question_answer

Charge +q and -q are placed at points A and B respectively which are a distance 2L apart, C is the midpoint between A and B.  The work done in moving a charge +Q along the semicircle CRD is

A) \[\frac{qQ}{2\pi {{\varepsilon }_{0}}L}\]

B) \[\frac{qQ}{3\pi {{\varepsilon }_{0}}L}\]

C) \[-\frac{qQ}{6\pi {{\varepsilon }_{0}}L}\]

D) \[\frac{qQ}{4\pi {{\varepsilon }_{0}}L}\]

Correct Answer: C

Solution :

[c] 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 \[{{V}_{D}}-{{V}_{C}}=\frac{-2}{3}\frac{kq}{L}=\frac{1}{4\pi {{\varepsilon }_{0}}}\left( -\frac{2}{3}.\frac{q}{L} \right)\] \[\Rightarrow \text{ Work done}=Q\left( {{V}_{D}}-{{V}_{C}} \right)\] \[=-\frac{2}{3}\times \frac{1}{4\pi {{\varepsilon }_{0}}}\frac{qQ}{L}=\frac{-qQ}{6\pi {{\varepsilon }_{0}}L}\]