An infinitely long uniform line charge distribution of charge per unit length $\mathrm{\lambda }$ lies parallel to the $\mathrm{y}$ -axis in the $\mathrm{y}$ - $\mathrm{z}$ plane at $\mathrm{z}=\frac{\sqrt{3}}{2}\mathrm{\alpha }$ (see figure). If the magnitude of the flux of the electric field through the rectangular surface $\mathrm{A}\mathrm{B}\mathrm{C}\mathrm{D}$ lying in the $\mathrm{x}$ - $\mathrm{y}$ plane with its centre at the origin is $\frac{\mathrm{\lambda }\mathrm{L}}{\mathrm{n}{\mathrm{\epsilon }}_0}$ ( ${\mathrm{\epsilon }}_0=$ permittivity of free space), then the value of $\mathrm{n}$ is

Solution

$\mathrm{d}\mathrm{\varphi }=\frac{\mathrm{\lambda }}{2\mathrm{\pi }{\mathrm{ϵ}}_0\sqrt{{\mathrm{d}}^2+{\mathrm{x}}^2}}.\frac{\mathrm{d}}{\sqrt{{\mathrm{d}}^2+{\mathrm{x}}^2}}\mathrm{L}\mathrm{d}\mathrm{x}$
$\mathrm{\varphi }=\frac{\mathrm{\lambda }\mathrm{d}\mathrm{L}}{2\mathrm{\pi }{\mathrm{ϵ}}_0}\underset{-\frac{\mathrm{a}}{2}}{\overset{\frac{\mathrm{a}}{2}}{\int }}\frac{\mathrm{d}\mathrm{x}}{{\mathrm{d}}^2+{\mathrm{x}}^2}$
$=\mathrm{ }\frac{\mathrm{\lambda }\left(\mathrm{d}\right)\left(\mathrm{L}\right)}{2\mathrm{\pi }{\mathrm{ϵ}}_0}\left(\frac{1}{\mathrm{d}}\right)2{\left[{\tan }^{-1}\frac{\mathrm{x}}{\mathrm{d}}\right]}_0^{\frac{\mathrm{a}}{2}}=\frac{\mathrm{\lambda }\mathrm{L}}{\mathrm{\pi }{\mathrm{ϵ}}_0}\left({\tan }^{-1}\frac{\mathrm{a}}{2.\frac{\mathrm{a}\sqrt{3}}{2}}\right)=\frac{\mathrm{\lambda }\mathrm{L}}{\mathrm{\pi }{\mathrm{ϵ}}_0}\left(\frac{\mathrm{\pi }}{6}\right)$
$\therefore \mathrm{n}=6$

Hence, the correct answer is 6.