In a particular Millikan oil-drop apparatus, the plates are 3.25 cm apart. The o

Question

In a particular Millikan oil-drop apparatus, the plates are 3.25 cm apart. The oil used has a density of 0.725 g/cm^3, and the atomizer that sprays the oil drops produces drops of diameter 1.00×103 mm.

A. What strength of electric field is needed to hold such a drop stationary against gravity if the drop contains four excess electrons?

B. What should be the potential difference across the plates to produce this electric field?

C. If another drop of the same oil requires a plate potential of 94.5 V to hold it stationary, how many excess electrons did it contain?

In a particular Millikan oil-drop apparatus, the plates are 3.25 cm apart. The oil used has a density of 0.725 g/cm^3, and the atomizer that sprays the oil drops produces drops of diameter 1.00×103 mm.

A. What strength of electric field is needed to hold such a drop stationary against gravity if the drop contains four excess electrons?

B. What should be the potential difference across the plates to produce this electric field?

C. If another drop of the same oil requires a plate potential of 94.5 V to hold it stationary, how many excess electrons did it contain?

Explanation / Answer

A) mg = qE

density*volume*g=qE

Putting all the values in SI units
(725)(4(5*10-7)3/3)(9.81) = (4)(1.6*10-19)(E)
E = 5818 N/C

(b)
For parallel plates:
V = E*x
V = (5818 N/C)(0.0325 m)
V = 189 V

(c)
mg = (ne)(V/x)
n = mgx/(eV)
n = (725 )*(4(5*10-7 )3/3)*(9.8)(0.0325 ) / [(1.6*10-19 )(94.5)]
n = 8

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