The electrical force between two charges is given by
[tex]F=k \frac{q_1 q_2}{r^2} [/tex]
where
k is the Coulomb's constant
q1 and q2 are the two charges
r is the distance between the two charges
We see that the magnitude of the force is inversely proportional to the square of the distance. Therefore, if the distance is increased to 3 times the original distance:
[tex]r'=3 r[/tex]
the magnitude of the force will change as
[tex]F' \sim \frac{1}{(3r)^2}= \frac{1}{9 r^2} = \frac{1}{9} F [/tex]
So, the correct answer is it will decrease to one-ninth the original force.
