A) \[\frac{Q}{2}\]
B) \[-\frac{Q}{2}\]
C) \[\frac{Q}{4}\]
D) \[-\frac{Q}{4}\]
Correct Answer: D
Solution :
When two charges\[{{q}_{1}}\]and\[{{q}_{2}}\]are situated in vacuum at a distance r metre, then the electric potential energy of the system is \[U=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{{{q}_{1}}{{q}_{2}}}{r}joule\] In order that system be in equilibrium, the potential energy of the system should be zero. \[\therefore \] \[\Sigma U=\frac{1}{4\pi {{\varepsilon }_{0}}}\left[ \frac{qQ}{r/2}+\frac{qQ}{r/2}+\frac{Q\times Q}{r} \right]\] \[0=\frac{2qQ}{r}+\frac{2qQ}{r}+\frac{{{Q}^{2}}}{r}\] \[\Rightarrow \] \[4q=-Q\] \[\Rightarrow \] \[q=-\frac{Q}{4}\]You need to login to perform this action.
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