Question

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 in Fig. If a single dielectric material is to be used to have the same capacitance C in this capacitor, then its dielectric constant K is given by

• A. $\frac{1}{K}=\frac{1}{K_1}+\frac{1}{K_2}+\frac{1}{2K_3}$
  
• B. $\frac{1}{K}=\frac{1}{K_1+K_2}+\frac{1}{2K_3}$
  
• C. $K=\frac{K_1{K}_2}{K_1+K_2}+2K_3$
  
• D. $K=K_1+K_2+K_3$

Answer 1

The correct option is B

$\frac{1}{K}=\frac{1}{K_1+K_2}+\frac{1}{2K_3}$

We have $C_1=\frac{{ϵ}_0\left(A/2\right)K_1}{\left(d/2\right)}=\frac{{ϵ}_{0}A{K}_1}{d}{C}_2=\frac{{ϵ}_0\left(A/2\right)K_2}{\left(d/2\right)}=\frac{{ϵ}_{0}A{K}_2}{d}and{C}_3=\frac{{ϵ}_0\left(A\right)K_3}{\left(d/2\right)}=\frac{2{ϵ}_{0}A{K}_3}{d}$
The capacitors $C_{1}and{C}_2$ are in parallel and their equivalent capacitance is
$C^{\prime }=C_1+C_2=\frac{{ϵ}_{0}A}{d}(K_1+K_2)$
This combination is in series with $C_3$. Hence, the net capacitance is
$\frac{1}{C^{\prime \prime }}=\frac{1}{C^{\prime }}+\frac{1}{C_3}=\frac{d}{{ϵ}_{0}A(K_1+K_2)}+\frac{d}{2{ϵ}_{0}A{K}_3}=\frac{d}{{ϵ}_{0}A}[\frac{1}{(K_1+K_2)}+\frac{1}{2K_3}]or{C}^{\prime \prime }=\frac{{ϵ}_{0}AK}{d}where\frac{1}{K}=\frac{1}{(K_1+K_2)}+\frac{1}{2K_3}$
Hence, the correct choice is (b).