91Ó°ÊÓ

When a wire is positioned in the external electric field of a source (in this case, the charged plane), it produces a surface charge in it, culminating in a field inside the conducting that is always zero. In this scenario, the lower surface charge density of the cable will be $-\frac{\mathrm{η}}{2}$, although the upper surface charge must be $\frac{\mathrm{η}}{2}$in need for the conductor to remain quiet.

As more than just a response, we can determine that the field emitted by each charged surface is

$E_{\mathrm{A}}=\frac{\mathrm{η}}{4{\mathrm{Î\mu }}_0}$

$E_{\mathrm{B}}=\frac{\mathrm{η}}{4{\mathrm{Î\mu }}_0}$

$E_{\mathrm{C}}=\frac{\mathrm{η}}{2{\mathrm{Î\mu }}_0}$

This originates in

$$\begin{gathered} {\stackrel{→}{E}}_1=\frac{\mathrm{η}}{2{\mathrm{Î\mu }}_0}\left(\frac{1}{2}-\frac{1}{2}+1\right)\hat{\mathbf{j}} \\ =\frac{\mathrm{η}}{2{\mathrm{Î\mu }}_0}\hat{\mathbf{j}} \end{gathered}$$

$$\begin{gathered} {\stackrel{→}{E}}_2=\frac{\mathrm{η}}{2{\mathrm{Î\mu }}_0}\left(-\frac{1}{2}-\frac{1}{2}+1\right)\hat{\mathbf{j}} \\ =\stackrel{→}{0} \end{gathered}$$

$$\begin{gathered} {\stackrel{→}{E}}_3=\frac{\mathrm{η}}{2{\mathrm{Î\mu }}_0}\left(-\frac{1}{2}+\frac{1}{2}+1\right)\hat{\mathbf{j}} \\ =\frac{\mathrm{η}}{2{\mathrm{Î\mu }}_0}\hat{\mathbf{j}} \end{gathered}$$

$$\begin{gathered} {\stackrel{→}{E}}_4=\frac{\mathrm{η}}{2{\mathrm{Î\mu }}_0}\left(-\frac{1}{2}+\frac{1}{2}-1\right) \\ =-\frac{\mathrm{η}}{2{\mathrm{Î\mu }}_0}\hat{\mathbf{j}} \end{gathered}$$

Cultivated area pasture is located in Zones 1 to 4.

${\stackrel{→}{E}}_1=\frac{\mathrm{η}}{2{\mathrm{Î\mu }}_0}\hat{\mathbf{j}}$

${\stackrel{→}{E}}_2=\stackrel{→}{0}$

${\stackrel{→}{E}}_3=\frac{\mathrm{η}}{2{\mathrm{Î\mu }}_0}\hat{\mathbf{j}}$

${\stackrel{→}{E}}_4=-\frac{\mathrm{η}}{2{\mathrm{Î\mu }}_0}\hat{\mathbf{j}}$