In Millikan's oil drop experiment,an oil drop | vedclass

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(B) In the balance condition,the electric force equals the gravitational force:$QE = mg$Since $E = \frac{V}{d}$ and $m = \frac{4}{3}\pi r^3 \rho$,we h

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$\frac{Q}{2}$

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(B) In the balance condition,the electric force equals the gravitational force:$QE = mg$Since $E = \frac{V}{d}$ and $m = \frac{4}{3}\pi r^3 ho$,we have:$Q \frac{V}{d} = \frac{4}{3}\pi r^3 ho g$This implies $Q \propto \frac{r^3}{V}$.For the two drops,we can write:$\frac{Q_1}{Q_2} = \left( \frac{r_1}{r_2} ight)^3 	imes \frac{V_2}{V_1}$Given $Q_1 = Q$,$r_1 = r$,$V_1 = 2400\,V$,$r_2 = \frac{r}{2}$,and $V_2 = 600\,V$:$\frac{Q}{Q_2} = \left( \frac{r}{r/2} ight)^3 	imes \frac{600}{2400}$$\frac{Q}{Q_2} = (2)^3 	imes \frac{1}{4} = 8 	imes \frac{1}{4} = 2$Therefore,$Q_2 = \frac{Q}{2}$.

In Millikan's oil drop experiment,an oil drop carrying a charge $Q$ is held stationary by a potential difference of $2400\,V$ 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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