A vertical electric field of magnitude 4 . 9 × 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 )

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Correct answer: B. 2 . 0 × 10 ^ - 9 C

A vertical electric field of magnitude 4 . 9 × 10 ^ 5 N C ^ - 1 just prevents a water droplet of a mass 0 . 1g from

To balance gravitational force, electric force on the liquid drop should act in upward direction.

For translational equilibrium along vertical direction,

$qE = mg$

$⇒ q = \dfrac{mg}{E} = \dfrac{0 . 1 \times {10}^{- 3} \times 9 . 8}{4 . 9 \times {10}^{5}} = 2 . 0 \times {10}^{- 9}C$

A vertical electric field of magnitude 4 . 9 × 10 ^ 5 N C ^ - 1...

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