JEE Main · 2022 · 25 Jun 2022 · Shift 2 · Capacitance

Two metallic plates form a parallel plate capacitor. The distance between the plates is 'd'. A metal sheet of thickness

\frac{d}{2}

and of area equal to area of each plate is introduced between the plates. What will be the ratio of the new capacitance to the original capacitance of the capacitor ?

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

Answer: (A)

Solution

Original capacitance:

C_{1} = \frac{ϵ _{0} A}{d}

. When a metal sheet of thickness

\frac{d}{2}

is inserted, the effective gap becomes

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

(the metal sheet is an equipotential, eliminating its thickness from the gap). New capacitance:

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

. Therefore,

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

, giving ratio

2 : 1

JEE Main · 2022 · 25 Jun 2022 · Shift 2 · Single correct option

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