Chapter 25, Problem 32PQ

Two long, thin rods each have linear charge density λ = 6.0 μC/m and lie parallel to each other, separated by 20.0 cm as shown in Figure P25.32. Determine the magnitude and direction of the net electric field at point P, a distance of 15.0 cm directly above the right rod.

Figure P25.32

To determine

The magnitude and direction of the net electric field at point $P$ a distance of $15.0\text{cm}$ directly above the right rod.

Answer to Problem 32PQ

The magnitude and direction of the net electric field at point $P$ a distance of $15.0\text{cm}$ directly above the right rod is $\underset{\_}{1.0\times {10}^6\text{N}/\text{C}}$ $\underset{\_}{71°}$ above the $x$ axis.

Explanation of Solution

The following figure gives the components of the electric field with direction.

Write the expression for the electric field due to the right rod.

    ${\overrightarrow{E}}_R=\frac{\lambda }{2\pi {\epsilon }_0{r}_R}\widehat{j}$                                                                                                         (I)

Here, $\lambda$ is the line charge density and $r_R$ is the distance of the point $P$ from the right rod.

Write the expression for the electric field due to the left rod.

    $E_L=\frac{\lambda }{2\pi {\epsilon }_0{r}_L}$                                                                                                          (II)

Here, $\lambda$ is the line charge density and $r_L$ is the distance of the point $P$ from the left rod.

Write the expression for the angle $\theta$ above the $x$ axis for $E_L$.

    $\theta ={\tan }^{-1}\frac{r_R}{r_{LR}}$                                                                                                         (III)

Here, $r_{LR}$ is the distance between the rods.

Write the expression for the vector field of the left rod.

    ${\overrightarrow{E}}_L=E_L\left(\cos \theta \widehat{i}+\sin \theta \widehat{j}\right)$                                                                                     (IV)

Write the expression for total electric field.

    ${\overrightarrow{E}}_{\text{tot}}={\overrightarrow{E}}_R+{\overrightarrow{E}}_L$                                                                                                        (V)

Write the expression for the magnitude of the electric field.

    $E=\sqrt{E_x^2+E_y^2}$                                                                                                      (VI)

Here, $E_x$ is the $x$ component of the total electric field and $E_y$ is the $y$ component of the total electric field.

Write the expression for the direction of the electric field.

    $\phi ={\tan }^{-1}\frac{E_y}{E_x}$                                                                                                       (VII)

Conclusion:

Substitute $6.0\times {10}^{-6}\text{C}/\text{m}$ for $\lambda$, $8.85\times {10}^{-12}{\text{C}}^2/\text{N}\cdot {\text{m}}^2$ for ${\epsilon }_0$ and $0.15\text{m}$ for $r_R$ in equation (I).

    $\begin{array}{c}{\overrightarrow{E}}_R=\frac{6.0\times {10}^{-6}\text{C}/\text{m}}{2\pi \left(8.85\times {10}^{-12}{\text{C}}^2/\text{N}\cdot {\text{m}}^2\right)\left(0.15\text{m}\right)}\widehat{j}\\ =7.2\times {10}^5\widehat{j}\text{N}/\text{C}\end{array}$

Substitute $6.0\times {10}^{-6}\text{C}/\text{m}$ for $\lambda$, $8.85\times {10}^{-12}{\text{C}}^2/\text{N}\cdot {\text{m}}^2$ for ${\epsilon }_0$ and $\sqrt{{\left(0.15\text{m}\right)}^2+{\left(0.20\text{m}\right)}^2}$ for $r_L$ in equation (II).

    $\begin{array}{c}{E}_L=\frac{6.0\times {10}^{-6}\text{C}/\text{m}}{2\pi \left(8.85\times {10}^{-12}{\text{C}}^2/\text{N}\cdot {\text{m}}^2\right)\sqrt{{\left(0.15\text{m}\right)}^2+{\left(0.20\text{m}\right)}^2}}\\ =4.3\times {10}^5\text{N}/\text{C}\end{array}$

Substitute $0.15\text{m}$ for $r_R$ and $0.20\text{m}$ for $r_{LR}$ in equation (III).

    $\begin{array}{c}\theta ={\tan }^{-1}\frac{0.15\text{m}}{0.20\text{m}}\\ =36.9°\end{array}$

Substitute $4.3\times {10}^5\text{N}/\text{C}$ for $E_L$ and $36.9°$ for $\theta$ in equation (IV).

    $\begin{array}{c}{\overrightarrow{E}}_L=4.3\times {10}^5\text{N}/\text{C}\left(\cos 36.9°\widehat{i}+\sin 36.9°\widehat{j}\right)\\ =\left(3.4\times {10}^5\widehat{i}+2.6\times {10}^5\widehat{j}\right)\text{N}/\text{C}\end{array}$

Substitute $\left(3.4\times {10}^5\widehat{i}+2.6\times {10}^5\widehat{j}\right)\text{N}/\text{C}$ for ${\overrightarrow{E}}_L$ and $7.2\times {10}^5\widehat{j}\text{N}/\text{C}$ for ${\overrightarrow{E}}_R$ in equation (V).

    $\begin{array}{c}{\overrightarrow{E}}_{\text{tot}}=7.2\times {10}^5\widehat{j}\text{N}/\text{C}+\left(3.4\times {10}^5\widehat{i}+2.6\times {10}^5\widehat{j}\right)\text{N}/\text{C}\\ =\left(3.4\times {10}^5\widehat{i}+9.8\times {10}^5\widehat{j}\right)\text{N}/\text{C}\end{array}$

Substitute $3.4\times {10}^5\text{N}/\text{C}$ for $E_x$ and $9.8\times {10}^5\text{N}/\text{C}$ for $E_y$ in equation (VI).

    $\begin{array}{c}E=\sqrt{{\left(3.4\times {10}^5\text{N}/\text{C}\right)}^2+{\left(9.8\times {10}^5\text{N}/\text{C}\right)}^2}\\ =1.0\times {10}^6\text{N}/\text{C}\end{array}$

Substitute $3.4\times {10}^5\text{N}/\text{C}$ for $E_x$ and $9.8\times {10}^5\text{N}/\text{C}$ for $E_y$ in equation (VII).

    $\begin{array}{c}\phi ={\tan }^{-1}\frac{9.8\times {10}^5\text{N}/\text{C}}{3.4\times {10}^5\text{N}/\text{C}}\\ =71°\end{array}$

Therefore, the magnitude and direction of the net electric field at point $P$ a distance of $15.0\text{cm}$ directly above the right rod is $\underset{\_}{1.0\times {10}^6\text{N}/\text{C}}$ $\underset{\_}{71°}$ above the $x$ axis.