A parallel-plate capacitor is constructed using three different dielectric materials as shown in the figure. What is the capacitance across P and Q?

(\frac{K_{1}}{2} + \frac{K_{2} K_{3}}{K_{2} + K_{3}}) \frac{ε_{0} A}{t}

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(K_{1} + \frac{K_{2} K_{3}}{K_{2} + K_{3}}) \frac{ε_{0} A}{t}

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(K_{1} + \frac{2 K_{2} K_{3}}{K_{2} + K_{3}}) \frac{ε_{0} A}{t}

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(K_{1} + \frac{K_{2} K_{3}}{2 (K_{2} + K_{3})}) \frac{ε_{0} A}{t}

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Solution

The correct option is A $(\frac{K_{1}}{2} + \frac{K_{2} K_{3}}{K_{2} + K_{3}}) \frac{ε_{0} A}{t}$
On redrawing the circuit we get,
$C_{1} = K_{1} \frac{ε_{0} A / 2}{t} = K_{1} \frac{ε_{0} A}{2 t}$

$C_{2} = K_{2} \frac{ε_{0} A / 2}{t / 2} = K_{2} \frac{ε_{0} A}{t}$

$C_{3} = K_{3} \frac{ε_{0} A / 2}{t / 2} = K_{3} \frac{ε_{0} A}{t}$

Now as $C_{2}$ , $C_{3}$ are in series and this combination is parallel with $C_{1}$ . Therefore the effective capacitance will be

$C_{e q} = \frac{ε_{0} A}{t} [\frac{K_{2} \times K_{3}}{K_{2} + K_{3}} + \frac{K_{1}}{2}]$

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