There are n electrons on a drop of oil. It is in equilibrium in an electric field of intensity E. If the density of the oil is ρ, then radius of the drop will be
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a
4πeE3nρg1/3
b
3nρg4πeE1/3
c
4πρg3neE1/3
d
3neE4πρg1/3
answer is D.
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Detailed Solution
In equilibrium, q E = m gThere are n electrons on the drop hence, q = n e∴ neE=43πr3ρgor r=3neE4πρg1/3