The electrostatic force between two charges located 2 meters apart is 40 N. What will the force be between these charges when they are located one-third meters apart?

Answer 1

F_a=kq₁q₂/r_a²

and F_b=kq₁q₂/r_b²

F_a = 40 N

r_a = 2 m

kq₁q₂ = constant

r_b = 0.33 m

F_ar_a² = kq₁q₂ = constant = F_br_b²

F_ar_a² = F_br_b²

40*2² = F_b*0.33²

F_b = 40 * 2² / 0.33² = 40 * 4 / 0.11 = 1,454 N