A point charge q is placed at a distance r from another point charge Q. If the distance between the charges is doubled then:

the force between the charges becomes one-fourth and the potential energy becomes half.

The correct answer is Option (2) → the force between the charges becomes one-fourth and the potential energy becomes half.

Using Coloumb's law,

$F=\frac{K|qQ|}{r^2}$

when $r'=2r$

$F'=\frac{K|qQ|}{(2r)^2}=\frac{F}{4}$

and,

$U'=\frac{KqQ}{2r}=\frac{U}{2}$ [U = Potential energy = $\frac{KqQ}{r}$]