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. 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. $\frac{1}{K}=\frac{K_1{K}_2}{K_1+K_2}+2K_3$
  
• D. $K=\frac{K_1{K}_3}{K_1+K_3}+\frac{K_2{K}_3}{K_2+K_3}$
  

Answer 1

Answer: (D)

Applying C $=\frac{{\epsilon }_{0}A}{d-t_1-t_2+\frac{t_1}{K_1}+\frac{t_2}{K_2}},$ we have
$\frac{{\epsilon }_0(A/2)}{d-d/2-d/2+\frac{d/2}{K_1}+\frac{d/2}{K_3}}$
$+\frac{{\epsilon }_0(A/2)}{d-d/2-d/2+\frac{d/2}{K_2}+\frac{d/2}{K_3}}=\frac{K{\epsilon }_{0}A}{d}$
Solving this equation, we get
$K=\frac{K_1{K}_3}{K_1+K_3}+\frac{K_2{K}_3}{K_2+K_3}$