JEE MAIN - Physics (2022 - 24th June Morning Shift - No. 8)
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)
1.6 $$\times$$ 10$$-$$9 C
2.0 $$\times$$ 10$$-$$9 C
3.2 $$\times$$ 10$$-$$9 C
0.5 $$\times$$ 10$$-$$9 C
Explanation
Since the droplet is at rest
$$\Rightarrow$$ Net force = 0
$$\Rightarrow$$ mg = qE
$$\Rightarrow$$ $$q = {{mg} \over E}$$ = 2 $$\times$$ 10$$-$$9 C
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