Chapter 25, Problem 35PQ

Two infinitely long, parallel lines of charge with linear charge densities 3.2 μC/m and −3.2 μC/m are separated by a distance of 0.50 m. What is the net electric field at points A, B, and C as shown in Figure P25.35?

FIGURE P25.35

To determine

The net electric field at points $A$, $B$ and $C$.

Answer to Problem 35PQ

The net electric field at points $A$, $B$ and $C$ are $\underset{\_}{-2.1\times {10}^5\widehat{i}\text{N}/\text{C}}$, $\underset{\_}{4.6\times {10}^5\widehat{i}\text{N}/\text{C}}$ and $\underset{\_}{-4.8\times {10}^5\widehat{i}\text{N}/\text{C}}$ respectively.

Explanation of Solution

Write the expression for finding the electric field due to first line charge.

    ${\overrightarrow{E}}_1=\frac{1}{2\pi {\epsilon }_0}\frac{{\lambda }_1}{r_1}\widehat{x}$                                                                                                 (I)

Here, ${\lambda }_1$ is the line charge density of the first line and $r_1$ is the distance between the line charge and the point.

Write the expression for finding the electric field due to second line charge.

    ${\overrightarrow{E}}_2=\frac{1}{2\pi {\epsilon }_0}\frac{{\lambda }_2}{r_2}\widehat{x}$                                                                                               (II)

Here, ${\lambda }_1$ is the line charge density of the second line and $r_2$ is the distance between the line charge and the point.

Write the expression for the net electric field.

    $\overrightarrow{E}={\overrightarrow{E}}_1+{\overrightarrow{E}}_2$                                                                                                   (III)

Here, $\overrightarrow{E}$ is the net charge at any point.

Conclusion:

The following figure gives the fields acting on the point $A$ due to the first and the second line charges.

Substitute $3.2\times {10}^{-6}\text{C}/\text{m}$ for ${\lambda }_1$, $0.70\text{m}$ for $r_1$, $\widehat{i}$ for $\widehat{x}$ and $8.85\times {10}^{-12}{\text{C}}^2/\text{N}\cdot {\text{m}}^2$ for ${\epsilon }_0$ in equation (I).

    $\begin{array}{c}{\overrightarrow{E}}_1=\frac{1}{2\pi \left(8.85\times {10}^{-12}{\text{C}}^2/\text{N}\cdot {\text{m}}^2\right)}\frac{3.2\times {10}^{-6}\text{C}/\text{m}}{0.70\text{m}}\widehat{i}\\ =8.2\times {10}^4\widehat{i}\text{N}/\text{C}\end{array}$

Substitute $3.2\times {10}^{-6}\text{C}/\text{m}$ for ${\lambda }_2$, $0.20\text{m}$ for $r_2$, $-\widehat{i}$ for $\widehat{x}$ and $8.85\times {10}^{-12}{\text{C}}^2/\text{N}\cdot {\text{m}}^2$ for ${\epsilon }_0$ in equation (II).

    $\begin{array}{c}{\overrightarrow{E}}_2=\frac{1}{2\pi \left(8.85\times {10}^{-12}{\text{C}}^2/\text{N}\cdot {\text{m}}^2\right)}\frac{3.2\times {10}^{-6}\text{C}/\text{m}}{0.20\text{m}}\left(-\widehat{i}\right)\\ =-2.9\times {10}^5\widehat{i}\text{N}/\text{C}\end{array}$

Substitute $8.2\times {10}^4\widehat{i}\text{N}/\text{C}$ for ${\overrightarrow{E}}_1$ and $-2.9\times {10}^5\widehat{i}\text{N}/\text{C}$ for ${\overrightarrow{E}}_2$ in equation (III).

    $\begin{array}{c}{\overrightarrow{E}}_A=8.2\times {10}^4\widehat{i}\text{N}/\text{C}+\left(-2.9\times {10}^5\widehat{i}\text{N}/\text{C}\right)\\ =-2.1\times {10}^5\widehat{i}\text{N}/\text{C}\end{array}$

The following figure gives the fields acting on the point $B$ due to the first and the second line charges.

Substitute $3.2\times {10}^{-6}\text{C}/\text{m}$ for ${\lambda }_1$, $0.25\text{m}$ for $r_1$, $\widehat{i}$ for $\widehat{x}$ and $8.85\times {10}^{-12}{\text{C}}^2/\text{N}\cdot {\text{m}}^2$ for ${\epsilon }_0$ in equation (I).

