A vertical electric field of magnitude 4.9 × 10^5 N/C just prevents a water droplet of a mass 0.1 g from falling.

Physics | JEE Mains 2022 | 24 - June Shift - 1
Chapter : Electric Charges and Fields | Topic : Electric Field and Electric Field Lines

A vertical electric field of magnitude 4.9 x \(\tt 10^{5}\) N/C just prevents a water droplet of a mass 0.1g from failing. The value charge on the droplet will be : Given g = 9.8 \(\tt m/s^{2}\)

\(\tt 1.6 \times 10^{-9}\) C

\(\tt 2.0 \times 10^{-9}\) C

\(\tt 3.2 \times 10^{-9}\) C

\(\tt 0.5 \times 10^{-9}\) C

Solution : B
Since the droplet is at rest
=> Net force = 0
=> mg = qE
=> q = \(\tt \frac{mg}{E}\)
=> \(\tt 2 \times 10^{-9}\) C