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An oil drop of 12 excess electrons is he...

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))`.

Text Solution

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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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