Each of two small non-conducting spheres is charged positively, the combined charge being 40 μC. When the two spheres are 50 cm apart, each sphere is repelled from the other by a force of magnitude 2.0 N. Determine the magnitude of the smaller of the two charges.
A) 1.4 μC
B) 1.1 μC
C) 2.0 μC
D) 3.3 μC
E) 17 μC

Question

Quizplus · Accepted Answer

The force between the spheres is given by Coulomb's law:$$F = k\frac{q_1q_2}{d^2}$$where k is Coulomb's constant, q1 and q2 are the charges on the spheres, and d is the distance between them.We are given that F = 2.0 N, d = 50 cm = 0.5 m, and the combined charge is 40 μC = 40 × 10^-6 C.Let q1 be the smaller charge and q2 be the larger charge. Then we have:$$2.0 N = k\frac{q_1q_2}{(0.5 m)^2}$$$$q_1q_2 = \frac{2.0 N 	imes (0.5 m)^2}{k}$$$$q_1q_2 = \frac{2.0 N 	imes 0.25 m^2}{9 	imes 10^9 N m^2 C^{-2}}$$$$q_1q_2 = 5.56 	imes 10^{-11} C^2$$We also know that the combined charge is 40 μC:$$q_1 + q_2 = 40 \mu C$$$$q_1 = 40 \mu C - q_2$$Substituting this into the previous equation, we get:$$(40 \mu C - q_2)q_2 = 5.56 	imes 10^{-11} C^2$$$$40 \mu C \ q_2 - q_2^2 = 5.56 	imes 10^{-11} C^2$$$$q_2^2 - 40 \mu C \ q_2 + 5.56 	imes 10^{-11} C^2 = 0$$Using the quadratic formula, we get:$$q_2 = \frac{40 \mu C \pm \sqrt{(40 \mu C)^2 - 4(5.56 	imes 10^{-11} C^2)}}{2}$$$$q_2 = \frac{40 \mu C \pm \sqrt{1600 \mu C^2 - 22.24 	imes 10^{-11} C^2}}{2}$$$$q_2 = \frac{40 \mu C \pm \sqrt{1577.76 	imes 10^{-11} C^2}}{2}$$$$q_2 = \frac{40 \mu C \pm 39.71 \mu C}{2}$$$$q_2 = 20.36 \mu C \quad or \quad q_2 = 1.29 \mu C$$Since q2 is the larger charge, we have:$$q_1 = 40 \mu C - q_2 = 40 \mu C - 20.36 \mu C = 19.64 \mu C$$Therefore, the magnitude of the smaller charge is 19.64 μC, which is closest to choice A, 1.4 μC.

Each of Two Small Non-Conducting Spheres Is Charged Positively, the Combined