Problem on Energy Density in Dielectric Medium

  • Thread starter genxium
  • Start date
  • #1
136
1
I'm reading this tutorial and having some difficulty in understanding its derivation.

I take as granted that electric energy within a volume [itex]\Omega[/itex] is defined by:

[tex]W = \int_\Omega \phi \cdot \rho \cdot d^3r[/tex]

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

[tex]U = \phi \cdot \rho[/tex]

To my understanding, the tutorial is trying to show that

[tex] U = \frac{1}{2} \textbf{E}\cdot \textbf{D} [/tex]

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

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

[itex]\delta W = \int_\Omega \phi \cdot (\delta\rho_f) \cdot d^3r \; -- \; (1)[/itex]

then by [itex]\nabla \textbf{D} = \rho_f[/itex] equation [itex](1)[/itex] reduces to

[itex]\int_\Omega \phi \cdot \nabla (\delta \textbf{D}) \cdot d^3r [/itex]

[itex]= \int_\Omega \nabla (\phi \cdot (\delta \textbf{D})) \cdot d^3r - \int_\Omega \nabla \phi \cdot (\delta \textbf{D}) \cdot d^3r [/itex]

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

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 [itex]\delta W = - \int_\Omega \nabla \phi \cdot (\delta \textbf{D}) \cdot d^3r[/itex]" which implies that [itex] \int_{\partial\Omega} \phi \cdot (\delta \textbf{D}) \cdot d\textbf{S} = 0[/itex].

This doesn't seem trivial to me. I consulted some of my friends majored in Physics but most of them just took [itex] U = \frac{1}{2} \textbf{E}\cdot \textbf{D} [/itex] 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 :)
 

Answers and Replies

  • #2
DrDu
Science Advisor
6,106
805
I suppose you have to consider the limit Omega to infinity.
 
  • #3
136
1
@DrDu, sorry I can't see why considering [itex]\Omega[/itex] 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 [itex]\delta W = - \int \nabla \phi \cdot (\delta \textbf{D}) \cdot d^3 r[/itex]". Would you further explain the idea of introducing infinity here?
 
  • #4
DrDu
Science Advisor
6,106
805
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
136
1
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
DrDu
Science Advisor
6,106
805
At least, E and D are local quantities, in contrast to phi.
 

Related Threads on Problem on Energy Density in Dielectric Medium

Replies
2
Views
14K
Replies
8
Views
1K
Replies
7
Views
542
Replies
3
Views
2K
Replies
6
Views
878
  • Last Post
Replies
1
Views
1K
Replies
1
Views
535
Replies
5
Views
10K
Top