Electric potential, hollow metalic cylinder

The potential inside such a cavity would be a constant V_0.

 

Why are you saying that it's constant V_0? The cylinder is not closed as far as I understand. It's like a toilet paper roll. If it were closed with lids, I'd see why, but in this case how would you explain?

The boundary conditions are V_0 on the roll, and V_1 in one side, and V_2 in the other, such that (V_2-V_1)/l = E. If V_1 = V_2 = V_0 as you say, this means that E = 0, which is the trivial case.

 

This is not perfectly valid, because it depends on the permitivity of the cylinder. According to Faraday a metall would shield away every electric field. But if you plot the field, you will see, that the field gets into the hollow space.