An oil drop having charge 2e is kept stationary between two parallel horizontal plates 2.0 cm apart when a potential difference of 12000 volts is applied between them. If the density of oil is 900 kg/m3, the radius of the drop will be

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2.0 x 10-6 m

1.7 x 10-6 m

1.4 x 10-6 m

1.1 x 10-6 m

answer is B.

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Detailed Solution

In equilibrium, QE = mg⇒Q·Vd=mg=43πr3ρg ⇒2×1.6×10-19×120002×10-2=43πr3×900×10 ⇒r=1.7×10-6 m

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