College Physics question about charged particles!! 5 stars!?

Question

Three positively charged particles are fixed on an x axis. Particles B and C are so close to each other that they can be considered to be at the same distance from particle A. The net force on particle A due to particles B and C is 2.014 x 10^-23 Newtons in the negative direction of the x axis.

In the next figure, particle B has been moved to the opposite side of A but is still at the same distance from it. The net force on A is now 2.877 x 10^-24 Newtons in the negative direction of the x axis. What is the ratio of (charge C / charge B)?

1. A------------------------------------B-C

2. B--------A---------C

Answer 1

I redrew the diagram for a better visual. Since the net force is in the negative direction, C > B.

1. A------- r ---------BC

2. B------- r ---------A------- r -----------C

F = k Q1Q2/r²

F1 = k Qa/r² (Qb + Qc)

F2 = k Qa/r² (Qb - Qc)

F1/F2 = (Qb + Qc)/(Qb - Qc)

2.014e-23/2.877e-24 = 7.000 (close enough)

7 = (Qb + Qc)/(Qb - Qc)

7(Qb - Qc) = (Qb + Qc)

7Qb - 7Qc = Qb + Qc

6Qb = 8Qc

Qb/Qc = .75

Edit: Oops, you wanted C/B:

Qc/Qb = 1.33