# Gauss Law with a hollow asymmetric surface

In this video, Walter Lewin argues that no charge will appear on the inside surface of a hollow conductor in electrostatic equilibrium. He uses a Gaussian surface contained entirely within the conductor (so that the flux through that surface is zero) to make this argument.

However, this argument only states that there is no net charge on the inside surface of a hollow asymmetrical object. Is it possible for some parts of this surface to have a positive charge, and other parts to have a negative charge, such that the total charge on the surface is zero, and Gauss Law is still satisfied? If not, then why?

This is still cheating a bit, because we're ignoring the influence of the outside surface. We can redo the argument more carefully: we keep the same boundary condition, but demand total charge $Q$ on the outside surface and zero total charge on the inner surface. Again, we already know there exists a solution: it is the charge distribution on the outside surface of a solid (non-hollow) conductor, which guarantees zero field everywhere inside. So this must be the unique solution.
• Never mind. Counter example is to take a arbitrarily small $dV$ in the presence of a point charge q. There is no net flux through the surface of dV, though there is a nonzero field. – Jonathan Wheeler Apr 18 '16 at 23:21