Electric field inside a non-uniformly charged conductor

Question

With Gauss's Law, if we take a uniformly charged hollow sphere for instance, the electric field inside the hollow sphere is 0 because there is no charge inside. However, what if this sphere is not uniform?

I got confused because I heard this from a physics professor that said that if the charges around the sphere were not uniformly distributed, then the electric field inside would not be zero. Why is that?

Just to add, obviously every conductor would distribute its charge in a uniform way, so this question would not be relevant because it would not be something seen in nature. However, suppose that it was possible to have it non-uniformly distributed, or that it is an insulator for instance. In my opinion, the electric field inside should still be 0 because if not, Gauss's Law would break! There is still no charge inside, so the electric field should be 0, is that correct? Maybe I am not taking in consideration something and I am wrong, but I would appreciate if someone can clear my confusion, thanks in advance.

Edit: I will elaborate a bit more in my question.

I can think of why the electric field inside could be non-zero, just that I don't understand why Gauss's Law says that the electric flux is still 0.

Suppose that you have an electric dipole (just two opposite charges in space, each equal in magnitude) separated by a distance 'd'. Then, it is obvious that in between these two charges there must be an electric field, flowing out of the positive charge onto the negative charge. However, if we make a Gaussian spherical surface in between the charges, with a radius 'r' smaller than 'd' (r < d, so that the imaginary sphere does not enclose any charge). Then, Gauss's Law says that if the charge inside is 0 then the electric flux is 0, but we know that there is an electric field in between these two charges! Maybe I have a misconception of Gauss's Law, and it would be great if someone could explain to me what is my mistake (because I know that I am wrong). Certainly the electric field in (E)(dA)cosx is not 0, and the area is neither 0 obviously, and 'x' is not 90º. The only explanation that I can think of is if the integral of that basically sums quantities and subtracts other quantities...

Answer 1

Firstly, in electrostatic problems charges are only found at the surface of conductors. Electric fields exert a force on charges ($\mathbf{F} = q \mathbf{E}$). Therefore, these must be absent ($\mathbf{E} = 0$) within a conducting material. Thus, electric field lines must terminate with charges at the surface of the conductor. You can see this by constructing a small Gaussian pillbox enclosing part of the surface of the conductor. Flux $\Phi_E = E_{\perp} A$ leaves the side outside the conductor and there is no flux through the other sides. Then Gauss's law gives charge $Q = \epsilon E_{\perp} A$ at the surface. For any Gaussian surface entirely within the conductor flux and net charge will both be zero.

The above indicates that the distribution of charges on the surface of the hollow conducting sphere will be uniform if $E_{\perp}$ is. However, if the electric field outside the sphere is not spherically symmetric the charge distribution at its surface won't be. This would arise if there were a non-spherically symmetric charge distribution outside the sphere.

Your edit indicates a misconception about electric flux. In your example you consider a spherical Gaussian surface between two opposite charges but not enclosing them. Clearly there is an electric field at each point on this surface. This is not inconsistent with Gauss's law. What the law tells us is that the net flux integrated over the surface is zero. While there will be positive contributions due to $E_{\perp}$ pointing out of the flux surface at some points these will be cancelled by negative contributions $E_{\perp}$ pointing inward at others. The intuitive way to think about this is with electric field lines, which can only begin on positive charges and end on negative ones. If a surface contains no net charge the number of field lines entering (positive flux contribution) is the same as the number leaving (negative flux contribution).