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 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 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.

$(a)$ Two insulated charged copper spheres $A$ and $B$ have their centres separated by a distance of $50 \; cm$. What is the mutual force of electrostatic repulsion if the charge on each is $6.5 \times 10^{-7} \; C$? The radii of $A$ and $B$ are negligible compared to the distance of separation.
$(b)$ What is the force of repulsion if each sphere is charged double the above amount,and the distance between them is halved?

The dimensional formula of $k$ (Coulomb's constant) is . . . . . . .
(Take $I$ as the dimension of current.)

$ABC$ is a right-angled triangle in which $AB = 3\,cm$ and $BC = 4\,cm$. And $\angle ABC = \pi / 2$. The three charges $+15\,e.s.u.$,$+12\,e.s.u.$,and $-20\,e.s.u.$ are placed respectively on $A$,$B$,and $C$. The force acting on $B$ is.......$dynes$.