Electromagnetism - Dielectric and Various Cavities

  • #1
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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 = [itex]\frac{PA}{∏εl^2}[/itex] - [itex]\frac{-PA}{∏εl^2}[/itex]

l is the length of the slit

E = [itex]\frac{2PA}{∏εl^2}[/itex] 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 = [itex]\frac{V}{d}[/itex], where d should be considered extremely small in part (b)
 
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  • #2
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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