A) \[\frac{qQ}{2\pi {{\varepsilon }_{0}}L}\]
B) \[\frac{qQ}{6\pi {{\varepsilon }_{0}}L}\]
C) \[-\frac{qQ}{6\pi {{\varepsilon }_{0}}L}\]
D) \[\frac{qQ}{4\pi {{\varepsilon }_{0}}L}\]
Correct Answer: C
Solution :
Potential at \[C={{V}_{C}}=0\] |
Potential at \[D={{V}_{D}}\]\[=K\left( \frac{-q}{L} \right)+\frac{Kq}{3L}=-\frac{2}{3}\frac{Kq}{L}\] |
Potential difference \[{{V}_{D}}-{{V}_{C}}=\frac{-2}{3}\frac{Kq}{L}=\frac{1}{4\pi {{\in }_{0}}}\left( -\frac{2}{3}.\frac{q}{L} \right)\] |
\[\Rightarrow \] Work done \[=Q({{V}_{D}}-{{V}_{C}})\]\[=-\frac{2}{3}\times \frac{1}{4\pi {{\varepsilon }_{0}}}\frac{qQ}{L}=\frac{-qQ}{6\pi {{\varepsilon }_{0}}L}\] |
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