Question

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’.

Answer 1

The charge on drop in term of "e" is 7.4 x 10^-22 C

Explanation:

Since, the radius is 1.64 mm

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

As we know, m=∂  x  v

(where ∂ is density and v is volume)

therefore,   m = 0.851 x 4/3-pi-r3

                   = 1.45 x 10^-17 kg

We know that qE= mg

                         q=mg/E

                           =1.45 x 10^-17 x  9.8/1.92 x 10^5

                            = 7.4 x 10^-22 C .

Thus the charge on drop in term of "e" is 7.4 x 10^-22 C

Also learn more

What is charge, mass and charge to mass ratio of an electron?

https://brainly.in/question/6705130