Question

Millikan Experiment 4

Consider an oil droplet of mass m andcharge q. We want to determine the charge on thedroplet in a Millikan-type experiment. We will do this in several steps. Assume, for simplicity, that the charge is positive and that the electric field betweenthe plates points upward.

Oil has a density rho of 860;{rm kg}/{rm m}^{3}. An oildroplet is suspended between two plates 1.0; rm cm apart by adjusting the potential difference between them to 1177; rm V. When the voltage is removed, the droplet falls and quickly reaches constant speed. It is timed with a stopwatch andfalls 3.00; rm mm in 7.33; rm s. The viscosity of air is 1.83 times 10^{ - 5};rm kg/ m cdot s. What is the droplet's charge q?Express the charge in coulombs to two significant figures. Take the free-fall acceleration to be 9.81; rm m/s^2.
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Answer #1
densityof oil ρ = 860kg/m3viscosity of air isη = 1.83*10-5kg/mspotenial difference V= 1174vseperation betweenthe plates d = 0.90*10-2mthevelocity of the oil drop isv = 0.00306m/7.40s=4.135*10-4m/s sothe electric field between the plates is E = V/d caculatefor Eacceleration due to gravity g = 9.8m/s2we havethe formula for the radius of the drop asa ˜ [9ηv/2gρ]1/2caculate foranow theformula for the charge on the drop isEq ˜ 4/3*πa3ρgor q = 4πa3ρg / 3Ecaculate for q we getthe droplet's charge
answered by: Dr Ajit Handa
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