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For the infinite line of charge, we know the more we move away, dependence of E is 1/r.

To get to how we derive 1/r, I go with the following way:

  1. We only have perpendicular components. Hence change for each point of charge is the following $ \delta E_\perp = \frac{x^2-2d^2}{r^5} \delta d $ (d is perpendicular distance, x horizontal distance)

  2. Now, to get the total change from all the charge, I integrated it from -infinity to infinity which gave me the following result. $ -2\delta d/(d^2) $.

Since now, it has d^2 and not d in the denominator, it got me thinking, how I get d and not d^2. So I thought about the following logic:

I think what we do here is integration actually gives us $ -2/d^2 $. Then we can transform to d. is on the same axis as d, so it can definitely be expressed as = z * d. Then we got -2 * z * d / d^2 = -2 * z / d. Hence we got 1/d dependence and not 1/d^2.

Q1: is my assumption correct and if so, we're now left with z(some number), but isn't this wrong ? or this is what people mean with 1/r dependence ? They only mean what it's inversely proportional and that's it ?

Q2: Since dependence is 1/r and not 1/d. Do they mean 1/r in which r is presumed to be d(perpendicular) ?

Qmechanic
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Giorgi
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3 Answers3

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The dependence of field strength vs. distance depends on the geometry of the situation. For a point charge, the field propagates as an expanding sphere. Since the area of a sphere is $A=4 \pi r^2$, doubling the distance from a point source causes the associated field to "paint" 4 times the area, resulting in a $1/r^2$ relationship. For an infinite line charge, the field propagates as an expanding cylinder. Since the area of the sides of that cylinder is $A=2 \pi r L$, doubling the distance from an infinite line source causes the associated field to "paint" 2 times the area, resulting in a $1/r$ relationship. In other words, there is no need for higher level mathematics if this problem is viewed from a geometric standpoint.

David White
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to get the total change from all the charge, I integrated it from -infinity to infinity which gave me the following result. $−2\delta d/(d^2)$.

This is the correct result. That integral should be proportional to $d^{-2}$. I don't understand your reasoning when you bring in $z$, but let me explain what this integral represents. $$ \delta E_\perp(x,d,\delta d) = \frac{x^2 - 2d^2}{r^5}\delta d $$ is the change in $E_\perp$ from the point at $x$ when you move from $d$ to $d+\delta d$ from the line of charge. This is a simplification because there's no $k$ or charge. We've been omitting $k$ so far, so let's continue to do that, and assume it's 1. Let's do the same with the charge density $\lambda$. Still, you need a $\mathrm dx$ to convert $\lambda$ into charge. So let's integrate: $$ \delta E_\perp(d,\delta d) = \int_{-\infty}^\infty\frac{x^2 - 2d^2}{r^5}\delta d\mathrm dx = 2\delta d\int_0^\infty\frac{x^2 - 2d^2}{(x^2 + d^2)^{5/2}}\mathrm dx = 2\delta d\left(-\frac{1}{d^2}\right) = -\frac{2\delta d}{d^2} $$ This is saying that if you're at $d$ and take a small step away from the line, the field decreases by $\frac{2\delta d}{d^2}$. Note: decreases by. This is not saying that the electric field is $-\frac{2\delta d}{d^2}$. What if we looked at it a different way? $$ \delta E(d,\delta d) = -\frac{2\delta d}{d^2} \longrightarrow \frac{\delta E(d,\delta d)}{\delta d} = -\frac{2}{d^2} \longrightarrow \frac{\partial E}{\partial d} = -\frac{2}{d^2} $$ Now to confirm, we can take the standard equation for $E$ near an infinite line of charge: $$ E(d) = \frac{2}{d}\longrightarrow \frac{\partial E}{\partial d} = -\frac{2}{d^2} $$

This confirms our result. Integrating $\delta E(x)$ yields the same result as differentiating $E(d)$.

Jacob Stuligross
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With the line of charge along $\hat z$, linear density $\lambda$: What is $\vec E(\rho, \phi, z)$?

By symmetry: there is no $\phi$ nor $z$ dependence, the leaves only $ \vec r$ for $\vec E$, so:

$$ \vec E = E(r)\hat r $$

Then: there are two distance scales, $\lambda^{-1}$ and $||r||$, so:

$$ E(r) \propto r^a/\lambda^b $$

with $a-b = -2$ by dimensional analysis (DA). DA also gives $E \propto \lambda$ so $b=-1$, thus $a=-1$.

Thus:

$$ E(r) = a \frac 1 {4\pi}\frac{\lambda} r $$

Now figure out $a$....$a=1$ is for spheres.

JEB
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