Figure shows the electric lines of force emerging from a charged body. If the electric field at $A$ and $B$ are ${E_A}$ and ${E_B}$ respectively and if the displacement between $A$ and $B$ is $r$ then

$Assertion\,(A):$ A charge $q$ is placed on a height $h / 4$ above the centre of a square of side b. The flux associated with the square is independent of side length.

$Reason\,(R):$ Gauss's law is independent of size of the Gaussian surface.

When the electric flux associated with closed surface becomes positive, zero or negative ?

The electric field in a region is given $\overrightarrow{ E }=\left(\frac{3}{5} E _{0} \hat{ i }+\frac{4}{5} E _{0} \hat{ j }\right) \frac{ N }{ C } .$ The ratio of flux of reported field through the rectangular surface of area $0.2\, m ^{2}$ (parallel to $y - z$ plane) to that of the surface of area $0.3\, m ^{2}$ (parallel to $x - z$ plane $)$ is $a : b ,$ where $a =$ .............

[Here $\hat{ i }, \hat{ j }$ and $\hat{ k }$ are unit vectors along $x , y$ and $z-$axes respectively]

Assertion : Four point charges $q_1,$ $q_2$, $q_3$ and $q_4$ are as shown in figure. The flux over the shown Gaussian surface depends only on charges $q_1$ and $q_2$.

Reason : Electric field at all points on Gaussian surface depends only on charges $q_1$ and $q_2$ .

Gauss’s law states that