JEE Main · 2026 · 23 Jan 2026 · Shift 2 · Capacitance

With the battery still connected the voltage is fixed, so the charge drawn is proportional to the capacitance. Inserting a mica sheet of thickness $t=2\text{ mm}$ and dielectric constant $K$ into the $5\text{ mm}$ gap leaves $3\text{ mm}$ of air, giving (thicknesses in mm) an equivalent capacitance $C^{\prime }=\frac{{\epsilon }_{0}A}{(d-t)+\frac{t}{K}}=\frac{{\epsilon }_{0}A}{3+\frac{2}{K}}$, while the original capacitance is $C=\frac{{\epsilon }_{0}A}{5}$. A $25\%$ increase in charge means $C^{\prime }=1.25C$: $\frac{{\epsilon }_{0}A}{3+\frac{2}{K}}=1.25\times \frac{{\epsilon }_{0}A}{5}=\frac{{\epsilon }_{0}A}{4}$. Hence $3+\frac{2}{K}=4$, so $\frac{2}{K}=1$ and $K=2$.