Two point charges of magnitudes $-8 \mu C$ and $+32 \mu C$ are separated by a distance of $15 \ cm$ in air. The position of the point from the $-8 \mu C$ charge at which the resultant electric field becomes zero is (in $cm$)

Write the equation for the electric field produced by a point charge. How does it depend on the distance from the charge?

There exists an electric field of magnitude $E$ in the $x$-direction. If the work done in moving a charge of $0.2 \, C$ through a distance of $2 \, m$ along a line making an angle $60^{\circ}$ with the $x$-axis is $4 \, J$,then the value of $E$ is ........ $N / C$.

$A$ circular wire loop of radius $10 \ cm$ carries a total charge of $10^{-5} \ C$ distributed uniformly over its length. $A$ small length of $3.14 \times 10^{-6} \ m$ of wire is cut off. The magnitude of the electric field at the centre due to the remaining wire is:
(Assume $\frac{1}{4 \pi \epsilon_{0}} = 9 \times 10^{9} \ SI$ units) (in $N \ C^{-1}$)

Two charges $+q$ and $-q$ are situated at a certain distance. At the point exactly midway between them:

Two point charges $-5 \mu C$ and $+5 \mu C$ are placed at $A$ and $B$ respectively on a straight line separated by $5 \ cm$. At a point $C$,if the resultant electric field is parallel to the line joining the points $A$ and $B$,then: