A) \[Q/2\]
B) \[-Q/2\]
C) \[Q/4\]
D) \[-Q/4\]
Correct Answer: D
Solution :
Let charge q be placed at mid-point of line AB as shown below. Also \[AB=x\] (say) \[\therefore \] \[AC=\frac{x}{2},BC=\frac{x}{2}\] For the system to be in equilibrium, \[{{F}_{Qq}}\,+{{F}_{OQ}}\,=0\] \[\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{Qq}{{{(x/2)}^{2}}}+\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{QQ}{{{x}^{2}}}=0\] \[\Rightarrow \] \[q=-\frac{Q}{4}\]You need to login to perform this action.
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