(a) Define electric flux. Is it a scalar or a vector quantity?
A point charge $q$ is at a distance of $d/2$ directly above the centre of a square of side $d$, as shown in the figure. Use Gauss' law to obtain the expression for the electric flux through the square.
(b) If the point charge is now moved to a distance ${}^{\prime }{d}^{\prime }$ from the centre of the square and the side of the square is doubled, explain how the electric flux will be affected.

$\text{Part (a)}$

$\text{Step 1: Definition of electric flux}$

$\text{Qualitative Definition:}$ The Electric flux is proportional to the number of field lines passing a given area in a unit of time. This definition cannot be used for calculating the exact value of flux, it is used only for the comparison of flux through two surfaces.

$\text{Quantitative Definition: }$ Magnitude of Electric flux is given by 

                      $\Delta \varphi =\int \vec{E}.\overrightarrow{dS}$                     

For plane surface and uniform Electric field, the above equation becomes:

                       $\Delta \varphi =\vec{E}.\vec{S}=ES\cos \theta$

It is a scalar quantity.

$\text{Step 2: Electric flux through square }$

Suppose a cube enclosing the charge q of side $d$. 

Hence according to Gauss's law, the electric flux through the cube is $\frac{q}{{ϵ}_o}$

The given case can be considered as one of the face of such a cube.

By symmetry, the flux through the face is $\frac{1}{6}th$ of the flux through the cube.

Hence flux through the square face = $\frac{q}{6{ϵ}_o}$

$\text{Part (b)}$

$\text{Electric flux through square when the side is doubled}$

Now, the charge is placed at a distance $d$ from the centre of square. Also, the side of the square is $2d$ in this case.

In this case, the charge will be at the center of a cube of side 2d and the square will be one of its face.

According to a similar analysis as shown in Step 2 the electric flux through the face is: $\frac{q}{6{ϵ}_o}$

Hence there will be no change in the electric flux.