A parallel plate capacitor of capacitance has spacing between two plates having area . The region between the plates is filled with dielectric layers, parallel to its plates, each with thickness The dielectric constant of the layer is For a very lar

Question

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 $N (> 10^{3})$ , the capacitance $C$ is $α (\frac{K ε_{0} A}{d \ln 2}) .$ The value of $α$ will be ________.

[ $ε_{0}$ is the permittivity of free space] [2019]

Smart Achievers · Accepted Answer

(1)

Let the region between the plates be filled with N dielectric layers.

$m =$ number of dielectric layers within $N$

$d =$ distance between the plates

Here, $m^{th}$

$d C = \frac{k (1 + \frac{m}{N}) ε_{0} A}{d x}$

$∴$ $\frac{1}{d C} = \frac{d x}{k (1 + \frac{m}{N}) ε_{0} A}$

$\frac{1}{C} = \int d (\frac{1}{C}) = \int_{0}^{d} \frac{d x}{k (1 + \frac{m}{N}) ε_{0} A}$

Using

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

$\frac{1}{C} = \int_{0}^{d} \frac{d x}{k (1 + \frac{x}{d}) ε_{0} A}$ $= \frac{d}{k ε_{0} A} \int_{0}^{d} \frac{d x}{d + x}$

$= \frac{d}{k ε_{0} A} \ln α$

or, $\frac{1}{C} = \frac{d}{k ε_{0} A} \ln α$ $\Rightarrow C = \frac{k ε_{0} A}{d \ln α}$

Comparing this equivalent capacitance

$C = \frac{k ε_{0} A}{d \ln α}$ with $α (\frac{k ε_{0} A}{d \ln α}),$

we get, $α = 1$

Solution

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