The law, governing the force between electric charges is known as

Two identical charged spheres suspended from a common point by two massless strings of lengths $l,$ are initially at a distance $d\;(d < < l)$ apart because of their mutual repulsion. The charges begin to leak from both the spheres at a constant rate. As a result, the spheres approach each other with a velocity $v.$ Then $v$ varies as a function of the distance $x$ between the spheres, as 

Two point charges $3 \times 10^{-6} \,C$ and  $8 \times 10^{-6} \, C$ repel each other by a force of  $6 \times 10^{-3} \, N$. If each of them is given an additional charge $-6 \times 10^{-6} \, C$, the force between them will be

Two small conducting spheres of equal radius have charges $ + 10\,\mu C$ and $ - 20\,\mu C$ respectively and placed at a distance $R$ from each other experience force ${F_1}$. If they are brought in contact and separated to the same distance, they experience force ${F_2}$. The ratio of ${F_1}$ to ${F_2}$ is

In a medium, the force of attraction between two point charges, distance $d$ apart, is $F$. What distance apart should these point charges be kept in the same medium, so that the force between them becomes $16\, F$ ?

In given diagram. Find distance of neutral point from particle of charge $e$ is......$cm$