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Homework Help: Different descriptions of electrostatic energy

  1. Jul 25, 2011 #1
    This is actually a question pertaining to a paper i'm trying to understand (PRB 73, 115407 (2006)), but I decided to put it here just to be safe.

    1. The problem statement, all variables and given/known data
    The paper I'm reading involves starting with an electrostatic energy contribution, and rewriting it with a green's function solution to the Poisson equation rather than the standard 1/r form to incorporate a dielectric. The step I'm missing, though, is in how the author formulates the energy contribution at first.

    2. Relevant equations
    The electrostatic energy contribution to the total energy is described like so:
    [itex]-\int \frac{\epsilon(r)}{8\pi}\left|\nabla V(r)\right|^{2}d^{3}r +\int \rho(r) V(r)d^{3}r [/itex].
    After incorporating the green's function solution (which is not relevant to my holdup), the term becomes the more familiar (to me at least):
    [itex]\frac{1}{2}\int \rho(r) V(r)d^{3}r [/itex]. (the sign on this term is ambiguous as it is defined as positive in the paper and negative in an online formulation written by the same author)

    3. The attempt at a solution
    Starting from the top equation..
    [itex]-\int \frac{\epsilon(r)}{8\pi}\left|\nabla V(r)\right|^{2}d^{3}r +\int \rho(r) V(r)d^{3}r[/itex]
    and plugging in the identity: [itex]\textbf{E} = - \nabla V(r)[/itex], I am left with
    [itex]\int \frac{\epsilon(r)}{8\pi}\nabla V(r) \cdot \textbf{E} d^{3}r+\int \rho(r) V(r)d^{3}r
    Then using Gauss' theorem yields:
    [itex]\int \frac{\epsilon(r)}{8\pi}\nabla \cdot (V(r)\textbf{E}) d^{3}r-\int \frac{\epsilon(r)}{8\pi} V(r) (\nabla \cdot \textbf{E}) d^{3}r+\int \rho(r) V(r)d^{3}r
    By the Divergence theorem, the first term is equivalent to [itex]\oint (V(r) \textbf{E}) \cdot d\textbf{A}[/itex] which is zero under the assumption that the potential vanishes at infinity. The last step is where I am struggling. At first I thought I would merely plug in the identity [itex]\nabla \cdot \textbf{E} = 4\pi \rho[/itex], but this still leaves me with a dangling permittivity within the integral. I tried going through everything again substituting in the electric displacement instead of the field, but then my last step will have a [itex]\rho_f[/itex] instead of the total charge.

    Anyone know where I'm going wrong with this?
  2. jcsd
  3. Jul 25, 2011 #2


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    Homework Helper

    The correct formula is [itex]\nabla \cdot \textbf{D} = 4\pi \rho [/itex]

  4. Jul 25, 2011 #3
    Shouldn't the form be [itex]\nabla \cdot \textbf{D} = \rho_{free}[/itex] which would have me missing a part of the total charge?

    Going through the steps using [itex]\textbf{D}=-\epsilon(r)\nabla V(r)[/itex] gives me the following result:
    [itex]-\int \frac{1}{2} V(r) \rho_{free}(r) d^{3}r + \int \rho_{total}(r) V(r) d^{3}r[/itex]

    I don't see how I can combine a total charge with just a free charge (coming from the dielectric displacement). This model is used where there will be dielectric areas which should accumulate bound charges...

    Thank you, ehild, for taking a look at this.
  5. Jul 25, 2011 #4


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    Homework Helper

    Are you sure the charge in the second integral is not free charge?

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