# Plate with positive potential and Gouy-Chapman model

I am thinking about the following mental experiment. Imagine I have an infinite flat plate at $$z=0$$ with electrostatic potential $$\phi_0$$, i.e., $$\phi(z=0)=\phi_0$$.

Now I embedded this plate within an ionic solution with cations and anions. Which, of the positive or negative charges will be attracted by the plate?

How can I show that? Thanks!

Follow up: If I solve the Gouy-Chapman model, I obtain the following densities of anions and cations, which matches with the picture you have already described (also in agreement with other references). However, the problem is the following: if you integrate $$\rho(z) = q_+ n_{+}(z) + q_- n_{-}(z)$$ where $$n_{\pm}(z)$$ is the density of positive/negative ($$q_{\pm}$$) charges, you will clearly obtain a value different from zero -- thus yielding non-zero net charge!

People from this field argue that there is no problem since a surface charged density was absorved by the plate (positive surface charged if $$\phi_0>0$$). However, how do we see this physically? I do not understand why positive charges were absorbed by the plate.

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In order to know with any certainty which direction the particles will move in let's break down some of the additional clarifications to this problem:

• Charged particles will only accelerate in a non-zero electric field $$\mathbf{E}$$
• An electric field $$\mathbf{E}$$ is defined not by any singular potential $$\phi(x,y,0)=\phi_0$$, but by potential difference $$\Delta\phi$$

In order for the problem to make more physical sense(Aside from it being an infinite plane which can have it's own problems), let's make a few assumptions:

• The potential $$\lim_{z\rightarrow\infty}\phi(x,y,z)=0$$, which is to say that a potential difference exists
• The electric fields created by the cations and anions are much smaller than the electric field due to the plate for simplicity

Now that we've established these things, we can look at the behavior of the electric field. The electric field in a one dimensional potential is defined $$E=-\frac{\partial\phi}{\partial z}$$

If $$\phi_0>0$$, then the electric field is some vector pointing away from the plate, so the negatively charged anions will move towards the plate

If $$\phi_0<0$$, then the electric field is some vector pointing towards the plate, so the positively charged cations will move towards the plate

If $$\phi_0=0$$, then there is no large scale electric field and any cation/anion movement is entirely due to their respective electric fields

Hopefully that answers your question, let me know if that makes sense!

• Thanks @Anthony! However, this result seems in contradiction with the ordinary Gouy-Chapman theory. Are you aware of this theory? It essentially describes the electrodynamics of combined Poisson's eq and Maxwell distribution See for instance Fig.1 aip.scitation.org/doi/10.1063/1.443542 You start with a neutral solution, and through the analyses you have done (or using Maxwell distribution), we are able to see that for \phi_0>0, there is accumulation of negative charges close to the plate. However, we started with a neutral, so how can be possible to obtain charged system? May 25, 2022 at 16:44
• The Gouy-Chapman theory is certainly much more involved than this analysis. Though I'm having trouble seeing what the contradiction is. For $\phi_0>0$ anions will accumulate on the plate, and cations will be both pushed away by the plate and attracted to the anions which have moved towards the plate. So the originally neutral system will be polarized because of the applied electric field in the fluid medium. May 25, 2022 at 16:59
• Anthony, I have added a more detailed description of what I think the conflict might be. Please take a look and let me know whether I am being more clear now. May 25, 2022 at 17:27
• Any thoughts around that? Jun 7, 2022 at 14:50
• Based on the additional description you provided. The anions are getting adsorbed onto the plate. Adsorption is a common process modelled in statistical mechanics where some particle gets bound to a surface. Once those anions are bound to the surface, they're no longer part of the fluid medium or moving around with the particles in the fluid, resulting in a change in the net charge of the fluid medium when you integrate the charge density across the fluid medium. Jun 7, 2022 at 20:32