91Ó°ÊÓ

volume density $\mathrm{ÒÏ}=1.2\mathrm{nC}/{\mathrm{m}}^3$

Using the concept of the electric field to determine the magnitude of the electric field

Formula:

$$\begin{gathered} \mathrm{σ}=\frac{\mathrm{q}}{\mathrm{A}} \\ =\frac{\mathrm{ÒÏV}}{\mathrm{A}}...................\left(1\right) \\ =\mathrm{ÒÏ}∆\mathrm{x} \end{gathered}$$

At x = 0.040 m, the net field has a rightward (+x) contribution, from the charge

lying between x = –0.050 m and x = 0.040 m, and a leftward (–x) contribution from the charge in the region from x = 0.040 m to x = 0.050 m.

Thus, using equation (1), in this situation, we have

$$\begin{gathered} \left|\underline{\mathrm{E}}\right|=\left(\frac{\mathrm{ÒÏ}(0.090\mathrm{m})}{2{\mathrm{Î\mu }}_0}\right)-\left(\frac{\mathrm{ÒÏ}(0.010\mathrm{m})}{2{\mathrm{Î\mu }}_0}\right) \\ =\frac{(8.0×{10}^{-9}\mathrm{C}/{\mathrm{m}}^3)(0.090\mathrm{m}-0.010\mathrm{m})}{2(8.85×{10}^{-12}{\mathrm{C}}^2/\mathrm{N}.{\mathrm{m}}^2)} \\ =5.4\mathrm{N}/\mathrm{C} \end{gathered}$$

The magnitude of the electric field at any point with the coordinate x=4.0 cm is $5.4\mathrm{N}/\mathrm{C}$

In this case, the field contributions from all layers of charge point rightward, and

we obtain

$$\begin{gathered} \left|\underline{\mathrm{E}}\right|=\frac{\mathrm{ÒÏ}(0.100\mathrm{m})}{2{\mathrm{Î\mu }}_{\mathrm{o}}} \\ =\frac{(8.0×{10}^{-9}\mathrm{C}/{\mathrm{m}}^3)(0.100\mathrm{m})}{2(8.85×{10}^{-12}{\mathrm{C}}^2/\mathrm{N}.{\mathrm{m}}^2)} \\ =6.8\mathrm{N}/\mathrm{C} \end{gathered}$$

The magnitude of the electric field at any point with the coordinate x = 6.0 cm is $6.8N/C.$