A parallel plate condenser is filled with two dielectrics as shown. Area of each plate is `A "metre"^(2)` and the separtion is `t` metre. The dielectric constants are `K_(1)` and `K_(2)`, respectively. Its capacitance in farad will be

Options

A. `(epsilon_(0)A)/(t) (K_(1)+K_(2))`
B. `(2epsilon_(0)A)/(t) (K_(1)+K_(2))`
C. `(epsilon_(0)A)/(t).(K_(1)+K_(2))/(2)`
D. `(epsilon_(0)A)/(t).(K_(1)+K_(2))/(2)`

Answer

C

Explanation

The given arrangement is equivalent to two capacitors in parallel. In each capacitor the area of the plate will
`:.` Equivalent capacitance `C' = C_(1) + C_(2)`
`= (K_(1) epsilon_(0)A//2)/(t) + (K_(2) epsilon_(0)A//2)/(t) = ((K_(1)+K_(2))epsilon_(0)A)/(2t)`