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 . . . . . . $cm^2$.

The flux of the electric field $\overrightarrow{E} = 24 \hat{i} + 30 \hat{j} + 28 \hat{k} \text{ NC}^{-1}$ through an area of $20 \text{ m}^2$ on the $yz$ plane is

$A$ hollow cylinder has a charge $q$ inside it. If $\phi$ is the electric flux associated with the curved surface $B$,the flux linked with the plane surface $A$ will be

$A$ circular disc of radius $R$ carries surface charge density $\sigma(r) = \sigma_0 \left(1 - \frac{r}{R}\right)$,where $\sigma_0$ is a constant and $r$ is the distance from the center of the disc. Electric flux through a large spherical surface that encloses the charged disc completely is $\phi_0$. Electric flux through another spherical surface of radius $\frac{R}{4}$ and concentric with the disc is $\phi$. Then the ratio $\frac{\phi_0}{\phi}$ is:

What kind of Gaussian surface is used to calculate the electric field due to the charge distribution?

The electric field in a region of space is given by $\vec{E} = (5\hat{i} + 2\hat{j}) \text{ N/C}$. Calculate the electric flux through a surface of area $2 \text{ m}^2$ lying in the $YZ$-plane.