Physics · Capacitors and R-C Circuits

JEE Main 2026 — 23 January, Morning Shift — Question 48

The space between the plates of a parallel-plate capacitor of capacitance C (without any dielectric) is now filled with three dielectric slabs of dielectric constants $\mathrm{K}_{1}=2, \mathrm{~K}_{2}=3$ , and $\mathrm{K}_{3}=5$ (as shown in the figure). If new capacitance is $\frac{\mathrm{n}}{3} \mathrm{C}$ then the value of $n$ is $\_\_\_\_$ .

Answer: 8

Numerical answer — enter this value.

Step-by-step solution

$\mathrm{C}_{1}=\frac{5 \epsilon_{0} \mathrm{~A} / 2}{\mathrm{~d} / 2}=\frac{5 \epsilon_{0} \mathrm{~A}}{\mathrm{~d}}=5 \mathrm{C}$ $\mathrm{C}_{2}=\frac{2 \epsilon_{0} \mathrm{~A} / 2}{\mathrm{~d} / 2}=\frac{2 \epsilon_{0} \mathrm{~A}}{\mathrm{~d}}=2 \mathrm{C}$ $\mathrm{C}_{1} \& \mathrm{C}_{2}$ in series. $\mathrm{C}^{\prime}=\frac{\mathrm{C}_{1} \mathrm{C}_{2}}{\mathrm{C}_{1}+\mathrm{C}_{2}}=\frac{(5 \mathrm{C})(2 \mathrm{C})}{7 \mathrm{C}}=\frac{10}{7} \mathrm{C}$ $\mathrm{C}_{3}=\frac{3 \epsilon_{0} \mathrm{~A} / 2}{\mathrm{~d} / 2}=3 \mathrm{C}$ $\mathrm{C}_{4}=\frac{2 \epsilon_{0} \mathrm{~A} / 2}{\mathrm{~d} / 2}=2 \mathrm{C}$ $\mathrm{C}_{4} \& \mathrm{C}_{3}$ in series; $\mathrm{C}^{\prime \prime}=\frac{(2 \mathrm{C})(3 \mathrm{C})}{5 \mathrm{C}}=\frac{6}{5} \mathrm{C}$ $\mathrm{C}^{\prime} \& \mathrm{C}^{\prime \prime}$ in parallel; So, $\mathrm{C}_{\mathrm{eq}}=\mathrm{C}\left(\frac{6}{5}+\frac{10}{7}\right)=\mathrm{C}\left(\frac{42+50}{35}\right)=\left(\frac{92}{35}\right) \mathrm{C}$ $\frac{92}{35} \mathrm{C}=\frac{\mathrm{nC}}{3}$ $\mathrm{n}=\frac{92 \times 3}{35}=7.9 \Rightarrow \mathrm{n} \simeq 8$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 23 January, Morning Shift
Subject: Physics
Chapter: Capacitors and R-C Circuits
Topic: Combination of Capacitors and Circuit Analysis