Result

Long line of charge

(E2.3.2) E2.3 · Application of Gauss's Law to Various Charge Distributions →

E = \frac{λ}{2 π ε _{0} r} = \frac{2 k _{e} λ}{r}

Statement

The field points straight out from the wire and falls off as 1/r.

Symbols and units

$λ$	charge per unit length	C/m
$r$	perpendicular distance from the wire	m
$k_{e}$	Coulomb constant	8.99×10⁹ N·m²/C² (=1/4πε₀)
$ε_{0}$	permittivity of free space	8.854×10⁻¹² C²/(N·m²)
$E$	field magnitude	N/C

When it applies

Close to a long wire, far from its ends.

Don't use when

An idealization valid close to a long wire, far from its ends — near the ends of a finite wire the field differs significantly.

Derivation

By symmetry E is radial and depends only on r, so use a coaxial cylinder of radius r and length ℓ as the gaussian surface.

No flux through the flat ends (E is parallel to them). On the curved side E is perpendicular and constant:

Φ_{E} = E (2 π r ℓ)

The enclosed charge is the density times the length:

q_{in} = λ ℓ

Gauss's law, then cancel ℓ:

E (2 π r ℓ) = \frac{λ ℓ}{ε _{0}} \Rightarrow E = \frac{λ}{2 π ε _{0} r}

Worked examples

Charged wire

λ = 5.0 nC/m at r = 2.0 cm:

E = \frac{2 ( 8.99 \times 1 0 ^{9} ) ( 5 \times 1 0 ^{- 9} )}{0.020} = 4.5 \times 1 0^{3} N/C

Watch out for

An idealisation, good near a long wire but wrong near its ends. The ℓ cancelling is the signal that the answer cannot depend on how much wire you chose to enclose.

Connects to

Gauss's law Sphere and cylinder Infinite sheet of charge