Answer:
Force between two charges is 0.02 N when distance is increased by factor of 3
Explanation:
As we know that the two charges will attract or repel each other when they are placed near to each other
Here we know that electrostatic force is given as
[tex]F = \frac{kq_1 q_2}{r^2}[/tex]
here we know that two charges exert 0.18 N force on each other when they are placed at distance "r" from each other
Now when same charges are placed distance "3r" then we have
[tex]F' = \frac{kq_1q_2}{(3r)^2}[/tex]
so we have
[tex]F' = \frac{F}{9}[/tex]
[tex]F' = \frac{0.18}{9}[/tex]
[tex]F' = 0.02 N[/tex]
