A parallel Plate capacitor of area A, plate separation d and capacitance C is filled with three 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 in the capacitor, then its dielectric constant k is given by

k = k_{1} + k_{2} + 2 k_{3}

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k = \frac{k_{1} k_{2}}{k_{1} + k_{2}} + 2 k_{3}

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\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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Solution

The correct option is D $\frac{1}{k} = \frac{1}{k_{1} + k_{2}} + \frac{1}{2 k_{3}}$
The capacitances, $C_{1} = \frac{k_{1} E_{0} A / 2}{[d / 2]} = \frac{k_{1} E_{0} A}{d}$
$C_{2} = \frac{k_{2} E_{0} A / 2}{d / 2} = \frac{k_{0} E_{0} A}{d}$
and $C_{3} = \frac{k_{3} E_{0} A}{[d / 2]} = \frac{2 k_{3} E_{0} A}{d}$
The equivalent capacitance is given by
$\Rightarrow \frac{1}{C_{e q}} = \frac{1}{C_{1} + C_{2}} + \frac{1}{C_{3}} = \frac{1}{\frac{E_{0} A}{d} (k_{1} + k_{2})} + \frac{1}{\frac{E_{0} A}{d} \times 2 k_{3}}$
$\Rightarrow \frac{1}{C_{e q}} = \frac{d}{E_{0} A} [\begin{matrix} \frac{1}{k_{1} + k_{2}} + \frac{1}{2 k_{3}} \end{matrix}]$
$\Rightarrow C_{e q} = {[\begin{matrix} \frac{1}{k_{1} + k_{2}} + \frac{1}{2 k_{3}} \end{matrix}]}^{- 1} . \frac{E \cdot A}{d}$
$∴ \frac{1}{k} = [\begin{matrix} \frac{1}{k_{1} + k_{2}} + \frac{1}{2 k_{3}} \end{matrix}]$ $(\begin{matrix} ∵ C = \frac{K E A}{d} \end{matrix})$

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