Question

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

Answer 1

Answer: 6

Electric flux though elemental part -
$d\varphi =EdAcos\theta$
$d\varphi =\frac{\lambda }{2\pi {\in }_0\sqrt{x^2+\frac{3a^2}{4}}}\times Ldx\times \frac{\frac{\sqrt{3a}}{2}}{\sqrt{x^2+\frac{3a^2}{4}}}$
$\int d\varphi =\frac{\sqrt{3}a\lambda L}{4\pi {\in }_0}\underset{x=-\frac{a}{2}\frac{dx}{\left(x^2+\frac{3a^2}{4}\right)}}{\overset{a/2}{\int }}$
$\varphi =\frac{\sqrt{3}a\lambda L}{4\pi {\in }_0}\times 2\underset{0}{\overset{a/2}{\int }}\frac{dx}{x^2+\frac{3a^2}{4}}$
$\varphi =\frac{\sqrt{3}a\lambda L}{2\pi {\in }_0}\times \frac{2}{\sqrt{3a}}{\left[ta{n}^{-1}\left(\frac{2x}{\sqrt{3}a}\right)\right]}_0^{a/2}$
$\varphi =\frac{\lambda L}{\pi {\in }_0}ta{n}^{-1}\left(\frac{1}{\sqrt{3}}\right)$
$\varphi =\frac{\lambda L}{6{\in }_0}$
On comparing $n=6$