JEE Main 2026 — Capacitance Question with Solution

The initial capacitance is $C=\frac{{ϵ}_{0}A}{d}$.
From the figure, the capacitor is divided into three parts.
The bottom half is divided into two capacitors $C_2$ and $C_3$ in parallel, each with area $A/2$ and thickness $d/2$.
$C_2=\frac{K_2{ϵ}_0(A/2)}{d/2}=K_2\frac{{ϵ}_{0}A}{d}=3C$
$C_3=\frac{K_3{ϵ}_0(A/2)}{d/2}=K_3\frac{{ϵ}_{0}A}{d}=5C$
The equivalent capacitance of the bottom part is $C_{eq,bottom}=C_2+C_3=3C+5C=8C$.
The top half is a single capacitor $C_1$ with area $A$ and thickness $d/2$.
$C_1=\frac{K_1{ϵ}_{0}A}{d/2}=2K_1\frac{{ϵ}_{0}A}{d}=2(2)C=4C$.
The total capacitance $C^{\prime }$ is the series combination of $C_1$ and $C_{eq,bottom}$.
$\frac{1}{C^{\prime }}=\frac{1}{C_1}+\frac{1}{C_{eq,bottom}}=\frac{1}{4C}+\frac{1}{8C}=\frac{2+1}{8C}=\frac{3}{8C}$
$C^{\prime }=\frac{8}{3}C$
Comparing with $C^{\prime }=\frac{n}{3}C$, we get $n=8$.