Problem on Energy Density in Dielectric Medium

• genxium
In summary, 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).

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 :)

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

@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?

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##.

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.

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

1. What is energy density in a dielectric medium?

The energy density in a dielectric medium is the amount of energy stored per unit volume in the medium. It is a measure of the energy that can be stored in the electric field within the medium.

2. How is energy density related to the dielectric constant?

The energy density is directly proportional to the square of the dielectric constant. This means that as the dielectric constant increases, the energy density also increases.

3. What is the equation for calculating energy density in a dielectric medium?

The equation for energy density in a dielectric medium is: U = 1/2 * ε * E^2, where U is the energy density, ε is the dielectric constant, and E is the electric field strength.

4. How does the energy density in a dielectric medium affect the capacitance of a capacitor?

The energy density in a dielectric medium is directly proportional to the capacitance of a capacitor. This means that as the energy density increases, the capacitance also increases.

5. What factors affect the energy density in a dielectric medium?

The energy density in a dielectric medium is affected by the dielectric constant, the electric field strength, and the volume of the medium. It can also be affected by the temperature and the type of dielectric material used.