Millikan measured the electron\'s charge by observing tiny charged oil drops in
ID: 1687019 • 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 1.6 X 10^-19 C. Find the charge on an oil drop weighing 1.57 X 10^-14 N and suspended in a downward field of magnitude 2.45 X 10^4 N/C.Explanation / Answer
F = E Q = E N e electric force where e is charge on electronF = mg = W weight of drop
N = W / (E e) = 1.57 * 10E-14 / (2.45 * 10E4 * 1.6 * 10E-19) = 4.00
4 electrons on oil drop