Chapter 25, Problem 67PQ

A solid plastic sphere of radius R₁ = 8.00 cm is concentric with an aluminum spherical shell with inner radius R₂ = 14.0 cm and outer radius R₃ = 17.0 cm (Fig. P25.67). Electric field measurements are made at two points: At a radial distance of 34.0 cm from the center, the electric field has magnitude 1.70 × 10³ N/C and is directed radially outward, and at a radial distance of 12.0 cm from the center, the electric field has magnitude 9.10 × 10⁴ N/C and is directed radially inward. What are the net charges on

1. a. the plastic sphere and
2. b. the aluminum spherical shell?
3. c. What are the charges on the inner and outer surfaces of the aluminum spherical shell?

FIGURE P25.67

To determine

The net charge on the plastic sphere.

Answer to Problem 67PQ

The net charge on the plastic sphere is $\underset{\_}{-1.46\times {10}^{-7}\text{C}}$.

Explanation of Solution

Write the expression for Gauss’s law.

    ${\displaystyle \oint \overrightarrow{E}\cdot d\overrightarrow{A}=\frac{q_{\text{in}}}{{\epsilon }_0}}$                                                                                                   (I)

Here, $q_{\text{in}}$ is the charge enclosed and ${\epsilon }_0$ is a constant.

Rewrite the above equation in terms of the volume charge density.

    $EA=\frac{q_{\text{in}}}{{\epsilon }_0}$

Here, $A$ is the area and $E$ is the electric filed.

Substitute $4\pi r^2$ for $A$ in the above equation.

    $E\left(4\pi r^2\right)=\frac{q_{\text{in}}}{{\epsilon }_0}$

Here, $r$ is the radius of the cylinder and $L$ is the length of the cylinder.

Conclusion:

Rearrange the above equation to find the charge in the sphere.

    $q_{\text{in}}=E\left(4\pi r^2\right){\epsilon }_0$                                                                                              (II)

Substitute $0.120\text{m}$ for $r$, $-9.10\times {10}^4\text{N}/\text{C}$ for $E$ and $8.85\times {10}^{-12}{\text{C}}^2/\text{N}\cdot {\text{m}}^2$ for ${\epsilon }_0$ in the above equation.

    $\begin{array}{c}{q}_{\text{in}}=-9.10\times {10}^4\text{N}/\text{C}\left(4\pi {\left(0.120\text{m}\right)}^2\right)8.85\times {10}^{-12}{\text{C}}^2/\text{N}\cdot {\text{m}}^2\\ =-1.46\times {10}^{-7}\text{C}\end{array}$

Therefore, the net charge on the plastic sphere is $\underset{\_}{-1.46\times {10}^{-7}\text{C}}$.

To determine

The net charge on the aluminum spherical shell.

Answer to Problem 67PQ

The net charge on the aluminum spherical shell is $\underset{\_}{1.68\times {10}^{-7}\text{C}}$.

Explanation of Solution

Write the expression for total charge enclosed.

    $q_{\text{in}}=E\left(4\pi r^2\right){\epsilon }_0$

Conclusion:

Substitute $q_{\text{sphere}}+q_{\text{shell}}$ for $q_{\text{in}}$, $0.340\text{m}$ for $r$, $+1.70\times {10}^3\text{N}/\text{C}$ for $E$ and $8.85\times {10}^{-12}{\text{C}}^2/\text{N}\cdot {\text{m}}^2$ for ${\epsilon }_0$ in the above equation.

    $\begin{array}{c}{q}_{\text{sphere}}+q_{\text{shell}}=\left(+1.70\times {10}^3\text{N}/\text{C}\right)\left(4\pi {\left(0.340\text{m}\right)}^2\right)\left(8.85\times {10}^{-12}{\text{C}}^2/\text{N}\cdot {\text{m}}^2\right)\\ =2.19\times {10}^{-8}\text{C}\end{array}$

Substitute $-1.46\times {10}^{-7}\text{C}$ for $q_{\text{sphere}}$ in the above equation and rearrange it to find the charge on the aluminum shell.

    $\begin{array}{c}-1.46\times {10}^{-7}\text{C}+q_{\text{shell}}=2.19\times {10}^{-8}\text{C}\\ q_{\text{shell}}=2.19\times {10}^{-8}\text{C}-\left(-1.46\times {10}^{-7}\text{C}\right)\\ =1.68\times {10}^{-7}\text{C}\end{array}$

Therefore, the net charge on the aluminum spherical shell is $\underset{\_}{1.68\times {10}^{-7}\text{C}}$.

To determine

The charge on the inner and outer surface of the aluminum spherical shell.

Answer to Problem 67PQ

The charge on the inner and outer surface of the aluminum spherical shell are $\underset{\_}{1.46\times {10}^{-7}\text{C}}$ and $\underset{\_}{2.2\times {10}^{-8}\text{C}}$ respectively.

Explanation of Solution

Write the expression for charge on the inner surface of the aluminum spherical shell.

    $Q_{\text{inner shell surface}}=-q_{\text{sphere}}$                                                                                      (III)

Write the expression for charge on the inner surface of the aluminum spherical shell.

    $Q_{\text{outer shell surface}}=q_{\text{shell}}-Q_{\text{inner shell surface}}$                                                                      (IV)

Conclusion:

Substitute $-1.46\times {10}^{-7}\text{C}$ for $q_{\text{sphere}}$ in equation (III).

    $\begin{array}{c}{Q}_{\text{inner shell surface}}=-\left(-1.46\times {10}^{-7}\text{C}\right)\\ =1.46\times {10}^{-7}\text{C}\end{array}$

Substitute $1.46\times {10}^{-7}\text{C}$ for $Q_{\text{inner shell surface}}$ and $1.68\times {10}^{-7}\text{C}$ for $q_{\text{shell}}$ in equation (IV).

    $\begin{array}{c}{Q}_{\text{outer shell surface}}=1.68\times {10}^{-7}\text{C}-1.46\times {10}^{-7}\text{C}\\ =2.2\times {10}^{-8}\text{C}\end{array}$

Therefore, the charge on the inner and outer surface of the aluminum spherical shell are $\underset{\_}{1.46\times {10}^{-7}\text{C}}$ and $\underset{\_}{2.2\times {10}^{-8}\text{C}}$ respectively.