Question

The capacitance of a parallel plate capacitor becomes $\frac{4}{3}$ times its original value. If a dielectric slab of thickness $t=\frac{d}{2}$ is inserted between the plates (where, $d$ is the distance of separation between the plates). What is the dielectric constant of the slab?

Answer 1

Answer: 2

Let $C$ be the original capacitance.
$\therefore$ New, capacitance, $C^{\prime }=\frac{4}{3}C$
Thickness, $t=\frac{d}{2}$
$\Rightarrow \frac{{\epsilon }_{0}A}{d-t\left(1-\frac{1}{K}\right)}=\frac{4}{3}\frac{{\epsilon }_{0}A}{d}$
$\Rightarrow \frac{2}{\left(1+\frac{1}{K}\right)}=\frac{4}{3}$
$\Rightarrow K=2$