An oil drop of 12 excess electrons is field stationary under constant clectric field of `2.55 xx 10^(4) NC^(-1)` in Millikan's oil drop experiment. Then density of the oil is 1.26g `cm^(-3)`. Estimate the radius of the drop. `(g=9.81 ms^(-2), e=1.60 xx 10^(-19)C)`.

Explanation

`n=12, e=1.6 xx 10^(-19)C, r=?, E=2.55 xx 10^(4) NC^(-1)`
`d=1.26 g cm^(-3) =1.26 xx 10^(3) kgm^(-3)`
Weight of the drop = Foree due to electric field
`4/3 pi r^(3) dg=ne E " "r^(3)=((3ne E)/(4pi dg)) therefore r=((3ne E)/(4pi dg))^(1/3)=[(3 xx 12 xx 1.6 xx 10^(-19) xx 2.55 xx 10^(+4))/(4 xx 3.14 xx 1.26 xx 10^(3) xx 9.8)]^(1/3)`
`r=9.81 xx 10^(-4) mm`