Question 97 of 311

A vertical electric field of magnitude $4.9\times {10}^{5}N{C}^{-1}$ just prevents a water droplet of a mass $0.1g$ from falling. The value of charge on the droplet will be : (Given $g=9.8m{s}^{-2}$)

A$1.6\times {10}^{-9}C$B$2.0\times {10}^{-9}C$C$3.2\times {10}^{-9}C$D$0.5\times {10}^{-9}C$

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

To balance gravitational force, electric force on the liquid drop should act in upward direction.
For translational equilibrium along vertical direction,
$qE=mg$
$\Rightarrow q=\dfrac{mg}{E}=\dfrac{0.1\times {10}^{-3}\times 9.8}{4.9\times {10}^{5}}=2.0\times {10}^{-9}C$