Chapter 25, Problem 44PQ

(a)

To determine

The magnitude of the electric field at a distance $0.100\text{m}$ from the center of the sphere.

Answer to Problem 44PQ

The magnitude of the electric field at a distance $0.100\text{m}$ from the center of the sphere is $\underset{\_}{5.24\times {10}^9\text{N}/\text{C}}$.

Explanation of Solution

Write the expression to find the charge enclosed.

    $q_{\text{in}}={\displaystyle \int \rho dV}$                                                                                                             (I)

Here, $\rho$ is the charge density and $dV$ is the enclosed elemental volume.

Write the expression for Gauss’s Law for a spherical symmetry.

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

Here,$q_{\text{in}}$ is the charge enclosed and $r$ is the radius of the sphere.

Conclusion:

Substitute $\frac{70.9\times {10}^{-3}\text{C}}{\left(4/3\right)\pi {\left(0.230\text{m}\right)}^3}$ for $\rho$ in equation (I).

    $\begin{array}{c}{q}_{\text{in}}={\displaystyle \int \frac{70.9\times {10}^{-3}\text{C}}{\left(4/3\right)\pi {\left(0.230\text{m}\right)}^3}dV}\\ =\frac{70.9\times {10}^{-3}\text{C}}{\left(4/3\right)\pi {\left(0.230\text{m}\right)}^3}{\displaystyle \int dV}\end{array}$

Substitute $\left(4/3\right)\pi r^3$ for ${\displaystyle \int dV}$ in the above equation.

    $\begin{array}{c}{q}_{\text{in}}={\displaystyle \int \frac{70.9\times {10}^{-3}\text{C}}{\left(4/3\right)\pi {\left(0.230\text{m}\right)}^3}dV}\\ =\frac{70.9\times {10}^{-3}\text{C}}{\left(4/3\right)\pi {\left(0.230\text{m}\right)}^3}\left(4/3\right)\pi r^3\end{array}$

Substitute $\frac{70.9\times {10}^{-3}\text{C}}{\left(4/3\right)\pi {\left(0.230\text{m}\right)}^3}\left(4/3\right)\pi r^3$ for $q_{\text{in}}$, $8.85\times {10}^{-12}{\text{C}}^2/\text{N}\cdot {\text{m}}^2$ for ${\epsilon }_0$ and $0.100\text{m}$ for $r$ in equation (II).

    $\begin{array}{c}E=\frac{1}{4\pi {\epsilon }_0{r}^2}\frac{70.9\times {10}^{-3}\text{C}}{\left(4/3\right)\pi {\left(0.230\text{m}\right)}^3}\left(4/3\right)\pi r^3\\ =\frac{1}{4\pi \left(8.85\times {10}^{-12}{\text{C}}^2/\text{N}\cdot {\text{m}}^2\right)}\frac{70.9\times {10}^{-3}\text{C}}{\left(4/3\right)\pi {\left(0.230\text{m}\right)}^3}\left(4/3\right)\pi \left(0.100\text{m}\right)\\ =5.24\times {10}^9\text{N}/\text{C}\end{array}$

Therefore, the magnitude of the electric field at a distance $0.100\text{m}$ from the center of the sphere is $\underset{\_}{5.24\times {10}^9\text{N}/\text{C}}$.

(b)

To determine

The magnitude of the electric field at a distance $0.230\text{m}$ from the center of the sphere.

Answer to Problem 44PQ

The magnitude of the electric field at a distance $0.230\text{m}$ from the center of the sphere is $\underset{\_}{1.21\times {10}^{10}\text{N}/\text{C}}$.

Explanation of Solution

Write the expression for Gauss’s Law for a spherical symmetry.

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

Here,$q_{\text{in}}$ is the charge enclosed and $r$ is the radius of the sphere.

Conclusion:

Substitute $\frac{70.9\times {10}^{-3}\text{C}}{\left(4/3\right)\pi {\left(0.230\text{m}\right)}^3}\left(4/3\right)\pi r^3$ for $q_{\text{in}}$, $8.85\times {10}^{-12}{\text{C}}^2/\text{N}\cdot {\text{m}}^2$ for ${\epsilon }_0$ and $0.230\text{m}$ for $r$ in the above equation.

    $\begin{array}{c}E=\frac{1}{4\pi {\epsilon }_0{r}^2}\frac{70.9\times {10}^{-3}\text{C}}{\left(4/3\right)\pi {\left(0.230\text{m}\right)}^3}\left(4/3\right)\pi r^3\\ =\frac{1}{4\pi \left(8.85\times {10}^{-12}{\text{C}}^2/\text{N}\cdot {\text{m}}^2\right)}\frac{70.9\times {10}^{-3}\text{C}}{\left(4/3\right)\pi {\left(0.230\text{m}\right)}^3}\left(4/3\right)\pi \left(0.230\text{m}\right)\\ =1.21\times {10}^{10}\text{N}/\text{C}\end{array}$

Therefore, the magnitude of the electric field at a distance $0.230\text{m}$ from the center of the sphere is $\underset{\_}{1.21\times {10}^{10}\text{N}/\text{C}}$.

(c)

To determine

The magnitude of the electric field at a distance $0.500\text{m}$ from the center of the sphere.

Answer to Problem 44PQ

The magnitude of the electric field at a distance $0.500\text{m}$ from the center of the sphere is $\underset{\_}{2.55\times {10}^9\text{N}/\text{C}}$.

Explanation of Solution

Write the expression for Gauss’s Law for a spherical symmetry.

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

Here,$q_{\text{in}}$ is the charge enclosed and $r$ is the radius of the sphere.

Conclusion:

Substitute $70.9\times {10}^{-3}\text{C}$ for $q_{\text{in}}$, $8.85\times {10}^{-12}{\text{C}}^2/\text{N}\cdot {\text{m}}^2$ for ${\epsilon }_0$ and $0.500\text{m}$ for $r$ in the above equation.

    $\begin{array}{c}E=\frac{70.9\times {10}^{-3}\text{C}}{4\pi {\epsilon }_0{\left(0.500\text{m}\right)}^2}\\ =\frac{70.9\times {10}^{-3}\text{C}}{4\pi \left(\left(8.85\times {10}^{-12}{\text{C}}^2/\text{N}\cdot {\text{m}}^2\right)\right){\left(0.500\text{m}\right)}^2}\\ =2.55\times {10}^9\text{N}/\text{C}\end{array}$

Therefore, the magnitude of the electric field at a distance $0.500\text{m}$ from the center of the sphere is $\underset{\_}{2.55\times {10}^9\text{N}/\text{C}}$.