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In Millikan's oil drop experiment, a cha...

In Millikan's oil drop experiment, a charged drop of mass `1.8 xx 10^(-14)` kg is stationary between its plates. The distance between the plates is 0.90 cm. The potential difference is 2.0 kV. The number of electrons on the drop is:

A

500

B

5

C

5

D

zero

Text Solution

Verified by Experts

The correct Answer is:
C
`E = (V)/(d) = (2000)/( 0. 9 xx 10 ^(-2)) = (2)/(9) xx 10 ^(6) V // m`
Since oil drop is stationary, therefore, (ne) E = mg
`n = ( mg)/(cE) =(1.8 xx 10 ^(-14)xx 10 xx 9)/( 1. 6 xx 10 ^(-19) xx 2 xx 10 ^(6))= 5`
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