Electrostatics – Vanishing Electric Field Inside a Conductor Implies Vanishing Charge Density but There Is a Puzzle

Question

If we place a metallic conductor in a static, external electric field, the free electrons inside the conductor will move opposite to the external field leaving the immobile positive ions (nucleus+core electrons) where they are inside the conductor. This will cause an internal field to build up opposite to the external field, and in static equilibrium, they cancel out. This is how the net electric field inside a conductor vanishes, in static equilibrium. By Gauss' law, $\nabla\cdot{\vec E}=\rho/\varepsilon_0$, it further implies that the charge density $\rho({\vec r})$ also vanishes at every point inside the conductor.

But after equilibrium is established, let us consider that part of the conductor which contains only immobile positive ions (or certainly a large excess of it). If we now consider a small volume $\Delta V$ in that part of the conductor which contains predominantly immobile positive ions, $\Delta V$ will contain a nonzero positive charge $\Delta q$. When we take the limit $\Delta V\to 0$, we should get, $$\rho(\vec r)=\lim_{\Delta V\to 0}\frac{\Delta q}{\Delta V}\neq 0.$$ The figure below is a rough cartoon of what I mean (after static equilibrium is established).

What is wrong with the argument in the second paragraph?

Answer 1

The excess of positive ions is only on the surface, just like the excess of electrons on the other side is only on the surface. The inside of the conductor remains neutral.