A parallel plate capacitor has capacitance `C`. If it is equally filled with parallel layers of materials of dielectric constants `K_(1)` and `K_(2)` its capacity becomes `C_(1)`. The ratio of `C_(1) to C` is

Options

A. `K_(1) + K_(2)`
B. `(K_(1)K_(2))/(K_(1) - K_(2))`
C. `(K_(1) + K_(2))/(K_(1)K_(2))`
D. `(2K_(1)K_(2))/(K_(1) + K_(2))`

Answer

D

Explanation

`rArr` Energy lost `= E - U = 11 xx 10^(-6) J`

`C_(A) = (K_(1)epsilon_(0)A)/(d//2), C_(B) = (K_(2)epsilon_(0)A)/(d//2)`
`C_(eq) = (C_(1))/(C_(2)) = (2K_(1)K_(2))/(K_(1) + K_(2))`
`= (C_(A)C_(B))/(C_(A) + C_(B)) = ((2K_(1)K_(2))/(K_(1) + K_(2))) (epsilon_(0)A)/(d)`