge anywhere inside the shell. Hence, the E field is 0.

But why is the E field 0 inside a nonconducting shell?

Thanks!

2. Feb 22, 2010

### collinsmark

3. Feb 22, 2010

### reising1

Gauss' Law states that Enot times the flux equals the charge enclosed

Using Gauss' Law, I get an equation which does not conclude that E is zero. Could you please help explain?

4. Feb 22, 2010

### collinsmark

[Edit:] Sorry, but when I originally posted this, I didn't read your question very carefully, and I thought you were asking something else. You might not find this post very pertinent. See my next post for a response more on the topic of your original question. The general idea is that the electric field in a nonconducting shell is not always zero.

Well, I would say,

$$\oint _S \vec E \cdot \vec{dA} = \frac{Q_{enc}}{\epsilon _0}$$

Think about what that means. By definition, the integral is over a closed surface. In relation to this problem, think of the surface of interest as being completely inside the shell. If you integrate the electric field lines parallel to the surface, over the entire closed surface, the result is proportional to $$Q_{enc}$$, the charge contained within the surface.

So there are three things of interest, the electric field, the Gaussian surface, and the enclosed charge. If $$Q_{enc}$$ is equal to zero, it means the left hand side of the equation is equal to zero too. Since the Gaussian surface is not necessarily zero, what does that tell you about E?

Last edited: Feb 22, 2010
5. Feb 22, 2010

### collinsmark

Oh, by the way, regarding your original question,

The electric field inside a nonconducting shell is zero too if there no charge inside the nonconducting shell itself, and the charge distribution on the shell itself is spherically symmetrical, and there are no other electric fields outside the shell.

The electrons in a conducting shell align themselves to force the electric field to be zero inside the conductor.

However, in a nonconducting shell, the electrons don't get to move around. So the electric field inside a non-conducting shell is not necessarily zero, if there is an electric field outside the shell, or if the charge distribution is not symmetric.

6. Feb 22, 2010

### reising1

So if there is a nonconducting shell of uniform surface charge density 5 Coloumbs, for example, is there an electric field on the inside of the shell?

7. Feb 22, 2010

### collinsmark

In that particular case, there is not an electric field inside the shell, if there are no other charges nearby. But in this case, the only thing keeping the electric field equal to zero is the spherical symmetry of the charge distribution. However, if you were to put a point charge somewhere, even outside the shell, there would be an electric field inside the shell.

Conducting shells act differently. If you were to put a point charge outside the shell, the charge distribution of the shell itself would automatically realign in just such a way as make the electric field on the interior 0.

8. Feb 22, 2010

### reising1

What do you mean by the spherical symmetry of the charge distribution?

9. Feb 22, 2010

### ideasrule

Suppose you're inside a nonconducting shell (as in, a hollow sphere). Choose a random (very small) area on the shell. As you get farther from that area, its electric influence diminishes as 1/r^2, but the solid angle subtended by it also diminishes as 1/r^2. This means that for any given small solid angle, the electric field caused by the area of shell subtended by that solid angle does not depend on the distance to the area. So no matter where you are in the shell, the electric field caused by the charges in front of you is cancelled out by the field caused by the charges behind you.

10. Feb 22, 2010

### collinsmark

Well, you mentioned "if there is a nonconducting shell of uniform surface charge density..." meaning the charge is distributed evenly on the surface of the shell.

That means you could rotate the shell anywhere around the center, at any angle, and it wouldn't change the situation at all. The charge distribution is spherically symetric.

Now imagine a non-conducting sphere where the charge distribution was not symmetric, such there was more charge on one side of the shell than the other side. That is another situation where there would be a non-zero electric field inside the non-conducting shell. (I'm assuming here that there are no other charges present besides the shell itself.)

That's where conducting spheres get a little weird. If you put a point charge near, but outside of a conducting shell, the charge on the surface of the sphere automatically rearranges itself such that the electric field inside the shell is zero. The situation here is
(a) a external electric field
(b) the charge distribution on the shell is not spherically symmetric.
But as it happens, the non-spherically symmetric charge distribution is arranged in exactly the sort of way that it cancels out the electric field caused by the external electric field, within the shell. That's the interesting thing about conductors.

11. Feb 22, 2010

### reising1

Thanks so much. One last question.

If there is a sphere with uniform charge (not specified if the sphere is a conductor or a nonconductor) of radius A concentric with a shell (So there is a sphere sitting inside a shell with an inner radius B and an outer radius C where C>B>A), why is the electric field on the inside of the sphere (at a distance from the center, for example, of A/2) not zero?

Is this because the uniform charge is distributed not only on the surface but on the inside as well, since the problem statement only specified "a sphere with uniform charge." Thus, there is charge on the inside.

