Given two stationary charged point particles -as in your problem-, the force they feel is expressed by Coulomb's Law, written as:
[tex]F= k\frac{q_1q_2}{r^2} [/tex]
In which [tex]q_1[/tex] is the charge of one of the particles, [tex]q_2[/tex] is the charge of the other particle, [tex]r[/tex] is the distance between the particles, and [tex]k[/tex] is simply a constant.
From Coulomb's Law, lets assume [tex]q_1[/tex] and [tex]r[/tex] are doubled, what happens to [tex]F[/tex]?
[tex]F=k\frac{q_1q_2}{r^2}[/tex]→
[tex]k\frac{2q_1q_2}{(2r)^2}=k\frac{2q_1q_2}{4r^2}= \frac{1}{2} k\frac{q_1q_2}{r^2}= \frac{1}{2}F [/tex]
So, the force is halved when you double the charge of one particle and the distance.
The answer is -2.
