JEE Main · 2020 · 08 Jan 2020 · Shift 2 · Capacitance

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A capacitor is made of two square plates each of side 'a' making a very small angle α between them, as shown in figure. The capacitance will be close to:

(A) $\frac{ϵ _{0} a ^{2}}{d} (1 + \frac{α a}{d})$
(B) $\frac{ϵ _{0} a ^{2}}{d} (1 - \frac{α a}{4 d})$
(C) $\frac{ϵ _{0} a ^{2}}{d} (1 - \frac{3 α a}{2 d})$
(D) $\frac{ϵ _{0} a ^{2}}{d} (1 - \frac{α a}{2 d})$ ✓ Correct

Answer: (D)

Solution

Because the gap widens across the plate, treat it as many thin strips in parallel. A strip of width

d x

at distance

x

from the narrow edge has gap

d + α x

and contributes

d C = \frac{ε _{0} a d x}{d + α x}

. Integrating across the plate,

C = \frac{ε _{0} a}{α} ln (1 + \frac{α a}{d})

. For the small angle

\frac{α a}{d} ≪ 1

, so

ln (1 + u) \approx u - \frac{u ^{2}}{2}

gives

C \approx \frac{ε _{0} a ^{2}}{d} (1 - \frac{α a}{2 d})

JEE Main · 2020 · 08 Jan 2020 · Shift 2 · Single correct option

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JEE Main · 2020 · 08 Jan 2020 · Shift 2 · Capacitance

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