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 ....... $N m^{2} C^{-1}$.

Consider a uniform electric field $E = 3 \times 10^{3} \hat{i} \; N/C$.
$(a)$ What is the flux of this field through a square of $10 \; cm$ on a side whose plane is parallel to the $yz$ plane?
$(b)$ What is the flux through the same square if the normal to its plane makes a $60^{\circ}$ angle with the $x$-axis?

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.

The electric flux passing through the cube for the given arrangement of charges placed at the corners of the cube (as shown in the figure) is:

$A$ line charge of length $\frac{a}{2}$ is kept at the center of an edge $BC$ of a cube $ABCDEFGH$ having edge length $a$ as shown in the figure. If the linear charge density is $\lambda \; C/m$, then the total electric flux through all the faces of the cube will be . . . . . . . (Take $\varepsilon_0$ as the free space permittivity)

$A$ point charge $q$ is placed at a distance $a/2$ directly above the center of a square of side $a$. The electric flux through the square is: