If two particles $A$ and $B$ of charges $1.6 \times 10^{-19} \ C$ and $3.2 \times 10^{-19} \ C$ respectively are separated by a distance of $3 \ cm$ in air,then the magnitude of electrostatic force on particle $A$ due to particle $B$ is

Two balls of same mass and carrying equal charge are hung from a fixed support of length $l$. At electrostatic equilibrium,assuming that the angle made by each thread is small,the separation $x$ between the balls is proportional to

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 identical charged conducting spheres $A$ and $B$ have their centres separated by a certain distance. Charge on each sphere is $q$ and the force of repulsion between them is $F$. $A$ third identical uncharged conducting sphere $C$ is brought in contact with sphere $A$ first and then with $B$ and finally removed from both. The new force of repulsion between spheres $A$ and $B$ (radii of $A$ and $B$ are negligible compared to the distance of separation so that for calculating force between them they can be considered as point charges) is best given as:

Two point charges $q_1$ and $q_2$ are $l$ distance apart. If one of the charges is doubled and the distance between them is halved, the magnitude of the force becomes $n$ times, where $n$ is:

Three point charges $q$,$-2q$,and $2q$ are placed on the $x$-axis at distances $x=0$,$x=\frac{3}{4}R$,and $x=R$ respectively from the origin as shown. If $q = 2 \times 10^{-6} \, C$ and $R = 2 \, cm$,the magnitude of the net force experienced by the charge $-2q$ is .......... $N$.