Question

There are 2 negative charges, both sides are -3.0 C with a force of 19.2 N, what is the distance ?

Answer 1

F=K×q1×q2/r^2 ,,, 19.2=9×10^9×3×3/r^2 then r^2=9×910^9/19.2 ,,, r=√(9×9×10^8/1.92 then r=9×10^4/1.38 then r=6.58×10^4m