An oil drop of 12 excess electrons is held stationaty under a constant electric field of `2.55xx10^(4) NC^(-1)` in Millikan's oil drop experi,ment. The density of the oil is `1.26 g cm^(-3)`. Estimate the radius of the drop. `(g = 9.81 ms^(-2) , e = 1.60xx10^(19) C)`

Explanation

The electric force on the oil drop balances the weight of the drop .
qE=mg
`rArr qE=4/3 pi r^3 rhog`
`rArr (12xx1.6xx10^(-19))(2.55xx10^4)`
`=4/3xx3.14xxr^3xx(1.26xx10^3)(9.81)`
`rArr r^3=(3(12xx1.6xx10^(-19))(2.55xx10^4))/(4xx3.14xx1.26xx10^3xx9.8)=944xx10^(-21)`
`rArr r=9.81xx10^(-7)m =9.81xx10^(-4) `mm