A tiny spherical oil drop carrying a net charge q is balanced in still air with a vertical uniform electric field of strength `(81pi)/(7)xx10^5Vm^-1`. When the field is switched off, the drop is observed to fall with terminal velocity `2xx10^-3ms^-1`. Given `g=9.8ms^-2`, viscoisty of the air `=1.8xx10^-5Nsm^-2` and the denisty of oil `=900kg m^-3`, the magnitude of q is

Explanation

When ball falls with terminal velocity
`W = F_(v)`
`mg = 6 pi eta r v`
`sigma . (4)/(3) pi r^(3) g = 6 pi eta rv`
`r^(2) = (9 eta v)/(2 sigma g) rArr = ((9 eta v)/(2 sigma g))^((1)/(2)) = ((9 xx 1.8 xx 10^(-5) xx 2 xx 10^(-3))/(2 xx 900 xx 9.8))^((1)/(2))`
`= (3)/(7) xx 10^(-5) m`
`qE = mg = 6 pi eta r v`
(since `mg = 6 pi eta r v`)
`q = (6 pi eta r v)/(E ) = ( 6pi xx 1.8 xx 2 xx 10^(-3))/((81 pi)/( 7) xx 10^(5)) xx (3)/(7) xx 10^(-5)`
`= 8 xx 10^(-19) C`