A parallel plate capacitor of area A, plate separation d and capacitance C is filled with three different dielectric materials having dielectric constants $K_{1}, K_{2}$ And $K_{3}$ as shown. 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

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

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

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

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

answer is D.

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Detailed Solution

Let $C = \frac{ε_{0} A}{d}; C_{1} = K_{1} \frac{ε_{0} \frac{A}{2}}{\frac{d}{2}} = K_{1} C$

$C_{2} = K_{2} \frac{ε_{0} \frac{A}{2}}{\frac{d}{2}} = K_{2} C;$

$C_{3} = \frac{K_{3} ε_{0} \frac{A}{2}}{\frac{d}{2}}, = K_{3};$

$C_{3^{1}} = \frac{K_{3} ε_{0} \frac{A}{2}}{\frac{d}{2}}, = K_{3} C$

The equivalent circuit as shown

$C_{e f f} = \frac{C_{1} C_{3}}{C_{1} + C_{3}} + \frac{C_{2} C_{3^{'}}}{C_{2} + C_{3^{'}}}$

$= \frac{(\frac{K_{1} ε A / 2}{d / 2}) (\frac{K_{3} ε A / 2}{d / 2})}{(\frac{K_{1} ε A / 2}{d / 2} + \frac{K_{3} ε A / 2}{d / 2})} + \frac{(\frac{K_{2} ε A / 2}{d / 2}) (\frac{K_{3} ε A / 2}{d / 2})}{(\frac{K_{2} ε A / 2}{d / 2} + \frac{K_{3} ε A / 2}{d / 2})}$