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:
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)
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.