If the Coulombian force acting between two protons separated by a distance $r$ is $F$, what would be the force acting between two alpha particles separated by a distance $2r$?

Two spheres carrying charges $+6 \mu C$ and $+9 \mu C$ are separated by a distance $d$ and experience a force of repulsion $F$. When a charge of $-3 \mu C$ is added to both spheres and they are kept at the same distance as before,the new force of repulsion is:

$A$ charge $Q$ is divided into two parts $q$ and $Q - q$. If the Coulomb repulsion between them when they are separated by a distance $r$ is to be maximum,the ratio of $\frac{Q}{q}$ should be:

The unit of permittivity of free space ${\varepsilon _0}$ is

Two small conducting balls of identical mass $20 \text{ g}$ and identical charge $10^{-10} \text{ C}$ hang from non-conducting threads of length $L = 300 \text{ cm}$. If the equilibrium separation of the balls is $x$ and $x \ll L$,then the magnitude of $x$ is (Assume $4 \pi \varepsilon_0 = \frac{1}{9 \times 10^9} \text{ F/m}$ and $g = 10 \text{ m/s}^2$):

$ABC$ is a right-angled triangle in which $AB = 3 \ cm$,$BC = 4 \ cm$ and the right angle is at $B$. Three charges $+15 \ \mu C$,$+12 \ \mu C$ and $-20 \ \mu C$ are placed respectively at $A$,$B$ and $C$. The force acting on the charge at $B$ is (in $N$)