Electromagnetism: Questions About Energy Density

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Homework Help Overview

The discussion revolves around concepts in electromagnetism, specifically focusing on energy density and the behavior of electric displacement fields in dielectric materials. Participants are examining a problem related to the surface integral being zero and the manipulation of terms involving electric fields.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to understand why the surface integral is considered zero and are discussing the implications of using the electric displacement vector D in the context of dielectric materials. There are questions about the mathematical manipulation involving the product rule and how terms are derived.

Discussion Status

The discussion is active, with participants providing insights and interpretations regarding the mathematical expressions and physical principles involved. Some guidance has been offered regarding the behavior of the electric displacement vector at infinity and the conditions under which certain integrals vanish.

Contextual Notes

There is mention of assumptions related to the behavior of fields at infinity and the treatment of free versus bound charges in dielectric materials. Participants are navigating these concepts without full consensus on the interpretations of the mathematical expressions presented.

Gavins
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Hey I am studying electromagnetism at the moment and one problem I'm having is the energy density.
Instead of typing everything out, this pdf has everything. http://www.physics.sfsu.edu/~lea/courses/ugrad/360notes11.pdf There are some things I don't understand. Why is the surface integral 0? He just says that it's 0 for the usual reasons. Then I'm not sure what he's doing with the (deltaD). E thing. I don't understand how he puts the whole thing in the delta and then a half pops out.

Thank you.
 
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I think he is just using the usual product rule to get the half:
[tex]\Delta (E \cdot E)=\Delta E \cdot E + E \cdot \Delta E = 2(\Delta E\cdot E)[/tex]
[tex]\Rightarrow \Delta E\cdot E = \frac{1}{2} \Delta (E \cdot E)[/tex]

As for the integral being zero, I'm not sure, perhaps it's zero in the limit.
 
You must be having so much fun, i love the book of Griffiths;
now, he used D because he is dealing with dielectric metarial, there you should not consider bounded charges since they are not cause but result.
div(D)=Pf but div(E)=P (here since i don't have rho, P is charge density) if you use E you can not undrestand if the thing you got is either free or bounded.
Zorba is right, USUALLY V is zero at the infinity, so first integral vanishes.
As you continue, you don't have to be afraid of confusing E or D because from now on you won't deal with charge any more, you can convert D to k€E then as you have same variable in delta or d, you can tell d(E^2)=2EdE, or the reverse.
After all if you like it this way you may not touch it and have some k€ there or you can convert it back and have ½ED
 
"Why is the surface integral 0? He just says that it's 0 for the usual reasons."

the electric displacement D vector as you go to infinity is proportional to 1/r^2 and the potential is a 1/r. the dA has a r^2 term in it. therefore making the whole term in the integral a 1/r, hence as you go out to infinity(integrating over all of space), the surface integral goes to zero
 

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