A water drop of mass m has charge 2e, where e is an electronic charge. To keep the water drop, just to remain suspended in air, the required electric field will be (g is acceleration due to gravity)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → mg/2e
The drop remains suspended when the upward electric force balances its weight.
\( \displaystyle qE = mg \)
\( \displaystyle E = \frac{mg}{q} \)
With \( q = 2e \):
\( \displaystyle E = \frac{mg}{2e} \)