Two charges,each of $1\;C$,are at a distance of $1\;km$ apart. The force between them is:

In a right-angled triangle $ABC$,$AB = 3 \ cm$,$BC = 4 \ cm$,and $\angle ABC = \frac{\pi}{2}$. Charges of $+15 \ e.s.u.$,$+12 \ e.s.u.$,and $-20 \ e.s.u.$ are placed at $A$,$B$,and $C$ respectively. What is the net force on the charge at $B$ in $dynes$?

The force between two point charges $A$ and $B$ is $F$. If $75\%$ of the charge of $A$ is transferred to $B$,then the new force between $A$ and $B$ is:

$N$ point charges are distributed into two groups and separated by a fixed distance. Then the ratio of maximum to minimum forces between the two groups is ($N$ is even and greater than $2$).

The ratio of gravitational force and electrostatic repulsive force between two electrons is approximately (gravitational constant $= 6.7 \times 10^{-11} \, Nm^2/kg^2$,mass of an electron $= 9.1 \times 10^{-31} \, kg$,charge on an electron $= 1.6 \times 10^{-19} \, C$).

Three charges $4q, Q$ and $q$ are placed at positions $0, \frac{l}{2}$ and $l$ respectively along a straight line. If the resultant force on $q$ is zero,then $Q$ is equal to