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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?

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  • $\begingroup$ You can use calculus to prove that (quite easily)... $\endgroup$ May 8 '20 at 20:59
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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.

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