# Electric field inside cavity is zero

If an isolated conductor without cavities is charged, its excess charge will distribute itself on its surface in order to guarantee that the electric field is zero on its interior.

If instead the conductor had an interior cavity, the charges would again distribute themselves on the outer surface in order to eliminate the electric field on the inside material of the conductor. (Free electrons cannot move in equilibrium.) Why is the electric field within the cavity zero as well? Why does the distribution of excess charge on the surface eliminate electric field within cavities as a result of eliminating electric field within the material itself? Is the distribution of charges on the outer surface independent of the interior shape?

• The proof follows from Gauss's law $\nabla \cdot \mathbf {E} = \frac{\rho}{\epsilon_0}$ and the uniqueness theorem. If the field at every point on the surface is zero, and there is no charge inside, then the field everywhere inside must also be zero (there can be no normal component of field on the wall when there is no charge inside, and the tangential component is zero because the wall is conducting). – Floris Aug 21 '17 at 21:58