Millikan measured the electron\'s charge by observing tiny charged oil drops in
ID: 1706631 • 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 3.18 10-14 N and suspended in a downward field of magnitude 3.31 104 N/C.Explanation / Answer
gravity = 3.18*10-14 N (downward)
magnitude of electric force = |q|*3.31*104 N/C
two forces are balanced, so electric force must be upward, E is downward, so q must be negative.
q = -3.18*10-14/3.31*104 = -9.61*10-19 C = -6e