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Chapter 22, Problem 039 In Millikans experiment, an oil drop of radius 1.70 μm and density 0.859 g/cm3 is suspended in chamber C (see the figure) when a downward electric field of 0.318 105 N/C is applied. Find the charge on the drop, in terms of e Oil A Iasalating chamber wall dre Microscop Number Units the tolerance is +/-2%
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Answer #1

Since drop is suspended, the force of gravity ( downward) is balanced by the electrostatic force ( upward implies the drop has negative charge)

So,

0= mg-|q|E

m = (4/3)\rhoπa3 = (4/3)(0.859)(3.14)(1.70×10-4)3= 17.67×10-9 kg

mg = qE

(17.67×10-9)9.81= q(0.318×105)

q = -5.45×10-13C

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