The gap between the plates of a parallel-plate capacitor is filled with isotropic dielectric whose relative permittivity $K$ varies linearly from $K_{1}$ to $K_{2}$ ( $K_{2} > K_{1}$ ) in the direction perpendicular to the plates. The area of each plate equals A, the separation between the plates is equal to d. Then the capacitance of the capacitor will be given by

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$\frac{(K_{2} - K_{1}) \in_{O} A}{2 d}$

$\frac{(K_{2} - K_{1}) \in_{O} A}{d}$

$\frac{(K_{2} - K_{1}) \in_{O} A}{d \ln (\frac{K_{2}}{K_{1}})}$

$\frac{2 (K_{2} - K_{1}) \in_{O} A}{d \ln (\frac{K_{2}}{K_{1}})}$

answer is A.

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Detailed Solution

$K (x) = \frac{K_{2} - K_{1}}{d} x + K_{1} \frac{d K}{d x} = (\frac{K_{2} - K_{1}}{d})$

Now, $d C = \frac{K ε_{0} A}{d x}$ $\Rightarrow \frac{1}{C_{e q}} = \int \frac{d x}{K ε_{0} A}$
   $\Rightarrow \frac{1}{C_{e q}} = \frac{d}{K_{2} - K_{1}} \int_{K_{1}}^{K_{2}} \frac{d x}{K ε_{0} A}$
   $\Rightarrow \frac{1}{C_{e q}} = \frac{d}{(K_{2} - K_{1}) ε_{0} A} \ln (\frac{K_{2}}{K_{1}})$
   $\Rightarrow \frac{1}{C_{e q}} = \frac{(K_{2} - K_{1}) ε_{0} A}{d \ln (\frac{K_{2}}{K_{1}})}$