Calculating Potential from Charge Density via Poisson Equation

[corroded_b]

Hi,
I'm doing MD-simulations in a capacitor-like system: 2 charged electrodes with a dense ionic liquid in between (non-diluted) with periodic boundaries in 2 dimensions (so for the electrodes I get infinite planes (xy) ,charged).

I want to get the potential $U(z)$ along the z-axis (witch is perpendicular to the electrodes). This is the superposition of the linear electrode pot. and the potential contribution of the ions.

So I calculate the charge density $\rho(z)$ and use the poisson equation $\nabla^2\Psi=-\frac{\rho}{\epsilon}$ to get the potential.

Now, in the http://pubs.acs.org/doi/suppl/10.1021/jp803440q/suppl_file/jp803440q_si_002.pdf" (page 3 on top) this potential (in gaussian units) is written as $\Psi(z)=-\frac{4 \pi}{\epsilon^*} \int_0^z (z-z') \rho(z')dz'$. Can someone explain/proof this expression?

Thanks,
corro

 

[corroded_b]

Ok, got think I got it now:

I calculated the charge density $\rho(z)$ by splitting the system in infinite sheets (or boxes in practical numerics). The surface charge on sheet $n$ is $\sigma_n$.

One sheet at position $z_n$ gives me the potential $\Psi(z)=-\frac{4 \pi}{\epsilon^*}\sigma_n (z-z_n)$.

For a continuous charge density I get the integral expression $\Psi(z)=-\frac{4 \pi}{\epsilon^*} \int_0^z \rho(z') (z-z') dz'$.

Correct?

 

[entphy]

I think you certainly can do it this way, by the linearity of the electric field contributed by each layer of charge of sheet density $\sigma$_n(z_n)=$\rho$(z)dz, and then sum up (integrated with respect to z) the contribution to the potential of all layer of $\rho$(z)dz from 0 to z.

The mathematical formality is as followed, but it is equivalent to the method you employed (if I understood you correctly as stated above), starting with Gauss's law over unit area Gaussian column extending from electrode at 0 to z,

-$\epsilon$$\nabla$$\psi$=$\int^{z}_{0}$$\rho$(z$^{'}$)dz$^{'}$
$\Rightarrow$$\int$$^{z}_{0}$$\nabla\psi$dz$^{'}$=-$\frac{1}{\epsilon}$$\int$$^{z}_{0}$$\int$$^{z^{'}}_{0}$$\rho$(z$^{''}$)dz$^{''}$dz$^{'}$
$\Rightarrow$$\int$$^{z}_{0}$d$\psi$=-$\frac{1}{\epsilon}${z$\int$$^{z}_{0}$$\rho$(z$^{'}$)dz$^{'}$-$\int$$^{z}_{0}$z$^{'}$$\rho$(z$^{'}$)dz$^{'}$}
$\Rightarrow$$\psi$(z)=-$\frac{1}{\epsilon}$$\int$$^{z}_{0}$(z-z$^{'}$)$\rho$(z$^{'}$)dz$^{'}$

Note : Integrate by parts the right hand side of 2nd eqn to arrive at the 3rd eqn.