Question

In Millikan's oil drop experiment, a charged oil drop is moving upwards with a terminal velocity V1 in certain electric field. After it acquires an additional charge, it moves up with a terminal velocity V2. In the absence of electric field the drop falls with a terminal velocity V, then the ratio of the chargess before and after acquiring charge is 1. V1/V2 2. (V1+V) / (V2+V) 3.(V1-V) / (V2-V) 4.V2/V1

Answer 1

Answer:

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Explanation:

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