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← JEE Advanced 2019 · Paper 1
PhysicsNumerical+3Question 16 of 54

JEE Advanced 2019 · Paper 1, Physics question 16

A parallel plate capacitor of capacitance CC has spacing dd between two plates having area AA. The region between the plates is filled with NN dielectric layers, parallel to its plates, each with thickness δ=dN\delta = \dfrac dN. The dielectric constant of the mthm^{th} layer is Km=K(1+mN)K_m = K\left(1 + \dfrac mN\right). For a very large N (>103)N\ (> 10^3), the capacitance CC is α(Kϵ0Adln⁡2)\alpha\left(\dfrac{K\epsilon_0A}{d\ln2}\right). The value of α\alpha will be ______.

[ϵ0\epsilon_0 is the permittivity of free space]

Answer and worked solution

Official answer: 0.99

Layers in series.
1C=∑m=1Nδϵ0AKm=dϵ0AK⋅1N∑m=1N11+m/N.\frac1C=\sum_{m=1}^N\frac{\delta}{\epsilon_0AK_m}=\frac d{\epsilon_0AK}\cdot\frac1N\sum_{m=1}^N\frac1{1+m/N}.

Large NN. The sum becomes an integral:
1N∑m=1N11+m/N→∫01dx1+x=ln⁡2.\frac1N\sum_{m=1}^N\frac1{1+m/N}\to\int_0^1\frac{dx}{1+x}=\ln2.

So C=Kϵ0Adln⁡2C=\dfrac{K\epsilon_0A}{d\ln2} and α=1\alpha=1.

Answer: 1.00 (key: 0.99 to 1.01)