A parallel plate capacitor of area A, plate separation d and capacitance C is filled with three different dielectric materials having dielectric constants `k_1`, `k_2` and `k_3` as shown. If a isngle dielectric material is to be used to have the same capacitance C in this capacitor, then its dielectic constant k is given by

Options

A. `1/(K_(1)+K_(2))+1/(2K_(3))`
B. `1/(K_(1)+K_(2))-1/(2K_(3))`
C. `1/(K_(1)-K_(2))+1/(2K_(3))`
D. `2/(K_(1)+K_(2))+1/(2K_(3))`

Answer

A

Explanation

The given capacitor may be supposed to be formed of three component capacitors `C_(1), C_(2) and C_(3)`. Let `K_(1), K_(2) and K_(3)` be dielectric constant of `C_(1), C_(2) and C_(3)` respectively. Condenser `C_(1) and C_(2)` are in parallel. Let `C_(p) = C_(1)+C_(2)`. Now, `C_(p) and C_(3)` are connected in series. There equivalent capacitance is given by `C_(s)`.
From figure, `C_(p)=C_(1)+C_(2)`
`C_(p)=(AK_(1)in_(0))/(2d/2)+(AK_(2)in_(0))/(2d/2)=(Ain_(0))/d(K_(1)+K_(2))`
Now `C_(p) and C_(3)` are in series
`1/C_(s)=1/C_(p)+1/C_(3)" "(C_(3)=(2AK_(3)in_(0))/d)`
`1/C_(s)=1/((Ain_(0))/d(K_(1)+K_(2)))+1/((2Ain_(0)K_(3))/d)`
`1/((AKin_(0))/d)=1/((Ain_(0))/d(K_(1)+K_(2)))+1/((Ain_(0))/d2k_(3))`
`therefore" "1/K=1/(K_(1)+K_(2))+1/(2K_(3))`