Question

Re: Lecture 5 and Faraday Cage

The argument that a charged hollow conductor contains no surface charge on the interior was made using Gauss's Law with the Gaussian surface on the interior of the conductor. However the conclusion seems to be a misapplication of Gauss's Law, which can only say the net charge inside the surface is zero. Is there another argument for non-symmetrical hollow conductors that disallows any charge on the interior surface other than, say, an equal amount of positive and negative charge on the interior?

Answer 1

There is a similar question in the closed area that I didn't see (sorry I am new and didn't spot immediately that past questions are added as you scroll down) and the closest answer I am looking for was given by chrissimo. At the interior surface, Gauss's Law can only lead to the conclusion that equal positive and negative charges lie on the interior and I think Prof Lewin's explanation is therefore a misapplication of Gauss's Law. However, chrissimo added something extra -- the metal is an equipotential solid. If a hollow conductor has excess +ve charge and the interior as areas with + and - charge (total charge = 0 to satisfy Gauss), then field lines must leave the -ve charge and terminate on a +charge. This other +ve charge must lie in the conductor. The presence of the field line means there is a potential difference, which contradicts the statement that the conductor is equipotential. That's an explanation, now I have to become comfortable with it or find something more intuitive. Time to think about external fields applied to this hollow conductor now :)

Thanks chrissimo