First slide

Capacitance

Question

A parallel plate condenser with plate area A and separation d is filled with dielectrics as shown in fig. The dielectric constants are K₁ and K₂ respectively. The capacitance will be

Moderate

A

$\frac{ε_{0} A}{d} (K_{1} + K_{2})$
B

$\frac{ε_{0} A}{d} (\frac{K_{1} + K_{2}}{K_{1} K_{2}})$
C

$\frac{2 ε_{0} A}{d} (\frac{K_{1} K_{2}}{K_{1} + K_{2}})$
D

$\frac{2 ε_{0} A}{d} (\frac{K_{1} + K_{2}}{K_{1} K_{2}})$

Solution

The combination is equivalent to two capacitors in series, each with plate area A and separation d / 2.
$\begin{aligned} C_{1} = \frac{K_{1} ε_{0} A}{(d / 2)} = \frac{2 K_{1} ε_{0} A}{d} \\ C_{2} = \frac{K_{2} ε_{0} A}{(d / 2)} = \frac{2 K_{2} ε_{0} A}{d} \end{aligned}$
The equivalent capacitance C is given by
$\begin{array}{l} \frac{1}{C} = \frac{1}{C_{1}} + \frac{1}{C_{2}} \\ ∴ \frac{1}{C} = \frac{d}{2 K_{1} ε_{0} A} + \frac{d}{2 K_{2} ε_{0} A} \\ = \frac{d}{2 ε_{0} A} [\frac{1}{K_{1}} + \frac{1}{K_{2}}] = \frac{d}{2 ε_{0} A} (\frac{K_{1} + K_{2}}{K_{1} K_{2}}) \end{array}$
or $C = \frac{2 ε_{0} A}{d} [\frac{K_{1} K_{2}}{K_{1} + K_{2}}]$

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