Question

At what distance of separation, r, must two 3.20 x 10-9 Coulomb charges be positioned in order for the repulsive force between them to be 83.4 N?

Answer 1

I'm not 100% sure but I think the answer is 60.4, because you multiply 3.20 by 10, to get 32 - 9 = 23, then subtract 23 from 83.4. Hope this helps you, and good luck!!!

Answer 2

F = q₁*q₂* C / r²

F =83.4
q₁ = q₂ = 3.2 * 10⁻⁹
C couloumb constant 8.9 * 10⁹

solve for r