A parallel plate capacitor has plates of area A separated by distance 'd' between them. It is filled with a dielectric which has a dielectric constant that varies as k(x) = K(1 + α x) where 'x' is the distance measured from one of the plates. If (ad) << 1, the total capacitance of the system is best given by the expression: (a) (Aε₀ K)/(d)(1 + ((α d)/(2))²) (b) (Aε₀ K)/(d)(1 + (α d)/(2)) (c) (Aε₀ K)/(d)(1 + (α² d²)/(2)) (d) (Aε₀ K)/(d)(1 + α d)

Answer: (b). Thin layers in series: 1/C = ∫₀d (dx)/(ε₀ K(1 + α x)A) = (ln (1 + α d))/(ε₀ K A α). For α d 1: ln (1 + α d) ≈ α d - (α² d²)/(2), so 1/C ≈ (d)/(ε₀ K A)(1 - (α d)/(2)). C ≈ (Aε₀ K)/(d)(1 + (α d)/(2)), so the answer is (b).

A parallel plate capacitor has plates of area A separated by distance… | JEE Main 2020 (7 Jan, Shift 1)

Q91JEE Main 2020 · 7 JAN · SHIFT 1Difficult

Dielectrics

A parallel plate capacitor has plates of area A separated by distance 'd' between them. It is filled with a dielectric which has a dielectric constant that varies as k(x) = K(1 + $\alpha$ x) where 'x' is the distance measured from one of the plates. If (ad) << 1, the total capacitance of the system is best given by the expression:

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Answer: (b)

Thin layers in series: $\frac{1}{C} = \int_0^d \frac{dx}{\epsilon_0 K(1 + \alpha x)A} = \frac{\ln(1 + \alpha d)}{\epsilon_0 K A \alpha}$ .
For $\alpha d \ll 1$ : $\ln(1 + \alpha d) \approx \alpha d - \frac{\alpha^2 d^2}{2}$ , so $\frac{1}{C} \approx \frac{d}{\epsilon_0 K A}\left(1 - \frac{\alpha d}{2}\right)$ .
$C \approx \frac{A\epsilon_0 K}{d}\left(1 + \frac{\alpha d}{2}\right)$ , so the answer is (b).

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Dielectrics question from JEE Main 2020 (7 Jan, Shift 1)

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