\(=\frac{1}{2 \varepsilon_0}\left\lfloor\frac{\lambda}{3}-\sigma\right\rfloor\)

\(E _{ B }=\frac{\lambda}{2 \pi \varepsilon_0 r _{ A }}-\frac{\sigma}{2 \varepsilon_0}\left\{ I _{ B }=\frac{4}{\pi}\right\}\)

\(=\frac{1}{2 \varepsilon_0}\left\lfloor\frac{\lambda}{4}-\sigma\right\rfloor\)

\(\frac{ E _{ A }}{ E _{ B }}=\frac{4}{3}\left(\frac{\lambda-3 \sigma}{\lambda-4 \sigma}\right)\)

\(=\frac{4}{3}\left\lfloor\frac{2 \sigma-3 \sigma}{2 \sigma-4 \sigma}\right\rfloor\)

\(=\frac{4}{3}\left\lfloor\frac{-\sigma}{-2 \sigma}\right\rfloor\)

\(=\frac{4}{6}\)