A) \[\frac{{{Q}^{2}}}{4\pi {{\varepsilon }_{0}}}\]
B) \[\frac{{{Q}^{2}}}{2\sqrt{2}\pi {{\varepsilon }_{0}}}\]
C) \[\frac{{{Q}^{2}}}{4\pi {{\varepsilon }_{0}}}\left( 1+\frac{1}{\sqrt{3}} \right)\]
D) \[\frac{{{Q}^{2}}}{4\pi {{\varepsilon }_{0}}}\left( 1+\frac{1}{\sqrt{5}} \right)\]
Correct Answer: D
Solution :
\[W=q\Delta V=q\,\left[ {{V}_{0}}-{{V}_{\infty }} \right]=q{{V}_{0}}\] Potential at origin \[=\text{ }{{V}_{0}}\] \[{{V}_{0}}\,=\,2.\frac{kQ}{2}+2.\frac{kQ}{2\sqrt{5}}\] \[=\,\,kQ\left( 1+\frac{1}{\sqrt{5}} \right)\] \[\therefore \] Word done \[=\text{ }kqQ\left( 1+\frac{1}{\sqrt{5}} \right)\] \[=\,\,\frac{{{Q}^{2}}}{4\pi {{\varepsilon }_{0}}}\,\left( 1+\frac{1}{\sqrt{5}} \right)\]
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