Question

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.

Comments

Vey good explanation  sarthaks.. Keep it up and Ur team is doing a great job

Answer 1

(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