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. `3.3 xx 10^(-18) C`
B. `3.2 xx 10^(-18) C`
C. `1.6 xx 10^(-18) C`
D. `4.8 xx 10^(-18) C`

Answer

A

Explanation

If the drop neither falls nor rises, i.e. it remains stationary, then
Force on the drop due to electric field = Weight of the drop qE = mg
or `q = (mg)/(E) = ((9.9 xx 10^(-15))xx10)/(3xx 10^(4)) = 3.3 xx 10^(-18)C`