JEE Mains · Physics · STD 12 - 1. Electric charges and fields
Two point charges \(Q\) each are placed at a distance \(d\) apart. A third point charge \(q\) is placed at a distance \(x\) from mid-point on the perpendicular bisector. The value of \(x\) at which charge \(q\) will experience the maximum \(Coulomb's force\) is ...............
- A \(x=d\)
- B \(x=\frac{d}{2}\)
- C \(x=\frac{d}{\sqrt{2}}\)
- D \(x=\frac{d}{2 \sqrt{2}}\)
Answer & Solution
Correct Answer
(D) \(x=\frac{d}{2 \sqrt{2}}\)
Step-by-step Solution
Detailed explanation
\(F =\frac{ KQq }{\left( x ^{2}+\frac{ d ^{2}}{4}\right)}\) Net force on \(g =2 F \cos \theta\) \(F _{\text {act }}=\frac{2 KQqx }{\left( x ^{2}+\frac{ d ^{2}}{4}\right)^{3 / 2}}\) For maximum \(F _{\text {act }}\) \(\frac{ d F _{\text {net }}}{ dx }=0\) we get…
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