Q. The intensity of the electric field required to keep a water drop of radius $10^{- 5}cm$ just suspended in air when charged with one electron approximately $(g = 10\ newton/kg,e = 1.6 \times 10^{- 19}\ coulomb)$
A) $260\ volt/cm$
B) $260\ newtons/coulomb$
C) $130\ volt/cm$
D) $130\ newton/coulomb$
Explanation
(b) For balance $mg = eE\ \Rightarrow E = \frac{mg}{e}$
Also $m = \frac{4}{3}\pi r^{3}d = \frac{4}{3} \times \frac{22}{7} \times \left( 10^{- 7} \right)^{3} \times 1000\ kg$
$\Rightarrow E = \ \frac{4/3 \times 22/7 \times \left( 10^{- 7} \right)^{3} \times 1000 \times 10}{1.6 \times 10^{- 19}} = 260\ N/C$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.