Chapter 24, Problem 24.42P

In a certain region of space, the electric field is $\stackrel{\to }{E}$ = 6.00 × 10³ x² $\hat{i}$, where $\stackrel{\to }{E}$ is in newtons per coulomb and x is in meters. Electric charges in this region are at rest and remain at rest. (a) Find the volume density of electric charge at x = 0.300 m. Suggestion: Apply Gauss’s law to a box between x = 0.300 m and x = 0.300 m + dx. (b) Could this region of space be inside a conductor?

To determine

The volume density of electric charge at $x=0.300\text{m}$.

Answer to Problem 24.42P

The volume density of electric charge at $x=0.300\text{m}$ is $3.19\times {10}^{-8}\text{C}/{\text{m}}^{\text{2}}$.

Explanation of Solution

Given info: The electric field vector in the region of space is $6.00\times {10}^3{x}^2\widehat{\text{i}}$.

Consider a Gaussian box of thickness $dx$ in the given region of space.

Formula to calculate the electric flux at $x$ is equal $3.00\text{m}$ is,

${\varphi }_1=E_{1}A\cos {\theta }_1$

Here,

$E_1$ is the electric field in the region of space at $x=0.300\text{m}$.

$A$ is the normal surface area to the electric field of the Gaussian box.

${\theta }_1$ is the angle between the electric field and normal surface area of the box.

Write the expression for the electric field at $x=0.300\text{m}$.

Substitute $0.300\text{m}$ for $x$ in above equation.

Thus, electric field in the region of space at $x=0.300\text{m}$ is $\text{540}\widehat{\text{i}}\text{N}/\text{C}$.

The electric field enter into the box at $x$ is equal $3.00\text{m}$ and the normal vector of the area of box is opposite to the direction of the electric field. So, the angle between the electric field and the normal area vector becomes $180°$.

Substitute $180°$ for ${\theta }_1$ and $\text{540}\text{N}/\text{C}$ for $E_1$ in above equation.

$\begin{array}{c}{\varphi }_1=\left(\text{540}\text{N}/\text{C}\right)A\cos 180°\\ =-\left(\text{540}\text{N}/\text{C}\right)A\end{array}$

Formula to calculate the electric flux at $x$ is equal $3.00\text{m}+dx$ is,

${\varphi }_2=E_{2}A\cos {\theta }_2$

Here,

$E_2$ is the electric field in the region of space at $x=0.300\text{m}+dx$.

${\theta }_2$ is the angle between the electric field and normal surface area of the box at $x=3.00\text{m}+dx$.

Write the expression for the electric field at $x=0.300\text{m}+dx$.

The term $d{x}^2$ is very small so it can be neglected in above expression.

The electric field enter into the box at $x$ is equal $3.00\text{m}+dx$ and the normal vector of the area of box is in the same the direction of the electric field. So, the angle between the electric field and the normal area vector becomes $0°$.

Substitute $0°$ for ${\theta }_2$ and $\left(6.00\times {10}^3\right)\left(0.09{\text{m}}^{\text{2}}+0.6\text{m}dx\right)\text{N}/\text{C}$ for $E_2$ in above equation.

$\begin{array}{c}{\varphi }_2=\left(\left(6.00\times {10}^3\right)\left(0.09{\text{m}}^{\text{2}}+0.6\text{m}dx\right)\text{N}/\text{C}\right)A\cos 0°\\ =\left(\left(6.00\times {10}^3\right)\left(0.09{\text{m}}^{\text{2}}+0.6\text{m}dx\right)\text{N}/\text{C}\right)A\end{array}$

Write the expression for the net electric flux through the box.

${\varphi }_{\text{net}}={\varphi }_1+{\varphi }_2$

Substitute $-\left(\text{540}\text{N}/\text{C}\right)A$ for ${\varphi }_1$ and $\left(\left(6.00\times {10}^3\right)\left(0.09{\text{m}}^{\text{2}}+0.6\text{m}dx\right)\text{N}/\text{C}\right)A$ for ${\varphi }_2$ in above equation.

$\begin{array}{c}{\varphi }_{\text{net}}=-\left(\text{540}\text{N}/\text{C}\right)A+\left(\left(6.00\times {10}^3\right)\left(0.09{\text{m}}^{\text{2}}+0.6\text{m}dx\right)\text{N}/\text{C}\right)A\\ =\left(3.6\times {10}^3\text{m}\right)\left(dx\right)A\text{N}/\text{C}\end{array}$ (1)

Write an alternate expression for the net electric flux from Gauss law.

${\varphi }_{\text{net}}=\frac{q}{{\epsilon }_0}$ (2)

Here,

$q$ is the average charge enclosed inside the Gaussian box.

${\epsilon }_0$ is the permittivity of free space.

Formula to calculate the volume of the Gaussian box is,

$V=Adx$

Here,

$h$ is the height of the rectangular Gaussian surface.

Formula to calculate the average charge enclosed inside the rectangular Gaussian surface is,

$q=\rho V$

Here,

$\rho$ is the average volume density.

$V$ is the volume of the rectangular Gaussian surface.

Replace $V$ by $Adx$ in above equation.

$q=\rho Adx$

Substitute $\rho Adx$ for $q$ in equation (2).

$\varphi =\frac{\rho Adx}{{\epsilon }_0}$ (3)

Equate the equation (2) and (1) for same value of electric flux.

$\begin{array}{c}\frac{\rho Adx}{{\epsilon }_0}=\left(3.6\times {10}^3\text{m}\right)\left(dx\right)A\text{N}/\text{C}\\ \rho =\left(3.6\times {10}^3\text{m}\right){\epsilon }_0\text{N}/\text{C}\end{array}$

Substitute $8.85\times {10}^{-12}{\text{C}}^{\text{2}}/\text{N}\cdot {\text{m}}^{\text{2}}$ for ${\epsilon }_0$.

$\begin{array}{c}\rho =\left(3.6\times {10}^3\text{N}/\text{C}\cdot \text{m}\right)\times 8.85\times {10}^{-12}{\text{C}}^{\text{2}}/\text{N}\cdot {\text{m}}^{\text{2}}\\ =3.186\times {10}^{-8}\text{C}/{\text{m}}^{\text{2}}\\ \approx 3.19\times {10}^{-8}\text{C}/{\text{m}}^{\text{2}}\end{array}$

Conclusion:

Therefore, the volume density of electric charge at $x=0.300\text{m}$ is $3.19\times {10}^{-8}\text{C}/{\text{m}}^{\text{2}}$.

To determine

Whether the given region of space can be inside a conductor.

Answer to Problem 24.42P

No, the given region of space could not be inside a conductor.

Explanation of Solution

Given info: The electric field vector in the region of space is $6.00\times {10}^3{x}^2\widehat{\text{i}}$.

According to the principle of Electromagnetism, in electrostatics free charges in a good conductor reside only on the surface. So the free charge inside the conductor is zero. So the field in it is caused by charges on the surface. Since charges are of the same nature and distribution is uniform, the electric fields cancel each other. No matter, what is the shape of the conductor as long as there is field inside it, electrons always rearrange themselves to make the net field zero.

But, the volume density of electric charge at $x=0.300\text{m}$ is $3.19\times {10}^{-8}\text{C}/{\text{m}}^{\text{2}}$, which is a non-zero value. So, the given region of space at $x=0.300\text{m}$ could not be inside a conductor.

Conclusion:

Therefore, the given region of space could not be inside a conductor.