91Ó°ÊÓ

1. Surface charge density $8.00\mathrm{nC}/{\mathrm{m}}^2$is distributed over an entire x-y plane.
2. Surface charge density $3.00\mathrm{nC}/{\mathrm{m}}^2$is distributed over the parallel plane defined by: z = 2.00 m

Using the concept of the electric field of a Gaussian surface, we can calculate the electric fields at the given z-coordinates. The net electric field, due to the presence of two surface charge densities σ, is used for the net electric field.

Formula:

The electric field of a non-conduction sheet,

$E=\frac{\mathrm{σ}}{2{\mathrm{Î\mu }}_0}$ (i)

Both sheets are horizontal (parallel to the x-y plane), producing vertical fields (parallel to the z-axis). At points above the z = 0 sheet (sheet A), its field points upward (toward +z); at points above the z = 2 sheet (sheet B), its field does likewise. However, below the z = 2 sheet, its field is oriented downward.

The magnitude of the net field in the region between the sheets is given using equation (i) as follows:

$$\begin{gathered} \left|E\right|=\frac{{\mathrm{σ}}_A}{2{\mathrm{Î\mu }}_0}-\frac{{{\mathrm{σ}}_A}^{\text{'}}}{2{\mathrm{Î\mu }}_0} \\ =\frac{8.00×{10}^{-9}\mathrm{C}/{\mathrm{m}}^2-3.00×{10}^{-9}\mathrm{C}/{\mathrm{m}}^2}{2(8.85×{10}^{-12}{\mathrm{C}}^2/{\mathrm{Nm}}^2)} \\ =2.82×{10}^2\mathrm{N}/\mathrm{C} \end{gathered}$$

Hence, the value of the electric field is $2.82×{10}^2\mathrm{N}/\mathrm{C}$.

The magnitude of the net field at points above both sheets is given using equation (i) as follows:

$$\begin{gathered} \left|E\right|=\frac{{\mathrm{σ}}_A}{2{\mathrm{Î\mu }}_0}+\frac{{{\mathrm{σ}}_A}^{\text{'}}}{2{\mathrm{Î\mu }}_0} \\ =\frac{8.00×{10}^{-9}\mathrm{C}/{\mathrm{m}}^2-3.00×{10}^{-9}\mathrm{C}/{\mathrm{m}}^2}{2(8.85×{10}^{-12}{\mathrm{C}}^2/{\mathrm{Nm}}^2)} \\ =6.21×{10}^2\mathrm{N}/\mathrm{C} \end{gathered}$$

Hence, the value of the electric field is $6.21×{10}^2\mathrm{N}/\mathrm{C}$.