Chapter 24, Problem 42P

(a)

To determine

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

Answer to Problem 42P

Volume density of electric charge at $x=0.300\text{m}$ is $31.9\text{C}/{\text{m}}^{\text{3}}$.

Explanation of Solution

Calculation:

Write the expression for electric flux.

    $\varphi =EA$                                                                                                                       (I)

Here, $\varphi$ is the electric flux, $A$ is the surface area and $E$ is the electric field.

As given, consider a box which extends from $x=0.300\text{m}$ to $x=0.300\text{m}+dx$ thickness. Then calculate the electric field at $x=0.300\text{m}$

Write the expression for net flux through the box.

    ${\varphi }_{\text{net}}={\varphi }_{\text{in}}+{\varphi }_{\text{out}}$                                                                                                           (II)

Here, ${\varphi }_{\text{net}}$ is the net flux.

Write the expression for the net flux from Gauss law.

    ${\varphi }_{\text{net}}=\frac{\rho Adx}{{\epsilon }_0}$

Here, $\rho$ is the volume charge density, ${\epsilon }_0$ is permittivity of free space.

Rewrite the above equation for $\rho$.

    $\begin{array}{l}{\varphi }_{\text{net}}=\frac{\rho Adx}{{\epsilon }_0}\\ \rho =\frac{{\epsilon }_0{\varphi }_{\text{net}}}{Adx}\end{array}$                                                                                                            (III)

Conclusion:

Substitute $0.300\text{m}$ for $x$ and $\left(6.00\times {10}^3\right)x^2\text{N}\cdot {\text{m}}^2/\text{C}$ for $E$ in Equation (I) to calculate ${\varphi }_{\text{in}}$.

    $\begin{array}{c}{\varphi }_{\text{in}}=-\left(\left(6.00\times {10}^3\right)x^2\text{N}\cdot {\text{m}}^2/\text{C}\right){\left(0.3\text{m}\right)}^{2}A\\ =\left(-540\text{N}/\text{C}\right)A\end{array}$

Substitute $\left(0.300\text{m}+dx\right)$ for $x$ and $\left(6.00\times {10}^3\right)x^2\text{N}\cdot {\text{m}}^2/\text{C}$ for $E$ in Equation (I) to calculate ${\varphi }_{\text{out}}$.

    $\begin{array}{c}{\varphi }_{\text{out}}=\left(\left(6.00\times {10}^3\right)\text{N}\cdot {\text{m}}^2/\text{C}\right){\left(0.3\text{m}+dx\right)}^{2}A\\ =\left(\left(6.00\times {10}^3\right)\text{N}\cdot {\text{m}}^2/\text{C}\right)\left({\left(0.3\text{m}\right)}^2+\left(0.6\text{m}\right)dx+{\left(dx\right)}^2\right)A\\ =\left(\left(6.00\times {10}^3\right)\text{N}\cdot {\text{m}}^2/\text{C}\right)\left({\left(0.3\text{m}\right)}^2+\left(0.6\text{m}\right)dx\right)A\\ =\left(540\text{N}/\text{C}\right)A+\left(3600\text{N}\cdot \text{m}/\text{C}\right)dxA\end{array}$

Substitute $-540\text{N}/\text{C}A$ for ${\varphi }_{\text{in}}$ and $\left[\left(540\text{N}/\text{C}\right)A+\left(3600\text{N}\cdot \text{m}/\text{C}\right)dxA\right]$ for ${\varphi }_{\text{out}}$ in Equation (II) to calculate ${\varphi }_{\text{net}}$.

    $\begin{array}{c}{\varphi }_{\text{net}}=-540\text{N}/\text{C}A+\left[\left(540\text{N}/\text{C}\right)A+\left(3600\text{N}\cdot \text{m}/\text{C}\right)dxA\right]\\ =\left(3600\text{N}\cdot \text{m}/\text{C}\right)dxA\end{array}$

Substitute $8.85\times {10}^{-12}{\text{C}}^{\text{2}}/\text{N}\cdot {\text{m}}^{\text{2}}$ for ${\epsilon }_0$, $\left(3600\text{N}\cdot \text{m}/\text{C}\right)dx\times A$ for ${\varphi }_{\text{net}}$ in Equation (III) to calculate $\rho$.

    $\begin{array}{c}\rho =\frac{\left(8.85\times {10}^{-12}{\text{C}}^{\text{2}}/\text{N}\cdot {\text{m}}^{\text{2}}\right)\left(\left(3600\text{N}\cdot \text{m}/\text{C}\right)dxA\right)}{Adx}\\ =\left(3600\text{N}\cdot \text{m}/\text{C}\right)\left(8.85\times {10}^{-12}{\text{C}}^{\text{2}}/\text{N}\cdot {\text{m}}^{\text{2}}\right)\\ =31.9\text{C}/{\text{m}}^{\text{3}}\end{array}$

Therefore volume charge density at $x=0.300\text{m}$ is $31.9\text{C}/{\text{m}}^{\text{3}}$.

(b)

To determine

Whether the region of space could be inside the conductor.

Answer to Problem 42P

The region of space is not inside the conductor.

Explanation of Solution

The electric field inside a conductor should be zero but the region from $x=0.300\text{m}$ to $x=0.300\text{m}+dx$ contains electric field of $\left(6\times {10}^3\right)x^2\widehat{\text{i}}\text{N}\cdot {\text{m}}^2/\text{C}$. Hence electric field can’t be zero for the value of $x=0.300\text{m}$ and all the above values as this region can’t be inside the conductor.

Conclusion:

Therefore, this region of space is not inside the conductor.