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A vertical electric field of magnitude \mathrm{4.9\times 10^{5}N/C} just prevents a water droplet of a mass \mathrm{0.1\, g} from falling. The value of charge on the droplet will be :

(\mathrm{Given\; g= 9.8\: m/s^{2}})

Option: 1

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


Option: 2

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


Option: 3

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


Option: 4

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


Answers (1)

best_answer


\mathrm{qE= mg}
\mathrm{q\left ( 4.9\times 10^{5} \right )= 10^{-4}\times 9.8}
\mathrm{q= 2\times 10^{-9}C}

The correct option is (2)

Posted by

Divya Prakash Singh

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