F = (kq1q2)/r^2
0.078 N = (9 * 10^9 N*m^2/C^2)(3.7 * 10^-6 C)(q2)/(1.3 m)^2
q2 = 0.078 N * (1.3 m)^2/((9 * 10^9 N*m^2/C^2)(3.7 * 10^-6 C))
q2 = _______ C
Two charges, separated by a distance of 1.3 m, attract each other with a force of 0.078
N. If one charge is +3.7 μC, what is the other charge?
2 answers
Also note that q2 will be negative because the force is attraction.