Charges \[+q\] and \[-q\] are placed at points A and B respectively which are a distance 2 L apart, C is the midpoint between A and S. The work done in moving a charge +Q along the semicircle CRD is:[AIPMT (S) 2007] |
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 :
[d] 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 \[=\Delta U={{U}_{2}}-{{U}_{1}}\] |
\[=\left( -\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}^{2}}}{2L}+\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{qQ}{3L}-\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{qQ}{L} \right)\] |
\[-\left( -\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}^{2}}}{2L}-\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{qQ}{L}+\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{qQ}{L} \right)\] |
\[=\frac{qQ}{4\pi {{\varepsilon }_{0}}}.\left[ \frac{1}{3L}-\frac{1}{L} \right]=\frac{qQ}{4\pi {{\varepsilon }_{0}}}\frac{(1-3)}{3L}\] |
\[=\frac{-2qQ}{12\pi {{\varepsilon }_{0}}L}=-\frac{qQ}{6\pi {{\varepsilon }_{0}}L}\] |
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