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### Kurs: Fizyka - 12 klasa (Indie)>Rozdział 1

Lekcja 6: Continuous charge distribution

# Natężenie pola elektrycznego pochodzącego od jednorodnie naładowanej, nieskończenie długiej, prostej nici

Natężenie pola elektrycznego pochodzącego od nieskończenie długiej, jednorodnie naładowanej, prostej nici. Stworzone przez Willy McAllister.

## Worked Example: Electric field near a line of charge

We derive an expression for the electric field near a line of charge.
The result will show the electric field near a line of charge falls off as $1/a$, where $a$ is the distance from the line.
Assume we have a long line of length $L$, with total charge $Q$. Assume the charge is distributed uniformly along the line. The total charge on the line is $Q$, so the charge density in coulombs/meter is,
$\mu =\frac{Q}{L}$
Assume a test charge $q$ is positioned opposite the center of the line, at a distance $a$.
What is the electric field at the location of $q$ due to (created by) the line of charge?
This derivation will lead to a general solution of the electric field for any length $L$, and any distance $a$. Using this general solution, we will solve a particularly useful case where the line is very long relative to the distance to the test charge, $L\gg a$.
First, create and name some variables to talk about.
• $a$ is the distance from the line to the location of our test charge, $q$.
• $dQ$ is a tiny amount of charge contained in a tiny section of the line, $dx$.
• $x$ is the distance from where $a$ touches the line to $dQ$.
• $r$ is the distance from $dQ$ to the location of the test charge.
• $\theta$ is the angle between $a$ and $r$.
The electric field surrounding some point charge, $Q$ is,
$E=\frac{1}{4\pi {ϵ}_{0}}\phantom{\rule{0.167em}{0ex}}\frac{Q}{{r}^{2}}$
The electric field at the location of test charge $q$ due to a small chunk of charge in the line, $dQ$ is,
$dE=\frac{1}{4\pi {ϵ}_{0}}\frac{dQ}{{r}^{2}}$
The amount of charge $dQ$ can be restated in terms of charge density, $dQ=\mu \phantom{\rule{0.167em}{0ex}}dx$,
$dE=\frac{1}{4\pi {ϵ}_{0}}\phantom{\rule{0.167em}{0ex}}\mu \phantom{\rule{0.167em}{0ex}}\frac{dx}{{r}^{2}}$
The most suitable independent variable for this problem is the angle $\theta$. The analysis is simplified by recasting the equation to sweep $d\theta$ through a range of angles instead of sweeping $dx$ along the line (this is a change of variable).
After the change of variables, we can redraw the diagram in terms of $d\theta$,
The change of variables allows us substitute $\frac{d\theta }{a}$ for $\frac{dx}{{r}^{2}}$ in the previous equation,
$dE=\frac{1}{4\pi {ϵ}_{0}}\phantom{\rule{0.167em}{0ex}}\mu \phantom{\rule{0.167em}{0ex}}\frac{d\theta }{a}$
Now we exploit the symmetry of the charge arrangement by figuring out the electric field in just the $y$ direction (the direction going straight from the line through $q$).
This means we scale the electric field $dE$ down by the cosine of the angle $\theta$,
$d{E}_{y}=\frac{1}{4\pi {ϵ}_{0}}\phantom{\rule{0.167em}{0ex}}\frac{\mu }{a}\phantom{\rule{0.167em}{0ex}}\mathrm{cos}\theta \phantom{\rule{0.167em}{0ex}}d\theta$
We are ready to integrate (add up) all the contributions from each $dQ$ to get the electric field,
${E}_{y}={\int }_{-\theta }^{+\theta }\frac{1}{4\pi {ϵ}_{0}}\frac{\mu }{a}\phantom{\rule{0.167em}{0ex}}\mathrm{cos}\theta \phantom{\rule{0.167em}{0ex}}d\theta$
This is the general solution for the electric field near any length of line, $L$, at any distance $a$ away from the line. The limits $±\theta$ are the angles to either end of the line.

### Useful case: long line of charge

Now we solve for the useful case where the line of charge is very long relative to the separation $a$, or $L\gg a$. If you stand at $q$ and turn your head to look in either direction towards each end of this very long line, your head turns (very nearly) $±\phantom{\rule{0.167em}{0ex}}{90}^{\circ }$ ($±\phantom{\rule{0.167em}{0ex}}\pi /2$ radians). These become the limits on our integration.
${E}_{y}={\int }_{-\pi /2}^{+\pi /2}\frac{1}{4\pi {ϵ}_{0}}\frac{\mu }{a}\phantom{\rule{0.167em}{0ex}}\mathrm{cos}\theta \phantom{\rule{0.167em}{0ex}}d\theta$
Move anything that doesn't depend on $\theta$ outside the integral.
${E}_{y}=\frac{1}{4\pi {ϵ}_{0}}\frac{\mu }{a}{\int }_{-\pi /2}^{+\pi /2}\phantom{\rule{0.167em}{0ex}}\mathrm{cos}\theta \phantom{\rule{0.167em}{0ex}}d\theta$
and evaluate the integral,
${E}_{y}=\frac{1}{4\pi {ϵ}_{0}}\frac{\mu }{a}\phantom{\rule{0.167em}{0ex}}\mathrm{sin}\phantom{\rule{0.167em}{0ex}}\theta \phantom{\rule{0.167em}{0ex}}{|}_{-\pi /2}^{+\pi /2}=\frac{1}{4\pi {ϵ}_{0}}\frac{\mu }{a}\phantom{\rule{0.167em}{0ex}}\left(+1--1\right)=\frac{2}{4\pi {ϵ}_{0}}\frac{\mu }{a}$
Finally, the electric field created by a long line of charge at a point $a$ away from the line is,
${E}_{y}=\frac{\mu }{2\pi {ϵ}_{0}}\frac{1}{a}$
Well done if you followed this all the way through. The important finding from this exercise is: in contrast to $1/{r}^{2}$ for a point charge, the field surrounding a line of charge falls off as $1/a$.
We did a lot of math to derive this result. It is worthwhile to take a moment to sit with this solution to let it soak in. Now that you have seen the math, does it make intuitive sense that distance has a different exponent, $1/a$, compared to a point charge, $1/{r}^{2}$?
As an amusing distraction, if you recall the fable of the butter gun from the Inverse Square Law article, can you design a new butter gun for a line of charge, that sprays in a $1/a$ pattern?

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