A vertical electric field of magnitude \( 4.9 \times 10^{5}\,\mathrm{NC^{-1}} \) just prevents a water droplet of mass 0.1 g from falling. What is the charge on the droplet ? Take acceleration due to gravity \( 9.8\,\mathrm{ms^{-2}} \).
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 2.0 \times 10^{-9}\,\mathrm{C} \)
The droplet is held at rest, so the upward electric force balances its weight.
\( \displaystyle qE = mg \;\Rightarrow\; q = \frac{mg}{E} \)
\( \displaystyle m = 0.1\,\mathrm{g} = 0.1\times10^{-3}\,\mathrm{kg} = 1\times10^{-4}\,\mathrm{kg} \)
\( \displaystyle q = \frac{1\times10^{-4} \times 9.8}{4.9\times10^{5}} = 2.0\times10^{-9}\,\mathrm{C} \)