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

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 $= 1.67 \times 10^{-27} \; kg$):

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

Three charges are placed at the vertices of an equilateral triangle of side '$a$' as shown in the figure. The force experienced by the charge placed at the vertex $A$ in a direction normal to $BC$ is

Two small spheres,each carrying a charge of $+q$,are connected by an insulating string of length $2a$. What is the tension in the string?

Two charges $q$ and $-3q$ are placed fixed on the $x$-axis separated by a distance $d$. Where should a third charge $2q$ be placed such that it will not experience any force?