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:

$A$ 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:

$A$ metallic sphere is kept in between two oppositely charged plates. The most appropriate representation of the field lines is

Draw the electric field lines of a positive point charge.

The charge distribution is shown in the figure below. The electric flux through the surface $S$ due to these charges is .........

$A$ metallic spherical shell has an inner radius $R_1$ and outer radius $R_2$. $A$ charge $Q$ is placed at the centre of the spherical cavity. What will be the surface charge density on the inner surface?