A parallel plate capacitor of capacitance C has spacing d between two plates having area A. The region between the plates is filled with N dielectric layers, parallel to its plates, each with thickness $\delta=\frac{\mathrm{d}}{\mathrm{N}}$. The dielectric constant of the $\mathrm{m}^{\text {th}}$ layer is $\mathrm{K}_{\mathrm{m}}=\mathrm{K}\left(1+\frac{\mathrm{m}}{\mathrm{N}}\right)$. For a very large $\mathrm{N}(>10^{3})$, the capacitance C is $\alpha\left(\frac{\mathrm{K} \epsilon_{0} \mathrm{~A}}{\mathrm{~d} \ell \mathrm{n} 2}\right)$. The value of $\alpha$ will be ____. [$\epsilon_{0}$ is the permittivity of free space]

JEE Advanced · 2019 · 27 May 2019 · Paper 1 · Capacitance

A parallel plate capacitor of capacitance C has spacing d between two plates having area A. The region between the plates is filled with N dielectric layers, parallel to its plates, each with thickness

δ = \frac{d}{N}

. The dielectric constant of the

m^{th}

layer is

K_{m} = K (1 + \frac{m}{N})

. For a very large

N (> 1 0^{3})

, the capacitance C is

α (\frac{K ϵ _{0} A}{d ℓ n 2})

. The value of

α

will be ____.
[

ϵ_{0}

is the permittivity of free space]

Answer: 1

Solution

For N layers in series, each with thickness

d x = \frac{d}{N}

δ = d x = \frac{d}{N}, \frac{m}{N} = \frac{x}{d}

K_{m} = K (1 + \frac{x}{d})

Capacitance of each layer:

C^{'} = \frac{K _{m} A ϵ _{0}}{d x}

For series combination:

\frac{1}{C _{e q}} = \int_{0}^{d} \frac{d x}{K _{m} A ϵ _{0}} = \frac{1}{K A ϵ _{0}} \int_{0}^{d} \frac{d x}{1 + \frac{x}{d}}

\frac{1}{C _{e q}} = \frac{d}{K A ϵ _{0}} [ln (1 + \frac{x}{d})]_{0}^{d}

\frac{1}{C _{e q}} = \frac{d}{K A ϵ _{0}} [ln 2 - ln 1] = \frac{d ln 2}{K A ϵ _{0}}

C_{e q} = \frac{K A ϵ _{0}}{d ln 2}

Comparing with

α (\frac{K ϵ _{0} A}{d l n 2})

α = 1

JEE Advanced · 2019 · 27 May 2019 · Paper 1 · Numerical answer

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JEE Advanced · 2019 · 27 May 2019 · Paper 1 · Capacitance

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