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 $\phi_0$ enter image description here

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?

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    $\begingroup$ The plate was charged in the first place. The imbalance of cations vs anions in the solution neutralizes that charge. $\endgroup$ May 26, 2022 at 15:48
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    $\begingroup$ The surface charge is the origin of the non-zero surface potential $\phi_0$. Since it is positive anions are more attracted to the surface than cations. $\endgroup$
    – Buck Thorn
    May 27, 2022 at 9:57
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    $\begingroup$ Look for a relation between the surface charge density and the z potential at the surface, and evaluate what the surface charge density corresponding to the potential you used in your calc might be. The net countercharge density in solution should be opposite this. $\endgroup$
    – Buck Thorn
    May 27, 2022 at 9:59
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    $\begingroup$ That's right: a positively charged plate is surrounded by the negatively charged solution. That's what the word "neutralize" means. $\endgroup$ Jun 7, 2022 at 15:00
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    $\begingroup$ The solution has as many negative charges as it had before, that is, exactly as many as positive ones. The layer around the plate has an excess of negative charges. See where this is going? $\endgroup$ Jun 7, 2022 at 15:32


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