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A spherical drop of radius $10^{- 6} m$ has absorbed 20 electrons. The energy required to give an additional electron to it is

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$4.6 \times 10^{- 21} J$

$5.7 \times 10^{- 21} J$

$9.21 \times 10^{- 23} J$

$5.96 \times 10^{24} J$

answer is A.

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Detailed Solution

$\begin{matrix} U = \frac{1}{4 π \in_{0}} \frac{q_{1} q_{2}}{r} \end{matrix}$

$= 9 \times 10^{9} \times \frac{20 \times {(1.6 \times 10^{- 19})}^{2}}{10^{- 6}} = 4 \cdot 6 \times 10^{- 21} J$

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