Gauss's law can help in easy calculation of electric field due to

A square surface of side $L$ metres is in the plane of the paper. A uniform electric field $\vec E(V/m) $, also in the plane of the paper, is limited only to the lower half of the square surface, (see figure). The electric flux in SI units associated with the surface is

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 .................. $\mathrm{Nm}^2 / \mathrm{C}$

A few electric field lines for a system of two charges $Q_1$ and $Q_2$ fixed at two  different points on the $x$ -axis are shown in the figure. These lines suggest that:-

What will be the total flux through the faces of the cube as in figure with side of length $a$ if a charge $q$ is placed at ?

$(a)$ $A$ $:$ a corner of the cube.

$(b)$ $B$ $:$ midpoint of an edge of the cube.

The figure shows some of the electric field lines corresponding to an electric field. The figure suggests