This question is a sequel of sorts to my earlier (resolved) question about a recent paper. In the paper, the authors performed molecular dynamics (MD) simulations of parallel-plate supercapacitors, in which liquid resides between the parallel-plate electrodes. The system has a "slab" geometry, so the authors are only interested in variations of the liquid structure along the $z$ direction.

In my previous question, I asked about how particle number density is computed. In this question, I would like to ask about how the electric potential is computed, given the charge density distribution.

Recall that in CGS (Gaussian) units, the Poisson equation is

$$\nabla^2 \Phi = -4\pi \rho$$

where $\Phi$ is the electric potential and $\rho$ is the charge density. So the charge density $\rho$ is proportional to the Laplacian of the potential.

Now suppose I want to find the potential $\Phi(z)$ along $z$, by integrating the Poisson equation. How can I do this?

In the paper, on page 254, the authors write down the average charge density $\bar{\rho}_{\alpha}(z)$ at $z$:

$$\bar{\rho}_{\alpha}(z) = A_0^{-1} \int_{-x_0}^{x_0} \int_{-y_0}^{y_0} dx^{\prime} \; dy^{\prime} \; \rho_{\alpha}(x^{\prime}, y^{\prime}, z)$$

where $\rho_{\alpha}(x, y, z)$ is the local charge density arising from the atomic charge distribution of ionic species $\alpha$, $\bar{\rho}_{\alpha}(z)$ is the average charge density at $z$ obtained by averaging $\rho_{\alpha}(x, y, z)$ over $x$ and $y$, and $\sum_{\alpha}$ denotes sum over ionic species.

The authors then integrate the Poisson equation to obtain $\Phi(z)$:

$$\Phi(z) = -4\pi \sum_{\alpha} \int_{-z_0}^z (z - z^{\prime}) \bar{\rho}_{\alpha}(z^{\prime}) \; dz^{\prime} \; \; \; \; \textbf{(eq. 2)}$$

My question is, how do I "integrate the Poisson equation" to obtain equation (2)? How do I go from $\nabla^2 \Phi = -4\pi \rho$ to equation (2)? In paricular, where does the $(z - z^{\prime})$ factor come from?

Thanks for your time.

  • $\begingroup$ Do you just know that Φ(z=−z0) = 0 ? What is the value of the potential on the second plane of the slab ? $\endgroup$
    – Shaktyai
    Aug 13, 2012 at 17:07
  • $\begingroup$ @Shaktyai Yes, I know that $\phi(z = -z_0) = 0$; that is the definition that is used (albeit implicitly) in the paper. I don't think I know the value of the potential on the second plane of the slab; I think that it depends on an integral of $\rho$. $\endgroup$
    – Andrew
    Aug 13, 2012 at 21:48
  • $\begingroup$ It's weird, the liquid is located within a capacitor. One should expect the potential to be known at the plates...A second order equation needs two boundary values, so the solution 2 is dubious. $\endgroup$
    – Shaktyai
    Aug 14, 2012 at 5:47

2 Answers 2


I don't know your level of knowledge, so let me start with the very basic fact that the electric field of a uniformly charged plate is $$ E=2\pi\sigma,\qquad\left( 1\right) $$ where $\sigma$ is the surface charge density. To derive this result you can utilize the Gauss formula: $$ \Phi=4\pi Q,\qquad\left( 2\right) $$ where $\Phi$ is the total flux of the electric field through a closed surface and $Q$ is the total charge in a space bounded by the surface. In the figure below I depicted charged plate as a blue plane and the closed surface as the box with green sides.

application of the Gauss formula

The flux is only non zero for these green rectangles $\Phi=2ES$, where $S$ is the area of the rectangles. The total charge inside the box is $Q=S\sigma$ hence $$ 2ES=4\pi S\sigma\quad\Rightarrow\quad E=2\pi\sigma. $$

Let's now approximate your system as the set of of plates with surface charge density $\sigma=\rho\left( z\right) \,dz$ where $\rho\left( z\right) $ is the $xy$-averaged charge density. Therefore, the total electric field in a point $z$ is the difference of the contributions of planes before $z$ and after $z$ (see figure below): $$ E\left( z\right) =E_{1}\left( z\right) -E_{2}\left( z\right) ,\qquad(3) $$ where $$ E_{1}\left( z\right) =2\pi\int_{-z_{0}}^{z}\rho\left( z^{\prime\prime }\right) \,dz^{\prime\prime},\qquad E_{2}\left( z\right) =2\pi\int _{z}^{z_{0}}\rho\left( z^{\prime\prime}\right) \,dz^{\prime\prime}. $$

