The distance between two plates of a capacitor is d and its capacitance is $\mathrm{C}_{1}$, when air is the medium between the plates. If a metal sheet of thickness $\frac{2 \mathrm{~d}}{3}$ and of same area as plate is introduced between the plates, the capacitance of the capacitor becomes $\mathrm{C}_{2}$. The ratio $\frac{\mathrm{C}_{2}}{\mathrm{C}_{1}}$ is:

JEE Main · 2023 · 10 Apr 2023 · Shift 2 · Capacitance

The distance between two plates of a capacitor is d and its capacitance is

C_{1}

, when air is the medium between the plates. If a metal sheet of thickness

\frac{2 d}{3}

and of same area as plate is introduced between the plates, the capacitance of the capacitor becomes

C_{2}

. The ratio

\frac{C _{2}}{C _{1}}

is:

(A) $2 : 1$
(B) $4 : 1$
(C) $3 : 1$ ✓ Correct
(D) $1 : 1$

Answer: (C)

Solution

Initially,

C_{1} = \frac{ϵ _{0} A}{d}

. When a metal sheet of thickness

t = \frac{2 d}{3}

is inserted, the effective separation becomes

d - t = d - \frac{2 d}{3} = \frac{d}{3}

(since the metal sheet acts as an equipotential and effectively reduces the gap). Therefore,

C_{2} = \frac{ϵ _{0} A}{d /3} = 3 \frac{ϵ _{0} A}{d} = 3 C_{1}

. The ratio

\frac{C _{2}}{C _{1}} = 3 : 1

JEE Main · 2023 · 10 Apr 2023 · Shift 2 · Single correct option

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