Force of attraction of repulsion between two charges bodies is given by Coulomb's law as,
[tex]F = \frac{kq_1q_2}{r^2}[/tex]
k = Coulomb's Constant
[tex]q_{1,2} =[/tex] Charge of the bodies
r = Distance
If the distance r is increased by 3 times then new force [tex]F_1[/tex] between charges is
[tex]F_1 = \frac{kq_1q_2}{(3r)^2}[/tex]
[tex]F_1 = \frac{kq_1q_2}{9r^2}[/tex]
[tex]F_1 = \frac{1}{9} \frac{kq_1q_2}{r^2}[/tex]
The first force is given, then if we again replace the force we have that,
[tex]F_1 = \frac{1}{9} F[/tex]
[tex]F_1 = \frac{1}{9} 5[/tex]
[tex]F_1 = 0.556N[/tex]
Therefore the final force will be 0.556N
