Force between $A$ and $B$ is $F$. If $75\%$ charge of $A$ is transferred to $B$ then force between $A$ and $B$ 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 $=1.67 \times 10^{-27}\,kg$ ):

${F_g}$ and ${F_e}$ represents gravitational and electrostatic force respectively between electrons situated at a distance $10\, cm$. The ratio of ${F_g}/{F_e}$ is of the order of

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

A paisa coin is made up of $\mathrm{Al - Mg}$ alloy and weighs $0.75\, g$ It is electrically neutral and contains equal amounts of positive and negative charge of magnitude $34.8$ $\mathrm{kC}$. Suppose that these equal charges were concentrated in two point charges separated by :

$(i)$ $1$ $\mathrm{cm}$ $(\sim \frac{1}{2} \times $ diagonal of the one paisa coin $)$

$(ii)$ $100\,\mathrm{m}$ $(\sim $ length of a long building $)$

$(iii)$ $10^6$ $\mathrm{m}$ (radius of the earth).

Find the force on each such point charge in each of the three cases. What do you conclude from these results ?

The law, governing the force between electric charges is known as