Chapter 25, Problem 47PQ

The infinite sheets in Figure P25.47 are both positively charged. The sheet on the left has a uniform surface charge density of 48.0 μC/m², and the one on the right has a uniform surface charge density of 24.0 μC/m².

1. a. What are the magnitude and direction of the net electric field at points A, B, and C?
2. b. What is the force exerted on an electron placed at points A, B, and C?

FIGURE P25.47

To determine

The magnitude and direction of the net electric fields at points $A$, $B$ and $C$.

Answer to Problem 47PQ

The magnitude and direction of the net electric fields at points $A$, $B$ and $C$ are $\underset{\_}{-4.07\times {10}^6\widehat{i}\text{N}/\text{C}}$, $\underset{\_}{+1.36\times {10}^6\widehat{i}\text{N}/\text{C}}$ and $\underset{\_}{+4.07\times {10}^6\widehat{i}\text{N}/\text{C}}$ respectively.

Explanation of Solution

Write the expression for finding the electric field due to first sheet.

    $E_1=\frac{{\sigma }_1}{2{\epsilon }_0}$                                                                                                         (I)

Here, ${\sigma }_1$ is the surface charge density of the first sheet and ${\epsilon }_0$ is a constant.

Write the expression for finding the electric field due to second sheet.

    $E_2=\frac{{\sigma }_2}{2{\epsilon }_0}$                                                                                                        (II)

Here, ${\sigma }_2$ is the surface charge density of the second sheet and ${\epsilon }_0$ is a constant.

The following figure shows the direction and the sum of the fields at points $A$, $B$ and $C$.

Write the expression to find the electric field at point $A$.

    ${\overrightarrow{E}}_A=-E_1\widehat{i}-E_2\widehat{i}$                                                                                             (III)

Write the expression to find the electric field at point $B$.

    ${\overrightarrow{E}}_B=+E_1\widehat{i}-E_2\widehat{i}$                                                                                             (IV)

Write the expression to find the electric field at point $C$.

    ${\overrightarrow{E}}_C=+E_1\widehat{i}+E_2\widehat{i}$                                                                                              (V)

Conclusion:

Substitute $48.0\times {10}^{-6}\text{C}/{\text{m}}^2$ for ${\sigma }_1$ and $8.85\times {10}^{-12}{\text{C}}^2/\text{N}\cdot {\text{m}}^2$ for ${\epsilon }_0$ in equation (I).

    $\begin{array}{c}{E}_1=\frac{48.0\times {10}^{-6}\text{C}/{\text{m}}^2}{2\left(8.85\times {10}^{-12}{\text{C}}^2/\text{N}\cdot {\text{m}}^2\right)}\\ =2.71\times {10}^6\text{N}/\text{C}\end{array}$

Substitute $24.0\times {10}^{-6}\text{C}/{\text{m}}^2$ for ${\sigma }_1$ and $8.85\times {10}^{-12}{\text{C}}^2/\text{N}\cdot {\text{m}}^2$ for ${\epsilon }_0$ in equation (II).

    $\begin{array}{c}{E}_2=\frac{24.0\times {10}^{-6}\text{C}/{\text{m}}^2}{2\left(8.85\times {10}^{-12}{\text{C}}^2/\text{N}\cdot {\text{m}}^2\right)}\\ =1.36\times {10}^6\text{N}/\text{C}\end{array}$

Substitute $2.71\times {10}^6\text{N}/\text{C}$ for $E_1$ and $1.36\times {10}^6\text{N}/\text{C}$ for $E_2$ in equation (III).

    $\begin{array}{c}{\overrightarrow{E}}_A=-2.71\times {10}^6\text{N}/\text{C}\widehat{i}-1.36\times {10}^6\text{N}/\text{C}\widehat{i}\\ =-4.07\times {10}^6\widehat{i}\text{N}/\text{C}\end{array}$

Substitute $2.71\times {10}^6\text{N}/\text{C}$ for $E_1$ and $1.36\times {10}^6\text{N}/\text{C}$ for $E_2$ in equation (IV).

