Different descriptions of electrostatic energy

[Mexicorn]

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.

Homework Statement

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.

Homework Equations

The electrostatic energy contribution to the total energy is described like so:
$-\int \frac{\epsilon(r)}{8\pi}\left|\nabla V(r)\right|^{2}d^{3}r +\int \rho(r) V(r)d^{3}r$.
After incorporating the green's function solution (which is not relevant to my holdup), the term becomes the more familiar (to me at least):
$\frac{1}{2}\int \rho(r) V(r)d^{3}r$. (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)

The Attempt at a Solution

Starting from the top equation..
$-\int \frac{\epsilon(r)}{8\pi}\left|\nabla V(r)\right|^{2}d^{3}r +\int \rho(r) V(r)d^{3}r$
and plugging in the identity: $\textbf{E} = - \nabla V(r)$, I am left with
$\int \frac{\epsilon(r)}{8\pi}\nabla V(r) \cdot \textbf{E} d^{3}r+\int \rho(r) V(r)d^{3}r$
Then using Gauss' theorem yields:
$\int \frac{\epsilon(r)}{8\pi}\nabla \cdot (V(r)\textbf{E}) d^{3}r-\int \frac{\epsilon(r)}{8\pi} V(r) (\nabla \cdot \textbf{E}) d^{3}r+\int \rho(r) V(r)d^{3}r$
By the Divergence theorem, the first term is equivalent to $\oint (V(r) \textbf{E}) \cdot d\textbf{A}$ 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 $\nabla \cdot \textbf{E} = 4\pi \rho$, 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 $\rho_f$ instead of the total charge.

Anyone know where I'm going wrong with this?

 

[ehild]

At first I thought I would merely plug in the identity $\nabla \cdot \textbf{E} = 4\pi \rho$,

The correct formula is $\nabla \cdot \textbf{D} = 4\pi \rho$

ehild

 

[Mexicorn]

Shouldn't the form be $\nabla \cdot \textbf{D} = \rho_{free}$ which would have me missing a part of the total charge?

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

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.

 

[ehild]

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

ehild