A) \[nV\]
B) \[\frac{V}{n}\]
C) \[{{n}^{2/3}}V\]
D) \[n{{V}^{2}}\]
Correct Answer: C
Solution :
Since \[n\] drops of radius r coalesced, hence \[\frac{4\pi {{R}^{3}}}{3}=n\frac{4}{3}\pi {{r}^{3}}\] \[R=r{{n}^{1/3}}\] The potential is given by the formula \[V=\frac{q}{4\pi {{\varepsilon }_{0}}r}\] Now, potential of bigger drop is \[\frac{nq}{4\pi {{\varepsilon }_{0}}R}=n\frac{q}{4\pi {{\varepsilon }_{0}}r{{n}^{1/3}}}\] \[={{n}^{1-\frac{1}{3}}}.V={{n}^{2/3}}V.\]
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