﻿ In Millikan’s oil drop experiment, an oil drop carrying a charge q falls with a terminal velocity v0 when there is no electric field between the plates. An electric field E is applied to keep it stationary. What additional charge should the oil drop acquire so that it begins to move upwards with a velocity 2v0 in the same electric field?      : Kaysons Education

# In Millikan’s Oil Drop Experiment, An Oil Drop Carrying A Charge q falls With A Terminal Velocity v0 when There Is No Electric Field Between The Plates. An Electric Field E is Applied To Keep It Stationary. What Additional Charge Should The Oil Drop Acquire So That It Begins To Move Upwards With A Velocity 2v0 in The Same Electric Field?

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## Question

### Solution

Correct option is

2q

When the drop is stationary, we have

let Q be the additional charge acquired by the drop so that its charge now is q’ = q + Q. When it moves upwards with a velocity v = 2v0, then we have

= 3q.

Thus,

.

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