Field Due to An Infinitely Long Straight Uniformly Charged Wire

Electric charge is distributed uniformly along an infinitely long, thin wire. The charge per unit length is (assumed positive). Find the electric field using Gauss’s law.

We found that the field E of a uniformly charged, infinite wire is radially outward if is positive and radially inward if $\lambda$ is negative, and that the field magnitude E depends only on the radial distance from the wire. This suggests that we use a cylindrical Gaussian surface, of radius r and arbitrary length l coaxial with the wire and with its ends perpendicular to the wire (Fig. 1).

The flux through the flat ends of our Gaussian surface is zero because the radial electric field is parallel to these ends, and so $\mathbf{E}.\hat{\mathbf{n}}=0$. On the cylindrical part of our surface we have $\mathbf{E}.\hat{\mathbf{n}}=E_{\perp}=E$ everywhere. (If $\lambda$ were negative, we would have $\mathbf{E}.\hat{\mathbf{n}}=E_{\perp}=-E$ The area of the cylindrical surface is $2\pi rl$ so the flux through it and hence the total flux $\Phi_E$ through the Gaussian surface is $EA=2\pi rlE$. The total enclosed charge is $Q_{encl}=\lambda l$ and so from Gauss’s law,

$\Phi_E=\oint \mathbf{E.S}=\frac{q_{encl}}{\epsilon_0}$

$\Phi_E=2\pi rlE=\frac{\lambda l}{\epsilon_0}$ and

$E=\frac{1}{2\pi \epsilon_0}\frac{\lambda l}{r}$ (field of an infinite line of charge)     (1)

If is negative, E is directed radially inward, and in the above expression for E we must interpret $\lambda$ as the absolute value of the charge per unit length.

Fig.1:A coaxial cylindrical Gaussian surface is used to find the electric field outside an infinitely long, charged wire.

We saw in the previous discussion that the entire charge on the wire contributes to the field at any point, and yet we consider only that part of the charge $Q_{encl}=\lambda l$ within the Gaussian surface when we apply Gauss’s law. There’s nothing inconsistent here; it takes the entire charge to give the field the properties that allow us to calculate $\Phi_E$ so easily, and Gauss’s law always applies to the enclosed charge only. If the wire is short, the symmetry of the infinite wire is lost, and E is not uniform over a coaxial, cylindrical Gaussian surface. Gauss’s law then cannot be used to find $\Phi_E$; we must solve the problem the hard way. 

We can use the Gaussian surface in Fig. 1 to show that the field outside a long, uniformly charged cylinder is the same as though all the charge were concentrated on a line along its axis. We can also calculate the electric field in the space between a charged cylinder and a coaxial hollow conducting cylinder surrounding it.

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