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A small oil drop of mass $10^{- 6}$ kg is hanging in at rest between two plates separated by 1 mm having a
potential difference of 500 V. The charge on the drop is ( $g = 10 {ms}^{- 2}$ )

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$2 \times 10^{- 9} C$

$2 \times 10^{- 11} C$

$2 \times 10^{- 6} C$

$2 \times 10^{- 7} C$

answer is B.

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Detailed Solution

As, F = mg = qE

$∴ E = \frac{mg}{q} or q = \frac{mg}{E}$

Also, $E = \frac{V}{d}$

$∴ q = \frac{mgd}{V} = \frac{(10^{- 6} \times 10 \times 1 \times 10^{- 3})}{500} = 2 \times 10^{- 11} C$

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