There are two metallic spheres of same radii but one is solid and the other is hollow, then

Force between two identical spheres charged with same charge is $F$. If $75\%$ charge of one sphere is transfered to the other sphere then the new force will be

Two charges, each equal to $q$, are kept at $x = -a$ and $x = a$ on the $x-$axis. A particle of mass $m$ and charge $q_0=\frac{q}{2}$ is placed at the origin. If charge $q_0$ is given a small displacement $(y < < a)$ along the $y-$axis, the net force acting on the particle is proportional to

Two charges $q$ and $-3q$ are placed fixed on $x-axis$ separated by distance $'d'$. Where should a third charge $2q$ be placed such that it will not experience any force ?

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 free point charges $+q$ and $+4q$ are a distance $R$ apart. $A$ third charge is so  placed that the entire system is in equilibrium. Then the third charge is :-