A vertical electric field of magnitude $4.9 × 105 NC^{-1}$ just prevents a water droplet of mass 0.1 g from falling. What is the charge on the droplet ? Take acceleration due to gravity $9.8 ms^{-2}$.

$2.0 × 10^{-9} C$

The correct answer is Option (1) → $2.0 × 10^{-9} C$

The gravitational force acting on droplet -

$F_{gravity}=m.g$

$=(0.1×10^{-3})×9.8$

$=9.8×10^{-4}N$

The electric force on charge droplet,

$F_{electric}=qE$

$=q×(4.9×10^5)N/C$

$∴F_{gravity}=F_{electric}$

$q=\frac{9.8×10^{-4}}{4.9×10^5}$

$=2.0×10^{-9}C$