Physics · Ch 1 — Electric Charges and Fields

Application of Gauss's Theorem: Field Due to an Infinitely Long Straight Charged Wire

1.15

Application of Gauss's Theorem: Field Due to an Infinitely Long Straight Charged Wire

The problem. Find the electric field at a perpendicular distance $r$ from an infinitely long, thin, straight wire carrying a uniform linear charge density $\lambda$ (Section 1.5).

Exploiting the symmetry. An infinite line of charge looks identical from every point along its own length, and identical under rotation about its own axis; the only physically meaningful field direction consistent with this cylindrical symmetry is a field that points radially outward (perpendicular to the wire) at every point, with a magnitude that depends only on the perpendicular distance $r$ from the wire, and not at all on position along the wire's length or around its circumference.

Choosing the Gaussian surface. This symmetry is matched exactly by choosing, as the Gaussian surface, a right circular cylinder of radius $r$ and length $l$ , coaxial with the wire (i.e. the wire runs exactly along the cylinder's central axis), as shown in the figure. This closed surface has three parts: the curved lateral surface, and two flat circular end caps.

Evaluating the flux, part by part. On the curved lateral surface, $\vec{E}$ is radial and therefore exactly parallel to the outward normal everywhere, and its magnitude $E$ is the same at every point of this surface (by the symmetry argument above), so the flux through it is simply $E$ times the curved surface's area:

\Phi_{\text{curved}} = E\times(2\pi rl)

On the two flat end caps, $\vec{E}$ is radial (perpendicular to the wire), which means it runs exactly parallel to these flat surfaces and perpendicular to their outward normals (which point along the cylinder's own axis), so the flux through each end cap is zero: $\Phi_{\text{ends}}=0$ .

Applying Gauss's theorem. The total flux is $\Phi=E(2\pi rl)+0$ . The charge enclosed within the cylinder is just the length of wire $l$ inside it, carrying charge $q_{\text{enc}}=\lambda l$ . Gauss's theorem then gives

E(2\pi rl) = \frac{\lambda l}{\epsilon_0}

\boxed{E = \frac{\lambda}{2\pi\epsilon_0 r} = \frac{2k\lambda}{r}}

directed radially outward from the wire for a positive $\lambda$ (radially inward for a negative $\lambda$ ). …

Figure 1Cylindrical Gaussian surface around a charged wire

What this figure shows. A long straight wire is drawn running vertically through the centre of the figure, marked with evenly spaced plus signs along its length to show a uniform positive linear charge density $\lambda$ . A cylinder is drawn coaxial with the wire (its central axis coinciding with the wire), with radius $r$ and length $l$ , shown as a semi-transparent tube so that the wire is visible running along its axis; this cylinder is the chosen Gaussian surface. Short outward-pointing arrows are drawn perpendicular to the cylinder's curved side surface at several points around and along it, representing the radial electric field $E$ , all of equal length to show that $E$ has the same magnitude everywhere on the curved surface. The cylinder's two flat circular end caps (top and bottom) are shaded differently from the curved side, with a note that no field-line arrow …

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