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In Milikan's oil drop experiment, a char...

In Milikan's oil drop experiment, a charged drop falls with terminal velocity V. If an electric field E is applied in vertically upward direction then it starts moving in upward direction with terminal velocity 2V. If magnitude of electric field is decreased to `(E)/(2)`, then terminal velocity will become

A

`V/2`

B

V

C

`3V/2`

D

2V

Text Solution

Verified by Experts

The correct Answer is:
A
`V=(QxxExxt)/(m)`
`V prop E`
So `V/2`
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