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

Explanation

Given charge on an oil drop
`q=12x=12xx1.60xx10^(-19)C`
i.e. `q=19.20xx10^(-19)C`
`E=2.55xx10^(4)NC^(-1)`
`rho=1.26xx10^(3)kgm^(-3)`
By Definition density `rho=("Mass"(m))/("volume"(V))`
i.e. mass `=m=(rho)` (volume)
i.e. `m=(4//3)pi r^(3)rho `
i.e. `m=4/3xx3.142xxr^(3)xx1.26xx10^(3)kg`
i.e. `m=5.278(r^(3))xx10^(3)kg`
given gravitational force `=` electrostatic force
i.e. `mg=qE`
i.e. `10^(3)xx5.278r^(3)g=19.2xx10^(-19)xx2.55xx10^(4)`
i.e. `r^(3)=(48.96xx10^(-15))/(10^(3)xx5.278xx9.81)=0.9455xx10^(-18)`
i.e. `r~~0.981xx10^(-6)m`
or `r=9.81xx10^(-7)m`
Thus the radius of the drop is `9.81xx10^(-7)m`