A parallel plate capacitor is of area $6 c m^{2}$ and a separation $3 m m$ . 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 equivalent capacitance would be :

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answer is C.

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Detailed Solution

Consider three dielectric connected in parallel combination

$C_{n e t} = C_{1} + C_{2} + C_{3} \Rightarrow \frac{k A \in_{0}}{d} = k_{1} (\frac{A}{3}) \frac{\in_{0}}{d} + k_{2} (\frac{A}{3}) \frac{\in_{0}}{d} + k_{3} (\frac{A}{3}) \frac{\in_{0}}{d} \Rightarrow \frac{k A \in_{0}}{d} = \frac{A \in_{0}}{3 d} (k_{1} + k_{2} + k_{3}) \Rightarrow k = \frac{k_{1} + k_{2} + k_{3}}{3} = 12$