91Ó°ÊÓ

Surface charge densities are ${\mathrm{σ}}_1=+2.00 \mathrm{μ°ä}/{\mathrm{m}}^2,{\mathrm{σ}}_2=+4.00 \mathrm{μ°ä}/{\mathrm{m}}^2,$and ${\mathrm{σ}}_3=-5.00 \mathrm{μ°ä}/{\mathrm{m}}^2$.

Distance is L = 1.50 cm.

Since the fields involved are uniform, the precise location of P is not relevant; what is important is it is above the three sheets, with the positively charged sheets contributing upward fields and the negatively charged sheet contributing a downward field, which conveniently conforms to usual conventions (of upward as positive and downward as negative). The net field is directed upward (+j)

$$\begin{gathered} \left|\stackrel{→}{E}\right|=\frac{{\mathrm{δ}}_1}{2{\mathrm{Î\mu }}_0}+\frac{{\mathrm{δ}}_2}{2{\mathrm{Î\mu }}_0}+\frac{{\mathrm{δ}}_3}{2{\mathrm{Î\mu }}_0} \\ =\frac{2\mathrm{μ}C/m^2}{2{\mathrm{Î\mu }}_0}+\frac{4\mathrm{μ}C/m^2}{2{\mathrm{Î\mu }}_0}+\frac{-5\mathrm{μ}C/m^2}{2{\mathrm{Î\mu }}_0} \\ =\frac{1.0×{10}^{-6}C/m^2}{2\left(8.85×{10}^{-12}{C}^2/N.m^2\right)} \\ =5.65×{10}^{4}a\mathrm{N}/\mathrm{C} \end{gathered}$$

In unit-vector notation, we have $\stackrel{→}{\mathrm{E}}=(5.65×{10}^4\mathrm{N}/\mathrm{C})\hat{\mathrm{j}}$. As positively charged sheets contributed to the upward electric field and negative in the downward field. The direction is still y direction here.

Thus, the electric field in vector form is $\stackrel{→}{\mathrm{E}}=(5.65×{10}^4\mathrm{N}/\mathrm{C})\hat{\mathrm{j}}$.