Published by:
CGP EDU Academic Team
Published on: September 12, 2026
The intensity of the electric field required to keep a water drop of radius $10^{-5} \, \mathrm{cm}$ just suspended in air when charged with one electron is approximately $(g = 10 newton/kg, e = 1.6 \times 10^{-19} coulomb)$
Text Solution
Verified by ExpertsThe correct answer is:
B
For balance mg = eE ⇒ ⇒ $E = \frac{m g}{e}$
Also $m = \frac{4}{3} \pi r^{3} d = \frac{4}{3} \times \frac{22}{7} \times (10^{-7})^{3} \times 1000kg$
⇒ ⇒ $E = \frac{4/3 \times 22/7 \times (10^{-7})^{3} \times 1000 \times 10}{1.6 \times 10^{-19}}$ = 260 N/C
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