A system of three charges are placed as shown in the figure : |
If \[D>>d,\]the potential energy of the system is best given by: [JEE Main 9-4-2019 Morning] |
A) \[\frac{1}{4\pi {{\varepsilon }_{0}}}\left[ -\frac{{{q}^{2}}}{d}-\frac{qQd}{2{{D}^{2}}} \right]\]
B) \[\frac{1}{4\pi {{\varepsilon }_{0}}}\left[ +\frac{{{q}^{2}}}{d}+\frac{qQd}{{{D}^{2}}} \right]\]
C) \[\frac{1}{4\pi {{\varepsilon }_{0}}}\left[ -\frac{{{q}^{2}}}{d}+\frac{2qQd}{{{D}^{2}}} \right]\]
D) \[\frac{1}{4\pi {{\varepsilon }_{0}}}\left[ -\frac{{{q}^{2}}}{d}-\frac{qQd}{{{D}^{2}}} \right]\]
Correct Answer: D
Solution :
[d] |
\[{{U}_{total}}={{U}_{self\,of\,dipole\,}}+{{U}_{\operatorname{int}eraction}}\] |
\[=-\frac{k{{q}^{2}}}{d}-\left( \frac{kQ}{{{D}^{2}}} \right)qd\] |
\[=-k\left[ \frac{{{q}^{2}}}{d}+\frac{qQd}{{{D}^{2}}} \right]\] |
Option |
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