Lorentz Cavity in Uniformly Polarised Dielectric: Exam Question

  • Context: Graduate 
  • Thread starter Thread starter manofphysics
  • Start date Start date
  • Tags Tags
    Atom Field Local
Click For Summary
SUMMARY

The discussion centers on the behavior of the electric field components in a Lorentz cavity within a uniformly polarized dielectric. The key equation presented is E = E_{ex} + E_{P} + E_{L} + E_{near}, where E_{P} represents the electric field in the dielectric before the cavity is formed. The main question raised is whether E_{P} changes when a sphere is cut out of the polarized dielectric, and the consensus is that while E_{P} does change, this change is often negligible, allowing for useful approximations in calculations. The adiabatic formation of the cavity is noted as a critical factor in maintaining the balance of electric fields.

PREREQUISITES
  • Understanding of electric fields in dielectrics
  • Familiarity with Lorentz cavity concepts
  • Knowledge of polarization in materials
  • Basic principles of electrostatics
NEXT STEPS
  • Study the effects of cavity formation on electric fields in dielectrics
  • Learn about the adiabatic process in electrostatics
  • Explore the mathematical modeling of polarized dielectrics
  • Investigate the implications of surface charge on electric fields
USEFUL FOR

Students preparing for exams in electromagnetism, physicists studying dielectric materials, and researchers interested in electrostatic field behavior in polarized media.

manofphysics
Messages
41
Reaction score
0
I just have a small question regarding lorentz cavity:

Refer to a small lorentz cavity in a uniformly polarised dielectric. as shown in fig.
E_{ex}: External electric field.
E_{P}: Electric field in the uniformly polarised dielectric (when sphere has NOT been cut out)
E_{L}:Electric field due to surface charge on cavity
E_{near}:Field due to dipoles inside cavity.

Now, E=E_{ex}+E_{P}+E_{L}+E_{near}

But, does not E_{P} change if we cut out a sphere from the polarised dielectric?
Is it that we are neglecting the small change in E_{P} due to cut out sphere and our result is an useful approximation, but not exact?

Please, any help will be appreciated. I need to understand this for my term exam.
 

Attachments

  • lorentz1.gif
    lorentz1.gif
    22.2 KB · Views: 740
Physics news on Phys.org
I'm not sure, but offhand: If you form the cavity adiabatically, then EL would compensate for the change in EP, wouldn't it? So I suspect that's the approximation involved.
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 11 ·
Replies
11
Views
4K