A) \[3\frac{V}{2}\]
B) \[\frac{V}{4}\]
C) \[\frac{V}{2}\]
D) V
Correct Answer: B
Solution :
\[\frac{V}{2}=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{q}{R}\text{(q=charge}\,\text{on}\,\text{smaller}\,\text{sphere)}\] \[\therefore \]\[q=(2\pi {{\varepsilon }_{0}}R)V\] \[PD+\frac{q}{4\pi {{\varepsilon }_{0}}}\left[ \frac{1}{R}-\frac{1}{2R} \right]=\frac{VR}{2}\left( \frac{1}{2R} \right)=\frac{V}{4}\] Hence, the correction option is [b].You need to login to perform this action.
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