A thin infinite sheet charge and an infinite | vedclass

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Question

$E _{ A }=\frac{\lambda}{2 \pi \varepsilon_0 r _{ A }}-\frac{\sigma}{2 \varepsilon_0}\left\{ r _{ A }=\frac{3}{\pi}\right\}$

$=\frac{1}{2 \varepsil

Vedclass · Accepted Answer

$6$

Vedclass · Answer

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

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