1.

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

Answer»

`10^(18)`
`10^(15)`
`10^(12)`
`10^(9)`

Solution :For the drop to be stationary,
FORCE on the drop DUE to electric field = Weight of the drop
qE = mg
`:. q = (1.6 xx 10^(-6) xx10)/(100) = 1.6 xx 10^(-7) C`
Number of electrons carried by the drop is
`N = (q)/(e) = (1.6 xx 10^(-7)C)/(1.6 xx 10^(-19)C) = 10^(12)`


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