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. `3.3xx10^(-18)C`
B. `3.2xx10^(-18)C`
C. `1.6xx10^(-18)C`
D. `4.8xx10^(-18)C`

Answer

A

Explanation

In steady state
electric force on drop

=weight of drop
`qE=mg`
`q=(mg)/E=(9.9xx10^(-15)xx10)/(3xx10^(4))=3.3xx10^(-18)C`