JEE Main 2026 — Capacitance Question with Solution

Initial capacitor has plate separation 5 mm with capacitance $C_1={ϵ}_{0}A/5$.
After inserting 2 mm mica sheet, the gap remaining is 3 mm of air. The effective separation becomes: $d_{eff}=3+2/k$ where $k$ is the dielectric constant of mica.
New capacitance: $C_2=\frac{{ϵ}_{0}A}{3+2/k}$
Since battery remains connected, voltage is constant. The charge is: $Q=CV$
Ratio of charges: $\frac{Q_2}{Q_1}=\frac{C_2}{C_1}=\frac{{ϵ}_{0}A/(3+2/k)}{{ϵ}_{0}A/5}=\frac{5}{3+2/k}$
Given that charge increases by 25%: $Q_2=1.25Q_1$
$\frac{5}{3+2/k}=1.25$
$5=1.25(3+2/k)$
$5=3.75+2.5/k$
$1.25=2.5/k$
$k=2$