If a dielectric is isotropic and uniform, then assuming it carries no free charge, we know that the potential $V$ inside the dielectric is harmonic: $\Delta V = 0$. Hence the maximum principle holds inside the dielectric: the maximum of $V$ inside the dielectric must lie on its boundary.
Now, what happens if the dielectric is still isotropic, but not uniform: $\epsilon = \epsilon(x,y,z)$. We may assume that $\epsilon$ is smooth.
For such a dielectric, we have $\nabla \cdot D = 0$ hence $\nabla\cdot(\epsilon E) = 0$; so $$ \Delta V = \nabla\cdot E = - {1\over \epsilon}\nabla \epsilon \cdot E = {1\over \epsilon} \nabla \epsilon \cdot \nabla V = \nabla (\ln \epsilon) \cdot \nabla V.$$
I feel that the maximum principle for $V$ continues to hold for dielectrics of the above form, without free charges. But I'm insufficiently skilled to prove that. Some help would be welcome.
