An oil drop, carrying six electronic charges and having a mass of `1.6xx10^(-12)g`, falls with some terminal velocity in a medium. What magnitude of vertical electric field is required to make the drop move upward with the same speed as it was formerly moving down ward with ? Ignore buoyancy.

Options

A. `10^(5)NC^(-1)`
B. `10^(4)NC^(-1)`
C. `3.3xx10^(4)NC^(-1)`
D. `3.3xx10^(5)NC^(-1)`

Answer

C

Explanation

For the first case, F=mg.
For the second case
`F+mg=6eE or 2 mg=6eE`
,
or, `E=(mg)/(3e)=(1.6xx10^(-15)xx10)/(3xx1.6xx10^(-19))=3.3xx10^(4)NC^(-1)`.