A capacitor is made of two square plates each of side ‘a’ making a very small angle  $\alpha$ between them, as shown in figure. The capacitance will be close to :

$\begin{array}{l}\Rightarrow Select in small strip of dx, at a dis\tan ce x from left.\text{}\\ \Rightarrow \tan \alpha =\frac{y}{x}\\ \Rightarrow y=x\tan \alpha =x\alpha \left(For small angles \tan \alpha =\alpha \right)\text{}\\ \Rightarrow dis\tan ce between plates at a dis\tan ce x=d+x\alpha \\ \Rightarrow Capaci\tan ce of this element can be written as, \text{}\\ \Rightarrow dC=\frac{{\epsilon }_{0}dA}{d+\alpha x}\\ \Rightarrow dC=\frac{{\epsilon }_{0}adx}{d+\alpha x} \\ \text{}\Rightarrow All such elements are connected in parallel, integration will give total capaci\tan ce \\ \Rightarrow {\int }_0^{a}dC={\int }_0^a\frac{{\epsilon }_{0}adx}{d+\alpha x} \\ \Rightarrow C= {\epsilon }_{0}a{\int }_0^a\frac{dx}{d+\alpha x}\text{}\\ \Rightarrow C=\frac{{\epsilon }_{0}a}{\alpha }{\left[\ln \left(d+\alpha x\right)\right]}_0^a\\ \left(\because \int \frac{1}{ax}dx=\frac{\log ax}{\frac{d}{dx}ax}+C \right)\\ \Rightarrow C=\frac{{\epsilon }_{0}a}{\alpha }\ln \left(1+\frac{\alpha a}{d}\right)\\ \text{}\Rightarrow C\simeq \frac{{\epsilon }_0{a}^2}{d}\left(1-\frac{\alpha a}{2d}\right) \left[\ln \left(1+x\right) =x\left(1-\frac{x}{2}\right)\right]\end{array}$