$A$ point charge $+Q$ is positioned at the center of the base of a square pyramid as shown. The flux through one of the four identical upper faces of the pyramid is:

$A$ square surface of side $L$ metres is in the plane of the paper. $A$ uniform electric field $\vec{E} \text{ (V/m)}$,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

Five point charges $\frac{1}{\pi}, \frac{2}{\pi}, \frac{3}{\pi}, \frac{4}{\pi}$ and $\frac{-5}{\pi} \ nC$ are located inside a pyramid. The total electric flux through the surface of the pyramid is

$A$ positive charge $q$ is kept at the center of a thick shell of inner radius $R_1$ and outer radius $R_2$ which is made up of conducting material. If $\phi_1$ is the flux through a closed Gaussian surface $S_1$ whose radius is just less than $R_1$ and $\phi_2$ is the flux through a closed Gaussian surface $S_2$ whose radius is just greater than $R_1$,then:

The electric field in a region is given by $\vec{E}=(2 \hat{i}+4 \hat{j}+6 \hat{k}) \times 10^3 \ N/C$. The flux of the field through a rectangular surface parallel to the $x-z$ plane is $6.0 \ N m^2 C^{-1}$. The area of the surface is . . . . . . $cm^2$.

Four closed surfaces and corresponding charge distributions are shown below. Let the respective electric fluxes through the surfaces be $\phi_1, \phi_2, \phi_3$ and $\phi_4$. Then: