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A vertical electric field of magnitude $4.9 \times 10^{5} N / C$ just prevents a water droplet of a mass $0.1 g$ from falling. The value of charge on the droplet will be : $($ Given $g = 9.8 m / s^{2})$

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$1.6 \times 10^{- 9} C$

$2.0 \times 10^{- 9} C$

$3.2 \times 10^{- 9} C$

$0.5 \times 10^{- 9} C$

answer is B.

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Detailed Solution

$Mg = qE$

$(0 .1 \times 10^{- 3}) (9 .8) = 4 .9 \times 10^{5} q$

$\Rightarrow q = \frac{2 \times 10^{- 4}}{10^{5}}$

$\Rightarrow q = 2 \times 10^{- 9} C$

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