In a particular Millikan oil-drop apparatus, the plates are 2.25 cm apart. The o

Question

In a particular Millikan oil-drop apparatus, the plates are 2.25 cm apart. The oil used has a density of 0.800 g/cm^3, and the atomizer that sprays the oil drops produces drops of diameter 1.00 In a particular Millikan oil-drop apparatus, the plates are 2.25 cm apart. The oil used has a density of 0.800 g/cm^3, and the atomizer that sprays the oil drops produces drops of diameter 1.00 In a particular Millikan oil-drop apparatus, the plates are 2.25 cm apart. The oil used has a density of 0.800 g/cm3, and the atomizer that sprays the oil drops produces drops of diameter 1.00 times 10?3 mm What strength of electric field is needed to hold such a drop stationary against gravity if the drop contains five excess electrons? What should be the potential difference across the plates to produce this electric field? If another drop of the same oil requires a plate potential of 72.2 V to hold it stationary, how many excess electrons did it contain? express answer as integer

Explanation / Answer

a F = Eq = mg so E = mg/q = 4/3 Pi (5 x 10^-7)^3 800* 9.8 / ( 5 x 1.6 x 10^-19) = 5128.66 V/m

b E = V/d so V = Ed = 5128.66 (0.0225 ) = 115.39 V

c 72.3/115.39 = 0.625 so there must be .623 x 5 electrons = 3.1 if the drop (unlikely) is the same size.

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