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In Millikan’s experiment, an oil drop of radius 1.64micrometers &dencity 0.851g/cc is suspended in chamber when a downward direction electric field of 1.92×10 5 N/C is applied,find charge on drop in terms of ‘e’.

In Millikan’s experiment, an oil drop of radius 1.64micrometers &dencity 0.851g/cc is suspended in chamber when a downward direction electric field of 1.92×105 N/C is applied,find charge on drop in terms of ‘e’.

Grade:12

1 Answers

Atal Tiwari
37 Points
9 years ago
Since, the radius is 1.64 mm
i.e. r = 1.64 x 10-6 m
As we know, m=∂*v
(where ∂ is density and v is volume)
therefore,   m = 0.851 x 4/3-pi-r3
                    = 1.45 x 10-17 kg
We have already studued that qE= mg
                                             q=mg/E
                                               =1.45 x 10-17 x  9.8/1.92x 105
                                               = 7.4 x 10-22 C .

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