An infinity long uniform 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   $\lambda L/\left(n{\in }_0\right)$$({\in }_0=permittivityoffreespace)$ , then the value of n is …….

 

$\begin{array}{l}{\varphi }_{rec\tan gle}=\varphi (\alpha /360)=(\lambda L/{\in }_0)(60/360)\\ =\lambda L/\left(6{\in }_0\right).\end{array}$The configuration is shown in the figure.

Consider a cylindrical Gaussian surface of radius  $a\sqrt{3}/2andlengthL$. The charge$\varphi =q_{enc}/{\in }_0=\lambda L/{\in }_0$ enclosed by Gaussian flux through Gaussian surface

The flux through the rectangular surface is equal to the flux through the cylindrical section with chord PQ (see figure). From geometry, the chord PQ subtends an angle $\alpha ={60}^o$  at the axis of the cylinder. Thus the flux through the rectangular surface is 

We encourage you to find the flux through the rectangular surface by using  
 $\varphi =\int \overrightarrow{E}.d\overrightarrow{S}$