JEE Mains · Physics · STD 12 - Electric charges and fields
- JEE Mains 2022
- Hard
- MCQ
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 ...............
Pick an option to check your answer.
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 \(x=\frac{d}{2 \sqrt{2}}\)
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