JEE Mains · Physics · STD 12 - Electric charges and fields
- JEE Mains 2023
- Hard
- Numerical
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 \(................\)
Pick an option to check your answer.
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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