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  • An oil drop which contains 20 excess electrons is held stationary under the influence of an electric field of magnitude<img alt="6.55 \times 10^5 \mathrm{~N} / \mathrm{C}." src="https://entrancecor

An oil drop which contains 20 excess electrons is held stationary under the influence of an electric field of magnitude6.55 \times 10^5 \mathrm{~N} / \mathrm{C}. Density of the given oil is 1.27 \mathrm{~g} / \mathrm{cm}^3. Based on given information, estimate the radius of the oil drop. (Take g=9.8 \mathrm{~m} / \mathrm{s}^2 ).

Option: 1

r=34.2 \times 10^{-7} \mathrm{~m}


Option: 2

r=3.42 \times 10^{-7} \mathrm{~m}


Option: 3

r=0.342 \times 10^{-7} \mathrm{~m}


Option: 4

r=34.2 \times 10^{-8} \mathrm{~m}


Answers (1)

best_answer

Electrostatic force on drop =\mathrm{qE}=\text { neE }

Gravitational force on drop =\mathrm{mg}=\text { Volume } \times \text { Density } \times \mathrm{g}

=\frac{4}{3} \pi r^3 \times \rho \times g

As drop is held stationary, hence net force would be zero.

This gives, Electrostatic force=Gravitational force

\begin{aligned} & n e E=\frac{4}{3} \pi r^3 \times \rho \times g \\ r^3= & \frac{3 n e E}{4 \pi \rho g} \end{aligned}

After putting the desired values and calculating them, r=34.2 \times 10^{-7} \mathrm{~m} is the answer.

Posted by

Irshad Anwar

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