The space between the plates of a parallel plate capacitor is filled with a 'dielectric' whose 'dielectric constant' varies with distance as per the relation:
$K (x) = K_{o} + λ x (λ = a constant)$
The capacitance C, of the capacitor, would be related to its vacuum capacitance $C_{0}$ by the relation:

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$C = \frac{λ d}{\ln (1 + K_{o} λ d)} C_{o}$

$C = \frac{λ}{d \cdot \ln (1 + K_{o} λ d)} C_{o}$

$C = \frac{λ d}{\ln (1 + λ d / K_{0})} C_{o}$

$C = \frac{λ}{d \cdot \ln (1 + K_{0} / λ d)} C_{o}$

answer is C.

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Detailed Solution

$\begin{array}{l} Capacitance of capacitor with dielectric material = C_{0} = \frac{ε_{0} A}{d} \\ d C = \frac{ε_{0} (K_{0} + λ x) A}{d x} \\ \Rightarrow \frac{1}{C} = \int \frac{1}{d C} \\ \Rightarrow \frac{1}{C} = \int_{0}^{d} \frac{d x}{ε_{0} (K_{0} + λ x) A} \\ \Rightarrow \frac{1}{C} = \frac{1}{A ε_{0} λ} \times {[\ln (K_{0} + λ x)]}_{0}^{d} \\ \Rightarrow \frac{1}{C} = \frac{1}{A ε_{0} λ} \times [\ln (K_{0} + λ d) - \ln (K_{0})] \\ \Rightarrow C = \frac{A ε_{0} λ}{\ln (1 + \frac{λ d}{K_{0}})} \\ \Rightarrow C = \frac{λ d}{\ln (1 + \frac{λ d}{K_{0}})} C_{0} \end{array}$