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?