Electrostatics> In Millikan’s experiment, an oil drop of ...

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’.

Anusha , 11 Years ago

Grade 12

1 Answers

Atal Tiwari

Since, the radius is 1.64 mm

i.e. r = 1.64 x 10^-6m

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^-17kg

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^-22C .

Last Activity: 11 Years ago

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Electrostatics> In Millikan’s experiment, an oil drop of ...

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’.

Anusha , 11 Years ago

Atal Tiwari

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