question_answer

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

A)  \[-\frac{Q}{4}\]

B)  \[-\frac{Q}{2}\]

C)  \[+\frac{Q}{2}\]

D)  \[+\frac{Q}{4}\]

Correct Answer: A

Solution :

 The force on Q at A due to Q at B is given by \[F=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{QQ}{{{a}^{2}}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{Q}^{2}}}{{{a}^{2}}}\] ?(1) the force on Q at C due to q is given by \[F=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{Qq}{{{\left( \frac{a}{2} \right)}^{2}}}\] ...(2) These forces should be equal and opposite for the equilibrium of charge Q at A. This. is only posible when q is negative \[\frac{{{Q}^{2}}}{{{a}^{2}}}=-\frac{Qq}{{{(a/2)}^{2}}}\]or\[q=-\frac{Q}{4}\] Thus\[q=-\frac{Q}{4}\]