# (a) Define electric flux. Is it a scalar or a vector quantity? A point charge q is at a distance of$\frac{d}{2}$ 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 ‘d’ from the centre of the square and the side of the square is doubled, explain how the electric flux will be affected.

a) Electric flux:- Electric flux through a given area is designed as the total number of electric lines of force passing normal to that area. ie it can be measure by taking the dot product of electric field  'E' and area 'ds' perpendicular to the field
That is the electric flux

$\phi = \int \vec{E}\cdot \overrightarrow{ds}$
electric flux is a scalar quantity
Construct a cube of 'd' and the charge 'q' is placed at the centre of cube

According to the Gauss’ theorem the electric flux $\phi =\frac{q}{\varepsilon _{0}}$
In the case of the cube, it is the total electric flux through all the six faces of the cube
Therefore flux through the one face of the cube is

$\phi =\frac{1}{6}\times \phi \, total$
$\phi =\frac{q}{6\varepsilon _{0}}$

b) if the charge is moved to a distance d from the centre of the square and the side of the square is double then the cube is constructed of side 2d. The charge is at the centre of the cube. Again flux through the square is $\frac{1}{6}$ times of total electric flux. The total flux does not  change so flux through the square will not change and it remains as $\frac{q}{6\varepsilon _{0}}$

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