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 $......... \text{ V cm}$.

$A$ charge $Q \; \mu C$ is placed at the centre of a cube. The flux coming out from any one of its faces will be:

An electric field $\overrightarrow{E} = 4x \hat{i} - (y^2 + 1) \hat{j} \text{ N/C}$ passes through the box shown in the figure. The flux of the electric field through surfaces $ABCD$ and $BCGF$ are marked as $\phi_I$ and $\phi_{II}$ respectively. The difference $(\phi_I - \phi_{II})$ is (in $\text{Nm}^2/C$):

Three charges $q_1 = 1\,\mu C, q_2 = 2\,\mu C$,and $q_3 = -3\,\mu C$ and four surfaces $S_1, S_2, S_3$,and $S_4$ are shown in the figure. The electric flux emerging through surface $S_2$ in $N\cdot m^2/C$ is:

Gauss's law states that

What is called a Gaussian surface?