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