In figure 21-28(a), three positively charged particles are fixed on an x axis. P
ID: 1954335 • Letter: I
Question
In figure 21-28(a), 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 5.09 × 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 5.77 × 10-24 N in the negative direction of the x axis. What is the ratio qC/qB?Explanation / Answer
Because the charge is still being pulled in the negative x direction after particle B has been moved, we can reason that particle C must have the greater positive charge. When B and C are on the same side, they work together to create a force of 5.09 x 10-23, but when B has moved, the work against one another.
qB + qC = 5.09 x 10-23
qC - qB = 5.77 x 10-24
We now have a system of equations. We could solve it out multiple ways, depending on whether we want to solve for qC or qB. Say you solve for qC, so that qC = 5.09 x 10-23 - qB. Then plug the new equation for qC into the second equation, and solve for qB. Plug the value for qB back into the either equation, and solve for qC. You can now take the exact values of qC and qB and find the ration qC/qB.