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