Question 23.35: Figure 23-46a shows three plastic sheets that are large, par......
Figure 23-46a shows three plastic sheets that are large, parallel, and uniformly charged. Figure 23-46b gives the component of the net electric field along an x axis through the sheets. The scale of the vertical axis is set by E_s=6.0 × 10^5 N/C. What is the ratio of the charge density on sheet 3 to that on sheet 2?
In the region between sheets 1 and 2, the net field is E_1-E_2+E_3=2.0 \times 10^5 \,N / C .
In the region between sheets 2 and 3, the net field is at its greatest value:
E_1+E_2+E_3=6.0 \times 10^5 \,N / C .
The net field vanishes in the region to the right of sheet 3, where E_1 + E_2 = E_3 . We note the implication that σ_3 is negative (and is the largest surface-density, in magnitude). These three conditions are sufficient for finding the fields:
E_1=1.0 \times 10^5 \,N / C , E_2=2.0 \times 10^5 \,N / C , E_3=3.0 \times 10^5 \,N / C \text {. }
From Eq. 23-13, we infer (from these values of E)
E=\frac{\sigma}{2 \varepsilon_0} \quad \text { (sheet of charge). } (23-13)
\frac{\left|\sigma_3\right|}{\left|\sigma_2\right|}=\frac{3.0 \times 10^5 \,N / C }{2.0 \times 10^5 \,N / C }=1.5 .
Recalling our observation, above, about σ_3, we conclude that \frac{\sigma_3}{\sigma_2}=-1.5 .