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 4.44 10-14 N and suspended in a downward field of magnitude 3.47 104 N/C.

Answer 1

Solution,

Force, F = mg = qE

Charge, q = (4.44 x 10^-14)/3.47 x 10^4

Charge, q = 1.28 x 10^-18 C

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