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An oil drop of 12 excess electrons is held stationary under a constant electric field of 2.55 xx 10^4 NC^(-1). (Millikan's oil drop experiment). The density of the oil is 1.26 g cm^(-3). Estimate the radius of the drop, (g = 9.81 ms^(-2), e = 1.60 xx 10^(-19 C)). |
Answer» SOLUTION : For given oil DROP of charge q, mass m, density p and radius R to remain under equilibrium, vertically upward electric force must be equal to vertically downward gravitational force acting on it. Thus, `Fe(uarr) =Fg(darr)` `therefore qE = 4/3 piR^(3)rho g` (From above figure) `therefore (12e)E = 4/3 pir^(3) rho g` `therefore =((9 xx 1.6 xx 10^(-19) xx 2.55 xx 10^(4))/(3.14 xx 1.26 xx 10^(3) xx 9.81))^(1/3)` `therefore R = 0.9817 xx 10^(-6)` `therefore R = 9.817 xx 10^(-7)` m |
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