How many electrons should be removed from a coin of mass 1.6 g, so that it may float in an electric field intensity of 109 N/C which is directed upwards?

Explanation

To make the coin float, the upward electric force must equal the downward gravitational force. The gravitational force can be calculated using Fgravity = mg, where m = 1.6 g = 0.0016 kg and g = 9.8 m/s2. The electric force is Felectric = QE, where Q is the charge and E = 109 N/C. Setting Fgravity equal to Felectric, we have mg = QE. Solving for Q gives Q = (mg)/E = (0.0016 kg × 9.8 m/s2) / (109 N/C) = 1.568 × 10-8 C. The charge of an electron is approximately 1.6 × 10-19 C, so the number of electrons is Q/e = (1.568 × 10-8 C) / (1.6 × 10-19 C/electron) ≈ 9.8 × 107 electrons. Therefore, the correct answer is 9.8 × 107. The other options incorrectly calculate the number of electrons, leading to an imbalance of forces.