A vertical electric field of magnitude 4.9 × 10^{5} N / C just prevents a water droplet of a mass 0

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1. Electric Charges and Fields

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

A

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

B

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

C

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

D

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

(JEE MAIN-2022)

Solution

$Mg = qE$

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

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

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

Standard 12

Physics

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(JEE MAIN-2019)