The intensity of the electric field required | vedclass

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Question

(B) For the water drop to remain suspended in air,the electric force must balance the gravitational force: $qE = mg$.Given $q = e = 1.6 	imes 10^{-19

Vedclass · Accepted Answer

$260 \, N/C$

Vedclass · Answer

(B) For the water drop to remain suspended in air,the electric force must balance the gravitational force: $qE = mg$.Given $q = e = 1.6 	imes 10^{-19} \, C$ and radius $r = 10^{-5} \, cm = 10^{-7} \, m$.The mass $m$ of the water drop is $m = 	ext{Volume} 	imes 	ext{Density} = \frac{4}{3} \pi r^3 ho$.Taking density $ho = 1000 \, kg/m^3$ and $g = 10 \, m/s^2$:$m = \frac{4}{3} 	imes \frac{22}{7} 	imes (10^{-7})^3 	imes 1000 = \frac{4}{3} 	imes \frac{22}{7} 	imes 10^{-21} 	imes 10^3 = 4.19 	imes 10^{-18} \, kg$.Now,$E = \frac{mg}{e} = \frac{4.19 	imes 10^{-18} 	imes 10}{1.6 	imes 10^{-19}} \approx 262 \, N/C$.Rounding to the nearest given option,the intensity is approximately $260 \, N/C$.

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 is approximately:

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