| Four equal point charges Q each are placed in the xy plane at |
| \[\left( 0,\text{ }2 \right),\text{ }\left( 4,\text{ }2 \right),\text{ }\left( 4,\,-2 \right)\text{ }and\text{ }\left( 0,-\,2 \right)\]. |
| The work required to put a fifth charge Q at the origin of the coordinate system will be-[JEE Main 10-Jan-2019 Evening] |
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 :
| [d] \[W=q\Delta V=q\,\left[ {{V}_{0}}-{{V}_{\infty }} \right]=q{{V}_{0}}\] |
| 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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