In Millikan's oil drop experiment an oil | vedclass

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Question

(b) In balance condition
$&rArr;$ $QE = mg$ $&rArr;$ $Q\frac{V}{d} = \left( {\frac{4}{3}\pi {r^3}\rho } \right)\,g$
$&rArr;$ $Q \propto \frac{{{r^3}

Vedclass · Accepted Answer

$\frac{Q}{2}$

Vedclass · Answer

(b) In balance condition
$&rArr;$ $QE = mg$ $&rArr;$ $Q\frac{V}{d} = \left( {\frac{4}{3}\pi {r^3}ho } ight)\,g$
$&rArr;$ $Q \propto \frac{{{r^3}}}{V}$$&rArr;$ $\frac{{{Q_1}}}{{{Q_2}}} = {\left( {\frac{{{r_1}}}{{{r_2}}}} ight)^3} 	imes \frac{{{V_2}}}{{{V_1}}}$
$&rArr;$ $\frac{Q}{{{Q_2}}} = {\left( {\frac{r}{{r/2}}} ight)^3} 	imes \frac{{600}}{{2400}} = 2$

$&rArr;$ $Q_2$ = $Q / 2$

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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