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 $E _{0}=4.0 \times 10^{3}\, \frac{ N }{ C } .$ The flux of this field through a rectangular surface area $0.4 \,m ^{2}$ parallel to the $Y - Z$ plane is ....... $Nm ^{2} C ^{-1}$

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. . . . . .

Linear charge density of wire is $8.85\,\mu C/m$ . Radius and height of the cylinder are $3\,m$ and $4\,m$ . Then find the flux passing through the cylinder

A few electric field lines for a system of two charges $Q_1$ and $Q_2$ fixed at two  different points on the $x$ -axis are shown in the figure. These lines suggest that:-

Discuss some points about Gauss’s law.

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