A parallel plate capacitor is of area $6 {cm}^{2}$ and a separation $3 mm$ . The gap is filled with three dielectric materials of equal thickness (see figure) with dielectric constants $K_{1} = 10, K_{2} = 12$ and $K_{3} = 14$ . The dielectric constant of a material which when fully inserted in above capacitor, gives same capacitance would be

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answer is C.

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Detailed Solution

$C = C_{1} + C_{2} + C_{3}$

$C = \frac{ε_{0} KA}{d} = \frac{ε_{0} K_{1} A}{3 d} + \frac{ε_{0} K_{2} A}{3 d} + \frac{ε_{0} K_{3} A}{3 d}$

$\Rightarrow K = \frac{K_{1} + K_{2} + K_{3}}{3} = \frac{10 + 12 + 14}{3} = 12$

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