Chapter 22, Problem 39P

(a)

To determine

Total charge on the sphere

Explanation of Solution

Given:

Radius of non-conducting sphere is $6.00\text{cm}$ and uniform volume charge density is $450\text{nC}/{\text{m}}^3$ .

Formula used:

Write the expression for the total charge on the sphere.

  $Q=\rho V$ ....... (1) Here, $Q$ is the total charge on the sphere, $\rho$ is volume charge density and $V$ is total volumeof sphere.

Write the expression for the volume

  $V=\frac{4}{3}\pi r^3$ ....... (2)

Here, $r$ is the radius of the sphere.

Calculation:

Substitute $6.00\text{cm}$ for $r$ in equation (2).

  $\begin{array}{c}V=\frac{4}{3}\pi {\left(6\text{cm}\left(\frac{\text{1}\text{m}}{100\text{cm}}\right)\right)}^3\\ =9.05\times {10}^{-4}{\text{m}}^3\end{array}$

Substitute $450\text{nC}/{\text{m}}^3$ for $\rho$ , and $9.05\times {10}^{-4}{\text{m}}^3$ for $V$ in equation (1).

  $\begin{array}{c}Q=\left(450{\text{nC/m}}^3\right)\left(9.05\times {10}^{-4}{\text{m}}^3\right)\\ =.4072\text{nC}\end{array}$

Conclusion:

Thus, the total charge on the shell is $.4072\text{nC}$ .

(b)

To determine

The electric field at a distance $2.00\text{cm}$ from the sphere’s center.

Explanation of Solution

Given:

Radius of non-conducting sphere is $6.00\text{cm}$ and uniform volume charge density is $450\text{nC}/{\text{m}}^3$ .

Formula used:

Write the expression for Gauss’s Law.

  ${\displaystyle \underset{S}{\oint }\overrightarrow{E}.d\overrightarrow{A}}=\frac{Q_{\text{enc}}}{{\epsilon }_0}$

Here, $\overrightarrow{E}$ is electric field, $S$ is imaginary spherical Gaussian surface, $Q_{\text{enc}}$ is net charge enclosed bythe surface and ${\epsilon }_0$ is electrical permittivity of vacuum.

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

  $E(4\pi r^2)=\frac{Q_{\text{enc}}}{{\epsilon }_0}$

Rearrange above expression for $E$ .

  $E=\frac{Q_{\text{enc}}}{4\pi {\epsilon }_0{r}^2}$ ....... (3)

Write the expression for charge enclosed by Gaussian surface for $r<R$ .

  $Q_{\text{enc}}=\rho \left(\frac{4}{3}\pi r^3\right)$

Substitute $\rho \left(\frac{4}{3}\pi r^3\right)$ for $Q_{\text{enc}}$ in equation (3).

  $E=\frac{\rho \left(\frac{4}{3}\pi r^3\right)}{4\pi {\epsilon }_0{r}^2}$

Simplify the above expression for $E$ .

  $E=\frac{\rho r}{3{\epsilon }_0}$ ....... (4)

Calculation:

Since, $2.00\text{cm}<\text{6}\text{.00}\text{cm}$ implies that $r<R$ therefore,

  $E=\frac{\rho r}{3{\epsilon }_0}$

Substitute $8.85\times {10}^{-12}\text{F/m}$ for ${\epsilon }_0$ , $2.00\text{ cm}$ for $r$ and $450\text{nC}/{\text{m}}^3$ for $\rho$ in equation (4).

  $\begin{array}{c}E=\frac{\left(450{\text{nC/m}}^3\left(\frac{1\text{C}}{{10}^9\text{nC}}\right)\right)\left(2.00\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}{3\left(8.85\times {10}^{-12}\text{F/m}\right)}\\ =338.98\text{N/C}\end{array}$

Conclusion:

Thus, the electric field is $338.98\text{N/C}$ at a distance $2.00\text{ cm}$ from the sphere’s center.

(c)

To determine

The electric field at a distance $5.90\text{cm}$ from the sphere’s center.

Explanation of Solution

Given:

Radius of non-conducting sphere is $6.00\text{cm}$ and uniform volume charge density is $450\text{nC}/{\text{m}}^3$ .

Formula used:

Write the expression for Gauss’s Law.

  ${\displaystyle \underset{S}{\oint }\overrightarrow{E}.d\overrightarrow{A}}=\frac{Q_{\text{enc}}}{{\epsilon }_0}$

Here, $\overrightarrow{E}$ is electric field, $S$ is imaginary spherical Gaussian surface, $Q_{\text{enc}}$ is net charge enclosed bythe surface and ${\epsilon }_0$ is electrical permittivity of vacuum.

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

  $E(4\pi r^2)=\frac{Q_{\text{enc}}}{{\epsilon }_0}$

Rearrange above expression for $E$ .

  $E=\frac{Q_{\text{enc}}}{4\pi {\epsilon }_0{r}^2}$

Write the expression for charge enclosed by Gaussian surface for $r<R$ .

  $Q_{\text{enc}}=\rho \left(\frac{4}{3}\pi r^3\right)$

Substitute $\rho \left(\frac{4}{3}\pi r^3\right)$ for $Q_{\text{enc}}$ in equation (3).

