question_answer

A charge q is placed at the centre of the line joining two equal charges Q. The system of the 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 :

 In equilibrium position \[{{F}_{AB}}+{{F}_{AC}}=0\] \[\Rightarrow \] \[\frac{1}{4\pi {{\varepsilon }_{0}}}\left[ \frac{Qq}{{{d}^{2}}/4}+\frac{{{Q}^{2}}}{{{d}^{2}}} \right]=0\] \[\Rightarrow \] \[\frac{4Qq}{{{d}^{2}}}=-\frac{{{Q}^{2}}}{{{d}^{2}}}\] \[\therefore \] \[q=-\frac{Q}{4}\]