An oil drop with 10 excess electrons is held stationary under a constant electric field of `3 xx 10^(4) NC^(-1)` in Millikan oil drop experiment. The density of the oil is `1.26 xx 10^(3) kg m^(-3)`. Calculate the radius of the drop.

Explanation

Data supplied,
r=radius of the droplet, d=density of oil drop `=1.26 xx 10^(3) kg m^(-3)`
`g=9.8 m//s^(2), ` n=no. of excess electrons =10
`e=1.6 xx 10^(-19)C" "E=3 xx 10^(4) NC^(-1)`
As the oil drop is stationary, weight of the droplet=force due to electric field
`i.e 4/3 pir^(3) dg=ne E, r^(3) =(3ne E)/(4pi dg) =(3 xx 10 xx 1.6 xx 10^(-19) xx 3 xx 10^(4))/(4 xx 3.14 xx 1.26 xx 10^(3) xx 9.8)= 0.9285 xx 10^(-18)`
`r=(0.9285 xx 10^(-18))^(1/3)=0.9756 xx 10^(-6)=9.76 xx 10^(-7) m`