Also, can someone explain why the Electric field at a distance r = 0 (at the center of the sphere) is 0? According to the equation of electric field, would that not make the electric field undefined, rather than 0. Since we are dividing by r^2, or 0?

Thank you so much. This has been very helpful.

12. Feb 23, 2010

### collinsmark

The easiest way to explain this using Gauss' law (in particular because the charge distribution is spherically symmetrical in your description). If we make a spherical Gaussian surface to be inside the sphere (for example, A/2), there is charge enclosed within the closed surface! And since the electric field is perpendicular to the Gaussian surface (due to the spherical symmetry of the charge distribution), there is an electric field inside the Gaussian surface! And since the surface and the electric field are always perpendicular and symmetric, the magnitude of the electric field can be pulled out from under the integral.

In this case, mostly in part due to the spherical symmetry discussed in the above paragraph,

$$\oint _S \vec E \cdot \vec{dA} = \frac{Q_{enc}}{\epsilon _0}$$

reduces to

$$E \int _{\theta = 0} ^{\pi} \int _{\phi = 0} ^{2\pi} r^2 sin\theta d\theta d\phi = \frac{Q_{enc}}{\epsilon _0}$$

$$E4\pi r^2 = \frac{Q_{enc}}{\epsilon _0}$$

or

$$\vec E = = \frac{Q_{enc}}{4\pi \epsilon _0} \frac{1}{r^2}\hat r$$

But also note that the enclosed charge is a function of the volume within the Gaussian surface, and the charge density. Noting the that volume of a sphere is $$(4/3)\pi r^3$$, and assuming a uniform volume charge density $$\rho$$,

$$Q_{enc} = \rho \frac{4}{3}\pi r^3$$

Combining these these equations gives us a relationship for the electric field within a solid, uniformly charged sphere.

(And of course, that's assuming that there are no other charges around, *unless* those other charges are spherically symmetric with respect to the Gaussian surface. In other words, concentric, uniformly charged shells on the outside have no effect on the results. On the other hand, a point charge somewhere off to the side will effect the results, so it is assumed that there are none of those.)

Yes, that's right. But it would be more clear if the problem states something like "uniform charge across the volume" for volume charge or "uniform charge distribution about the surface" for surface charge.

If the problem states that the non-conducting sphere is solid and has a "uniform charge," you can probably assume that the question means uniform volume charge distribution. But it wouldn't hurt the authors of the problem statement to be more clear.

Go ahead and combine the above equations, and you will find that the electric field does in fact go to zero as r goes to zero.

Last edited: Feb 23, 2010
13. Feb 25, 2010

### reising1

Wow! Thank you so much. I understand now. What you wrote was very helpful.

14. Jun 9, 2011

i have a question . electric field inside a charged non conducting shell is zero while electric
filed inside a non conducting sphere is not zero.why?

15. Jun 9, 2011

inside a charged non conducting sphere charged is uniformly distributed inside the sphere but what in case of charged non conducting shell ?
is charge lie on the outer surface of shell or distributed on the small solid part of shell as n case of non conducting sphere

16. Jun 10, 2011

### collinsmark

Let's take each case independently.

Non-conducting, spherical shell:

If the charge distribution on the shell is uniform and there are no other charges around (neither inside nor outside of the shell), then the electric field within the shell is zero.

But if you place some other charge in the proximity of the shell (for example a point charge outside the shell), the electric field inside the shell is what it would be if the shell was not there, but the other charge was. The electric field attributed to the other charge is still there.

Also, if the non-conducting, spherical shell's charge distribution is not uniform (not spherically symmetric), then there will be an electric field inside the shell even if the shell is the only charge around.

Conducting, spherical shell:

If there are no charges around, the charge distribution on the conducting, spherical shell will be spherically symmetric. This will happen naturally. And like the non-conducting case, the electric field within the conducting shell is zero. (Assuming there is no other charge somewhere inside the shell.)

Now consider bringing some other charge close by, yet still outside the shell. In the conducting case, the charge distribution on the conducting shell will rearrange itself in just such a way to cancel out the electric field caused by any exterior charges. The charge distribution on the conducting shell is no longer uniform (no longer spherically symmetric) in this case, but the electric field within the conducting shell is still zero.

Like charges repel each other. So the charge distribution of a solid conductor will all be on the outer surface of the conductor (of any shape), and the charge distribution will be exactly that to create 0 electric field inside the solid conductor.

The same applies to hollow conductor (of any shape), as long as there are no other charges contained somewhere within the hollow part.

I'll leave it to you to consider a thick conductor (of some arbitrary shape), containing an inner cavity of arbitrary shape within, and with an isolated charge inside the hollow part (inside the cavity). It's a little more complicated, but not too complicated. Just remember the following:

• The electric field within a solid conducting material is zero. That also applies to any solid section of the conductor discussed here.
• Gauss' law can help out a lot.

17. Jun 11, 2011