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 ........ \(\times 10^{-9} \;C\) \(\left(\right.\) Given \(\left.g =9.8 m / s ^{2}\right)\)

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Answer & Solution

Correct Answer

(B) \(2.0 \times 10^{-9} C\)

Step-by-step Solution

Detailed explanation

\(Mg = qE\)

\(\left(0.1 \times 10^{-3}\right)(9.8)=4.9 \times 10^{5} q\)

\(\frac{2 \times 10^{-4}}{10^{5}}= q\)

\(q=2 \times 10^{-9} C\)

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