A) \[+0.5\mu C\]
B) \[-0.5\mu C\]
C) \[1\mu C\]
D) \[-1\mu C\]
E) none of above
Correct Answer: B
Solution :
From Coulombs law \[F=\frac{1}{4\pi {{\varepsilon }_{0}}}\,\frac{{{Q}^{2}}}{2{{a}^{2}}}\] \[{{F}_{1}}\]and\[{{F}_{2}}\]will be directed as shown, for this both q should be negative \[{{F}_{1}}={{F}_{2}}=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{Qq}{{{a}^{2}}}\] \[{{F}_{12}}=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{\sqrt{2}Qq}{{{a}^{2}}}\] For equilibrium of Q, we have \[F=-{{F}_{12}}\] \[\frac{{{Q}^{2}}}{2{{a}^{2}}}=\frac{\sqrt{2}Qq}{{{a}^{2}}}\] \[\Rightarrow \] \[q=\frac{Q}{2\sqrt{2}}\] Given, \[Q=+\sqrt{2}\mu C\] \[\therefore \] \[q=-\frac{\sqrt{2}\mu C}{2\sqrt{2}}=-0.5\,\mu C\]You need to login to perform this action.
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