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A negatively charged oil drop is prevent...

A negatively charged oil drop is prevented from falling under gravity by applying a vertical electric field of `100Vm^(-1)`. If the mass of the drop is `1.6xx10^(3)g` the number of electrons carried by the drop is `(g=10ms^(-2))`

A

`10^(18)`

B

`10^(15)`

C

`10^(6)`

D

`10^(12)`

Text Solution

Verified by Experts

The correct Answer is:
D
`qE=mg implies q = (1.6 xx10^(-6) xx10)/(100) =1.6 xx10^(-7)C`
`n=q/e = (1.6 xx10^(-7))/(1.6 xx10^(-19))=10^(12)`
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