JEE Main 2025 — Capacitance Question with Solution

Area of plate is $A$. then
$\begin{aligned}
& \mathrm{C}=\frac{\varepsilon_2 \varepsilon_0 \mathrm{~A}}{\mathrm{~d} / 2}=\frac{2 \varepsilon_2 \varepsilon_0 \mathrm{~A}}{\mathrm{~d}} \\ & \mathrm{C}^{\prime}=\frac{\varepsilon_1 \varepsilon_0 \mathrm{~A}}{\mathrm{~d} / 2}=\frac{2 \varepsilon_1 \varepsilon_0 \mathrm{~A}}{\mathrm{~d}}
\end{aligned}$<br/>Let$\mathrm{C}_0=\frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d}}$<br/>$\begin{aligned}
& \mathrm{C}=2 \varepsilon_2 \mathrm{C}_0 \\ & \mathrm{C}^{\prime}=2 \varepsilon_1 \mathrm{C}_0
\end{aligned}$<br/>$\mathrm{C} \& \mathrm{C}^{\prime}$areinseries<br/>$\mathrm{C}_1=\frac{\mathrm{CC}^{\prime}}{\mathrm{C}+\mathrm{C}^{\prime}}=\frac{4 \varepsilon_2 \varepsilon_1 \mathrm{C}_0^2}{2 \mathrm{C}_0\left(\varepsilon_2+\varepsilon_1\right)}$<br/>$=\frac{2 \varepsilon_2 \varepsilon_1 \mathrm{C}_0}{\left(\varepsilon_2+\varepsilon_1\right)}$<imgsrc="https://firebasestorage.googleapis.com/v0/b/prepsharp-c91fd.firebasestorage.app/o/je{e}_{m}ains$\mathrm{C}=\frac{\varepsilon_1 \varepsilon_0 \mathrm{~A}}{2 \mathrm{~d}}=\frac{\varepsilon_1 \mathrm{C}_0}{2}$<br/>$\mathrm{C}^{\prime}=\frac{\varepsilon_2 \mathrm{C}_0}{2}$<br/>$\mathrm{C} \& \mathrm{C}^{\prime}$areinparallel<br/>$\mathrm{C}_2=\mathrm{C}^{\prime}+\mathrm{C}=\left(\varepsilon_1+\varepsilon_2\right) \frac{\mathrm{C}_0}{2}$<br/>Thus$\frac{\mathrm{C}_1}{\mathrm{C}_2}=\frac{2 \varepsilon_2 \varepsilon_1 \mathrm{C}_0}{\left(\varepsilon_2+\varepsilon_1\right)} \times \frac{2}{\left(\varepsilon_1+\varepsilon_2\right) \mathrm{C}_0}$<br/>$=\frac{4 \varepsilon_2 \varepsilon_1}{\left(\varepsilon_2+\varepsilon_1\right)^2}$