A charge of $10^{- 9} C$ is placed on each of the 64 identical drops of radius 2 cm. They are then combined to form a bigger drop. Find its potential

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$7 .2 \times 10^{3} V$

$7 .2 \times 10^{2} V$

$1 .44 \times 10^{2} V$

$1 .44 \times 10^{3} V$

answer is A.

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Detailed Solution

$\begin{array}{l} V = n^{2 / 3} v \\ \Rightarrow V = (64)^{2 / 3} \times 9 \times 10^{9} \times \frac{10^{- 9}}{(2 \times 10^{- 2})} \\ = 7 .2 \times 10^{3} V \end{array}$

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