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Electrostatic Energy in Dielectric Media

In Section 1.11 we discussed the energy of a system of charges in free space. The result obtained there, $$W=\frac{1}{2}\int \rho(\mathbf{x})\Phi(\mathbf{x})\text{ d}^3x \tag{4.83}$$ for the energy due to a charge density $\rho(\mathbf{x})$ and a potential $\Phi(\mathbf{x})$ cannot in general be taken over as it stands in our macroscopic description of dielectric media. The reason becomes clear when we recall how (4.83) was obtained. We thought of the final configuration of charge as being created by assembling bit by bit the elemental charges, bringing each one in from infinitely far away against the action of the then existing electric field. The total work done was given by (4.83). With dielectric media, work is done not only to bring real (macroscopic) charges into position, but also to produce a certain state of polarisation in the medium. If $\rho$ and $\Phi$ in (4.83) represent macroscopic variables, it is certainly not evident that (4.83) represents the total work, including that done on the dielectric.

To be general in our description of dielectrics, we will not initially make any assumptions about linearity, uniformity, etc., of the response of a dielectric to an applied field. Rather, let us consider a small change in the energy $\delta W$ due to some sort of change $\delta \rho$ in the macroscopic charge density $\rho$ existing in all space. The work done to accomplish this change is $$\delta W=\int \delta \rho (\mathbf{x})\Phi (\mathbf{x})\text{ d}^3x \tag{4.84}$$ where $\Phi(\mathbf{x})$ is the potential due to the charge density $\rho (\mathbf{x})$ already present. Since $\nabla \cdot D=\rho$, we can relate the change $\delta \rho$ to a change in the displacement of $\delta D$: $$\delta\rho=\nabla\cdot (\delta D) \tag{4.85}$$

I cannot reconcile (4.84) with (4.83). I think I'm taking the wrong approach here: deriving (4.84) from (4.83). Maybe (4.84) is more fundamental? My first few attempts were to vary (4.83) as a functional (in the spirit of calculus of variations) and come up with (4.84). This failed. (4.84) feels like it's double counting, since the integral is over all space. I would appreciate any help on this!

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The potential is also dependent on the charge density. You can show the result by replacing the potential with its expression dependent on the charge density and calculating the variation on the product of densities, assuming no variation in positions of the charges. Better yet, just saw the explanation on page 168 (3rd edition).

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  • $\begingroup$ Yep, that does it. Thanks! $\endgroup$ – Spinor Apr 19 '20 at 2:53

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