If the distance between two equal point charges is doubled,what would happen to the force between them?

Two copper balls,each weighing $10\,g$,are kept in air $10\,cm$ apart. If one electron from every $10^6$ atoms is transferred from one ball to the other,the Coulomb force between them is (atomic weight of copper is $63.5$).

$ABC$ 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.......$dynes$.

Four charges,each of magnitude $+Q$,are placed at the corners of a square of side $a$. What is the net force on one of the charges?

Two similar tiny balls of mass $m$,each carrying charge $q$,are hung from silk threads of length $l$ as shown in the figure. These are separated by a distance $x$ and the angle $2\theta$ is small. Then,for equilibrium:

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: