Using Gauss's law, I find that the expression for this field is :
$$E(r) = \frac{\lambda}{2\pi r \epsilon_0}$$
where $\lambda$ is the line charge density.
However, when I try to use the direct formula :
$$E(r) = \frac{1}{4\pi\epsilon_0} \int_{-\infty}^{+\infty}\frac{\lambda}{r^2}dl$$
The integral diverges…
Is it not possible to find the expression of this field without using Gauss' law ?
Why doesn't the formula just work, like it is supposed to ?
EDIT : I forgot to mention that the wire is infinitely thin
Best Answer
Coulomb's law says
$$ \vec{E(\vec{r})} = \frac{1}{4 \pi \epsilon_0} \int \frac{\rho(\vec{r}') (\vec{r} - \vec{r}') \, d^3 r'}{|\vec{r} - \vec{r}'|^3} .$$
The charge density is a constant $\lambda$ and is linear. We'll fix that the wire runs along the $Y$ axis and we're calculating the field along the $X$ axis. This being the case, let us fixe $\vec{r}=R\,\hat{x}$ as the horizontal distance $R$ from the wire to where we're measuring the field, and $\vec{r}'= z\, \hat{z}$ as the vertical coordinate ranging from $-\infty$ to $\infty$ along the wire. This brings us to
$$ \vec{E} = \frac{\lambda \hat{x}}{4 \pi \epsilon_0} \int_{-\infty}^{\infty} \frac{R \, dz}{(R^2 + z^2)^{3/2}} ,$$
where we've canceled the $Y$ component due to symmetry considerations. Integrating this expression leads us to the proper result.
This problem is usually solved using an angle as a parameter, but this integral here is so much fun. You can of course simplify by exchanging to angle variables, but do play a little bit proving that it converges, evaluating it and then taking the limit. It is also necessary to prove that this limit exists (try using l'Hôpital), and to evaluate it I'd advise you to use a geometric series in $z$.
EDIT: We're calculating the field on circle with radius $R$, of course. Maybe that didn't come out very clear.