The Millikan Oil-Drop Experiment The charge of an electron was first measured by the American physicist Robert Millikan during 1909-1913.
(a) Show that an oil drop of radius r at rest between the plates will remain at rest if the magnitude of its charge is q = 4π/3 pr3gd/ V AB
Where p is the density of the oil (ignore the buoyant force of the air.) By adjusting V AB to keep a given drop at rest, the charge on that drop can be determined, provided its radius is known.
(b) Millikans oil drops were much too small to measure their radii directly.
Instead, Millikan determined r by cutting off the electric field and measuring the terminal speed v, of the drop as it fell. (We discussed the concept of terminal speed in Section 5.3.) The viscous force F on a sphere of radius r moving with speed u through a fluid with viscosity η is given by Stokes's law: F = 6πηru. When the drop is falling at uv the viscous force just balances the weight w = mg of the drop. Show that the magnitude of the charge on the drop is
Within the limits of their experimental error, every one of the thousands of drops that Millikan and his coworkers measured had a charge equal to some small integer multiple of a basic charge e. That is, they found drops with charges of ±2e, ±5e, and so on, but none with values such as 0.76e or 2.4ge. A drop with charge e bas acquired one extra electron; if its charge is - 2e, it bas acquired two extra electrons, and so on.
(c) A charged oil drop in a Millikan oil-drop apparatus is observed to fall 1.00 mm at constant speed in 39.3 s if VAB = 0. The same drop can be held at rest between two plates separated by 1.00 mm if VAB = 9.16 V. How many excess electrons bas the drop acquired, and what is the radius of the drop? The viscosity of air is 1.81 X 10-5 N¢ s/m2 and the density of the oil is 824 kg/m3
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Question Posted: March 26, 2010 05:45:10