Two identical particles of mass $m$ and charge $q$ are shot at each other from a very great distance with an initial speed $v$. The distance of closest approach of these charges is

Three point charges of magnitude $5 \mu C$,$0.16 \mu C$,and $0.3 \mu C$ are located at the vertices $A$,$B$,and $C$ of a right-angled triangle whose sides are $AB = 3 \, cm$,$BC = 3 \sqrt{2} \, cm$,and $CA = 3 \, cm$. Point $A$ is the right-angle corner. Calculate the magnitude of the net electrostatic force (in $N$) experienced by the charge at point $A$ due to the other two charges.

Three charges $4q, Q$ and $q$ are in a straight line along the $x$-axis at the positions $x=0 \ m$,$x=\frac{1}{2} \ m$,and $x=1 \ m$ respectively. The resultant force on $q$ will be zero if $Q$ is equal to:

$A$ charge of $1 \mu C$ is divided into two parts such that their charges are in the ratio of $2: 3$. These two charges are kept at a distance of $1 \ m$ apart in vacuum. Then,the electric force between them (in $N$) is

$A$ proton of mass $m$ and charge $e$ is projected from a very large distance towards an $\alpha$-particle with velocity $v$. Initially,the $\alpha$-particle is at rest,but it is free to move. If gravity is neglected,then the minimum separation along the straight line of their motion will be:

$F$ is the force between two identical charged particles placed at a distance $Y$ from each other. If the distance between the charges is reduced to half the previous distance,then the force between them becomes: