Electric field

1. Jul 23, 2009

bodensee9

Hello
I have a conceptual question about the following. Suppose I have a spherical shell with charge density $$\sigma$$ that is uniformly distributed throughout its surface. My shell has radius a. Then I cut a little circular piece of radius b with b << a. Then I know that my electric field at the center of this little hole is given by $$2\pi\sigma$$, because I can treat my shell with hole as a perfect shell, with $$E = 4\pi\sigma$$ going outwards and then a little piece with $$2\pi\sigma$$ going inwards (and then I add the two). So then at the center of this sphere, I would have a field of magnitude
$$2\pi\sigma$$ going inwards because inside the shell I have no field? Thanks.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jul 23, 2009

turin

There are several errors here, not the least of which is the assumption of the field of the small disk of radius b. Hint: the disk of radius b is neither a conductor nor an infinite plane. Also, there seems to be extra π's and missing ε's, but that may be a characteristic of your unit system.

3. Jul 23, 2009

bodensee9

thanks, but actually, could you let me know what I'm doing wrong with the superposition principle? by Gauss' law, the field on one side of the disk - field on other side of the disk = 4pi(sigma). So don't we have 2E*A (area) = 4pi(sigma)*A, so E on each side = 2pi(sigma). Actually, I'm pretty sure that at the center of the little hole this is the right answer, but I'm just wondering about the center of the spherical shell. I'm using CGS units. In these units k = 1, so I don't have any epsilons or pi's. Thanks.

4. Jul 23, 2009

turin

OK, I think that explains your π's and ε's. So, I think that you found the correct field value at the center of the hole.

However, you cannot assume that the field from the negative disk is independent of position in general. The E=2πσ expression (for an infinite, uniform surface charge) works at the center of the disk because, at that distance, the disk is approximately infinite (compared to the distance from the disk), and the disk is symmetric about the center. If you either move away from the center, or move away from the disk, then this approximation does not hold. In particular, the center of the shell is at a distance d>>b away from the disk, so E=2πσ is a terrible approximation there. However, can you think of a different geometry that might approximate the disk well, when you are so far away that it looks very tiny, like a single ...

5. Jul 23, 2009

bodensee9

so if I'm very far away from the disk, would it be like a point charge? so then does that mean I will only have E = q/r^2? where q is the charge on the disk?
just to follow up, so if I'm not at the center of the disk, then I will need to integrate? In this case, wouldn't I be better off integrating over the whole sphere with the missing hole then? For example, will I have
$$dE = \frac{dA\sigma}{r^{2}}cos\alpha$$
where
$$\alpha$$ is the angle approximated by my distance l from the center of the disk over R (radius of shell?)
Thanks.

6. Jul 23, 2009

turin

That would be my approximation.

I don't know. Do you need to consider this case as well?