Electromagnetism - Dielectric and Various Cavities

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SUMMARY

This discussion focuses on the behavior of electric fields within dielectric materials containing small cavities. Specifically, it establishes that inside a needle-like cavity parallel to the polarization vector P, the electric field E remains equivalent to that in the surrounding dielectric. Conversely, within a thin crack perpendicular to P, the electric field E is εr times greater than in the dielectric. Additionally, it is demonstrated that at the center of a small spherical cavity, the electric field E is calculated as E = P/(3μ0).

PREREQUISITES
  • Understanding of dielectric materials and their polarization
  • Familiarity with electric field equations, specifically E = -∇V
  • Knowledge of the relationship between electric displacement D and electric field E
  • Basic principles of electrostatics, including capacitor behavior
NEXT STEPS
  • Study the concept of electric displacement fields in dielectrics
  • Learn about the mathematical derivation of electric fields in cavities within dielectric materials
  • Explore the implications of boundary conditions at dielectric interfaces
  • Investigate the effects of varying cavity shapes on electric field distribution
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Students and professionals in physics, electrical engineering, and materials science who are studying electromagnetism and the behavior of electric fields in dielectric materials.

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Homework Statement


A large block of dielectric contains small cavities of various shapes whose sizes are much smaller than the block. The dielectric has uniform polarization P except those cavities which are assumed not to disturb the polarization of dielectric.
(a) Show that, inside a needle-like cavity parallel to P, the electric field E(inside the cavity) is the same as in the dielectric.
(b) Show that, inside a thin crack perpendicular to P, the electric field E(inside the cavity) is εr times larger than in the dielectric.
(c) Show that, at the center of a small spherical cavity, E=P/3*μ0.

Attempt at Solution:

a) q = σbA = P*A
Working this out E = \frac{PA}{∏εl^2} - \frac{-PA}{∏εl^2}

l is the length of the slit

E = \frac{2PA}{∏εl^2} Z

However as A is very small in that direction, E needle is close to zero. So E0 is the E-field in the cavity.

b) This is where I'm stuck.
I assumed that the slit acted like a plate capacitor and attempted to go from there with σ+ on the top and σ- on the bottom. However, I get the E field of the perpendicular cavity is E0 + P/ε, which is not a multiple of εr

Homework Equations



E = - ∇V

E = \frac{V}{d}, where d should be considered extremely small in part (b)
 
Last edited:
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For a) and b) I would use continuity rules of E and D fields parallel and perpendicular to an air-dielectric interface. What are they? You might want to be in a position to defend them if called upon ...

Hints:
a) what would discontinuity of the E field imply that we know is incorrect?
b) what would discontinuity of D imply that we know is incorrect? Realize there is no free charge at the air-dielectric interface. (D = epsilon*E for both air and dielectric).
c) got to think about that one.
 

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