An electric charge $q$ is placed at the centre of a cube of side $a$. The electric flux through one of its faces will be

$A$ square of side $20 \ cm$ is enclosed by a spherical surface of radius $80 \ cm$. The centers of the square and the sphere are the same. Four charges $2 \times 10^{-6} \ C, -5 \times 10^{-6} \ C, -3 \times 10^{-6} \ C$,and $6 \times 10^{-6} \ C$ are placed at the four corners of the square. The total flux coming out of the spherical surface in $N \cdot m^2/C$ is:

$A$ charge $q$ is located at the centre of a cube. The electric flux through any face is

In the figure,a $+Q$ charge is located at one of the corners of the cube. What is the electric flux through the cube due to the $+Q$ charge?

$A$ disk of radius $a/4$ having a uniformly distributed charge $6 \text{ C}$ is placed in the $x-y$ plane with its centre at $(-a/2, 0, 0)$. $A$ rod of length $a$ carrying a uniformly distributed charge $8 \text{ C}$ is placed on the $x$-axis from $x = a/4$ to $x = 5a/4$. Two point charges $-7 \text{ C}$ and $3 \text{ C}$ are placed at $(a/4, -a/4, 0)$ and $(-3a/4, 3a/4, 0)$,respectively. Consider a cubical surface formed by six surfaces $x = \pm a/2, y = \pm a/2, z = \pm a/2$. The electric flux through this cubical surface is

The expression for an electric field is given by $\vec{E} = 4000 x^2 \hat{i} \text{ V/m}$. The electric flux through the cube of side $20 \text{ cm}$ when placed in the electric field (as shown in the figure) is $......... \text{ V cm}$.