Question

A parallel plate capacitor of plate area $A$, plate separation $d$ and capacitance $C$ is filled with three different dielectric materials having dielectric constants $K_1,K_2$ and $K_3$. If a single dielectric material is to be used to having 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. $\frac{1}{K}=\frac{K_1{K}_2}{K_1+K_2}+\frac{K_2{K}_3}{K_2+K_3}$

Answer 1

The correct option is B $\frac{1}{K}=\frac{1}{K_1+K_2}+\frac{1}{2K_3}$

For single dielectric system the capacitance ,$C=\frac{AK{ϵ}_0}{d}=K{C}_0$ where $C_0=\frac{A{ϵ}_0}{d}$

For three dielectric there are 3 capacitors.

$C_1=\frac{\left(A/2\right)K_1{ϵ}_0}{d/2}=\left(K_1{C}_0\right)$,

$C_2=\frac{\left(A/2\right)K_2{ϵ}_0}{d/2}=\left(K_2{C}_0\right)$ and

$C_3=\frac{A{K}_3{ϵ}_0}{d/2}=2K_3{C}_0$.

Here $C_1,C_2$ are in parallel and this group and $C_3$ are in series.

thus, $\frac{1}{C_{eq}}=\frac{1}{(C_1+C_2)}+\frac{1}{C_3}=\frac{1}{(K_1{C}_0+K_2{C}_0)}+\frac{1}{2K_3{C}_0}$

As$C=C_{eq},\frac{1}{C}=\frac{1}{C_{eq}}$

$\frac{1}{K{C}_0}=\frac{1}{(K_1{C}_0+K_2{C}_0)}+\frac{1}{2K_3{C}_0}$

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