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I was doing a physics problem in Purcell's E&M book when I encountered a problem that asked to find the work needed per unit length to assemble an infinite wire charge of radius $a$, by bringing in infinitesimal charges from a radius $R$. However, after solving the problem, I get that the work required per unit length to assemble the wire this way from a radius $R$ is $$\frac{\lambda^2}{4\pi\epsilon_0}\left(\frac14 + \ln\left(\frac Ra\right)\right).$$

Taking the limit as $ R\rightarrow \infty$, we get that this approaches infinity. Wouldn't this mean it would take infinite work per unit length to assemble this, and that the infinitely long wire has an infinite potential energy per unit length?

I'm having trouble physically and intuitively understanding this result.

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  • $\begingroup$ To charge an infinite capacitor takes infinite charge and infinite energy to pump it all in. $\endgroup$
    – Jon Custer
    Jul 13 at 19:56
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The E field of an infinite line charge goes like $\frac{1}{r}$. The E field of an infinite line charge with a small gap is very close to the same except in the immediate neighborhood of the gap. So to get the work to bring in the little rod of charge that will fill the gap from infinity, you are basically integrating $\frac{1}{r}$ out to infinity, which diverges.

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  • $\begingroup$ Thank you, that makes sense. But I still find it strange that the potential energy per unit length is infinite. Does this just mean it takes extremely much work to assemble a long wire like in the problem? How are real wires like this constructed, then? $\endgroup$ Jul 13 at 22:16
  • $\begingroup$ While the integral diverges, it does so very slowly. Say your “wire” has a radius $a$ of 1 meter. Then for a particular charge density and/or particular test charge, the work, in Joules, required to bring the test charge from distance $R$ to the surface of the wire is $ln(R)$. That’s infinity if $R$ is infinite, but if $R=10^{27}$m (size of the known universe), it’s 62 J. $\endgroup$
    – Ben51
    Jul 13 at 22:34
  • $\begingroup$ Thank you @Ben51, that helped very much. It now makes lots of sense. I guess I didn't really realize how slowly the log diverges and it makes much more sense with some numbers plugged in. $\endgroup$ Jul 13 at 22:43
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Even if you want to just compute the potential energy of an infinite line of point charges, seperated by a distance $a$, then, counting from the "center" point, symmetrically, the energy is:

$$\begin{align} E &= \frac{2kq^{2}}{a} + \frac{2kq^{2}}{2a} + \frac{2kq^{2}}{3a} + ...\\ &= \frac{2kq^{2}}{a}\left(1 + \frac{1}{2} + \frac{1}{3} + ...\right)\\ &= \frac{2kq^{2}}{a}\sum\frac{1}{n} \\ &= \infty \end{align}$$

So, it's not surprising that the continuum limit of this setup diverges.

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  • $\begingroup$ That's a bit disingenuous. The infinity in the question would remain for a finite section of wire, for which the specific calculation here is not a good analogy. $\endgroup$ Jul 13 at 20:22
  • $\begingroup$ I agree with @EmilioPisanty here. 1) That is not the same scenario; your example deals with discrete point charges in one line, while mine deals with a uniform distribution over an infinitely long cylinder, which has a width, unlike your line. 2) The potential energy is infinite for even a finite section of wire in my case (as Emilio mentioned) $\endgroup$ Jul 13 at 21:31

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