$A$ negatively charged oil drop is prevented from falling under gravity by applying a vertical electric field of $100 \ V m^{-1}$. If the mass of the drop is $1.6 \times 10^{-3} \ g$,the number of electrons contained in the drop 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:

The force between two identical spheres charged with the same charge $q$ is $F$. If $50\%$ of the charge from one sphere is transferred to the second sphere,then the new force will be:

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 charges $-q_1$,$+q_2$,and $-q_3$ are placed as shown in the figure. The $x$-component of the force on $-q_1$ is proportional to

Two point charges of equal magnitude but opposite sign are placed at a certain distance. The force between them is $F$. If $75\%$ of the charge from one is transferred to the other,what will be the new force between them?