Question

Charge of uniform volume density$\mathbf{r}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{2}\mathbf{nC}\mathbf{/}{\mathbf{m}}^{\mathbf{3}}$fills an infinite slab between role="math" localid="1657340713406" $\mathbf{x}\mathbf{=}\mathbf{-}\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{cm}$and role="math" localid="1657340708898" $\mathbf{x}\mathbf{=}\mathbf{+}\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{cm}\mathbf{}\mathbf{.}$What is the magnitude of the electric field at any point with the coordinate (a) $\mathbf{x}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{cm}$and (b)$\mathbf{x}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{cm}$?

Short answer

a)The magnitude of the electric fieldat any point with the co-ordinate x = 4.0 cm is$5.4\mathrm{N}/\mathrm{C}$

b) The magnitude of the electric field at any point with the co-ordinate x = 6.0 cm is $6.8\mathrm{N}/\mathrm{C}$

Step-by-step solution

Listing the given quantities

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

Understanding the concept of the magnitude of the electric field

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

Formula:

$$\begin{gathered} \mathrm{\sigma }=\frac{\mathrm{q}}{\mathrm{A}} \\ =\frac{\mathrm{\rho V}}{\mathrm{A}}...................\left(1\right) \\ =\mathrm{\rho }∆\mathrm{x} \end{gathered}$$

(a) Calculations for the magnitude of the electric field at any point with the coordinate x = 4.0 cm

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

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

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

(b) Calculations for the magnitude of the electric field at any point with the coordinate x= 6.0 cm

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

we obtain

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

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