In a Millikan-type oil-drop experiment, the plates are `8 mm` apart. An oil drop is found to remain at rest when the upper plate is at a potential `135 V` higher than that of the lower one. When the electric field is switched off, the drop is found to fall a distance of `2.0 mm` in 36 seconds with a uniform speed. Find (a) the charge on the drop and (b) the number of electrons attached to this drop. Density of oil `= 880 kg m^(-3)` and coefficient of viscosity of `air = 180 mupoise`.

Explanation

(a) The charge on the drop is
`q = (18 pi)/(E) sqrt((eta^3 v^3)/(2(rho - sigma)g))`. ..(i)`
Here `E = (136 V)/(8 xx 10^(-3) m) = 1.7 xx 10^4 V m^(-1)`
`eta = 180 mupoise = 1.8 xx 10^(-5) N sm^(-2)`
`v = (2.0 mm)/(36 s) = 1/18 xx 10^(-5) N sm^(-2)`
and `rho = 880 kg m^(-3)`.
The density of air `sigma (1.29 kg m^(-3)` may be neglected in comparision to that of the oil. Putting values in (i),
`q = 7.93 xx 10^(-19) C`.
(b) The number of electrons attached to the drop is, ` n = (7.93 xx 10^(-19) C)/(1.6 xx 10^(19) C) = 4.96`.
It is clear that 5 electrons are attached to the drop.