A vertical electric field of magnitude 4 . 9 × 10 5 N C - 1 just prevents a water droplet of a mass 0 . 1 g from falling. The value of charge on the droplet will be : (Given g = 9 . 8 m s - 2 )

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TestHub · Accepted Answer

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

For translational equilibrium along vertical direction,

$q E = m g$

$\Rightarrow q = \frac{m g}{E} = \frac{0.1 \times 10^{- 3} \times 9.8}{4.9 \times 10^{5}} = 2.0 \times 10^{- 9} C$

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