First slide

Insertion of dielectric in capacitor

Question

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 $k (x) = K_{0} (1 + α x)$ where ‘x’ is the distance measured from one of the plates. If $(α d) < < 1$ , the total capacitance of the system is best given by the expression

Difficult

A

$\frac{A \in_{0} K_{0}}{d} [1 + {(\frac{α d}{2})}^{2}]$
B

$\frac{A \in_{0} K_{0}}{d} [1 + \frac{α^{2} d^{2}}{2}]$
C

$\frac{A K_{0} \in_{0}}{d} (1 + α d)$
D

$\frac{A K_{0} \in_{0}}{d} [1 + \frac{α d}{2}]$

Solution

Capacitance of “dx” element $C' = \frac{K ε_{0} A}{d x}$
Capacitance of “dx” element $C' = \frac{K_{0} (1 + α x) ε_{0} A}{d x}$
All such elements are connected in series. Hence, net capacitance is given by,
$\sum \frac{1}{C'} = \int_{0}^{d} \frac{d x}{K_{0} ε_{0} A (1 + α x)}$
$\Rightarrow \frac{1}{C} = \frac{1}{K_{0} ε_{0} A α} {[\ln (1 + α x)]}_{0}^{b} (∵ \int \frac{1}{a x} d x = \frac{\log a x}{a} + c) \Rightarrow \frac{1}{C} = \frac{1}{K_{0} ε_{0} A α} \ln (1 + α d)$
Given $α d < < 1$
   $\Rightarrow \frac{1}{C} = \frac{1}{K_{0} ε_{0} A α} (α d - \frac{α^{2} d^{2}}{2}) [u \sin g \ln (1 + x) = x - \frac{x^{2}}{2} + . . .]$
$\Rightarrow \frac{1}{C} = \frac{d}{K_{0} ε_{0} A} (1 - \frac{α d}{2})$
$\Rightarrow C = \frac{K_{0} ε_{0} A}{d} {(1 - \frac{α d}{2})}^{- 1} \Rightarrow C = \frac{K_{0} ε_{0} A}{d} (1 + \frac{α d}{2}) [u \sin g {(1 + x)}^{n} = 1 + n x]$

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