Question

Figure 23-51 shows a cross-section through a very large non-conducting slab of thickness$\mathit{d}\mathbf{=}\mathbf{9}\mathbf{.}\mathbf{40}\mathit{m}\mathit{m}$and uniform volume charge density $\mathbf{p}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{80}\mathbf{fC}\mathbf{/}{\mathbf{m}}^{\mathbf{3}}$ . The origin of an x-axis is at the slab’s center. What is the magnitude of the slab’s electric field at an xcoordinate of (a) $\mathbf{0}$ , (b) $\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{mm}$ , (c) $\mathbf{4}\mathbf{.}\mathbf{70}\mathbf{mm}$ , and (d) $\mathbf{26}\mathbf{.}\mathbf{0}\mathbf{mm}$?

Short answer

a) The magnitude of the slab’s electric field at $\mathrm{x}=0\mathrm{mm}\mathrm{is}0\mathrm{N}/\mathrm{C}.$

b) The magnitude of the slab’s electric field at $\mathrm{x}=2\mathrm{mm}\mathrm{is}1.31\times {10}^{-6}\mathrm{N}/\mathrm{C}$.

c) The magnitude of the slab’s electric field at $\mathrm{x}=4.70\mathrm{mm}\mathrm{is}3.08\times {10}^{-6}\mathrm{N}/\mathrm{C}$ .

d) The magnitude of the slab’s electric field at $x=26mmis3.08\times {10}^{-6}$$\mathrm{N}/\mathrm{C}$.

Step-by-step solution

The given data

a) Thickness of the slab,$\mathrm{d}=9.40\mathrm{mm}$

b) Uniform volume charge density,$\mathrm{p}=5.80\mathrm{fC}/{\mathrm{m}}^3$

Understanding the concept of the electric field

Using the concept of the electric flux of Gauss's flux theorem, we get the value of the electric field using the given data and substituting the charge value of the volume charge density.

Formulae:

The volume charge density of a body, $\mathrm{p}=\frac{\mathrm{q}}{\mathrm{V}}$ (1)

The electric field of a Gaussian surface, $\mathrm{\varphi }=\oint \mathrm{E}.\mathrm{dA}=\frac{\mathrm{q}}{{\mathrm{\epsilon }}_0}$ (2)

a) Calculation of the electric field at x = 0 mm

We use a Gaussian surface in the form of a box with rectangular sides. It is centered at the central plane of the slab, so the left and right faces are each a distance x from the central plane. We take the thickness of the rectangular solid to be a, the same as its length, so the left and right faces are squares.

The electric field is normal to the left and right faces and is uniform over them. Since $\mathrm{p}=5.80\mathrm{fC}/{\mathrm{m}}^3$ is positive, it points outward at both faces: toward the left at the left face and toward the right at the right face. Furthermore, the magnitude is the same at both faces. The electric flux through each of these faces from equation (1) is given as:

${\mathrm{\varphi }}_{\mathrm{E}}={\mathrm{Ea}}^2$

The field is parallel to the other faces of the Gaussian surface and the flux through them is zero.

So, the total flux through the Gaussian surface is given as

$\mathrm{\varphi }={\mathrm{Ea}}^2$......................(3)

The volume enclosed by the Gaussian surface is given by:

volume = (Area)(distance)

=$2{\mathrm{a}}^2\mathrm{x}$

So, the charge contained within it using equation (1) and the above value is given as:

$\mathrm{q}=2{\mathrm{a}}^2\mathrm{xp}$…………………….(4)

Using the given data of equations (a) and (b) in equation (i), we get the equation for the electric field as given:

$2{\mathrm{\epsilon }}_0{\mathrm{Ea}}^2=2{\mathrm{a}}^2\mathrm{xp}$

$\mathrm{E}=\frac{\mathrm{px}}{{\mathrm{\epsilon }}_0}$…………………(5)

For x=0 mm , the value of the electric field using the given data in equation (5) as given: $\mathrm{E}=0\mathrm{N}/\mathrm{C}$

.

Hence, the value of the electric field is$0\mathrm{N}/\mathrm{C}$

b) Calculation of the electric field at x = 2.0 mm

For x = 2.0 mm , the value of the electric field using the given data in equation (5) as given:

$$\begin{gathered} \mathrm{E}=\frac{(5.80\times {10}^{-15}\mathrm{C}/{\mathrm{m}}^3)(2.00\times {10}^{-3}\mathrm{m})}{8.85\times {10}^{-12}{\mathrm{C}}^2/\mathrm{N}·{\mathrm{m}}^2} \\ =1.31\times {10}^{-6}\mathrm{N}/\mathrm{C} \end{gathered}$$

Hence, the value of the electric field is $=1.31\times {10}^{-6}\mathrm{N}/\mathrm{C}$.

c) Calculation of the electric field at x = 4.70 mm

For x = 4.70 mm , the value of the electric field using the given data in equation (5) as given:

$$\begin{gathered} \mathrm{E}=\frac{(5.80\times {10}^{-15}\mathrm{C}/{\mathrm{m}}^3)(4.70\times {10}^{-3}\mathrm{m})}{8.85\times {10}^{-12}{\mathrm{C}}^2/\mathrm{N}·{\mathrm{m}}^2} \\ =3.08\times {10}^{-6}\mathrm{N}/\mathrm{C} \end{gathered}$$

Hence, the value of the electric field is$\mathbf{3}\mathbf{.}\mathbf{08}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{6}}\mathbf{N}\mathbf{/}\mathbf{C}$

d) Calculation of the electric field at x = 26.0 mm

For x= 26.00mm , we take a Gaussian surface of the same shape and orientation, but with$\mathrm{x}>\frac{\mathrm{d}}{2}$ , so the left and right faces are outside the slab.

The total flux through the surface is again $\mathrm{\varphi }=2{\mathrm{Ea}}^2$……………………(6)

But, the charge enclosed is using equation (1) now,, $\mathrm{q}={\mathrm{a}}^2\mathrm{dp}$ ………………..(7)

Thus, substituting the given values of equations (6) and (7) in equation (1), we get the electric field equation as given:

$$\begin{gathered} 2{\mathrm{\epsilon }}_0{\mathrm{Ea}}^2={\mathrm{a}}^2\mathrm{dp} \\ \mathrm{E}=\frac{\mathrm{pd}}{2{\mathrm{\epsilon }}_0} \end{gathered}$$

Using the given data in the above equation and we get the value of the electric field as given:

role="math" localid="1657347742834" $$\begin{gathered} \mathrm{E}=\frac{(5.80\times {10}^{-15}\mathrm{C}/{\mathrm{m}}^3)(9.40\times {10}^{-3}\mathrm{m})}{2(8.85\times {10}^{-12}{\mathrm{C}}^2/\mathrm{N}·{\mathrm{m}}^2)} \\ =3.08\times {10}^{-6}\mathrm{N}/\mathrm{C} \end{gathered}$$

Hence, the value of the electric field is $=3.08\times {10}^{-6}\mathrm{N}/\mathrm{C}$.