Coulomb's Law Question: Please Help?

Question

A 0.50 kg uniform bar with a length of 1.00 m has a fulcrum placed 40.0 cm from the left end of the bar. Also located from the left end of the bar are a 1.0 kg mass (m1) at 10.0 cm and a 2.0 kg mass (m2) at the 90.0 cm mark. Mass m2 also posses a charge (q2) of 4C. A charge q3 is placed 20.0 cm below the beam and holds the bar in rotational equilibrium parallel to the ground. Ignoring the polarization effects of the charges on the bar, what is the charge of q3?

I got that Fg=FE so I solved for q1: 8.7*(10^-5)C

But I'm stuck now..please help

Answer 1

Assume that q3 is directly under m2 (or q2)

Equilibrium equation

r1m1g -r2m2g+r2f(electric from q3 onto q2)

Coulumb's Law

f(electric from q3 onto q2)=kq2q3/(R^2)

form eq1 and 2

kq2q3/(R^2) = (r2m2g - r1m1g )/r2
q3=(r2m2 - r1m1 ) g R^2/(r2 k q2)
q3=(.50 x 2.0 - 30 x 1.0) 9.81x (.2)^2 /(.50 x 8.988 x e+9 x 4 e-6)
q3= 15.3 uC