A parallel-plate capacitor of area $A$ , plate separation $d$ and capacitance $C$ is filled with four dielectric materials having dielectric constants $k_{1}, k_{2}, k_{3}$ and $k_{4}$ as shown in the figure below. 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}{k_{3}} + \frac{3}{2 k_{4}}

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k = k_{1} + k_{2} + k_{3} + 3 k_{4}

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k = \frac{2}{3} (k_{1} + k_{2} + k_{3}) + 2 k_{4}

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\frac{1}{k} = \frac{3}{2 (k_{1} + k_{2} + k_{3})} + \frac{1}{k_{4}}

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Solution

The correct option is C
$k = \frac{2}{3} (k_{1} + k_{2} + k_{3}) + 2 k_{4}$
Here the capacitance, $C = \frac{A k ϵ_{0}}{d}$
$C_{K_{1}}, C_{K_{2}}, C_{K_{3}}$ are in parallel combination and they are connected in series with $C_{K_{4}}$

Here, $C_{K_{1}} = \frac{(A / 3) k_{1} ϵ_{0}}{d / 2} = \frac{2 k_{1}}{3} C$ ,
$C_{K_{2}} = \frac{(A / 3) k_{2} ϵ_{0}}{d / 2} = \frac{2 k_{2}}{3} C$ ,

$C_{K_{3}} = \frac{(A / 3) k_{3} ϵ_{0}}{d / 2} = \frac{2 k_{3}}{3} C$ ,

$C_{K_{4}} = \frac{(A) k_{4} ϵ_{0}}{d / 2} = 2 k_{4} C$
Now the equivalent capacitance for the combination of four capacitors is
$\frac{1}{C_{e q}} = \frac{1}{(C_{k_{1}} + C_{k_{2}} + C_{k_{3}})} + \frac{1}{C_{K_{4}}}$
$C_{e q} = k C$

$\frac{1}{k C} = \frac{3}{2 C} [\frac{1}{k_{1} + k_{2} + k_{3}}] + \frac{1}{2 k_{4} C}$

or $\frac{1}{k} = \frac{3}{2 (k_{1} + k_{2} + k_{3})} + \frac{1}{2 (k_{4})}$
or
$k = \frac{2 (k_{1} + k_{2} + k_{3})}{3} + 2 (k_{4})$

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