Additional Exercises

1.25 An oil drop of 12 excess electrons is held stationary under a constant electric field of 2.55×10-4 N/C2.55×104 N/C (Millikan’s oil drop experiment). The density of the oil is 1.26 gm/cm31.26 gm/cm3. Estimate the radius of the drop. 
(
g = 9.81 m/s2m/s2e = 1.60×10-19 C1.60×1019 C).


 

Let the radius of the oil drop =r

Force (F) due to electric field E is equal to the weight of the oil drop (W).

Fe=WFe=W
qE=mgqE=mg
neE=mg=43πr3ρgneE=mg=43πr3ρg

r=[3neE4πρg]13=[3×2.55×104×12×1.6×10194×3.14×1.26×103×9.81]13=[946.09×1021]13=9.82×107mmr=[3neE4πρg]13=[3×2.55×104×12×1.6×10194×3.14×1.26×103×9.81]13=[946.09×1021]13=9.82×107mm

The radius of the oil drop is 9.82 x 10-7107 mm.