question_answer

Three charges +Q, q, +Q are placed respectively, at distance. 0. d/2 and d from the origin, on the x-axis. If the net force experienced by +Q, placed at  \[x=0,\]is zero. then value of q is:

A) \[+Q\text{/2}\]

B) \[-\text{Q/2}\]

C) \[-\text{Q/4}\]

D) \[+Q\text{/4}\]

Correct Answer: C

Solution :

For equilibrium,  \[{{\text{\vec{F}}}_{a}}+{{\text{\vec{F}}}_{b}}\,=\,0\]

\[{{\vec{F}}_{a}}\,=\,-{{\vec{F}}_{b}}\]

\[\frac{\text{KQQ}}{{{d}^{2}}}\,=\,\frac{\text{KQq}}{{{\text{(d/2)}}^{\text{2}}}}\]

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