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+1 vote
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in Physics by (75.0k points)

The gap between the plates of a parallel-plate capacitor is filled with isotropic dielectric whose permittivity ε varies linearly from ε1 to ε2 (ε2 > ε1) in the direction perpendicular to the plates. The area of each plate equals S, the separation between the plates is equal to d. Find: 

(a) the capacitance of the capacitor;

(b) the space density of the bound charges as a function of ε if the charge of the capacitor is q and the field E in it is directed toward the growing ε values.

by (10 points)
Vey good explanation  sarthaks.. Keep it up and Ur team is doing a great job

1 Answer

+2 votes
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Best answer

(a) We point the x-axis towards right and place the origin on the left hand side plate. The left plate is assumed to be positively charged. 

Since ε varies linearly, we can write, 

ε(x) = a + bx 

where a and b can be determined from the boundary condition. We have

Now potential difference between the plates

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