question_answer

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 \[=\Delta U={{U}_{2}}-{{U}_{1}}\]

\[=\left( -\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}^{2}}}{2L}+\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{qQ}{3L}-\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{qQ}{L} \right)\]

\[-\left( -\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}^{2}}}{2L}-\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{qQ}{L}+\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{qQ}{L} \right)\]

\[=\frac{qQ}{4\pi {{\varepsilon }_{0}}}.\left[ \frac{1}{3L}-\frac{1}{L} \right]=\frac{qQ}{4\pi {{\varepsilon }_{0}}}\frac{(1-3)}{3L}\]

\[=\frac{-2qQ}{12\pi {{\varepsilon }_{0}}L}=-\frac{qQ}{6\pi {{\varepsilon }_{0}}L}\]