A parallel plate capacitor of area A, plat separation d and capacitance C is filled with four dielectric material having dielectric constant $K_{1}$ $K_{2}$ $K_{2}$ and $K_{4}$ as shown in the figure. If a single dielectric material is to be used to have the sam capacitance in this capacitor. then its diectric constant K is given by:-

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

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\frac{1}{K} = \frac{1}{K_{1}} + \frac{1}{K_{2}} + \frac{1}{K_{3}} + \frac{3}{2 K_{4}}

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K = \frac{3 K_{1} K_{2} K_{3}}{2 (K_{2} K_{3}^{'} + K_{1} + K_{2} + K_{3} K_{1})} + \frac{K_{4}}{2}

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K = \frac{2}{3} (K_{1} + K_{2} + K_{3}) + 2 K_{4}

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Solution

The correct option is D $\frac{1}{K} = \frac{1}{K_{1}} + \frac{1}{K_{2}} + \frac{1}{K_{3}} + \frac{3}{2 K_{4}}$
$\begin{matrix} C_{e q} = \frac{\frac{K_{3} ε_{o} A / 2}{d / 3} . \frac{K_{4} ε_{o} A / 2}{d / 3}}{(K_{3} + K_{4}) \frac{ε_{o} A / 2}{d / 3}} + \frac{\frac{K_{2} ε_{o} A / 2}{d / 3} . \frac{K_{4} ε_{o} A / 2}{d / 3}}{(K_{2} + K_{4}) \frac{ε_{o} A / 2}{d / 3}} + \frac{\frac{K_{1} ε_{o} A / 2}{d / 3} . \frac{K_{4} ε_{o} A / 2}{d / 3}}{(K_{1} + K_{4}) \frac{ε_{o} A / 2}{d / 3}} = (\frac{K_{3} . K_{4}}{K_{3} + K_{4}} + \frac{K_{2} K_{4}}{K_{2} + K_{4}} + \frac{K_{1} K_{4}}{K_{1} + K_{4}}) \frac{3 ε_{o} A}{2 d} \Rightarrow \frac{K ε_{o} A}{d} = \frac{3 ε_{o} A}{2 d} (\frac{K_{3} . K_{4}}{K_{3} + K_{4}} + \frac{K_{2} K_{4}}{K_{2} + K_{4}} + \frac{K_{1} K_{4}}{K_{1} + K_{4}}) \Rightarrow \frac{2 K}{3} = K_{4} (\frac{K_{3}}{K_{3} + K_{4}} + \frac{K_{2}}{K_{2} + K_{4}} + \frac{K_{1}}{K_{1} + K_{4}}) \Rightarrow \frac{2 K}{3} = \frac{2 K_{4} (K_{1} K_{2} + K_{2} K_{3} + K_{3} K_{4} + K_{4} K_{1})}{(K_{1} + K_{4}) (K_{2} + K_{4}) (K_{3} + K_{4})} H e n c e, o p t i o n B i s c o r r e c t a n s w e r . \end{matrix}$

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