JEE Mains · Physics · STD 12 - 1. Electric charges and fields
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
- A \(-\frac Q2\)
- B \(+\frac Q2\)
- C \(+\frac Q4\)
- D \(-\frac Q4\)
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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