The classic Millikan oil drop experiment was the first to obtain an accurate mea

Question

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 gravity by a vertical electric field. Assume the oil drop to be 1.40 mu m in radius and have a density of 895 kg/m^3. (a) Find the weight of the drop. 10 074e-15 How do you calculate an object's weight? N (b) If the drop has a single excess electron, find the magnitude of the electric field strength needed to balance its weight. 6.29e4 If the oil drop hovers in place how can you relate the weight to the electric force? How does the electric force relate to the electric field? N/C

Explanation / Answer

a) Weight of the drop=mass*g=volume*density*g=(4/3)*pi*r3*density*g=(4/3)*pi*(1.40*10-6)3*895*9.8= 1.0081*10-13 N

b) Since weight is balanced by force of electric field

eE=weight

=>E=weight/e=1.0081*10-13 /1.6*10-19= 630060 N/C

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