A capacitor is made of two square plates each of side ‘2a’making a very small angle a between them, as shown in figure. The capacitance will be close to: 

Consider an infinitesimal strip of capacitor of thickness dx at a distance x as shown. Capacitance of parallel plate capacitor of area A is given  by  $C=\frac{{\epsilon }_{0}A}{t}$
[Here  $t=$seperation between plates] 
So, capacitance of thickness $dx$ will be  $\therefore dC=\frac{{\epsilon }_{0}2adx}{d+x\text{tan}\alpha }$
Total capacitance of system can be obtained by integrating with limits $x=0$  to  $x=a$
$\begin{array}{rr}\therefore C_{eq}=\int dC& =2a{\epsilon }_0{\int }_{x=0}^{x=2a}\frac{dx}{x\text{tan}\alpha +d}\\ & =2a{\epsilon }_0{\int }_{x=0}^{x=2a}\frac{dx}{d(1+\frac{x}{d}\text{tan}\alpha )}\end{array}$
  $\begin{array}{l} \Rightarrow C_{eq}=2\frac{a{\epsilon }_0}{d}{\int }_0^{2a}(1-\frac{x \text{tan}\alpha }{d})dx=\frac{2a{\epsilon }_0}{d}{(x-\frac{x^2\text{tan}\alpha }{2d})}_0^{2a}\\ \Rightarrow C_{eq}=\frac{4a^2{\epsilon }_0}{d}=(1-\frac{2a \text{tan}\alpha }{2d})=\frac{{\epsilon }_{0}2a^2}{d}(1-\frac{\alpha a}{d})\end{array}$