Question

A parallel plate capacitor of area $A$, plate separation $d$ and capacitance $C$ is filled with three different dielectric materials having dielectric constant $k_1,k_2$ 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. $k=\frac{{k_{1}k}_2}{k_1+k_2}+2k_3$
• D. $k=k_1+k_2+2k_3$

Answer 1

Answer: (B)

$\frac{1}{k}=\frac{1}{k_1+k_2}+\frac{1}{{2k}_3}$

solution:

Let $C_1,C_2,C_3$ be the capacitance with dielectric $K_1,K_2,K_3$

and by using the formula

$C=K\frac{ϵA}{d}$ we get

Let $C_{eq}$ be $C_1+C_2=\frac{ϵA}{d}(K_1+K_2)$

$\frac{1}{C}=\frac{1}{C_{eq}}+\frac{1}{C_3}$

$\frac{d}{KAϵ}=\frac{d}{(k_1+k_2)Aϵ}+\frac{d}{k_{3}2Aϵ}$

We get

$\frac{1}{K}=\frac{1}{(k_1+k_2)}+\frac{1}{2k_3}$

hence the correct opt: B