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
Medium
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(b) In balance condition
$⇒$ $QE = mg$ $⇒$ $Q\frac{V}{d} = \left( {\frac{4}{3}\pi {r^3}\rho } \right)\,g$
$⇒$ $Q \propto \frac{{{r^3}}}{V}$$⇒$ $\frac{{{Q_1}}}{{{Q_2}}} = {\left( {\frac{{{r_1}}}{{{r_2}}}} \right)^3} \times \frac{{{V_2}}}{{{V_1}}}$
$⇒$ $\frac{Q}{{{Q_2}}} = {\left( {\frac{r}{{r/2}}} \right)^3} \times \frac{{600}}{{2400}} = 2$
$⇒$ $QE = mg$ $⇒$ $Q\frac{V}{d} = \left( {\frac{4}{3}\pi {r^3}\rho } \right)\,g$
$⇒$ $Q \propto \frac{{{r^3}}}{V}$$⇒$ $\frac{{{Q_1}}}{{{Q_2}}} = {\left( {\frac{{{r_1}}}{{{r_2}}}} \right)^3} \times \frac{{{V_2}}}{{{V_1}}}$
$⇒$ $\frac{Q}{{{Q_2}}} = {\left( {\frac{r}{{r/2}}} \right)^3} \times \frac{{600}}{{2400}} = 2$
$⇒$ $Q_2$ = $Q / 2$
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