An electric field,$\overrightarrow{E} = \frac{2 \hat{i} + 6 \hat{j} + 8 \hat{k}}{\sqrt{6}} \ V/m$,passes through a surface of $4 \ m^2$ area having a unit normal vector $\hat{n} = \left( \frac{2 \hat{i} + \hat{j} + \hat{k}}{\sqrt{6}} \right)$. The electric flux through that surface is:

The linear charge density of a wire is $8.85 \ \mu C/m$. The radius and height of the cylinder are $3 \ m$ and $4 \ m$ respectively. Find the electric flux passing through the cylinder.

$A$ cubical Gaussian surface has a side of length $a = 10 \,cm$. Electric field lines are parallel to the $X$-axis as shown in the figure. The magnitudes of the electric fields through surfaces $ABCD$ and $EFGH$ are $6 \,kNC^{-1}$ and $9 \,kNC^{-1}$ respectively. Then,the total charge enclosed by the cube is (Take $\varepsilon_0 = 9 \times 10^{-12} \,Fm^{-1}$): (in $\,nC$)

Shown below is a distribution of charges. The flux of electric field due to these charges through the surface $S$ is

$A$ point charge $q$ is placed at the center $O$ of a cube of side length $L$. Another identical charge $q$ is placed at a distance $L$ from $O$ as shown in the figure. The electric flux through the face $ABCD$ is:

The figure shows the electric field lines. The spacing between the lines is parallel to the paper at every point. If the magnitude of the field at $A$ is $40 \ N/C$,then the approximate magnitude of the field at $B$ is ....... $N/C$.