The electric flux linked with the closed surface in $N m^2 C^{-1}$ is given by:
$\left(\varepsilon_0 = 8.85 \times 10^{-12} C^2 N^{-1} m^{-2}\right)$

The figure shows some of the electric field lines corresponding to an electric field. The figure suggests

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

$A$ metallic solid sphere is placed in a uniform electric field. The lines of force follow the path$(s)$ shown in the figure as:

Eight dipoles of charges of magnitude $e$ are placed inside a cube. The total electric flux coming out of the cube will be

The electric field in a region is given by $\overrightarrow{E} = \frac{2}{5} E_{0} \hat{i} + \frac{3}{5} E_{0} \hat{j}$ with $E_{0} = 4.0 \times 10^{3} \, N/C$. The flux of this field through a rectangular surface area $0.4 \, m^{2}$ parallel to the $Y-Z$ plane is ....... $N m^{2} C^{-1}$.