A vertical electric field of magnitude 4.9 $\times$ 105 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/s2)
Solution
<p>Since the droplet is at rest</p>
<p>$\Rightarrow$ Net force = 0</p>
<p>$\Rightarrow$ mg = qE</p>
<p>$\Rightarrow$ $q = {{mg} \over E}$ = 2 $\times$ 10<sup>$-$9</sup> C</p>
About this question
Subject: Physics · Chapter: Electrostatics · Topic: Electric Field and Field Lines
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