JEE Mains · Physics · STD 12 - Electric charges and fields
- JEE Mains 2023
- Hard
- Numerical
A thin infinite sheet charge and an infinite line charge of respective charge densities \(+\sigma\) and \(+\lambda\) are placed parallel at \(5\,m\) distance from each other. Points \(P\) and \(Q\), are at \(\frac{3}{\pi} m\) and \(\frac{4}{\pi} m\) perpendicular distance from line charge towards sheet charge, respectively. \(E_P\) and \(E_Q\) are the magnitudes of resultant electric field intensities at point \(P\) and \(Q\), respectively. If \(\frac{E_p}{E_Q}=\frac{4}{a}\) for \(2|\sigma|=|\lambda|\). Then the value of \(a\) is ...........
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(6\)
Step-by-step Solution
Detailed explanation
\(E _{ A }=\frac{\lambda}{2 \pi \varepsilon_0 r _{ A }}-\frac{\sigma}{2 \varepsilon_0}\left\{ r _{ A }=\frac{3}{\pi}\right\}\)
\(=\frac{1}{2 \varepsilon_0}\left\lfloor\frac{\lambda}{3}-\sigma\right\rfloor\)
\(E _{ B }=\frac{\lambda}{2 \pi \varepsilon_0 r _{ A }}-\frac{\sigma}{2 \varepsilon_0}\left\{ I _{ B }=\frac{4}{\pi}\right\}\)
\(=\frac{1}{2 \varepsilon_0}\left\lfloor\frac{\lambda}{4}-\sigma\right\rfloor\)
\(\frac{ E _{ A }}{ E _{ B }}=\frac{4}{3}\left(\frac{\lambda-3 \sigma}{\lambda-4 \sigma}\right)\)
\(=\frac{4}{3}\left\lfloor\frac{2 \sigma-3 \sigma}{2 \sigma-4 \sigma}\right\rfloor\)
\(=\frac{4}{3}\left\lfloor\frac{-\sigma}{-2 \sigma}\right\rfloor\)
\(=\frac{4}{6}\)
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