# Electric potential of infinite line from direct integration

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.

• 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. Aug 18, 2021 at 13:15

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 $$2\log(\frac{r_0}{r})$$ !