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  • The intensity of the electric field required to keep a water drop of radius <img alt="10^{-5} \mathrm{~cm}" src="https://entrancecorner.oncodecogs.com/gif.latex?10%5E%7B-5%7D%20%5Cmathrm%7B%7Ecm%7D

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 -

Option: 1

260 \mathrm{~V} / \mathrm{m}


Option: 2

260 \mathrm{~N} / \mathrm{C}


Option: 3

130 \: \mathrm{V} / \mathrm{cm}


Option: 4

130 \mathrm{~N} / \mathrm{C}


Answers (1)

best_answer

From figure ,


e E=m g

E=\frac{m g}{e}

=\frac{\frac{4}{3} \pi r^{3} \rho \cdot g}{e}
\text { but, density }=\frac{\text { mass }}{\text { volume }}

E=\frac{\frac{4}{3} \times \frac{22}{7} \times\left(10^{-5} \times 10^{-2}\right)^{3} \times 1 \times 10}{1.6 \times 10^{-19}}
\rho=\frac{m}{\frac{4}{3} \pi r^{3}}
m=\frac{4}{3} \pi r^{3} \rho

E=260 \mathrm{~N} / \mathrm{C} Ans.

Posted by

rishi.raj

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