Energy density of an electromagnetic field

AI Thread Summary
The discussion focuses on the energy density of an electromagnetic field in a linear dielectric, highlighting the distinction between free and bound charges. It clarifies that the energy density can be accurately expressed using macroscopic electrodynamics equations, specifically u = E · D / 2. The integration approach discussed raises questions about whether bound charge energy is neglected and why treating bound charge as independent fails in dielectric contexts. The conversation references Griffith's Electrodynamics for further insights on energy in dielectric systems. Overall, the interaction emphasizes the complexities of integrating macroscopic and microscopic electrodynamics.
PhysicsKT
Messages
17
Reaction score
0
The energy density of an electromagnetic field with a linear dielectric is often expressed as
1994848e8909b58aaa7dfa748264681c15b04cdb.png
. It is also known that energy can be found by
90ce12f273329132bc0a22e77cabd6fadd9317ec.png
. Using the latter, the energy density is found to be
2b95635ceca0346d915aadc5ef5f3f8047d12dd6.png
, as is well known. If you integrate the latter only over free charge and ignore bound charge, you write
a3e754ebda6b4b4d609f6ac85bb3d8b3f6fa3516.png
, use integration by parts, and obtain the first result. Does the first result neglect the energy from bound charge? If not, why does
2b95635ceca0346d915aadc5ef5f3f8047d12dd6.png
break down (I.e. why can’t one find the energy with a dielectric by treating the bound charge as its own independent charge arrangement and using formulae for a vacuum?)
 

Attachments

  • 1994848e8909b58aaa7dfa748264681c15b04cdb.png
    1994848e8909b58aaa7dfa748264681c15b04cdb.png
    741 bytes · Views: 674
  • 90ce12f273329132bc0a22e77cabd6fadd9317ec.png
    90ce12f273329132bc0a22e77cabd6fadd9317ec.png
    1,020 bytes · Views: 591
  • 2b95635ceca0346d915aadc5ef5f3f8047d12dd6.png
    2b95635ceca0346d915aadc5ef5f3f8047d12dd6.png
    826 bytes · Views: 649
  • a3e754ebda6b4b4d609f6ac85bb3d8b3f6fa3516.png
    a3e754ebda6b4b4d609f6ac85bb3d8b3f6fa3516.png
    700 bytes · Views: 531
  • 2b95635ceca0346d915aadc5ef5f3f8047d12dd6.png
    2b95635ceca0346d915aadc5ef5f3f8047d12dd6.png
    826 bytes · Views: 536
Physics news on Phys.org
You are mixing macroscopic with microscopic electrodynamics. For macroscopic electrodynamics ##u=\vec{E} \cdot \vec{D}/2## is correct. Within Markovian linear-response theory ##\vec{D}=\epsilon \vec{E}=\epsilon_0 \epsilon_r \vec{E}##.
 
For some discussion, see Griffith's Electrodynamics.
https://www.amazon.com/gp/product/1108420419/?tag=pfamazon01-20
View the table of contents and click on section 4.4.3 "Energy in Dielectric Systems". See especially the bottom of page 198 and the top of 199.
 
  • Like
Likes Delta2
I see. Makes perfect sense now. Thanks!
 
I was using the Smith chart to determine the input impedance of a transmission line that has a reflection from the load. One can do this if one knows the characteristic impedance Zo, the degree of mismatch of the load ZL and the length of the transmission line in wavelengths. However, my question is: Consider the input impedance of a wave which appears back at the source after reflection from the load and has traveled for some fraction of a wavelength. The impedance of this wave as it...
Back
Top