According to Ohm's law $$\textbf{J}=\sigma\textbf{E}$$ where $\textbf{J}$ is the current, $\sigma$ is the electric conductivity, and $\textbf{E}$ is the electric field. Now from the continuity equation, we get $$\frac{\partial \rho}{\partial t}=-\nabla\cdot\textbf{J}=-\sigma\nabla\cdot\textbf{E}=-\frac{\sigma}{\epsilon}\rho$$ where $\epsilon$ is the electric permittivity and $\rho$ is the charge density. Its solution is $$\rho(x,t)=\rho_0(x)\exp(-\frac{\sigma}{\epsilon}t)$$ For good conductors, the charge density goes to zero rapidly. Since they must go somewhere, they accumulate on the surface of the conductor.
Now in a lecture, the instructor says that the derivation fails on the surface. Why is that?