Gauss’s law should be invalid if

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.

It is not convenient to use a spherical Gaussian surface to find the electric field due to an electric dipole using Gauss’s theorem because

Two surfaces $S_1$ and $S_2$ are shown in figure. Flux associated with $S_1$ is ${\phi _1}$ and $S_2$ is ${\phi _2}$. Which is correct ?

An electric field is given by $(6 \hat{i}+5 \hat{j}+3 \hat{k}) \ N / C$.

The electric flux through a surface area $30 \hat{\mathrm{i}}\; m^2$ lying in $YZ-$plane (in SI unit) is

A square surface of side $L$ meter in the plane of the paper is placed in a uniform electric field $E(volt/m)$ acting along the same plane at an angle $\theta$ with the horizontal side of the square as shown in figure.The electric flux linked to the surface, in units of $volt \;m $