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 single 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_(1)+1/K_(2)+(1)/(2K_(3))`
B. `1/K=(1)/(K_(1)+K_(2))+(1)/(2K_(3))`
C. `1/K=(K_(1)+K_(2))/(K_(1)+K_(2))+2K_(3)`
D. `K=(K_(1)K_(3))/(K_(1)+K_(3))+(K_(2)K_(3))/(K_(2)+K_(3))`

Answer

D

Explanation

Applying `C=(epsilon_0A)/(d-t_1-t_2+(t_1)/(K_1)+(t_2)/(K_2))` we have
`(epsilon_0(A//2))/(d-d//2-d//2+(d//2)/(K_1)+(d//2)/(K_3))+(epsilon_0(A//2))/(d-d//2-d//2+(d//2)/(K_2)+(d//2)/(K_3))=(Kepsilon_0A)/d`
Solving the equation we get ,`K=(K_1K_3)/(K_1+K_3)+(K_2K_3)/(K_2+K_3)`