Field of an infinite wire

An infinite straight wire with linear charge density $\lambda$ (coulombs per metre) possesses cylindrical symmetry: the field is radial with respect to the axis and depends only on the distance $d$ from the wire, not on the position along it.

Calculation with Gauss. As the Gaussian surface we choose a cylinder coaxial with the wire, of radius $d$ and length $L$. The field is parallel to the two bases of the cylinder (which therefore give no contribution to the flux) and perpendicular to the lateral surface, of area $2\pi d L$. The enclosed charge is $\lambda L$, and Gauss’s theorem gives:

$E \cdot 2\pi d L = \frac{\lambda L}{\varepsilon_0} \quad\Rightarrow\quad \ev{E = \frac{\lambda}{2\pi\varepsilon_0\,d}}$

Key formula

Infinite wire with linear density $\lambda$: $\ev{E = \frac{\lambda}{2\pi\varepsilon_0\,d}}$ The field decreases as $1/d$ (not as $1/d^2$ for a point charge).

Infinite wire: view from above (cross-section). The Gaussian surface is a coaxial cylinder of radius $d$. The field lines are radial due to cylindrical symmetry. The flux is non-zero only on the lateral surface; on the bases the field is parallel to the normal and gives no contribution.

Summary — Fields from Gaussian symmetries

The fields of the main charge distributions, obtained by exploiting symmetry with Gauss’s theorem:

Distribution	Electric field
Sphere, $r>R$	$E = kQ/r^2$
Uniform sphere, $r<R$	$E = kQr/R^3$
Spherical shell, $r<R$	$E = 0$
Infinite plane	$E = \sigma/(2\varepsilon_0)$
Infinite wire	$E = \lambda/(2\pi\varepsilon_0\,d)$
Parallel-plate capacitor	$E = \sigma/\varepsilon_0$

Collegamenti

Argomenti: Campo elettrico e potenziale Concetti: Teorema di Gauss Competenze: Applicazione del teorema di Gauss Oggetti: Filo rettilineo infinito

Esercizi collegati: Which graph of E · From the field to the spherical radius · Speculative physics: Coulomb 1 over r cubed