Recently I have been investigating the problem of a neutral solutions (with equal number of anions and cations) interacting with an infinite surface $z=0$ with electrostatic potential $\phi(z=0)=\phi_0>0$.
If we assume only one type of cation and anion, with opposite charges, we obtain \begin{equation} \nabla^2\phi\left(\textbf{r}\right)=-\frac{1}{\varepsilon}\sum_{i=1}^{2}q_{i}n_{i}\left(\textbf{r}\right) \end{equation} with $q_1=-q_2=q$. If we assume that the density of charges is in thermal equlibrium, we have through Maxwell-Boltzmann distribution \begin{equation} n_{i}\left(\textbf{r}\right)=n_{i}^{0}e^{-\frac{q_{i}\phi\left(\textbf{r}\right)}{k_{B}T}}. \end{equation}
After manipulating a bit, we obtain
\begin{equation}
\nabla^2\phi\left(\textbf{r}\right)=2qn_{0}\sinh\left[\frac{q\phi\left(\textbf{r}\right)}{k_{B}T}\right],
\end{equation}
which describes the very famous Gouy-Chapman model
with analytical solution given by
\begin{equation}
\phi\left(z\right)=2\frac{k_{B}T}{q}\ln\left[\frac{1+\tanh\left(\frac{qW}{4k_{B}T}\right)e^{-z/\lambda_{D}}}{1-\tanh\left(\frac{qW}{4k_{B}T}\right)e^{-z/\lambda_{D}}}\right]
\end{equation}
and anion (-), cation (+) densities given by
\begin{equation}
n_{\pm}\left(z\right)=n_{0}\left[\frac{1+\tanh\left(\frac{qW}{4k_{B}T}\right)e^{-z/\lambda_{D}}}{1-\tanh\left(\frac{qW}{4k_{B}T}\right)e^{-z/\lambda_{D}}}\right]^{\mp2}
\end{equation}, where I plot for different positive values of $\phi_0$
Contradiction: As we can clearly see from figure above, the neutrality condition does not hold anymore -- see imbalance between positive and negative charges. People from this field argue that there is no problem since a surface charged density was absorbed by the plate (positive surface charged if $𝜙_0>0$). However, how do we see this physically? I do not understand why positive charges were absorbed by the plate. Can we somehow show this mathematically? If we integrate the total density $\rho(z)$, we will not obtain zero, implying there is no charge neutrality within the problem.
Any thoughts on that?
