A charged oil drop falls with terminal velocity `v_(0)` in the absence of electric field. An electric field E keeps it stationary. The drop acquires charge 3q, it starts moving upwards with velocity `v_(0)`. The initial charge on the drop is

Options

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

Answer

C

Explanation

When drop is stationary, then `q_(1)E=6pi etarv_(0)`
or `q_(1)=6pieta rv_(0)//E`
when drop moves upwards, then
`3q=(6pietar(v_(0)+v_(0)))/(E)=2xx((6pietarv_(0))/(E))=2q_(1)`
`implies q_(1)=(3)/(2)q`