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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 ...........
$3$
$9$
$6$
$12$
Solution
$E _{ A }=\frac{\lambda}{2 \pi \varepsilon_0 r _{ A }}-\frac{\sigma}{2 \varepsilon_0}\left\{ r _{ A }=\frac{3}{\pi}\right\}$
$=\frac{1}{2 \varepsilon_0}\left\lfloor\frac{\lambda}{3}-\sigma\right\rfloor$
$E _{ B }=\frac{\lambda}{2 \pi \varepsilon_0 r _{ A }}-\frac{\sigma}{2 \varepsilon_0}\left\{ I _{ B }=\frac{4}{\pi}\right\}$
$=\frac{1}{2 \varepsilon_0}\left\lfloor\frac{\lambda}{4}-\sigma\right\rfloor$
$\frac{ E _{ A }}{ E _{ B }}=\frac{4}{3}\left(\frac{\lambda-3 \sigma}{\lambda-4 \sigma}\right)$
$=\frac{4}{3}\left\lfloor\frac{2 \sigma-3 \sigma}{2 \sigma-4 \sigma}\right\rfloor$
$=\frac{4}{3}\left\lfloor\frac{-\sigma}{-2 \sigma}\right\rfloor$
$=\frac{4}{6}$
