In figure (a) below, three positively charged particles arefixed on an x axis. P
ID: 1663070 • Letter: I
Question
In figure (a) below, three positively charged particles arefixed on an x axis. Particles B and Care so close to each other that they can be considered to be at thesame distance from particle A. The net force on particleA due to particles B and C is 5.00× 10-23 N in the negative direction of thex axis. In figure (b) below, particle B has beenmoved to the opposite side of A but is still at the samedistance from it. The net force on A is now 3.70 ×10-24 N in the negative direction of the xaxis. What is the ratioqC/qB?
Explanation / Answer
F1 = 5.00 × 10-23 N, F2= 3.70 × 10-24 N F1 = kqA(qB +qC)/r2 F2 = kqA(qB -qC)/r2 (qB + qC)/(qB - qC) =F1/F2 (1 + qC/qB)/(1 - qC/qB) = F1/F2let x =qC/qB, a =F1/F2 (1 + x)/(1 - x) = ax = (a - 1)/(a+ 1) = 0.862 Since there are no figures, this value may beqC/qBor qB/qC,