Laplace's equation vs. Poisson's equation for electric field in hollow conductor

Question

I've been reading in Purcell's Electricity and Magnetism about why the electric field in a hollow conductor is zero, irregardless of external applied fields. The explanation seems to follow the following ideas:

1. the electrostatic potential at the boundary of the hollow cavity is constant, say $V_i$, because conductors redistribute charges to eliminate internal electric fields (I understand this)

2. Laplace's equation of $\nabla^2 (\phi)= 0$ has only one solution $\phi(x,y,z)$ for the potential inside the cavity, so the trivial solution of $\phi(x,y,z) = V_i$, satisfies the boundary condition and also the space inside the cavity.

My question is why this logic breaks down/cannot be applied to a NON-EMPTY cavity within a conductor that encloses a point charge, applying the logic to an empty annulus or shell of space (no charge density), whose outer boundary is the internal wall of the conductor. The inner edge of the shell would be some distance from the point charge, but not touching it. Gauss's law dictates that there must be flux through this shell, so that the $\phi$ value changes with radius. Yet since the charge density is 0 in the space of the shell, $\phi(x,y,z) = V_i$, satisfies the boundary condition and also the space inside the shell.

Answer 1

Let's consider for simplicity the spherically symmetric case, where we have a spherical cavity and a point particle in the middle. In the presence of non-zero charge density we need to consider the Poisson equation, $\nabla^2 \phi = - 4\pi \rho$. Here $\rho(x) = q \delta(x)$ is the charge density of the point particle which is placed at $x=0$ (the middle of the cavity). Notice that there is a singularity at $x=0$, so the situation is not the same as with an empty cavity: we cannot assume that the solution is smooth at $x=0$. This is why a constant potential is not a solution in this case.

One way to treat the singularity is to 'cut out' a small ball $B$ of radius $r$ around the point charge from the space, and then to treat the surface of this ball as another boundary (in addition to the boundary that is the conducting shell). The new space exists between the outer conducting and the small sphere that surrounds the charge. This way, the solution will be smooth everywhere in the new space. But we now need to specify the boundary conditions on this inner surface, which should somehow be related to the charge we removed. Let's work out these boundary conditions. Integrating the Laplace equation over the ball $B$, we have \begin{align} \int_B \nabla^2 \phi = -4\pi \int_B \rho = -4\pi q \,. \end{align} Using Stoke's theorem, the integral on the left becomes \begin{align} \int_{\partial B} ds \, \hat{n} \cdot \vec{\nabla} \phi = 4\pi r^2 \partial_r \phi \,. \end{align} Here $\partial B$ is the boundary of the ball, namely the small shell of radius $r$. $\hat{n}=\hat{r}$ is the normal vector, and $ds$ is an area element on the shell. We find the boundary condition \begin{align} \partial_r \phi(r) = - \frac{q}{r^2} \,. \end{align} Therefore, the potential should have a non-vanishing gradient close to the point charge (as expected), and this rules out a constant solution in the cavity. Of course, we know the solution in this case is simply $\phi(r) = q/r$, which is consistent with the boundary conditions on the conductor and close to the point charge.

Answer 2

The logic breaks down in that the region you've described has two boundaries: the outer boundary which is at some potential $V_i$, and the inner boundary which is at some different (not necessarily constant) potential. You have to satisfy all the boundary conditions, and $\phi=V_i$ violates the inner boundary condition.

Answer 3

In the case of a charge enclosed by a conducting cavity, a charge of opposite sign and equal magnitude would be induced in the inner cavity of the conductor. This leads to a vanishing flux in the conductor, consistent with a vanishing electric field (in the conductor) and the potential being constant through it.