The law governing the force between electric charges is known as

Two identical conducting spheres carry identical charges. If the spheres are set at a certain distance apart,they repel each other with a force $F$. $A$ third conducting sphere identical to the other two,but initially uncharged,is touched to one sphere and then to the other before being removed. The force between the original two spheres is now

Two charges of equal magnitudes and at a distance $r$ exert a force $F$ on each other. If the charges are halved and the distance between them is doubled,then the new force acting on each charge is

The ratio of Coulomb's electrostatic force to the gravitational force between an electron and a proton separated by some distance is $2.4 \times 10^{39}$. The ratio of the proportionality constant $K = \frac{1}{4 \pi \varepsilon_0}$ to the gravitational constant $G$ is nearly (Given that the charge of the proton and electron each $= 1.6 \times 10^{-19} \; C$,the mass of the electron $= 9.11 \times 10^{-31} \; kg$,the mass of the proton $= 1.67 \times 10^{-27} \; kg$):

Three point charges of magnitude $5 \mu C$,$0.16 \mu C$,and $0.3 \mu C$ are located at the vertices $A$,$B$,and $C$ of a right-angled triangle whose sides are $AB = 3 \, cm$,$BC = 3 \sqrt{2} \, cm$,and $CA = 3 \, cm$. Point $A$ is the right-angle corner. Calculate the magnitude of the net electrostatic force (in $N$) experienced by the charge at point $A$ due to the other two charges.

Four charges equal to $-Q$ are placed at the four corners of a square of side length $a$,and a charge $q$ is at its centre. If the system is in equilibrium,the value of $q$ is: