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 ....... $Nm ^{2} C ^{-1}$
$624$
$661$
$620$
$640$
If $\oint_s \vec{E} \cdot \overrightarrow{d S}=0$ over a surface, then:
An electric field is given by $(6 \hat{i}+5 \hat{j}+3 \hat{k}) \ N / C$.
The electric flux through a surface area $30 \hat{\mathrm{i}}\; m^2$ lying in $YZ-$plane (in SI unit) is
An electric charge $q$ is placed at the centre of a cube of side $\alpha $. The electric flux on one of its faces will be
If a spherical conductor comes out from the closed surface of the sphere then total flux emitted from the surface will be
Draw electric field lines when two positive charges are near.