Two square plates of side ‘a’ are arranged as shown in the figure. The minimum separation between plates is ‘d’ and one of the plate is inclined at small angle $\alpha$  with plane parallel to another plate. The capacitance of capacitor is (given $\alpha$ is very small)

For small strip of width dx  and separation $d+\alpha x$ , differential capacitance can be given as

$$\begin{gathered} dC=\frac{{\in }_{0}adx}{d+\alpha x} \\ All such strips are in parallel, so we will integrate the equation \\ C.=\frac{{\in }_{0}a}{\alpha }{\left[In\left(d+\alpha x\right)\right]}_0^a \\ =\frac{{\in }_{0}a}{\alpha }In\left(1+\frac{\alpha a}{d}\right) \\ Use, In\left(1+x\right) \simeq x - \frac{x^2}{2} (for small values of x) \\ \approx \frac{{\in }_{0}a}{\alpha }\left[\frac{\alpha a}{d}-\frac{{\alpha }^2{a}^2}{2d^2}\right] \\ \approx \frac{{\in }_0{a}^2}{d}\left(1-\frac{\alpha a}{2d}\right) \end{gathered}$$