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 xx10^(-11)`
B. `16xx10^(-9)`
C. `16xx10^(-11)`
D. `16 xx10^(-13)`

Answer

C

Explanation

`qE = mg`
`implies q=(16 xx10^(-6) xx10)/10^6 = 16xx10^(-11) C`