The plates of a parallel plate capacitor are separated by a distance d with air as medium between the plates. In order to increase its capacitance by 60%, a slab of dielectric constant 5 having the same area as that of plates is introduced between the plates. The thickness of the dielectric slab is:

$\frac{15}{32}d$

The correct answer is Option (4) → $\frac{15}{32}d$

Capacitance of air capacitor: $C_0=\frac{\varepsilon_0 A}{d}$

With dielectric slab of thickness $t$ and $K=5$:

$\frac{1}{C}=\frac{d-t}{\varepsilon_0 A}+\frac{t}{5\varepsilon_0 A} =\frac{1}{\varepsilon_0 A}\Big(d-t+\frac{t}{5}\Big)$

$\Rightarrow C=\frac{\varepsilon_0 A}{d-\frac{4}{5}t}$

Given $C=1.6C_0=\frac{1.6\varepsilon_0 A}{d}$

$\frac{\varepsilon_0 A}{d-\frac{4}{5}t}=\frac{1.6\varepsilon_0 A}{d}$

$d-\frac{4}{5}t=\frac{d}{1.6}=\frac{5}{8}d$

$d-\frac{5}{8}d=\frac{4}{5}t$

$\frac{3}{8}d=\frac{4}{5}t$

$t=\frac{15}{32}d$

Answer: $t=\frac{15}{32}d$