Question

A parallel plate capacitor is of area \(6 ~\text{cm}^2\) and a separation \(3 ~\text{mm}\). The gap is filled with three dielectric materials of equal thickness (see figure) with dielectric constants \(K_1 = 10, K_2 =12\) and \(K_3 = 14\) The dielectric constant of a material which when fully inserted in above capacitor, gives same capacitance would be

\( F _{1} \) \( K _{2} \) \( K _{ 3 } \) \( 3 mm \)

• A. $14$
• B. $12$
• C. $4$
• D. $36$

Answer 1

The correct option is B $12$

Given:
Plate area $=6{\text{cm}}^2$
Separation between plates $=3\text{mm}$.
$K_1=10,K_2=12$
$K_3=14$


Let dielectric constant of material used be $K$,

$\Rightarrow \frac{K_1{ϵ}_0{A}_1}{d}+\frac{K_2{ϵ}_0{A}_2}{d}+\frac{K_3{ϵ}_0{A}_3}{d}=\frac{K{ϵ}_{0}A}{d}$

$\Rightarrow \frac{10{ϵ}_{0}A/3}{d}+\frac{12{ϵ}_{0}A/3}{d}+\frac{14{ϵ}_{0}A/3}{d}=\frac{K{ϵ}_{0}A}{d}(\because A_1=A_2=A_3=A/3)$

$\Rightarrow \frac{{ϵ}_{0}A}{d}(\frac{10}{3}+\frac{12}{3}+\frac{14}{3})=\frac{K{ϵ}_{0}A}{d}$

$\therefore K=12$

Hence, option $\left(C\right)$ is correct.