Question

The 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. d is the separation between the plates. The dielectric constant of the dielectric slab is :

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

Answer 1

Answer: (A)

Now $C_{eff}$ is series combination of $\frac{l_{0}A}{x},\frac{l_{0}A}{d-t-x}and\frac{k{l}_{0}A}{t}$

$\frac{1}{C_{eff}}=\frac{x}{l_{0}A}+\frac{d-t-x}{l_{0}A}+\frac{t}{k{l}_{0}A}$

$\Rightarrow C_{eff}=\frac{k{l}_{0}A}{(d-t)k+t}$

$\Rightarrow$ given $C_{eff}=\frac{7}{6}\cdot \frac{l_{0}A}{d}$

$\frac{7}{6}\cdot \frac{l_{0}A}{d}=\frac{k{l}_{0}A}{(d-\frac{2}{3}d)k+\frac{2}{3}d}$

$\Rightarrow \frac{7}{6}=\frac{3k}{\left(\frac{1}{3}\right)k+\frac{2}{3}}$

$\Rightarrow 7k+14=18k$

$\Rightarrow k=\frac{14}{11}$