A parallel plate capacitor of area A, plate separation d and capacitance C is filled with three different dielectric materials having dielectric constants $K_{1}, K_{2} a n d K_{3}$ as shown in Fig. If a single dielectric material is to be used to have the same capacitance C in this capacitor, then its dielectric constant K is given by

$\frac{1}{K} = \frac{1}{K_{1}} + \frac{1}{K_{2}} + \frac{1}{2 K_{3}}$

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$\frac{1}{K} = \frac{1}{K_{1} + K_{2}} + \frac{1}{2 K_{3}}$

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$K = \frac{K_{1} K_{2}}{K_{1} + K_{2}} + 2 K_{3}$

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$K = K_{1} + K_{2} + K_{3}$

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Solution

The correct option is B
$\frac{1}{K} = \frac{1}{K_{1} + K_{2}} + \frac{1}{2 K_{3}}$

We have $C_{1} = \frac{ϵ_{0} (A / 2) K_{1}}{(d / 2)} = \frac{ϵ_{0} A K_{1}}{d} C_{2} = \frac{ϵ_{0} (A / 2) K_{2}}{(d / 2)} = \frac{ϵ_{0} A K_{2}}{d} a n d C_{3} = \frac{ϵ_{0} (A) K_{3}}{(d / 2)} = \frac{2 ϵ_{0} A K_{3}}{d}$
The capacitors $C_{1} a n d C_{2}$ are in parallel and their equivalent capacitance is
$C^{'} = C_{1} + C_{2} = \frac{ϵ_{0} A}{d} (K_{1} + K_{2})$
This combination is in series with $C_{3}$ . Hence, the net capacitance is
$\frac{1}{C^{''}} = \frac{1}{C^{'}} + \frac{1}{C_{3}} = \frac{d}{ϵ_{0} A (K_{1} + K_{2})} + \frac{d}{2 ϵ_{0} A K_{3}} = \frac{d}{ϵ_{0} A} [\frac{1}{(K_{1} + K_{2})} + \frac{1}{2 K_{3}}] o r C^{''} = \frac{ϵ_{0} A K}{d} w h e r e \frac{1}{K} = \frac{1}{(K_{1} + K_{2})} + \frac{1}{2 K_{3}}$
Hence, the correct choice is (b).

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