Calculate the radius of a charged water drop which remains just suspended in equilibrium in the earth's electric field. Charge in the water drop is equal to that of an electron. Magnitude of the earth's electric field is `10^(-2)"stat"V.cm^(-1)`. `[e=4.805xx10^(-10)` esu of charge, `g=980cm.s^(-2)`]

Explanation

Let the radius of the water drop be r cm and mass mg.
Here intensity of electric field,
`E = 10^(-2)` stat`V · cm^(-1)`
In equilibrium, clectrical force on the charged water drop =weight of the water drop.
or, eE=mg
or, `eE4/3pir(3)pg` [density of water, `p=1g.cm^(-3)`]
or, `r^(3)=(3eE)/(4pig)=(3xx4.805xx10^(-10)xx10^(-2))/(4xxpixx980)=1.171xx10^(-15)`
`thereforer=1.054xx10^(-5)cm`