  $E=\frac{\rho \left(\frac{4}{3}\pi r^3\right)}{4\pi {\epsilon }_0{r}^2}$

Simplify the above expression for $E$ .

  $E=\frac{\rho r}{3{\epsilon }_0}$

Calculation:

Since, $5.90\text{cm}<\text{6}\text{.00}\text{cm}$, implies that $r<R$ therefore:

  $E=\frac{\rho r}{3{\epsilon }_0}$ .

Substitute $8.85\times {10}^{-12}\text{F/m}$ for ${\epsilon }_0$ , $5.90\text{ cm}$ for $r$ and $450\text{nC}/{\text{m}}^3$ for $\rho$ in equation (4).

  $\begin{array}{c}E=\frac{\left(\frac{450\text{nC}}{1{\text{m}}^3}\left(\frac{1\text{C}}{{10}^9\text{nC}}\right)\right)\left(5.90\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}{3\left(8.85\times {10}^{-12}\text{F/m}\right)}\\ =1000\text{N/C}\end{array}$

Conclusion:

Thus, the electric field is $1000\text{N/C}$ at a distance $5.90\text{ cm}$ from the sphere’s center.

(d)

To determine

The electric field at a distance $6.10\text{cm}$ from the sphere’s center.

Explanation of Solution

Given:

Radius of non-conducting sphere is $6.00\text{cm}$ and uniform volume charge density is $450\text{nC}/{\text{m}}^3$ .

Formula used:

Write the expression for Gauss’s Law.

  ${\displaystyle \underset{S}{\oint }\overrightarrow{E}.d\overrightarrow{A}}=\frac{Q_{\text{enc}}}{{\epsilon }_0}$

Here, $\overrightarrow{E}$ is electric field, $S$ is imaginary spherical Gaussian surface, $Q_{\text{enc}}$ is net charge enclosed bythe surface and ${\epsilon }_0$ is electrical permittivity of vacuum.

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

  $E(4\pi r^2)=\frac{Q_{\text{enc}}}{{\epsilon }_0}$

Rearrange above expression for $E$ .

  $E=\frac{Q_{\text{enc}}}{4\pi {\epsilon }_0{r}^2}$

Since the total charge on the sphere is enclosed by the Gaussian surface for $r\ge R$ :

  $Q_{\text{enc}}=Q$

Since, $6.10\text{cm}>\text{6}\text{.00}\text{cm}$,implies that $r>R$ therefore,

  $Q_{\text{enc}}=Q$ .

Calculation:

Substitute $8.85\times {10}^{-12}\text{F/m}$ for ${\epsilon }_0$ , $6.10\text{ cm}$ for $r$ and $4072.5\text{nC}$ for $Q_{\text{enc}}$ in equation (4).

  $\begin{array}{c}E=\frac{\left(.4072\text{nC}\left(\frac{1\text{C}}{{10}^9\text{nC}}\right)\right)}{4\pi \left(8.85\times {10}^{-12}\text{F/m}\right){\left(6.10\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}^2}\\ =980\text{N/C}\end{array}$

Conclusion:

Thus, the electric field is $980\text{N/C}$ at a distance $6.10\text{ cm}$ from the sphere’s center.

(e)

To determine

The electric field at a distance $10.0\text{cm}$ from the sphere’s center.

Explanation of Solution

Given:

Radius of non-conducting sphere is $6.00\text{cm}$ and uniform volume charge density is $450\text{nC}/{\text{m}}^3$ .

Formula used:

Write the expression for Gauss’s Law.

  ${\displaystyle \underset{S}{\oint }\overrightarrow{E}.d\overrightarrow{A}}=\frac{Q_{\text{enc}}}{{\epsilon }_0}$

Here, $\overrightarrow{E}$ is electric field, $S$ is imaginary spherical Gaussian surface, $Q_{\text{enc}}$ is net charge enclosed bythe surface and ${\epsilon }_0$ is electrical permittivity of vacuum.

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

  $E(4\pi r^2)=\frac{Q_{\text{enc}}}{{\epsilon }_0}$

Rearrange above expression for $E$ .

  $E=\frac{Q_{\text{enc}}}{4\pi {\epsilon }_0{r}^2}$

Since the total charge on the sphere is enclosed by the Gaussian surface for $r\ge R$ :

  $Q_{\text{enc}}=Q$

Since, $10.0\text{cm}>\text{6}\text{.00}\text{cm}$, implies that $r>R$ therefore,

  $Q_{\text{enc}}=Q$ .

Calculation:

Substitute $8.85\times {10}^{-12}\text{F/m}$ for ${\epsilon }_0$ , $10.0\text{ cm}$ for $r$ and $4072.5\text{nC}$ for $Q_{\text{enc}}$ in equation (4).

  $\begin{array}{c}E=\frac{\left(.4072\text{nC}\left(\frac{1\text{C}}{{10}^9\text{nC}}\right)\right)}{4\pi \left(8.85\times {10}^{-12}\text{F/m}\right){\left(10.0\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}^2}\\ =370\text{N/C}\end{array}$

Conclusion:

Thus, the electric field is $370\text{N/C}$ at a distance $10.0\text{ cm}$ from the sphere’s center.