A capacitor is made of two square plates each of side  $\text{'}a\text{'}$ making a very small angle  $\alpha$  between them, as shown in figure. The capacitance will be close to:  $C=\frac{{\in }_o{a}^2}{d}(1-\frac{\alpha a}{nd})$ then the value of $n$  is _______

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  $t=$  separation between plates]
So, capacitance of thickness $dx$  will be  $\therefore dC=\frac{{\epsilon }_{0}adx}{d+x\tan \alpha }$
Total capacitance of system can be obtained by integrating with limits  $x=0$  to  $x=a$
 $\therefore C_{eq}=\int dC=a{\epsilon }_0\underset{x=0}{\overset{x=a}{\int }}\frac{dx}{x\tan \alpha +d}$
 $=a{\epsilon }_0\underset{x=0}{\overset{x=a}{\int }}\frac{dx}{d(1+\frac{x}{d}\tan \alpha )}$
 [By Binomial expansion]
 $\Rightarrow C_{eq}=\frac{a{\epsilon }_0}{d}\underset{0}{\overset{a}{\int }}(1-\frac{x\tan \alpha }{d})dx=\frac{a{\epsilon }_0}{d}{(x-\frac{x^2\tan \alpha }{2d})}_0^a$
 $\Rightarrow C_{eq}=\frac{a^2{\epsilon }_0}{d}=(1-\frac{a\tan \alpha }{2d})=\frac{{\epsilon }_0{a}^2}{d}(1-\frac{\alpha a}{2d})$