A charged oil drop is suspended in a uniform filed of `3xx10^4 v//m` so that it neither falls nor rises. The charge on the drop will be (Take the mass of the charge `=9.9xx10^-15kg` and `g=10m//s^2`)

Options

A. (a) `1.6xx10^-18C`
B. (b) `3.2xx10^-18C`
C. (c) `3.3xx10^-18C`
D. (d) `4.8xx10^-18C`

Answer

C

Explanation

At equilibrium, electric force on drop balances weight of drop.
`qE=mgimpliesq=(mg)/(E)=(9.9xx10^-15xx10)/(3xx10^4)=3.3xx10^(-18)C`