When the distance between the charged particles is halved,the force between them becomes

The law governing the force between electric charges is known as

Two unit negative charges are placed on a straight line. $A$ positive charge $q$ is placed exactly at the midpoint between these unit charges. If the system of these three charges is in equilibrium,the value of $q$ (in $C$) is

Two pith balls carrying equal charges are suspended from a common point by strings of equal length. The equilibrium separation between them is $r$. Now,the strings are rigidly clamped at half the height. The equilibrium separation between the balls now becomes:

Three equal charges $+q$ are placed at the three vertices of an equilateral triangle centered at the origin. They are held in equilibrium by a restoring force of magnitude $F(r) = k r$ directed towards the origin,where $k$ is a constant. What is the distance of the three charges from the origin?

The distance between charges $5 \times 10^{-11} \, C$ and $-2.7 \times 10^{-11} \, C$ is $0.2 \, m$. The distance at which a third charge should be placed in order that it will not experience any force along the line joining the two charges is . . . . . . $m$.