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?


Good question! I watched Lewin's lectures a few years ago and distinctly remember this explanation as being unsatisfactory. It's one of the very few places where he "cheats" and skips a few steps.

The missing step is the existence/uniqueness theorem. We know, from Lewin's argument, that the total charge on the inside surface is zero, and the boundary condition is that there can be no field at the Gaussian surface itself. Moreover, we already know there exists a solution to this problem (i.e. zero charge density everywhere on the inside). So that has to be the unique solution.

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.

As a nice corollary, this implies that the outside of a conductor never knows anything about holes in the inside: in ideal electrostatics, there is no way of telling if a conductor is hollow.

  • $\begingroup$ Couldn't you also say that the Gaussian surfaces all are zero for arbitrarily small volumes within the zero charge region? This seems much easier and intuitive to me $\endgroup$ – Jonathan Wheeler Apr 18 '16 at 23:17
  • $\begingroup$ @JonathanWheeler I'm not sure how that works. Can you expand that into an answer? $\endgroup$ – knzhou Apr 18 '16 at 23:19
  • $\begingroup$ 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. $\endgroup$ – Jonathan Wheeler Apr 18 '16 at 23:21

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