Electromagnetic Fields and Waves: Notes, Solved Problems and Interactive Lab

Problem 3.2.1

Infinitely long line charge

Find the electric field at a normal distance of $ρ$ from an infinitely long line charge with a uniform charge density of $ρ_{L}$ by using Gauss's law.

Source: D. K. Cheng, Field and Wave Electromagnetics, 2nd ed., Example 3-5.

Try it yourself first.

Strategy

An infinite line has cylindrical symmetry, so the field is radial and depends on $ρ$ only. A coaxial cylinder through P is then a Gaussian surface: $E$ is normal to its side with constant magnitude, and tangential to its caps.

Symmetry

The infinite line is unchanged by a translation along its own axis ( $z$ ) and by a rotation about it, so the field cannot depend on $z$ or $ϕ$ : it depends on $ρ$ only. It is also unchanged by the reflection $z \to - z$ through the plane perpendicular to the line that contains the field point P, which would reverse a $z$ component, hence $E_{z} = 0$ . Finally, it is unchanged by the reflection through the plane that contains the line and P, which would reverse a $ϕ$ component, hence $E_{ϕ} = 0$ . The field is radial and its magnitude depends on $ρ$ only:

E = E_{ρ} (ρ) a_{ρ} .

Gaussian surface

For the left-hand side of Eq. (3.2.13) we take a cylindrical surface of radius $ρ$ , coaxial with the line, from $z = - L /2$ to $z = L /2$ ( $L$ to be taken to infinity later). It consists of three parts (Figure 3.2.16a):

the side, where $d S = ρ d ϕ d z a_{ρ}$ and $E_{ρ}$ is constant, so that $E \cdot d S = E_{ρ} ρ d ϕ d z$ , and
the top and bottom caps, where $d S = \pm ρ d ρ d ϕ a_{z}$ is perpendicular to $E$ , so that $E \cdot d S = 0$ .

Left-hand side

Only the side contributes:

ε_{0} \oint_{S} E \cdot d S = ε_{0} \int_{z = - L /2}^{L /2} \int_{ϕ = 0}^{2 π} E_{ρ} ρ d ϕ d z = ε_{0} E_{ρ} ρ ϕ (2 π) z (L) = 2 π L ε_{0} ρ E_{ρ} .

Enclosed charge

The cylinder contains a length $L$ of the line:

total charge inside S = ρ_{L} L .

Result

Equating both sides,

\Rightarrow 2 π L ε_{0} ρ E_{ρ} = ρ_{L} L E_{ρ} = \frac{ρ _{L}}{2 π ε _{0} ρ} .

The length $L$ has cancelled, so letting $L \to \infty$ has no effect on the result, and we find

Answer

D E = \frac{ρ _{L}}{2 π ρ} a_{ρ}, = \frac{ρ _{L}}{2 π ε _{0} ρ} a_{ρ} [V/m]

(a) The bookkeeping on the Gaussian cylinder.

(b) D_\rho falls as 1/\rho, but the flux per unit length is constant. Problem 3.2.1. (a) On the side of the coaxial cylinder (orange) \vec D is parallel to d\vec S and has the same length all round. On the caps (grey) it is tangential. (b) D_\rho versus \rho: the product 2\pi\rho D_\rho, the flux per unit length through a coaxial cylinder, is \rho_L for every radius. — **Figure 3.2.16.** (b) $D_{ρ}$ falls as $1/ ρ$ , but the flux per unit length is constant. Problem 3.2.1. (a) On the side of the coaxial cylinder (orange) $D$ is parallel to $d S$ and has the same length all round. On the caps (grey) it is tangential. (b) $D_{ρ}$ versus $ρ$ : the product $2 π ρ D_{ρ}$ , the flux per unit length through a coaxial cylinder, is $ρ_{L}$ for every radius.

Check

Units: $ρ_{L} / (ε_{0} ρ)$ is $(C/m) / ((F/m) m) = C/ (F m) = V/m$ . This is the result of Problem 3.1.1, obtained there by integrating $d E$ along the whole line. Here symmetry does the work: the flux per unit length through every coaxial cylinder is $ρ_{L}$ , and since the area per unit length of the cylinder grows as $2 π ρ$ , the field falls as $1/ ρ$ . Note that the infinite length was essential: for a finite line the field is not purely radial and is not constant on the side of a cylinder, and Gauss's law cannot be used to compute it.

Problem 3.2.1: Infinitely long line charge