Q.

(a) Electric flux:
Electric flux through an area is defined as the product of electric field strength E and area dS
perpendicular to the field.
It represents the field lines crossing the area.
It is a scalar quantity.
Imagine a cube of edge d, enclosing the charge.
The square surface is one of the six faces of this cube. According to Gauss’ theorem in
electrostatics,
Total electric flux through the cube = q / ε₀
This is the total flux through all six surface
∴ Electric flux through the square surface = q / σε₀

(b) On moving the charge to distance d from the center of square and making side of square
2d, does not change the flux at all because flux is independent of side of square or distance of
charge in this case.

OR
 

(a) Electric field $\overrightarrow{\mathrm{E}}$ due to a straight uniformly charged infinite line of charge density λ:
Consider a cylindrical Gaussian surface of radius r and length l coaxial with line charge. The
cylindrical Gaussian surface may be divided into three parts :
(i) curved surface S₁
(ii) flat surface S₂ and
(iii) flat surface S₃.
By symmetry, the electric field has the same magnitude E at each point of curved surface Sj
and is directed radially outward. We consider small elements of surfaces S₁, S₂ and S₃.
     The surface element vector d$\overrightarrow{\mathrm{S}}$ is directed along the direction of electric field (i.e., angle between $\overrightarrow{\mathrm{E}}$and d$\overrightarrow{\mathrm{S}}$₁  is 0°); the elements d${\overrightarrow{\mathrm{S}}}_2$ and d${\overrightarrow{\mathrm{S}}}_3$  are directed perpendicular to field vector $\overrightarrow{\mathrm{E}}$(i.e., angle between d${\overrightarrow{\mathrm{S}}}_2$ and $\overrightarrow{\mathrm{E}}$, and d${\overrightarrow{\mathrm{S}}}_3$ and $\overrightarrow{\mathrm{E}}$ is 90. Electric flux through the cylindrical surface, 
${\text{∮}}_{\mathrm{S}}\overrightarrow{\mathrm{E}}\cdot \mathrm{d}\overrightarrow{\mathrm{S}}={\text{∮}}_{{\mathbf{s}}_1}\overrightarrow{\mathrm{E}}\cdot \mathrm{d}{\overrightarrow{\mathrm{S}}}_1+{\text{∮}}_{{\mathrm{s}}_2}\overrightarrow{\mathrm{E}}\cdot \mathrm{d}{\overrightarrow{\mathrm{S}}}_2+{\text{∮}}_{{\mathrm{S}}_2}\overrightarrow{\mathrm{E}}\cdot \mathrm{d}{\overrightarrow{\mathrm{S}}}_3$
$=\underset{{\mathrm{S}}_1}{\text{∮}\mathrm{Fd}}{\mathrm{dS}}_1\cos 0^{\circ }+\underset{{\mathrm{S}}_2}{\text{∮}}{\mathrm{F}}_2{\mathrm{dS}}_2\cos {90}^{\circ }+{\text{∮}}_{{\mathrm{S}}_3}{\mathrm{FdS}}_3\cos {90}^{\circ }$
$\doteq \mathrm{E}\int {\mathrm{dS}}_1+0+0$
= E x area of curved surface = E x 2$\mathrm{\pi rl}$
Total charges enclosed, q=$\mathrm{\lambda l}$
By Gauss' theorem,
${\text{∮}}_{\mathrm{E}}=\frac{\mathrm{\lambda l}}{{\mathrm{\epsilon }}_0}=\frac{\mathrm{\lambda l}}{{\mathrm{\epsilon }}_0}$
$\mathrm{or} \mathrm{E}=\frac{\mathrm{\lambda }}{2{\mathrm{\pi \epsilon }}_0\mathrm{r}}$

b) Graph showing variation of E with perpendicular distance from line of charge:
The electric field is inversely proportional to distance V from line of charge.