JEE Mains · Physics · STD 12 - 1. Electric charges and fields
Two equal positive point charges are separated by a distance \(2 a\). The distance of a point from the centre of the line joining two charges on the equatorial line (perpendicular bisector) at which force experienced by a test charge \(q_0\) becomes maximum is \(\frac{a}{\sqrt{x}}\). The value of \(x\) is \(................\)
- A \(4\)
- B \(2\)
- C \(8\)
- D \(10\)
Answer & Solution
Correct Answer
(B) \(2\)
Step-by-step Solution
Detailed explanation
\(F=\frac{2 K q q_0 x}{\left(x^2+a^2\right)^{3 / 2}}\) For \(F\) to be maximum \(\frac{ d F}{ d x}=0\) \(x=\frac{a}{\sqrt{2}}\)
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