    $\begin{array}{c}{\overrightarrow{E}}_1=\frac{1}{2\pi \left(8.85\times {10}^{-12}{\text{C}}^2/\text{N}\cdot {\text{m}}^2\right)}\frac{3.2\times {10}^{-6}\text{C}/\text{m}}{0.25\text{m}}\widehat{i}\\ =2.3\times {10}^5\widehat{i}\text{N}/\text{C}\end{array}$

Substitute $3.2\times {10}^{-6}\text{C}/\text{m}$ for ${\lambda }_2$, $0.25\text{m}$ for $r_2$, $\widehat{i}$ for $\widehat{x}$ and $8.85\times {10}^{-12}{\text{C}}^2/\text{N}\cdot {\text{m}}^2$ for ${\epsilon }_0$ in equation (II).

    $\begin{array}{c}{\overrightarrow{E}}_2=\frac{1}{2\pi \left(8.85\times {10}^{-12}{\text{C}}^2/\text{N}\cdot {\text{m}}^2\right)}\frac{3.2\times {10}^{-6}\text{C}/\text{m}}{0.25\text{m}}\left(\widehat{i}\right)\\ =2.3\times {10}^5\widehat{i}\text{N}/\text{C}\end{array}$

Substitute $2.3\times {10}^5\widehat{i}\text{N}/\text{C}$ for ${\overrightarrow{E}}_1$ and $2.3\times {10}^5\widehat{i}\text{N}/\text{C}$ for ${\overrightarrow{E}}_2$ in equation (III).

    $\begin{array}{c}{\overrightarrow{E}}_B=2.3\times {10}^5\widehat{i}\text{N}/\text{C}+2.3\times {10}^5\widehat{i}\text{N}/\text{C}\\ =4.6\times {10}^5\widehat{i}\text{N}/\text{C}\end{array}$

The following figure gives the fields acting on the point $C$ due to the first and the second line charges.

Substitute $3.2\times {10}^{-6}\text{C}/\text{m}$ for ${\lambda }_1$, $0.10\text{m}$ for $r_1$, $-\widehat{i}$ for $\widehat{x}$ and $8.85\times {10}^{-12}{\text{C}}^2/\text{N}\cdot {\text{m}}^2$ for ${\epsilon }_0$ in equation (I).

    $\begin{array}{c}{\overrightarrow{E}}_1=\frac{1}{2\pi \left(8.85\times {10}^{-12}{\text{C}}^2/\text{N}\cdot {\text{m}}^2\right)}\frac{3.2\times {10}^{-6}\text{C}/\text{m}}{0.10\text{m}}\left(-\widehat{i}\right)\\ =-5.8\times {10}^5\widehat{i}\text{N}/\text{C}\end{array}$

Substitute $3.2\times {10}^{-6}\text{C}/\text{m}$ for ${\lambda }_2$, $0.60\text{m}$ for $r_2$, $\widehat{i}$ for $\widehat{x}$ and $8.85\times {10}^{-12}{\text{C}}^2/\text{N}\cdot {\text{m}}^2$ for ${\epsilon }_0$ in equation (II).

    $\begin{array}{c}{\overrightarrow{E}}_2=\frac{1}{2\pi \left(8.85\times {10}^{-12}{\text{C}}^2/\text{N}\cdot {\text{m}}^2\right)}\frac{3.2\times {10}^{-6}\text{C}/\text{m}}{0.60\text{m}}\left(\widehat{i}\right)\\ =9.6\times {10}^4\widehat{i}\text{N}/\text{C}\end{array}$

Substitute $-5.8\times {10}^5\widehat{i}\text{N}/\text{C}$ for ${\overrightarrow{E}}_1$ and $9.6\times {10}^4\widehat{i}\text{N}/\text{C}$ for ${\overrightarrow{E}}_2$ in equation (III).

    $\begin{array}{c}{\overrightarrow{E}}_C=-5.8\times {10}^5\widehat{i}\text{N}/\text{C}+9.6\times {10}^4\widehat{i}\text{N}/\text{C}\\ =-4.8\times {10}^5\widehat{i}\text{N}/\text{C}\end{array}$

Therefore, the net electric field at points $A$, $B$ and $C$ are $\underset{\_}{-2.1\times {10}^5\widehat{i}\text{N}/\text{C}}$, $\underset{\_}{4.6\times {10}^5\widehat{i}\text{N}/\text{C}}$ and $\underset{\_}{-4.8\times {10}^5\widehat{i}\text{N}/\text{C}}$ respectively.