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) (ε₀ a²)/(d)(1+(α a)/(d)) (b) (ε₀ a²)/(d)(1-(α a)/(4d)) (c) (ε₀ a²)/(d)(1-(α a)/(2d)) (d) (ε₀ a²)/(d)(1-(3α a)/(2d))

Answer: (c). At distance x from the narrow edge the gap is d + α x; strips of width dx are in parallel. C = ∫₀a (ε₀ a dx)/(d + α x) = (ε₀ a)/(α)ln (1 + (α a)/(d)) Using ln (1+x) ≈ x - x²/2: C ≈ (ε₀ a)/(α)((α a)/(d) - (α²a²)/(2d²)) So the answer is (c) (ε₀ a²)/(d)(1 - (α a)/(2d)).

A capacitor is made of two square plates each of side 'a' making a… | JEE Main 2020 (8 Jan, Shift 2)

Q15JEE Main 2020 · 8 JAN · SHIFT 2Difficult

Parallel Plate Capacitor

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

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Answer: (c)

At distance $x$ from the narrow edge the gap is $d + \alpha x$ ; strips of width $dx$ are in parallel.
$C = \int_0^a \frac{\varepsilon_0 a\,dx}{d + \alpha x} = \frac{\varepsilon_0 a}{\alpha}\ln\left(1 + \frac{\alpha a}{d}\right)$
Using $\ln(1+x) \approx x - \frac{x^2}{2}$ : $C \approx \frac{\varepsilon_0 a}{\alpha}\left(\frac{\alpha a}{d} - \frac{\alpha^2a^2}{2d^2}\right)$
So the answer is (c) $\frac{\varepsilon_0 a^2}{d}\left(1 - \frac{\alpha a}{2d}\right)$ .

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Parallel Plate Capacitor question from JEE Main 2020 (8 Jan, Shift 2)

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