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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 give same capacitance when fully inserted in above capacitor, would be

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answer is C.

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Detailed Solution

In the given arrangement capacitor can be viewed as three-different capacitors connected in parallel as shown below

So, equivalent capacity of the system is

$\begin{array}{l} C_{e q} = C_{1} + C_{2} + C_{3} \\ \Rightarrow \frac{K ε_{0} A}{d} = \frac{K_{1} ε_{0} A / 3}{d} + \frac{K_{2} ε_{0} A / 3}{d} + \frac{K_{3} ε_{0} A / 3}{d} \end{array}$

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

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

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