In Millikan's oil drop experiment, an oil drop mass `60 xx 10^(-6) kg` is balanced by an electric field of `10^(6) V//m`. The charge in coulomb on the drop, assuming `g = 10 m//s^(2)` is

Options

A. `6.2 xx 10^(-11)`
B. `16 xx 10^(-9)`
C. `16 xx 10^(-11)`
D. `16 xx 10^(-13)`

Answer

C

Explanation

`eE = mg`
`rArr e = (mg)/(E) = (16 xx 10^(-6) xx 10)/(10^(6)) = 16 xx 10^(-11) C`.