JEE Mains · Physics · STD 12 - Electric charges and fields
- JEE Mains 2019
- Hard
- MCQ
Three charges \(+Q, q, +Q\) are placed respectively, at distance, \(0, \frac d2\) 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
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Answer & Solution
Correct Answer
(D) \(-\frac Q4\)
Step-by-step Solution
Detailed explanation
For equilibrium,
\(\vec{F}_{a}+\vec{F}_{B}=0\)
\(\vec{F}_{a}=-\overrightarrow{F_{B}}\)
\(\frac{k Q Q}{d^{2}}=-\frac{k Q q}{(d / 2)^{2}}\)
\(\Rightarrow q=-\frac{Q}{4}\)
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