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  Solved  Paper-2002

A) \[Q/2\]      

B)                           \[-Q/2\]

C) \[Q/4\]                   

D)           \[-Q/4\]

Correct Answer: D

Solution :

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