Question

Two point-charges exert a 5 N force on each other. What will the force become if the distance between them is increased by a factor of three?

Answer 1

Force of attraction of repulsion between two charges bodies is given by Coulomb's law as,

`F = (kq_1q_2)/(r^2)`

k = Coulomb's Constant

`q_(1,2) =` Charge of the bodies

r = Distance

If the distance r is increased by 3 times then new force
`F_1` between charges is

`F_1 = (kq_1q_2)/((3r)^2)`

`F_1 = (kq_1q_2)/(9r^2)`

`F_1 = (1)/(9) (kq_1q_2)/(r^2)`

The first force is given, then if we again replace the force we have that,

`F_1 = (1)/(9) F`

`F_1 = (1)/(9) 5`

`F_1 = 0.556N`

Therefore the final force will be 0.556N