Practicing Success
If ‘n’ identical drops each charged to a potential V coalesce to a single bigger drop, the potential of the bigger drop is |
$n^{1/3} V$ $n^{2/3} V$ $n^{4/3} V$ $n^{5/3} V$ |
$n^{2/3} V$ |
$\text{ If charge on each drop is q then total charge on bigger drop is } nq.$ $\text{Since Volume remain constant then } n\times \frac{4\pi}{3} r^3 = \frac{4\pi}{3} R^3 $ $ \Rightarrow R = n^{1/3} r$ $ \text{Potential of bigger drop is }V' = \frac{k.nq}{R} = n^{2/3} \frac{kq}{r} = n^{2/3}V$
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