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\psidz^{'}=-\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.