question_answer

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

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

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

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

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

Correct Answer: D

Solution :

Key Idea Work done is equal to change in potential energy. In Ist case, when charge \[+Q\] is situated at\[C\]. Electric potential energy of system \[{{U}_{1}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{(q)(-q)}{2L}+\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{(-q)Q}{L}+\frac{1}{4\pi {{\varepsilon }_{0}}}\cdot \frac{qQ}{L}\] In IInd case, when charge \[+Q\] is moved from \[C\] to \[D\]. Electric potential energy of system in that case                 \[{{U}_{2}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\cdot \frac{(q)(-q)}{2L}+\frac{1}{4\pi {{\varepsilon }_{0}}}\cdot \frac{qQ}{3L}\]                                                 \[+\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{(-q)(Q)}{L}\] Work done\[=\Delta U={{U}_{2}}-{{U}_{1}}\] \[=\left( -\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}^{2}}}{2L}-\frac{1}{4\pi {{\varepsilon }_{0}}}\cdot \frac{qQ}{L}+\frac{1}{4\pi {{\varepsilon }_{0}}}\cdot \frac{qQ}{L} \right)\] \[-\left( -\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}^{2}}}{2L}-\frac{1}{4\pi {{\varepsilon }_{0}}}\cdot \frac{qQ}{L}+\frac{1}{4\pi {{\varepsilon }_{0}}}+\cdot \frac{qQ}{L} \right)\] \[=\frac{qQ}{4\pi {{\varepsilon }_{0}}}\cdot \left[ \frac{1}{3L}-\frac{1}{L} \right]=\frac{qQ}{4\pi {{\varepsilon }_{0}}}\frac{(1-3)}{3L}\] \[=\frac{-2qQ}{12\pi {{\varepsilon }_{0}}L}=-\frac{qQ}{6\pi {{\varepsilon }_{0}}L}\]