Q. A drop of $10^{- 6}kg$ water carries $10^{- 6}C$ charge. What electric field should be applied to balance its weight (assume $g = 10m/s^{2}$)
A) $10\ V/m$ upward
B) $10\ V/m\ $ downward
C) $0.1\ V/m$ downward
D) $0.1\ V/m$ upward
Explanation
(a) By using $QE = mg$
$\Rightarrow E = \frac{mg}{Q} = \frac{10^{- 6} \times 10}{10^{- 6}} = 10\ V/m;$ upward because charge is positive
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.