Electrostatics Question 372

Question: In Millikan’s oil drop experiment an oil drop carrying a charge Q is held stationary by a potential difference 2400V between the plates. To keep a drop of half the radius stationary the potential difference had to be made 600V . What is the charge on the second drop [MP PET 1997]

Options:

A) Q4

B) Q2

C) Q

D) 3Q2

Show Answer

Answer:

Correct Answer: B

Solution:

In balance condition therefore QE=mg

therefore QVd=(43πr3ρ)g

therefore Qr3V

therefore Q1Q2=(r1r2)3×V2V1

therefore QQ2=(rr/2)3×6002400=2

therefore Q2 = Q / 2



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