In Millikan's oil drop experiment an oil drop carrying a charge Q is held stationary by a potential difference 2400 V between the plates. To keep a drop of half the radius stationary the potential difference had to be made 600 V. What is the charge on the second drop
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a
Q4
b
Q2
c
Q
d
3Q2
answer is B.
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Detailed Solution
In balance condition ⇒QE=mg⇒QVd=43πr3ρg⇒Q∝r3V⇒Q1Q2=r1r23×V2V1⇒QQ2=rr/23×6002400=2⇒Q2=Q/2