Question Detail
Question
How many excess electrons must be placed on each of two small spheres spaced 3 cm apart, if the repulsive force between the spheres is to be 10-18 N? (e = 1.6 x 10-19 C, k = 9 x 109 Nm2C-2)
Correct Answer
Option B is the correct answer.
Detailed Explanation
Coulomb's Law states F = k * q1q2 / r2. Since the spheres are identical and have excess electrons, q1 = q2 = q.So, F = k * q2 / r2.We need to find q first: q2 = F * r2 / k.Given F = 10-18 N, r = 3 cm = 0.03 m, k = 9 x 109 Nm2C-2.q2 = (10-18 N) * (0.03 m)2 / (9 x 109 Nm2C-2)q2 = (10-18) * (9 x 10-4) / (9 x 109)q2 = 10-18 * 10-4 * 10-9 = 10-31 C2.q = √(10-31) = √(10 * 10-32) = 3.162 x 10-16 C.The number of excess electrons (n) is q / e, where e = 1.6 x 10-19 C.n = (3.162 x 10-16 C) / (1.6 x 10-19 C)n = (3.162 / 1.6) * 10(-16 - (-19))n = 1.976 * 103 electrons.This is approximately 1.98 x 103 electrons.
Hint
First, use Coulomb's Law to find the total charge on each sphere, then divide the total charge by the charge of a single electron to find the number of excess electrons.