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The intensity of the electric field required to keep a water drop of radius \small 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

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


Answers (1)

best_answer

From figure, 

                                           

\begin{aligned} e E & =m g \\ E & =\frac{m g}{e} \\ & =\frac{\frac{4}{3} \pi r^3 \rho \cdot g}{e} \end{aligned}                                                                         \text { but, density }=\frac{\text { mass }}{\text { volume }}

\begin{aligned} & \rho=\frac{m}{\frac{4}{3} \pi r^3} \\ & m=\frac{4}{3} \pi r^3 \rho \end{aligned}                                          

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}}        

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

Posted by

chirag

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