Two-point charges +2 C and +6 C separated by distance ' r ' repel each other with a force of 18 N . If negative charge of 4 C is added to each of these charges, then the force experienced by them is

The initial charges are $q_1 = +2\text{ C}$ and $q_2 = +6\text{ C}$, separated by distance $r$. The initial repulsive force is $F_1 = 18\text{ N}$.

Using Coulomb's Law, $F_1 = k \frac{|q_1 q_2|}{r^2} = k \frac{12}{r^2}$, giving $18 = k \frac{12}{r^2}$.

Adding $-4\text{ C}$ to each charge yields $q_1' = -2\text{ C}$ and $q_2' = +2\text{ C}$. The new force is $F_2 = k \frac{|(-2)(2)|}{r^2} = k \frac{4}{r^2}$.

Taking the ratio of the forces, $\frac{F_2}{18} = \frac{4}{12} = \frac{1}{3}$, so $F_2 = \frac{18}{3} = 6\text{ N}$. The charges are opposite in sign, resulting in an attractive force.