Millikan measured the electron?s charge by observing tiny charged oil drops in a
ID: 2148599 • Letter: M
Question
Millikan measured the electron?s charge by observing tiny charged oil drops in an electric field. Each drop had a charge imbalance of only a few electrons. The strength of the electric field was adjusted so that the electric and gravitational forces on a drop would balance and the drop would be suspended in air. In this way the charge on the drop could be calculated. The charge was always found to be a small multiple of 1.60E-19 C. Find the charge on an oil drop weighing 4.00E-14 N and suspended in a downward field of magnitude 2.08E4 N/C.
Explanation / Answer
The Milikan experiment problem: weight of oil = 4.00 e-14 N therefore the upward force due to the electric field would equal the same strength: F = 4.00 e-14 N F = Eq q = F/E q = 1.923 e-18 C Hanging masses problem: m1 = m2 = 0.500g m1 = m2 = 0.5 e-3 kg the electric force = the x component of the tension y-direction: Tsin70 = mg T = .00521 N electric force = Tcos70 electric force = 0.001783 N b) electric force = kq^2/r^2 0.001783 = 9e9(q^2)/.02^2 q = 8.9e-9 C c) number of elrectrons = q/e = 55625000000 (one loses this many while the other gains this many as both are oppositely charged due to the fact that they are attracted to each other) BOL