Question

Figure 23-59 shows, in cross section, three infinitely large nonconducting sheets on which charge is uniformly spread. The surface charge densities are ${\mathbf{\sigma }}_{\mathbf{1}}\mathbf{=}\mathbf{+}\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{\hspace{0.33em}}\mathbf{\mu C}\mathbf{/}{\mathbf{m}}^{\mathbf{2}}\mathbf{,}{\mathbf{\sigma }}_{\mathbf{2}}\mathbf{=}\mathbf{+}\mathbf{4}\mathbf{.}\mathbf{00}\mathbf{\hspace{0.33em}}\mathbf{\mu C}\mathbf{/}{\mathbf{m}}^{\mathbf{2}}\mathbf{,}$and ${\mathbf{\sigma }}_{\mathbf{3}}\mathbf{=}\mathbf{-}\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{\hspace{0.33em}}\mathbf{\mu C}\mathbf{/}{\mathbf{m}}^{\mathbf{2}}$, and $\mathit{L}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{50}\mathbf{\hspace{0.33em}}\mathit{c}\mathit{m}$distance . In unit vector notation, what is the net electric field at point P?

Short answer

The surface net electric field at a point P is$5.65\times {10}^4\hat{\mathrm{j}}\hspace{0.33em}\mathrm{N}/\mathrm{C}$

Step-by-step solution

Listing the given quantities

Surface charge densities are ${\mathrm{\sigma }}_1=+2.00\hspace{0.33em}\mathrm{\mu C}/{\mathrm{m}}^2,{\mathrm{\sigma }}_2=+4.00\hspace{0.33em}\mathrm{\mu C}/{\mathrm{m}}^2,$and ${\mathrm{\sigma }}_3=-5.00\hspace{0.33em}\mathrm{\mu C}/{\mathrm{m}}^2$.

Distance is L = 1.50 cm.

Understanding the concept of electric field

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)

Step 3: Net electric field at P

$$\begin{gathered} \left|\overrightarrow{E}\right|=\frac{{\delta }_1}{2{\epsilon }_0}+\frac{{\delta }_2}{2{\epsilon }_0}+\frac{{\delta }_3}{2{\epsilon }_0} \\ =\frac{2\mu C/m^2}{2{\epsilon }_0}+\frac{4\mu C/m^2}{2{\epsilon }_0}+\frac{-5\mu C/m^2}{2{\epsilon }_0} \\ =\frac{1.0\times {10}^{-6}C/m^2}{2\left(8.85\times {10}^{-12}{C}^2/N.m^2\right)} \\ =5.65\times {10}^{4}a\mathrm{N}/\mathrm{C} \end{gathered}$$

In unit-vector notation, we have $\overrightarrow{\mathrm{E}}=(5.65\times {10}^4\mathrm{N}/\mathrm{C})\hat{\mathrm{j}}$. 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 $\overrightarrow{\mathrm{E}}=(5.65\times {10}^4\mathrm{N}/\mathrm{C})\hat{\mathrm{j}}$.