Gauss's law and electric field inside spheres and shells

Question

I have a question regarding the electric field inside a sphere and shell and I know the result but don't really understand why it is what it is.

Lets say there is a shell with radius $R$ and that has charge $Q$ homogeneously distributed on its surface. If I apply a spherical Gaussian surface with radius $r$ ($r < R$) I can use Gauss's law to determine the electric field which tells me that the electric field must be $0$ since the enclosed charge is zero.

But how can this be? I know that the enclosed charge is 0 but what about the electric field due to the charge on the shell? If I try to google an answer I get something along the line of symmetry. But this doesn't really clear up how symmetry makes the field contribution from the outside charge zero.

A situation where we want to find the field inside an insulating sphere with charge $Q$ (uniformly distributed throughout) we apply a Gaussian surface as before and find a result. But how do we take the charge outside this Gaussian surface in to account? Because if the Gaussian surface has radius $r < R$ then I don't understand how the shell with thickness $R-r$ is included in the result.

I think both cases have to do something to do with symmetry. Can someone explain me what it is what I am missing or what the details are why symmetry tells us we can 'ignore' the outside charge?

Answer 1

I think you are essentially asking about The Shell Theorem. That article references gravity, but the math is the same for electromagnetism.

You can simplify the problem to two dimensions with a ring of charge which in turn allows you to analyze the field along any diameter of the ring.

As I recall, the integral you build up to calculate the field is complicated, but has some properties that can simplify it. Of course because of symmetry, in the exact center, the field is zero. Use a parameter to represent getting the field off center. I think you can prove that the derivative of the integral with respect to that parameter is 0 and therefore has the same value as at the center, 0. Alternatively, it can be shown the value of the integral off center is it's own negative, also suggesting its value is zero. Since the value is zero everywhere on the diameter, and by symmetry the result applies to every dimeter of the circle, it follows that the field is 0 everywhere within the circle.

Answer 2

The answer does have to do with symmetry (as Gauss's law only applies during cases of symmetrical charge distribution). Try using calculus to solve it. Pick a random point inside a sphere with charge evenly distributed along the outer surface. Take the integral of $\frac{k\cdot dq}{r^2}$ along the entire sphere and you'll see that it is indeed 0.