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The classic Millikan oil drop experiment was the first to obtain an accurate measurement of the charge on an electron. In it, oil drops were suspended against the gravitational force by a vertical electric field.

Part (a) Given that the oil drop is 0.85 ?m in radius and has a density of 910 kg/m3, find the weight of the drop in N

.Part (b) If the drop has a single excess electron, find the electric field strength needed to balance its weight.

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Added Thu, 29 Oct '15
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a)

Volume of oil drop

V=(4/3)pir3 =(4/3)*pi*(0.85*10-6)3

V=2.57*10-18 m3

Weight of oil drop

W=mg=pVg=(2.57*10-18)*920*9.8

W=2.32*10-14 N

b)

E=W/q =2.32*10-14/(1.6*10-19)

E=144,957 N/C

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Added Thu, 29 Oct '15
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