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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

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    $\begingroup$ One thing that you have neglected is the vector nature of Couloumb's Law. Your second equation does not give the electric field correctly. $\endgroup$
    – garyp
    Commented Mar 10, 2015 at 20:06
  • $\begingroup$ What do you mean ? The symmetry tells us that the field is radial, its direction is given by the sign of $\lambda$ and its norm is given by the formula. $\endgroup$
    – mwa1
    Commented Mar 10, 2015 at 21:11

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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.

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  • $\begingroup$ I forgot to mention that the wire is supposedly infinitely thin, so if understand your formula, the $z$ parameter is always zero and we're back to $\vec{E(r)} = \frac{\lambda}{4\pi\epsilon_0 r^2}\vec{e_r}\int_{-\infty}^{\infty}dl$ $\endgroup$
    – mwa1
    Commented Mar 10, 2015 at 21:56
  • $\begingroup$ No, $z$ is never $0$. $z$ is mapping the wire. The fact that the wire is thin is already included. $\endgroup$ Commented Mar 10, 2015 at 22:10
  • $\begingroup$ Oh, ok I get it. $\endgroup$
    – mwa1
    Commented Mar 10, 2015 at 22:28

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