the planes before and after $z$ contribute to the field with opposite signs

Thus, the potential $\phi\left( z\right) $ has the form: $$ \phi\left( z\right) =-\int_{-z_{0}}^{z}dz^{\prime}E\left( z^{\prime }\right) ,\qquad(4) $$ with the boundary value $\phi\left( -z_{0}\right) =0$. The expression (4) is the potential required. Let's now simplify it. First of all, I simplify the expression for the field: $$ E\left( z\right) =2\pi\int_{-z_{0}}^{z}\rho\left( z^{\prime\prime}\right) \,dz^{\prime\prime}-2\pi\int_{z}^{z_{0}}\rho\left( z^{\prime\prime}\right) \,dz^{\prime\prime}=4\pi\int_{-z_{0}}^{z}\rho\left( z^{\prime\prime}\right) \,dz^{\prime\prime}-2\pi\int_{-z_{0}}^{z_{0}}\rho\left( z^{\prime\prime}\right) \,dz^{\prime\prime}. $$ Therefore the potential takes the form: $$ \phi\left( z\right) =-\int_{-z_{0}}^{z}dz^{\prime}E\left( z^{\prime }\right) =-4\pi\int_{-z_{0}}^{z}dz^{\prime}\int_{-z_{0}}^{z^{\prime}} \rho\left( z^{\prime\prime}\right) \,dz^{\prime\prime}-2\pi\left( z+z_{0}\right) \int_{-z_{0}}^{z_{0}}\rho\left( z^{\prime}\right) \,dz^{\prime}. $$ To simplify the first term I change the order of integrations (integration domain is presented in the figure below): $$ \int_{-z_{0}}^{z}dz^{\prime}\int_{-z_{0}}^{z^{\prime}}\rho\left( z^{\prime\prime}\right) \,dz^{\prime\prime}=\int_{-z_{0}}^{z}dz^{\prime \prime}\int_{z^{\prime\prime}}^{z}\rho\left( z^{\prime\prime}\right) \,dz^{\prime}=\int_{-z_{0}}^{z}\left( z-z^{\prime\prime}\right) \rho\left( z^{\prime\prime}\right) dz^{\prime\prime}. $$

integration domain

Finally, we obtain the following result for the potential: $$ \phi\left( z\right) =-4\pi\int_{-z_{0}}^{z}\left( z-z^{\prime}\right) \rho\left( z^{\prime}\right) dz^{\prime}-2\pi\left( z+z_{0}\right) \int_{-z_{0}}^{z_{0}}\rho\left( z^{\prime}\right) \,dz^{\prime}. $$ One can see that the result you presented is valid only for a neutral liquid: $$ \int_{-z_{0}}^{z_{0}}\rho\left( z^{\prime}\right) \,dz^{\prime}=0. $$

  • $\begingroup$ Thanks so much for your time. Can you please elaborate on why "the total electric field in a point $z$ is the difference of the contributions of planes before $z$ and after $z$"? How do you obtain expressions for $E_1(z)$ and $E_2(z)$ (in particular, how do you determine the limits of integration in the expressions for $E_1(z)$ and $E_2(z)$?)? Also, why is $E(z) = E_1(z) - E_2(z)$? $\endgroup$
    – Andrew
    Aug 13, 2012 at 21:44
  • $\begingroup$ As you can see from the first picture the directions of electric field on the different sides of the plate are opposite. You can split the full bulk of liquid into plates $dz$ thick, which have the charge surface density $\sigma=\rho(z) dz$. $\endgroup$ Aug 13, 2012 at 22:21
  • $\begingroup$ If you consider a certain point with coordinate $z$ then all plates, which have coordinates $z'$ such that $-z_{0}<z'<z$, yield the contribution $E_{1}$, while all others with coordinates $z<z'<z_{0}$ give the contribution with opposite sign $E_{2}$ (because the point $z$ is on the opposite side for them and an electric field is a vector quantity). $\endgroup$ Aug 13, 2012 at 22:21
  • $\begingroup$ I depicted the directions of the fields $E_{1,2}$ in the second picture. The expressions for the fields are no more than the summation of $2\pi\sigma$ over all plates (integration=summation). $\endgroup$ Aug 13, 2012 at 22:22

I) OP notes that $\bar{\rho}_{\alpha}$ is the $xy$-averaged charge density
$$ \bar{\rho}_{\alpha}(z) ~:=~ A_0^{-1} \int_{-x_0}^{x_0} \int_{-y_0}^{y_0} dx^{\prime} \; dy^{\prime} \; \rho_{\alpha}(x^{\prime}, y^{\prime}, z), \qquad A_0~:=~4x_0y_0.$$

II) Similarly, eq. (2) must be for the $xy$-averaged potential

$$ \bar{\Phi}(z) ~:=~ A_0^{-1} \int_{-x_0}^{x_0} \int_{-y_0}^{y_0} dx^{\prime} \; dy^{\prime} \; \Phi(x^{\prime}, y^{\prime}, z),$$

$$\tag{eq. 2}\bar{\Phi}(z) ~=~ -4\pi \sum_{\alpha} \int_{-z_0}^z (z - z^{\prime}) \bar{\rho}_{\alpha}(z^{\prime}) \; dz^{\prime} .$$

Moreover, eq. (2) satisfies the boundary condition


Now differentiate eq. (2) twice wrt. $z$,

$$ \bar{\Phi}^{\prime\prime}(z) ~=~ -4\pi \sum_{\alpha} \bar{\rho}_{\alpha}(z).$$

This is consistent with the Poisson equation

$$ \nabla^2 \Phi ~=~ -4\pi \rho, \qquad \rho~=~\sum_{\alpha}\rho_{\alpha},$$

if there is no net electric flux going through the $4$ walls of the box that are parallel to the $z$-directions.


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