JEE Main 2022 · 24 Jun · Shift 1 · Physics Q38

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PhysicsSingle correct+4 / -1Question 38 of 90

A vertical electric field of magnitude $4.9\times10^5$ 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 $g=9.8\ \mathrm{m/s^2}$ )

1.6\times10^{-9}

2.0\times10^{-9}

3.2\times10^{-9}

0.5\times10^{-9}

Answer and worked solution

Official answer: B

Balance. $qE=mg$ , so $q=\dfrac{mg}E=\dfrac{10^{-4}\times9.8}{4.9\times10^5}=2.0\times10^{-9}$ C.

(B)

Topics:Electric field

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JEE Main 2022 · 24 Jun · Shift 1, Physics question 38