The figure here shows a uniform infinite charged solid cylinder of radius $5R$. Also shown is an imaginary square surface PQRS of side $8R$, having its sides QR and SP along its curved surface while sides PQ and RS along its cross-sections perpendicular to its axis. If minimum electric field on surface PQRS is $E_0$ , electric flux through the surface PQRS is $n{E}_0{R}^2$ .Find $n$.
 

To find flux through PQRS, join PS and QR with axis O as shown forming a closed surface.

As field lines are crossing only PQRS, flux through it will be
$\varphi =\frac{q_{inside}}{{\epsilon }_0}$ (as per Gauss’ Law) …………..(i)
OT = 3R
So, Volume of closed surface is
$V=\frac{1}{2}\times PQ\times OT\times PS$

$=96R^3$

$\Rightarrow q_{inside}=\rho V=96\rho R^3 \dots \dots \dots \dots .\left(ii\right)$

From Eqs. (i) and (ii), We get $\varphi =\frac{q_{inside}}{{\epsilon }_0}=\frac{96\rho R^3}{{\epsilon }_0}$  ………….(iii)
Now, electric field inside a uniform solid cylinder at a distance r from axis is given by, $E=\frac{\rho r}{2{\epsilon }_0}$ .
It will be minimum for r equal to 3R for surface PQRS,

$\Rightarrow E_0=\frac{\rho \left(3R\right)}{2{\epsilon }_0}$

$\Rightarrow \frac{\rho R}{{\epsilon }_0}=\frac{2{\epsilon }_0}{3} \dots \dots \dots \dots .\left(iv\right)$

From Eqs. (iii) and (iv), We get,

$$\begin{gathered} \varphi =96R^2.\frac{\rho R}{{\epsilon }_0}=96R^2\left(\frac{2E_0}{3}\right) \\ =64E_0{R}^2 \end{gathered}$$