JEE Main 2026 — Physics Electromagnetism

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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 $K_{1}=2, K_{2}=3$ and $K_{3}=5$ (as shown in figure). If new capacitance is $\frac{n}{3} C$ then the value of $n$ is $\_\_\_\_$ .

JEE Main 2026 Physics, Capacitance — question figure

Official previous-year question

Held on 23 Jan 2026 · Verified 6 Jul 2026.

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Solution

The initial capacitance is $C = \frac{\epsilon_0 A}{d}$ .

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 $d/2$ .

$C_2 = \frac{K_2 \epsilon_0 (A/2)}{d/2} = K_2 \frac{\epsilon_0 A}{d} = 3C$

$C_3 = \frac{K_3 \epsilon_0 (A/2)}{d/2} = K_3 \frac{\epsilon_0 A}{d} = 5C$

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 $d/2$ .

$C_1 = \frac{K_1 \epsilon_0 A}{d/2} = 2K_1 \frac{\epsilon_0 A}{d} = 2(2)C = 4C$ .

The total capacitance $C'$ is the series combination of $C_1$ and $C_{eq, bottom}$ .

$\frac{1}{C'} = \frac{1}{C_1} + \frac{1}{C_{eq, bottom}} = \frac{1}{4C} + \frac{1}{8C} = \frac{2+1}{8C} = \frac{3}{8C}$

$C' = \frac{8}{3}C$

Comparing with $C' = \frac{n}{3}C$ , we get $n = 8$ .

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