An oil drop with charge $n e$ is held stationary between two plates with an external potential difference of 500 $v o l t$ . If the size of the drop is doubled without any change of charge, the potential difference required to keep the drop stationary will be

500 V

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1000 V

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2000 V

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4000 V

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Solution

The correct option is D $4000 V$
When the oil drop is stationary we have
$m g = q E$

$⟹ \frac{4}{3} π r^{3} ρ g = \frac{q V}{d}$

$⟹ V \propto r^{3}$

Thus, $V_{2} = V_{1} (\frac{r_{2}}{r_{1}})^{3} = 500 \times (\frac{2}{1})^{3} = 4000 V$

So, the answer is option (D).

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An oil drop with charge ne is held stationary between two plates with an external potential difference of 500 volt. If the size of the drop is doubled without any change of charge, the potential difference required to keep the drop stationary will be

The correct option is D 4000VWhen the oil drop is stationary we havemg=qE⟹43πr3ρg=qVd⟹V∝r3Thus, V2=V1(r2r1)3=500×(21)3=4000VSo, the answer is option (D).

An oil drop with charge $n e$ is held stationary between two plates with an external potential difference of 500 $v o l t$ . If the size of the drop is doubled without any change of charge, the potential difference required to keep the drop stationary will be

The correct option is D $4000 V$
When the oil drop is stationary we have
$m g = q E$

$⟹ \frac{4}{3} π r^{3} ρ g = \frac{q V}{d}$

$⟹ V \propto r^{3}$

Thus, $V_{2} = V_{1} (\frac{r_{2}}{r_{1}})^{3} = 500 \times (\frac{2}{1})^{3} = 4000 V$

So, the answer is option (D).