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 ______.

JEE Main · 2026 · 23 Jan 2026 · Shift 1 · Capacitance

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 the figure). If new capacitance is

\frac{n}{3} C

then the value of

n

is ______.

Answer: 8

Solution

Let

C = \frac{ε _{0} A}{d}

be the empty capacitance. The upper slab

K_{1}

fills the full plate area over thickness

\frac{d}{2}

C_{top} = \frac{K _{1} ε _{0} A}{d /2} = 2 K_{1} C = 4 C

. The lower region holds

K_{2}

and

K_{3}

side by side, each of area

\frac{A}{2}

and thickness

\frac{d}{2}

, acting in parallel:

C_{bot} = \frac{K _{2} ε _{0} ( A /2 )}{d /2} + \frac{K _{3} ε _{0} ( A /2 )}{d /2} = (K_{2} + K_{3}) C = 8 C

. The top and bottom layers are in series:

C_{eq} = \frac{C _{top} C _{bot}}{C _{top} + C _{bot}} = \frac{4 C \cdot 8 C}{12 C} = \frac{8 C}{3}

. Comparing with

\frac{n}{3} C

gives

n = 8

JEE Main · 2026 · 23 Jan 2026 · Shift 1 · Integer answer

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JEE Main · 2026 · 23 Jan 2026 · Shift 1 · Capacitance

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