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?

In Millikan's oil drop experiment,an oil drop carrying a charge $Q$ is held stationary by a potential difference of $2400\,V$ between the plates. To keep a drop of half the radius stationary,the potential difference had to be made $600\,V$. What is the charge on the second drop?

Two electric charges are arranged as shown in the figure. Where should a third charge be placed so that it remains in equilibrium?

$A$ point charge $q_1 = 4 q_0$ is placed at the origin. Another point charge $q_2 = -q_0$ is placed at $x = 12 \, cm$. The charge of a proton is $q_0$. $A$ proton is placed on the $x$-axis such that the net electrostatic force on it is zero. In this situation,the position of the proton from the origin is $.......... \, cm$.

Two charges each of charge $+10 \mu C$ are kept on $Y$-axis at $y=-a$ and $y=+a$, respectively. Another point charge $-20 \mu C$ is placed at the origin and given a small displacement $x$ $(x \ll a)$ along $X$-axis. The force acting on the point charge is ($x$ and $a$ are in metres, $\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \text{ N m}^2 \text{ C}^{-2}$)

$A$ charge $Q$ is to be divided between two objects. The values of the charges on the objects so that the electrostatic force between them will be maximum is