In 1909, Robert Millikan (1868–1953) used his famous 1909 oil drop experiment to
ID: 2002734 • Letter: I
Question
In 1909, Robert Millikan (1868–1953) used his famous 1909 oil drop experiment to estimate the elementary charge. In this experiment, small drops of oil were charged, and an electric field was adjusted to help the drops stay suspended. From the required field, the charge on various drops was calculated—they were all whole-number ratios of the elementary charge. The actual analysis has to take into account buoyant forces and other details, but we will simplify it for this problem. Assume that the density of the oil used in the experiment is 880 kg/m3, a drop has a radius of 1.25 m, and a drop has a charge of +7e. What applied electric field is necessary to keep the drop suspended against gravity?
Magnitude in kN/C = ???
direction=??
Explanation / Answer
Suppose the required electric field is E virtically up. Then Fe = 7e*E.
The volume of drop = (4/3)*pi*(1.25*10^-6)^3 = 8.18*10^-18 m^3
The density = 880 kg/m^3
the mass = 7.198*10^-15 kg
The force due to gravity = mg = m*9.8 = 7.05*10^-14 N
Fe = Fg
7*1.6*10^-19*E = 7.05*10^-14
E = 7.05*10^-14/(7*1.6*10^-19) = 62946.4286 N/C upward.