A charged oil drop fall with a terminal velocity `V` in the absence of elcectric field . An electric field `E` keep keep the oil drop stationary in it . When the drop acquire a charge 'q' it moves up with same velocity. Find the initial charge on the drop.

Options

A. `q//2`
B. `q`
C. `3q//2`
D. `2q`

Answer

C

Explanation

When drop is stationary, then
`q_(1)E=6pietarv_(0)` or `q_(1)=6pietarv_(0)//E`
When drop moves upwards, then
`3q=(6pietar(v_(0)+v_(0)))/(E)`
`=2xx((6pietarv_(0))/(V))`
`=2q_(1)" "{because q_(1)=6pietarV_(0)//E}`
`therefore q_(1)=(3)/(2)q`.