Physics · Capacitors and R-C Circuits

JEE Main 2025 — 3 April, Morning Shift — Question 49

A parallel plate capacitor is filled equally (half) with two dielectrics of dielectric constants $\varepsilon_{1}$ and $\varepsilon_{2}$ , as shown in figures. The distance between the plates is $d$ and area of each plate is $A$ . If capacitance in first configuration and second configuration are $C_{1}$ and $C_{2}$ respectively, then $\frac{C_{1}}{C_{2}}$ is

Option A:
$\frac{4 \varepsilon_{1} \varepsilon_{2}}{\left(\varepsilon_{1}+\varepsilon_{2}\right)^{2}}$
Correct
Option B:
$\frac{\varepsilon_{0}\left(\varepsilon_{1}+\varepsilon_{2}\right)}{2}$
Option C:
$\frac{\varepsilon_{1} \varepsilon_{2}}{\varepsilon_{1}+\varepsilon_{2}}$
Option D:
$\frac{\varepsilon_{1} \varepsilon_{2}^{2}}{\left(\varepsilon_{1}+\varepsilon_{2}\right)^{2}}$

Answer: A

Step-by-step solution

$C=\frac{k \varepsilon_{0} A}{d}$

\begin{aligned} & C_{1}=\frac{\varepsilon_{0} A}{\frac{d}{2 \varepsilon_{1}}+\frac{d}{2 \varepsilon_{2}}}=\frac{\varepsilon_{0} A}{d}\left\{\frac{1}{\frac{\varepsilon_{2}+\varepsilon_{1}}{2 \varepsilon_{1} \varepsilon_{2}}}\right\} \\ & C_{2}=\frac{\varepsilon_{0}}{d}\left(\varepsilon_{1} \frac{A}{2}+\varepsilon_{2} \frac{A}{2}\right)=\frac{\varepsilon_{0} A}{d}\left\{\frac{\varepsilon_{1}}{2}+\frac{\varepsilon_{2}}{2}\right\} \\ & \frac{C_{1}}{C_{2}}=\frac{2 \varepsilon_{1} \varepsilon_{2}}{\left(\varepsilon_{2}+\varepsilon_{1}\right) \frac{\left(\varepsilon_{1}+\varepsilon_{2}\right)}{2}}=\frac{4 \varepsilon_{1} \varepsilon_{2}}{\left(\varepsilon_{1}+\varepsilon_{2}\right)^{2}} \end{aligned}

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 3 April, Morning Shift
Subject: Physics
Chapter: Capacitors and R-C Circuits
Topic: Effect of Dielectrics