A parallel plate capacitor has plates of area $A$ separated by distance $d$ between them. It is filled with a dielectric which has a dielectric constant that varies as $K(x) = K_0(1 + \alpha x)$, where $x$ is the distance measured from one of the plates. If $(\alpha d) \ll 1$, the total capacitance of the system is best given by the expression:
Answer: (A) $\dfrac{AK_0\varepsilon_0}{d}\left(1 + \dfrac{\alpha d}{2}\right)$
Thin slices of thickness $dx$ are in series:
$$\frac1C = \int_0^d \frac{dx}{K_0(1+\alpha x)\varepsilon_0A} = \frac{\ln(1 + \alpha d)}{K_0\varepsilon_0A\alpha}$$
For $\alpha d \ll 1$, $\ln(1+\alpha d) \approx \alpha d - \dfrac{\alpha^2d^2}{2}$:
$$\frac1C \approx \frac{d}{K_0\varepsilon_0A}\left(1 - \frac{\alpha d}{2}\right) \Rightarrow C \approx \frac{AK_0\varepsilon_0}{d}\left(1 + \frac{\alpha d}{2}\right)$$
Solution by Sreeraj P, M.Sc Physics