Millikan measured the electron\'s charge by observing tiny charged oil drops in
ID: 1628728 • 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 2.86 times 10^-14 N and suspended in a downward field of magnitude 3.5 times 10^4 N/C. eExplanation / Answer
The upward electrostatic force on the drop is equal to the downward gravitational force.
qE = mg
=> q[3.57 x 104] = 2.86 x 10-14
=> q = 8.011 x 10-19 C.
and q = ne
8.011 x 10-19 = n[1.6 x 10-19]
n = 5.
therefore, q = 5e.