The force between two-point charges is $100N$ in air. Calculate the force if the distance between them is increased by $50\%$.

Hint: Force between two charges is given by coulomb’s law. According to coulomb’s law, force between charges is inversely proportional to the square of the distance between them.
Force between charges can be calculated using this relation.
Formula used:
Force between two charges, $F\propto \dfrac{1}{{{r}^{2}}}$

Complete answer:
Force between two-point charges is given by Coulomb's law. Coulomb’s law states that force between two charges is directly proportional to product of the magnitude of the charges and inversely proportional to square of distance between them. Mathematically, it can be written as
$F\propto \dfrac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}$
On removing the proportionality sign,
$F=k\dfrac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}$
Where $k=\dfrac{1}{4\pi {{\epsilon }_{0}}}=9\times {{10}^{9}}N{{m}^{2}}/{{C}^{2}}$ if charges are placed in air.
Given that force between the charges is $100N$. If distance between the charges is increased by $50\%$, then the new distance becomes $r'=r+0.5r=1.5r$
Since, force between the charges is inversely proportional to separation between them, the force after the distance between them is increase by $50\%$ can be given by
$\dfrac{F'}{F}=\dfrac{{{r}^{2}}}{r{{'}^{2}}}=\dfrac{{{r}^{2}}}{{{(1.5r)}^{2}}}=\dfrac{4}{9}$

This implies that
$F'=\dfrac{4}{9}F$

Additional Information:
Electric charges are of two types: positive and negative. If both charges are of opposite nature then the coulomb force between them is attractive. Coulomb force between them is repulsive if charges are of same nature i.e. positive-positive or negative-negative.

Note:
Electric charges are of two types: positive and negative. Coulomb force between two-point charges is attractive if both charges are of opposite sign and is repulsive if both charges are of same sign. If force is negative, this implies that the force is attractive.