In a region, the intensity of an electric field is given by $E = 2i + 3j + k$ in $NC^{-1}$. The electric flux through a surface $S = 10i \ m^2$ in the region is

The electric field in a region is given by $\overrightarrow{E} = (\frac{3}{5} E_{0} \hat{i} + \frac{4}{5} E_{0} \hat{j}) \, N/C$. The ratio of the flux of this field through a rectangular surface of area $0.2 \, m^{2}$ (parallel to the $y-z$ plane) to that through a surface of area $0.3 \, m^{2}$ (parallel to the $x-z$ plane) is $a : b$,where $a = \dots$ [Here $\hat{i}, \hat{j}$ and $\hat{k}$ are unit vectors along the $x, y$ and $z$-axes respectively].

$A$ charge $q$ is placed inside a hollow cylinder. The electric flux associated with the curved surface $B$ is $\phi$. The flux associated with a plane surface (either $A$ or $C$) will be:

$A$ sphere encloses an electric dipole with charges $\pm 3 \times 10^{-6} \; C$. What is the total electric flux across the sphere in $\text{N m}^2 / \text{C}$?

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

$A$ square surface of side $L$ meters is in the plane of the paper. $A$ uniform electric field $\vec{E} \text{ (V/m)}$ is also in the plane of the paper,limited only to the lower half of the square surface. The electric flux associated with the surface in $SI$ units is: