question_answer

A charge q is placed at the centre of line joining two equal charges Q. The system of three charges will be in equilibrium, if q is equal to:

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, \[{{F}_{Qq}}+{{F}_{QQ}}=0\] \[=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{Qq}{{{(x/2)}^{2}}}+\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{QQ}{{{x}^{2}}}=0\] \[\Rightarrow \] \[q=-\frac{q}{4}\]