The space between the plates of a parallel-plate capacitor is filled consecutively with two dielectric layers 1 and 2 having the thickness of `d_(1)` and `d_(2)` and the permittivities `espilon_(1)` and `espilon_(2)` respectively. The area of each plate is equal to `S`. Find :

Explanation

(a) As it is series combination of two capacitors,
`(1)/(C ) = (d_(1))/(epsilon_(0) epsilon_(1) S) + (d_(2))/(epsilon_(0) epsilon_(2) S)` or, `C = (epsilon_(0) S)/((d_(1)//epsilon_(1)) + (d_(2)//epsilon_(2)))`
(b) Let, `sigma` be the initial surface charge density, then density of bound charge on the boundary plane,
`sigma' = sigma (1 - (1)/(epsilon_(1))) - sigma (1 - (1)/(epsilon_(2))) = sigma ((1)/(epsilon_(2)) - (1)/(epsilon_(1)))`
But, `sigma = (q)/(S) = (CV)/(S) = (epsilon_(0) S epsilon_(1) epsilon_(2))/(epsilon_(2) d_(1) + epsilon_(1) d_(2)) (V)/(S)`
So, `sigma' = (epsilon_(0) V (epsilon_(1) - epsilon_(2)))/(epsilon_(2) d_(1) + epsilon_(1) d_(2))`