    $\begin{array}{c}{\overrightarrow{E}}_B=+2.71\times {10}^6\text{N}/\text{C}\widehat{i}-1.36\times {10}^6\text{N}/\text{C}\widehat{i}\\ =+1.36\times {10}^6\widehat{i}\text{N}/\text{C}\end{array}$

Substitute $2.71\times {10}^6\text{N}/\text{C}$ for $E_1$ and $1.36\times {10}^6\text{N}/\text{C}$ for $E_2$ in equation (IV).

    $\begin{array}{c}{\overrightarrow{E}}_C=+2.71\times {10}^6\text{N}/\text{C}\widehat{i}+1.36\times {10}^6\text{N}/\text{C}\widehat{i}\\ =+4.07\times {10}^6\widehat{i}\text{N}/\text{C}\end{array}$

Therefore, the magnitude and direction of the net electric fields at points $A$, $B$ and $C$ are $\underset{\_}{-4.07\times {10}^6\widehat{i}\text{N}/\text{C}}$, $\underset{\_}{+1.36\times {10}^6\widehat{i}\text{N}/\text{C}}$ and $\underset{\_}{+4.07\times {10}^6\widehat{i}\text{N}/\text{C}}$ respectively.

To determine

The force exerted on an electron placed at points $A$, $B$ and $C$.

Answer to Problem 47PQ

The force exerted on an electron placed at points $A$, $B$ and $C$ are $\underset{\_}{6.51\times {10}^{-13}\widehat{i}\text{N}}$, $\underset{\_}{-2.17\times {10}^{-13}\widehat{i}\text{N}}$ and $\underset{\_}{-6.51\times {10}^{-13}\widehat{i}\text{N}}$ respectively.

Explanation of Solution

Write the expression for the force exerted on the electron.

    $\overrightarrow{F}=-e\overrightarrow{E}$                                                                                                         (VI)

Here, $e$ is the charge of the electron.

Conclusion:

Substitute $-4.07\times {10}^6\widehat{i}\text{N}/\text{C}$ for ${\overrightarrow{E}}_A$ and $1.6\times {10}^{-19}\text{C}$ for $e$ in equation (VI).

    $\begin{array}{c}{\overrightarrow{F}}_A=-e{\overrightarrow{E}}_A\\ =-1.6\times {10}^{-19}\text{C}\left(-4.07\times {10}^6\widehat{i}\text{N}/\text{C}\right)\\ =6.51\times {10}^{-13}\widehat{i}\text{N}\end{array}$

Substitute $+1.36\times {10}^6\widehat{i}\text{N}/\text{C}$ for ${\overrightarrow{E}}_B$ and $1.6\times {10}^{-19}\text{C}$ for $e$ in equation (VI).

    $\begin{array}{c}{\overrightarrow{F}}_B=-e{\overrightarrow{E}}_B\\ =-1.6\times {10}^{-19}\text{C}\left(+1.36\times {10}^6\widehat{i}\text{N}/\text{C}\right)\\ =-2.17\times {10}^{-13}\widehat{i}\text{N}\end{array}$

Substitute $+4.07\times {10}^6\widehat{i}\text{N}/\text{C}$ for ${\overrightarrow{E}}_C$ and $1.6\times {10}^{-19}\text{C}$ for $e$ in equation (VI).

    $\begin{array}{c}{\overrightarrow{F}}_C=-e{\overrightarrow{E}}_C\\ =-1.6\times {10}^{-19}\text{C}\left(+4.07\times {10}^6\widehat{i}\text{N}/\text{C}\right)\\ =-6.51\times {10}^{-13}\widehat{i}\text{N}\end{array}$

Therefore, the force exerted on an electron placed at points $A$, $B$ and $C$ are $\underset{\_}{6.51\times {10}^{-13}\widehat{i}\text{N}}$, $\underset{\_}{-2.17\times {10}^{-13}\widehat{i}\text{N}}$ and $\underset{\_}{-6.51\times {10}^{-13}\widehat{i}\text{N}}$ respectively.