Question

Capacitance of a capacitor becomes $\frac{7}{6}$ times its original value, if a dielectric slab of thickness $t=\frac{2}{3}d$ is introduced in between the plates. The dielectic constant of the dielectric slab is :

• A. $\frac{14}{11}$
• B. $\frac{11}{14}$
• C. $\frac{7}{11}$
• D. $\frac{11}{7}$

Answer 1

The correct option is A $\frac{14}{11}$

The capacitance is given by $C_1=\frac{{ϵ}_{0}A}{d}$ ...(i)
If electric slab of constant k of thickness $t=\frac{2}{3}d$ is introduced, then capacitance becomes
$C_2=\frac{{ϵ}_{0}A}{d-t(1-\frac{1}{k})}$
$=\frac{{ϵ}_{0}A}{d-\frac{2}{3}d(1-\frac{1}{k})}$ ...(ii)
Eq. (i) becomes
$C_2=\frac{{ϵ}_{0}A}{d\left[\begin{array}{c}(1-\frac{2}{3})+\frac{2}{3k}\end{array}\right]}$
$\Rightarrow C_2=\frac{C_1}{(\frac{1}{3}+\frac{2}{3k})}$
$\Rightarrow \frac{7}{6}{C}_1=\frac{C_1}{\frac{1}{3}+\frac{2}{3k}}$
$\Rightarrow \frac{1}{3}+\frac{2}{3k}=\frac{6}{7}$
$\Rightarrow \frac{2}{3k}=\frac{6}{7}-\frac{1}{3}=\frac{11}{21}$
or $33k=21\times 2$
$k=\frac{42}{33}=\frac{14}{11}$