The space between the plates of a parallel plate capacitor is filled with a 'dielectric' whose 'dielectric constant' varies with distance as per the relation: K(x) = K0 + λ x ( λ = a constant) The capacitance C, of this capacitor, would be related to its 'vacuum' capacitance C0 as per relation: (a) C = (λ d)/(ln (1+K₀λ d)) C0 (b) C = (λ)/(d · ln (1+K₀λ d)) C0 (c) C = (λ d)/(ln (1+λ d/K₀)) C0 (d) C = (λ)/(d · ln (1+K₀/λ d)) C0

Answer: (c). Layers in series: 1/C = ∫₀d (dx)/(ε₀ A(K₀ + λ x)) = (1)/(ε₀ Aλ)ln (1 + (λ d)/(K₀)). C = (ε₀ A λ)/(ln (1 + λ d/K₀)) = (λ d)/(ln (1 + λ d/K₀))C₀ with C₀ = (ε₀ A)/(d) So the answer is (c).

The space between the plates of a parallel plate capacitor is filled… | JEE Main 2014 (12 Apr, Shift 1)

Q107JEE Main 2014 · 12 APR · SHIFT 1Difficult

Dielectrics

The space between the plates of a parallel plate capacitor is filled with a 'dielectric' whose 'dielectric constant' varies with distance as per the relation:

K(x) = K0 + $\lambda$ x ( $\lambda$ = a constant)

The capacitance C, of this capacitor, would be related to its 'vacuum' capacitance C0 as per relation:

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Answer: (c)

Layers in series: $\frac{1}{C} = \int_0^d \frac{dx}{\epsilon_0 A(K_0 + \lambda x)} = \frac{1}{\epsilon_0 A\lambda}\ln\left(1 + \frac{\lambda d}{K_0}\right)$ .
$C = \frac{\epsilon_0 A \lambda}{\ln(1 + \lambda d/K_0)} = \frac{\lambda d}{\ln(1 + \lambda d/K_0)}C_0$ with $C_0 = \frac{\epsilon_0 A}{d}$
So the answer is (c).

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Dielectrics question from JEE Main 2014 (12 Apr, Shift 1)

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