8.02X Electricity and Magnetism

8.02x – Problem Set 1 SolutionsProblem 1 (5 points)a) The ratio of the electric force to the gravitational force for an electronand a proton in a hydrogen atom is∣ F e ∣∣∣∣ =F g=k eer 2= k e 2= (1)G mempG mr 2 e m p9.0 × 10 9 N · m 2 /C 2 (1.6 × 10 −19 C) 26.67 × 10 −11 N · m 2 /kg 2 (9.1 × 10 −31 kg)(1.67 × 10 −27 kg) ≈ 2 × 1039 .b) In a), we have seen that the ratio of the forces does not depend on thedistance, because both Newton’s gravitational law and Coulomb’s law have r −2dependence. Therefore, the ratio F e /F g doesn’t change with distance.c) The universe is charge neutral on large scales. Therefore, there is nonet electromagnetic force on astronomical distances. However, gravitationalmass adds up, and gravitation is only an attractive (and long-distance) force, itdominates.Problem 2 (5 points)a) The ratio of the forces would be∣ F e ∣∣∣∣ = k Q E Q M. (2)F g G M E M MWe need to estimate the number of protons (and electrons) in Earth and inthe Moon. We assume protons contribute to roughly half the mass (neutronsare the other part, and electrons are 1000 times lighter). Therefore, the Earthhas approximately N E ≈ (M E /m p )/2 protons. On the other hand, the Moonhas approximately N M ≈ (M M /m p )/2 protons. Assuming the electrons’ chargeis (1 − 10 −11 ) times the proton charge, the Earth would carry a charge Q E ≈eN E × 10 −11 , and the moon would have charge Q M ≈ eN M × 10 −11 . The ratioof the forces would then be( ) ( )∣ F e ∣∣∣∣ = k ME2m pe × 10 −11 MM2m pe × 10 −11 = k e 2× 10 −22 = (3)F g G M E M MG=4m 2 p9.0 × 10 9 N · m 2 /C 26.67 × 10 −11 N · m 2 /kg 2 (1.6 × 10 −19 C) 2(1.67 × 10 −27 kg) 2 × 10−22 ≈ 1 × 10 14 .1