An electric field $\overrightarrow{\mathrm{E}}=(2 \mathrm{xi}) \mathrm{NC}^{-1}$ exists in space. $\mathrm{A}$ cube of side $2 \mathrm{~m}$ is placed in the space as per figure given below. The electric flux through the cube is .................. $\mathrm{Nm}^2 / \mathrm{C}$

The $S.I.$ unit of electric flux is

Eight dipoles of charges of magnitude $e$ are placed inside a cube. The total electric flux coming out of the cube will be

For a closed surface $\oint {\overrightarrow {E \cdot } } \,\overrightarrow {ds} \,\, = \,\,0$, then 

Given below are two statement: one is labelled as Assertion $A$ and the other is labelled as Reason $R$.

Assertion $A:$ If an electric dipole of dipole moment $30 \times 10^{-5}\,Cm$ is enclosed by a closed surface, the net flux coming out of the surface will be zero.

Reason $R$ : Electric dipole consists of two equal and opposite charges.

In the light of above, statements, choose the correct answer from the options given below:

An infinitely long uniform line charge distribution of charge per unit length $\lambda$ lies parallel to the $y$-axis in the $y-z$ plane at $z=\frac{\sqrt{3}}{2} a$ (see figure). If the magnitude of the flux of the electric field through the rectangular surface $A B C D$ lying in the $x-y$ plane with its center at the origin is $\frac{\lambda L }{ n \varepsilon_0}\left(\varepsilon_0=\right.$ permittivity of free space $)$, then the value of $n$ is