A) \[V\]
B) \[V/N\]
C) \[V\times N\]
D) \[V\times {{N}^{2/3}}\]
Correct Answer: D
Solution :
If the drops are conducting, then \[\frac{4}{3}\pi {{R}^{3}}=N\,\left( \frac{4}{3}\pi {{r}^{3}} \right)\] Þ\[R={{N}^{1/3}}r\]. Final charge Q = Nq So final potential \[V=\frac{Q}{R}\] \[=\frac{Nq}{{{N}^{1/3}}r}=V\times {{N}^{2/3}}\]
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