The Electric Field Due to a Line of Charge — Question 10

Question 10

The null between parallel line charges reflects inverse-distance rather than inverse-square decay

Two infinite straight lines parallel to z cross the xy plane at $x=-a$ and $x=a$. Their positive densities are $\lambda$ and $2\lambda$. On the x axis use the line-field magnitude $2k\lambda/r$.

1. Find the field null between the lines.

2. Generalize the relation between a null location $x_0$ and the ratio $\lambda_R/\lambda_L$. Explain why the point-charge cancellation formula cannot be substituted.

Question 10 - Solution

The null between parallel line charges reflects inverse-distance rather than inverse-square decay

(a) Method and calculation

$$E_x=\frac{2k\lambda}{x+a}-\frac{4k\lambda}{a-x}=0.$$$$a-x=2(x+a)\Rightarrow x=-a/3.$$

(b) Interpretation and further calculation

$$\frac{\lambda_L}{x_0+a}=\frac{\lambda_R}{a-x_0}\Rightarrow\frac{\lambda_R}{\lambda_L}=\frac{a-x_0}{a+x_0}.$$$$E_{\mathrm{line}}\propto r^{-1},\quad E_{\mathrm{point}}\propto r^{-2}.$$The null is closer to the weaker line, but the distance ratio follows a different power law.

Physical check

The circles show cross-sections of infinite lines normal to the page, not individual point charges.