Question

There is an attractive force 3.5N between two point charges.what will the force be if the distance is increased to 5 times its oeiginal value?

Answer 1

0.14 N

Explanation:

Force acting initially is to `{(1/K)*(q1*q2)/r^2}`=3.5=F
On making the distance 5 times new force becomes
`{(1/K)*(q1*q2)/(5r)^2}`
i.e `1/25*F`
i.e `3.5/25` N or 0.14 N