Question

A capacitors has square plates each of side $l$ making an angle a with each other as shown in the figure. Then, for small value of a the capacitance C is given by :

• A. $\frac{{\in }_0{I}^2}{d}(1-\frac{al}{2d})$
• B. $\frac{{\in }_0{I}^2}{2d}(1-\frac{al}{d})$
• C. $\frac{{\in }_0{I}^2}{d}(1-\frac{al}{d})$
• D. $\frac{{\in }_0{I}^2}{2d}(1+\frac{al}{d})$

Answer 1

The correct option is A $\frac{{\in }_0{I}^2}{d}(1-\frac{al}{2d})$

Suppose the capacitor is made of various elementary capacitors connected in parallel.
Consider one such elementary capacitor of width dx at a distance x from left end.
The area of each plate of the elementary capacitor , $dA=ldx$ and separation between the plates , $BC=d+x\tan a=d+xa$ as a is very small, $\tan a=a$
The capacitance for elementary capacitor is $dC=\frac{{ϵ}_{0}dA}{BC}=\frac{{ϵ}_{0}ldx}{d+xa}=\frac{{ϵ}_{0}ldx}{d}(1+xa/d{)}^{-1}=\frac{{ϵ}_{0}ldx}{d}(1-xa/d)$ (taking only 1st term of binomial expn as a is very small)
Thus total capacitance, $C={\int }_0^l\frac{{ϵ}_{0}ldx}{d}(1-xa/d)=\frac{{ϵ}_0{l}^2}{d}(1-la/2d)$