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