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 electric field everywhere in the conductor must be zero. This means that the electric field is zero everywhere on the wall of the cavity - and that in turn implies there can be no field in the cavity.

It follows that the distribution of charge on the surface is independent of the shape of internal cavities: if you consider a "skin" of conductor (thin layer) right outside the cavity, then the electric field at that point can be considered a boundary condition of the problem "where is the charge". And that boundary condition is unchanged whether the cavity is filled with more conductor, or with nothing.

  • $\begingroup$ Can you elaborate as to why zero field on the wall implies zero field inside the cavity? Is there a rigorous proof of this fact? $\endgroup$ – sunjos20 Aug 21 '17 at 21:48
  • $\begingroup$ 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). $\endgroup$ – Floris Aug 21 '17 at 21:58
  • 2
    $\begingroup$ @sunjos20, (1) the inner surface is an equipotential surface. (2) Since this is the electrostatic case and there is no charge within the cavity, the potential inside the cavity must satisfy Laplace's equation and, thus, cannot have a local extremum (maximum or minimum within the cavity). (3) Thus, the potential within the cavity must be constant, i.e., the electric field within the cavity must be zero. $\endgroup$ – Alfred Centauri Aug 21 '17 at 23:47

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