Potential inside a hollow sphere (spherical shell) given potential at surface Suppose that we have a hollow sphere (spherical shell) whose surface is held at some constant potential V0. What is the potential inside the sphere?
I had an argument with my physics professor over this. He claims that the potential inside depends on how far you are from the center and becomes zero at the center ("so that it doesn't blow up").
Wouldn't the potential at ANY point inside the sphere just be V0? Would the answer matter depending on whether the surface is a conductor on insulator, even? 
 A: If there is no charge inside the sphere, the potential must be the solution of the equation
$$
\nabla^2 \phi =0
$$
with boundary condition $\phi=\phi_0$ on the surface. The function $\phi=\phi_0$ inside the sphere is a solution, and it is unique. If there are charges inside the sphere the potential is different, and can be constructed, for example, using the image charges method. As you are explicitly assigning the potential on the boundary, this is independent from the fact that the surface is conducting or not.
A: If the sphere is conductive, then there is no electric field inside.
This means that you do no work to move a charge from one point to another - which is the definition of "constant potential". So yes - you are right. The potential anywhere inside will be the same as the potential on the surface.
A: If the hollow sphere is conducting, then potential inside hollow sphere is constant and outside the sphere, the potential is inversely proportional to distance from the center of sphere.
If it is a insulator, then we cannot say that electric field inside the sphere is zero. then the potential will be different.
A: According to Gauss's Law, since there is no charge inside of an hollow object, the electric field inside would be zero, even if it were an insulator. Therefore, the potential difference would be zero. Therefore, the potential (Vo, not to be confused with potential energy) would indeed be constant inside the hollow object unless there is a net charge inside the object. 
Note: Gauss's Law states that the enclosed integral of the electric field times the portion of the entire area the field is going through dA equals to the enclosed charge divided by the electric constant (approximately 8.85*10^-12). 
