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 $900 \,kg/m^3$, the radius of the drop will be
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(b) In equilibrium $QE = mg$ $==>$ $Q.\frac{V}{d} = mg = \left( {\frac{4}{3}\pi {r^3}\rho } \right)\,g$
$==>$ $2 \times 1.6 \times {10^{ - 19}} \times \frac{{12000}}{{2 \times {{10}^{ - 2}}}} = \frac{4}{3}\pi {r^3} \times 900 \times 10$
$ r = 1.7 × 10^{-6} \,m$
$==>$ $2 \times 1.6 \times {10^{ - 19}} \times \frac{{12000}}{{2 \times {{10}^{ - 2}}}} = \frac{4}{3}\pi {r^3} \times 900 \times 10$
$ r = 1.7 × 10^{-6} \,m$
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