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 ....... $Nm ^{2} C ^{-1}$

Careful measurement of the electric field at the surface of a black box indicates that the net outward flux through the surface of the box is $8.0 \times 10^{3} \;Nm ^{2} / C .$

$(a)$ What is the net charge inside the box?

$(b)$ If the net outward flux through the surface of the box were zero, could you conclude that there were no charges inside the box? Why or Why not?

Five charges $+q,+5 q,-2 q,+3 q$ and $-4 q$ are situated as shown in the figure.

The electric flux due to this configuration through the surface $S$ is

Discuss some points about Gauss’s law.

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 

What is the direction of electric field intensity ?