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An oil drop carrying a charge Q is held stationary by a potential difference of 2400V between two horizontal parallel plates. To keep a drop of half the radius stationary, the potential difference had to be made 600 volt. The charge on the second drop is

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$\frac{Q}{4}$

$\frac{Q}{2}$

$Q$

$\frac{3 Q}{2}$

answer is B.

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Detailed Solution

$m g = E q$

$m g = \frac{V}{d} q \Rightarrow r^{3} \propto V q$ mass is proportional to volume and volume is proportional to r³

$\frac{q_{1}}{q_{2}} = {(\frac{r_{1}}{r_{2}})}^{3} (\frac{V_{2}}{V_{1}}) = {(\frac{2}{1})}^{3} (\frac{600}{2400}) q_{2} = \frac{q_{1}}{2} = \frac{Q}{2}$

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