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The electric field at a distance $r$ from am infinite line uniformly charged is $$\vec{E}=\frac{\lambda}{2\pi r}\hat r.$$ This can be shown from Gauss's law.

If we want a potential whose negative gradient is $\vec{E}$, then clearly $V(r)=-\frac{\lambda}{2\pi}\log(r)$ will do the job.

However, if I wanted to obtain this potential directly from Coulomb's law and the superposition principle, I would write the integral $$\int_{-\infty}^\infty \frac{\lambda dx}{\sqrt{r^2+x^2}},$$ and this integral is divergent.

So the question is how to obtain the potential from Coulomb's law and the superposition principle, without computing the electric field first.

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    $\begingroup$ The potential cannot be just $\ln r$. Remember that the argument of a function like a sine or a logarithm must be dimensionless. The potential must be $\ln \mu r$ for some constant $\mu$ with dimension of inverse length. As you can't get a $\mu$ from an integral which does not contain $\mu$, there is no non-divergent expression that gives the potential. Instead you need to use a renormalization proceedure: subtract the expression for the potential at two different points. the integral in the difference is then convergent. $\endgroup$
    – mike stone
    Aug 18, 2021 at 13:15

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You can do this on condition of not imposing a zero potential at infinity. If you impose a zero potential on the distance $r_0$ of the wire, the elementary potential will be of the form $dV=\frac{\lambda}{4\pi\epsilon}\left(\frac{1}{\sqrt{x^2+r^2}}-\frac{1}{\sqrt{x^2+{r_0}^2}}\right)$

It remains to integrate $\int_{-A}^{B}\left(\frac{1}{\sqrt{x^2+r^2}}-\frac{1}{\sqrt{x^2+{r_0}^2}}\right)dx=\hbox{arcsinh}\left(\frac{B}{r}\right)-\hbox{arcsinh}\left(\frac{B}{r_0}\right)+\hbox{arcsinh}\left(\frac{A}{r}\right)-\hbox{arcsinh}\left(\frac{A}{r_0}\right)$

Then to pass to the limit when $A$ and $B$ tend towards infinity $\hbox{arcsinh}\left(\frac{B}{r}\right)-\hbox{arcsinh}\left(\frac{B}{r_0}\right)\ -> \log(\frac{r_0}{r})$

We do indeed find the famous $2\log(\frac{r_0}{r})$ !

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This really is sensible. An infinite wire contains an infinite charge. It takes an infinite amount of energy to bring a charge from an infinite distance away up close to an infinite charge. The same idea might be clearer with an infinite plane, where the E field is constant all the way to infinity.

The force is never infinite (unless you hit the wire). Even though some of those charges are arbitrarily far off to the side, there is enough force at large distances that it takes an infinite energy to travel an infinite distance through the field.

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