Question

In Millikan's oil drop experiment, an oil drop carrying a charge q falls with a terminal velocity v 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 2v in the same electric field?

A. q′=q
B. q′=2q
C. q ′= 2q
D. q′= 2/3q

Answer 1

In Millikan's oil drop experiment, an oil drop carrying a charge q falls with a terminal velocity v when there is no electric field between the plates. When an electric field E is applied to keep the oil drop stationary, the electric force on the drop balances the gravitational force on the drop.

To make the oil drop move upwards with a velocity 2v in the same electric field, the additional charge the oil drop should acquire is q′ = 2q.

Step-by-step explanation:

In Millikan's oil drop experiment, an oil drop carrying a charge q falls with a terminal velocity v when there is no electric field between the plates. When an electric field E is applied to keep the oil drop stationary, the electric force on the drop balances the gravitational force on the drop:

qE = mg

To make the oil drop move upwards with a velocity 2v in the same electric field, the electrical force should balance the gravitational force in the opposite direction:

q′E = 2mg

where q′ is the additional charge the oil drop should acquire.

Divide the second equation by the first equation:

(q′E)/(qE) = (2mg)/(mg)

q′/q = 2

Therefore, the additional charge the oil drop should acquire is q′ = 2q. So, the answer is option C.