A vertical electric field of magnitude 4.9 x 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: (Given g = 9.8 m/s^2)

Question

Accepted Answer

Correct answer: B. (b) Mg= qE (0.1 10^-3) (9.8 )=4.9 10^8q 2 10^-410^5=q q=2 10^-9C

$A vertical electric field of magnitude 4.9 x 1 0^{5} N/C just prevents a water droplet of a mass 0.1 g from falling. The value of charge on the droplet will be: (Given g = 9.8 m/ s^{2})$

Solution

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