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

Two identical tennis balls each having mass $m$ and charge $q$ are suspended from a fixed point by threads of length $l$. What is the equilibrium separation when each thread makes a small angle $\theta$ with the vertical?

$A$ charge $q$ is placed at the centre of the line joining two equal charges $Q$. The system of the three charges will be in equilibrium if $q$ is equal to:

Two electrons are moving towards each other,each with a velocity of $10^6 \, m/s$. What will be the closest distance of approach between them in $m$?

Two identical charged spheres are suspended by strings of equal lengths. The strings make an angle of $37^{\circ}$ with each other. When suspended in a liquid of density $0.7 \text{ g/cm}^3$,the angle remains the same. If the density of the material of the sphere is $1.4 \text{ g/cm}^3$,the dielectric constant of the liquid is . . . . . . $\left(\tan 37^{\circ} = \frac{3}{4}\right)$.

If two charges $q_1$ and $q_2$ are separated by a distance '$d$' and placed in a medium of dielectric constant $K$,what will be the equivalent distance between the charges in air for the same electrostatic force?