Question

If the distance between two positive point charges is tripled, then the strength of the electrostatic repulsion between them will decrease by a factor of of how much?

Answer 1

Coulomb's Law:

`F=(kq_1q_2)/(r^2)`
k is a constant
q1 and q2 are charges
r is the distance



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

the distance is tripled:

`F_2=(kq_1q_2)/((3r)^2)=(kq_1q_2)/(9r^2)`



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

The strength of the electrostatic repulsion will decrease by a factor of 9.