question_answer

If a charge q is placed at the centre of the line joining two equal charges Q such that the system is in equilibrium, then the value of q is [AIEEE 2002]

A) \[Q/2\]

B) \[-Q/2\]

C) \[Q/4\]

D) \[-Q/4\]

Correct Answer: D

Solution :

[d] Let charge q be placed at mid-point of line AB as shown below.

Also                           \[AB=x\]                          (say)

\[\therefore \]      \[AC=\frac{x}{2},BC=\frac{x}{2}\]

For the system to be in equilibrium,

\[{{F}_{Qq}}+{{F}_{QQ}}=0\]

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

\[\Rightarrow \]                           \[q=-\frac{Q}{4}\]