Millikan measured the electron\'s charge by observing tiny charged oil drops in
ID: 1687166 • 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 1.6e-19 C. Find the charge on an oil drop weighing 1.68 10-14 N and suspended in a downward field of magnitude 3.51 104 N/C.Explanation / Answer
weight of the oil drop W = 1.68*10^-14 N Electric field strength E = 3.51*10^4 N/C Charge on each electron e = 1.602*10^-19 C Force acting on the charge F = qE According to the given data W = qE q = W/E = (1.68*10^-14)/(3.51*10^4) = 0.4876*10^-18 C = 4.876*10^-19 C = 3.04e