Question 23.80: Charge of uniform surface density 8.00 nC/m² is distributed ......
Charge of uniform surface density 8.00 nC/m² is distributed over an entire xy plane; charge of uniform surface density 3.00 nC/m² is distributed over the parallel plane defined by z = 2.00 m. Determine the magnitude of the electric field at any point having a z coordinate of (a) 1.00 m and (b) 3.00 m.
The field due to a sheet of charge is given by Eq. 23-13. Both sheets are horizontal (parallel to the xy 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 = 2.0 sheet (sheet B), its field does likewise. However, below the z = 2.0 sheet, its field is oriented downward.
E=\frac{\sigma}{2 \varepsilon_0} \quad \text { (sheet of charge). } (23-13)
(a) The magnitude of the net field in the region between the sheets is
|\vec{E}|=\frac{\sigma_A}{2 \varepsilon_0}-\frac{\sigma_B}{2 \varepsilon_0}=\frac{8.00 \times 10^{-9}\, C / m ^2-3.00 \times 10^{-9} \,C / m ^2}{2\left(8.85 \times 10^{-12} \,C ^2 / N \cdot m ^2\right)}=2.82 \times 10^2 \,N / C.
(b) The magnitude of the net field at points above both sheets is
|\vec{E}|=\frac{\sigma_A}{2 \varepsilon_0}+\frac{\sigma_B}{2 \varepsilon_0}=\frac{8.00 \times 10^{-9}\, C / m ^2+3.00 \times 10^{-9} \,C / m ^2}{2\left(8.85 \times 10^{-12} \,C ^2 / N \cdot m ^2\right)}=6.21 \times 10^2 \,N / C .