# Problem on Energy Density in Dielectric Medium

1. Dec 21, 2014

### genxium

I'm reading this tutorial and having some difficulty in understanding its derivation.

I take as granted that electric energy within a volume $\Omega$ is defined by:

$$W = \int_\Omega \phi \cdot \rho \cdot d^3r$$

where $\phi = \phi(\textbf{r})$ is the eletric potential, $\rho = \rho(\textbf{r})$ is the charge density and $d^3r \stackrel{\Delta}{=}$ volume element. Now that the energy density is defined by

$$U = \phi \cdot \rho$$

To my understanding, the tutorial is trying to show that

$$U = \frac{1}{2} \textbf{E}\cdot \textbf{D}$$

where $\textbf{E} = \textbf{E}(\textbf{r})$ is eletric field strength and $\textbf{D} = \epsilon_0 \textbf{E} + \textbf{P}$ is the electric displacement (FYI: definition of electric displacement if needed).

Now that the tutorial begines with introducing a change of free charge density $(\delta\rho_f)$ and yielding a change of total energy (within the volume I SUPPOSE):

$\delta W = \int_\Omega \phi \cdot (\delta\rho_f) \cdot d^3r \; -- \; (1)$

then by $\nabla \textbf{D} = \rho_f$ equation $(1)$ reduces to

$\int_\Omega \phi \cdot \nabla (\delta \textbf{D}) \cdot d^3r$

$= \int_\Omega \nabla (\phi \cdot (\delta \textbf{D})) \cdot d^3r - \int_\Omega \nabla \phi \cdot (\delta \textbf{D}) \cdot d^3r$

$= \int_{\partial\Omega} \phi \cdot (\delta \textbf{D}) \cdot d\textbf{S} - \int_\Omega \nabla \phi \cdot (\delta \textbf{D}) \cdot d^3r \; -- \; (2)$

where use has been made of Integration by Parts and Divergence Theorem. I'm fine with the derivation by far.

Here comes the part that I don't understand. The tutorial says "If the dielectric medium is of finite spatial extent then we can neglect the surface term to give $\delta W = - \int_\Omega \nabla \phi \cdot (\delta \textbf{D}) \cdot d^3r$" which implies that $\int_{\partial\Omega} \phi \cdot (\delta \textbf{D}) \cdot d\textbf{S} = 0$.

This doesn't seem trivial to me. I consulted some of my friends majored in Physics but most of them just took $U = \frac{1}{2} \textbf{E}\cdot \textbf{D}$ as granted when using it and some are still trying to help.

Hope I can get luck in this forum, any help will be appreciated :)

2. Dec 21, 2014

### DrDu

I suppose you have to consider the limit Omega to infinity.

3. Dec 22, 2014

### genxium

@DrDu, sorry I can't see why considering $\Omega$ to infinity will explain the equation, and as I quoted from the tutorial "If the dielectric medium is of finite spatial extent then we can neglect the surface term to give $\delta W = - \int \nabla \phi \cdot (\delta \textbf{D}) \cdot d^3 r$". Would you further explain the idea of introducing infinity here?

4. Dec 22, 2014

### DrDu

I think if the sources of $\delta D$ are in a finite region then the gradient of the potential and $\delta D$ will decay rapidly enough at infinity so that the surface integral vanishes in the limit $r \to \infty$.

5. Dec 22, 2014

### genxium

Oh I see, just checked the tutorial again and it does say that the integration is taken over all space. However another problem comes if $\Omega$ defined like so -- does it mean that $U = \frac{1}{2} \textbf{E} \cdot \textbf{D}$ is only correct when the definition of "electric energy density" is with respect to all space? I suppose that energy density is a term describing a local quantity which is irrelevant to the micro space extent.

6. Dec 22, 2014

### DrDu

At least, E and D are local quantities, in contrast to phi.