But Officer, You Haven't Charged me with Anything! - Electrostatics

P. 496 - 497: Q1, P1, 2, 8, 11, 20

Q1.

Problems:
1. 1.88 x 10¹⁴ electrons

2. F2 = 2.4 N

8. number of excess electrons = 3.8 x 10¹⁴ electrons, 3.4 x 10^-16 kg increase (since you know the increase of electrons, and the mass of one electron is 9.11 x 10^-31 kg)

11. the three net forces on each F1 = 1.2 x 10²N to the left, F2 = 5.3 x 10²N to the right, F3 = 3.9 x 10² N.

20. If one charge is Q1 , the other charge will be Q2 = Q – Q1 . For the force to be repulsive, the two charges must have the same sign. Because the total charge is positive, each charge will be positive. We account for this by considering the force to be positive:

F = kQ1Q2/r2 = kQ1(Q – Q1)/r2;

You know F = 12 N and r = 1.06 meters. This gives you a quadratic equation that can be solved numerically (on the calculator!) for Q1 = 50.0 x 10^-6 C, Q2 = 30 x 10^-6 C