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Explanation of potential equation (caculated from charge density via PoissonEquation)

  1. Jul 21, 2011 #1
    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 [itex]U(z)[/itex] 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 [itex]\rho(z)[/itex] and use the poisson equation [itex]\nabla^2\Psi=-\frac{\rho}{\epsilon}[/itex] to get the potential.

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

    Thanks,
    corro
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Jul 29, 2011 #2
    Re: Explanation of potential equation (caculated from charge density via PoissonEquat

    Ok, got think I got it now:

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

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

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

    Correct?
     
  4. Jul 29, 2011 #3
    Re: Explanation of potential equation (caculated from charge density via PoissonEquat

    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 [itex]\sigma[/itex]n(zn)=[itex]\rho[/itex](z)dz, and then sum up (integrated with respect to z) the contribution to the potential of all layer of [itex]\rho[/itex](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,

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

    Note : Integrate by parts the right hand side of 2nd eqn to arrive at the 3rd eqn.
     
    Last edited: Jul 29, 2011
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