Physics · Capacitors and R-C Circuits

JEE Main 2025 — 8 April, Evening Shift — Question 66

Space between the plates of a parallel plate capacitor of plate area $4 \mathrm{~cm}^{2}$ and separation of

$(d) 1.77 \mathrm{~mm}$ , is filled with uniform dielectric materials with dielectric constants (3 and 5)as

shown in figure. Another capacitor of capacitance 7.5 pF is connected in parallel with it. The effective

capacitance of this combination is $\qquad$ pF . (Given $\epsilon_{o}=8.85 \times 10^{-12} \mathrm{~F} / \mathrm{m}$ )

Answer: 15

Numerical answer — enter this value.

Step-by-step solution

$\frac{1}{C}=\frac{1}{C_{1}}+\frac{1}{C_{2}}=\frac{d / 2}{A k_{1} \varepsilon_{0}}+\frac{d / 2}{A k_{2} \varepsilon_{0}}$ $=\left(\frac{1}{k_{1}}+\frac{1}{k_{2}}\right) \frac{d}{2 A \varepsilon_{0}}$

$=\left(\frac{1}{3}+\frac{1}{5}\right) \frac{d}{2 A \varepsilon_{0}}$

$\frac{1}{C}=\frac{4}{15} \frac{d}{A \varepsilon_{0}}$

$C=\frac{15}{4} \times \frac{A \varepsilon_{0}}{d}=\frac{15}{4} \times \frac{4 \times 10^{-4} \times 8.85 \times 10^{-12}}{1.77 \times 10^{-3}}$

$=7.5 \mathrm{pF}$

$C+C_{3}=(7.5+7.5) \mathrm{pF}=15 \mathrm{pF}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 8 April, Evening Shift
Subject: Physics
Chapter: Capacitors and R-C Circuits
Topic: Combination of Capacitors and Circuit Analysis