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

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