$A$ cube of side $l$ is placed in a uniform electric field $E$,where $E = E\hat{i}$. The net electric flux through the cube is:

An electrostatic field line leaves at an angle $\alpha$ from a point charge $q_{1}$ and connects with a point charge $-q_{2}$ at an angle $\beta$ ($q_{1}$ and $q_{2}$ are positive). See the figure below. If $q_{2} = \frac{3}{2} q_{1}$ and $\alpha = 30^{\circ}$,then:

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

The electric field in a region is uniform and is given by $\vec{E} = a \hat{i} + b \hat{j} + c \hat{k}$. The electric flux associated with a surface of area $\vec{A} = \pi R^2 \hat{i}$ is:

Electric charge is uniformly distributed along a long straight wire of radius $1\, mm$. The charge per $cm$ length of the wire is $Q$ $coulomb$. Another cylindrical surface of radius $50\, cm$ and length $1\, m$ symmetrically encloses the wire as shown in the figure. The total electric flux passing through the cylindrical surface is

Find the electric flux through the shaded face $BCGF$ of the cube,as shown in the figure.