three positively charged particles are fixed on an x axis. Particles B and C are

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 3.78 × 10-23 N in the negative direction of the x axis. In figure 21-28(b), 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 4.15 × 10-24 N in the negative direction of the x axis. What is the ratio qC/qB?


http://edugen.wileyplus.com/edugen/courses/crs4957/art/qb/qu/c21/pict_21_16.gif

Explanation / Answer

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

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