Chapter 22, Problem 22.17E

A very long uniform line of charge has charge per unit length 4.80 μC/m and lies along the x-axis. A second long uniform line of charge has charge per unit length −2.40 μC/m and is parallel to the x-axis at y = 0.400 m. What is the net electric field (magnitude and direction) at the following points on the y-axis: (a) y = 0.200 m and (b) y = 0.600 m?

22.17.    IDENTIFY: Add the vector electric fields due to each line of charge. E(r) for a line of charge is given by Example 22.6 and is directed toward a negative line of charge and away from a positive line.

SET UP: The two lines of charge are shown in Figure 22.17.

$\text{E = }\frac{1}{2\pi {\epsilon }_0}\frac{\text{λ}}{\text{r}}.$

Figure 22.17

EXECUTE: (a) At point a, $\stackrel{\to }{{\text{E}}_1}$ and $\stackrel{\to }{{\text{E}}_2}$ are in the +y-direction (toward negative charge, away from positive charge).

$\begin{array}{l}{\text{E}}_1=(1/2\pi {\epsilon }_0)[(4.80\times {10}^{-6}C/m)/(0.200m)]\text{ = }4.314\times {10}^{5}N/C\text{.}\\ {\text{E}}_2=(1/2\pi {\epsilon }_0)[(2.40\times {10}^{-6}C/m)/(0.200m)]\text{ = }2.157\times {10}^{5}N/C.\\ \text{E}={\text{ E}}_1{\text{ + E}}_2\text{ = }6.47\times {10}^{5}N/C,inthe\text{ y}-direction\text{.}\end{array}$

(b) At point b, $\stackrel{\to }{{\text{E}}_1}$ is in the +y-direction and $\stackrel{\to }{{\text{E}}_2}$ is in the −y-direction.

${\text{E}}_1=(1/2\pi {\epsilon }_0)[(4.80\times {10}^{-6}C/m)/(0.600m)]\text{ = }1.438\times {10}^{5}N/C\text{.}$

${\text{E}}_2=(1/2\pi {\epsilon }_0)[(2.40\times {10}^{-6}C/m)/(0.200m)]\text{ = }2.157\times {10}^{5}N/C\text{.}$

E = E₂ − E₁ = 7.2×10⁴ N/C, in the −y-direction.