A negatively charged oil drop is prevented from falling under gravity by applying a vertical electric field `100 V m^(-1)`. If the mass of the drop is `1.6 xx 10^(-3)g`, the number of electrons carried by the drop is `(g = 10 ms^(-2))`

Options

A. `10^(18)`
B. `10^(15)`
C. `10^(12)`
D. `10^(9)`

Answer

C

Explanation

For the drop to be stationary,
Force on the drop due to electric field = Weight of the drop
qE = mg
`:. q = (1.6 xx 10^(-6) xx10)/(100) = 1.6 xx 10^(-7) C`
Number of electrons carried by the drop is
`N = (q)/(e) = (1.6 xx 10^(-7)C)/(1.6 xx 10^(-19)C) = 10^(12)`