 # Electric Field Due to an Infinite Line Charge using Gauss’ Law

Section 5.5 explains one application of Gauss’ Law, which is to find the electric field due to a charged particle. In this section, we present another application – the electric field due to an infinite line of charge. The result serves as a useful “building block” in a number of other problems, including determination of the capacitance of coaxial cable (Section 5.24). Although this problem can be solved using the “direct” approach described in Section 5.4 (and it is an excellent exercise to do so), the Gauss’ Law approach demonstrated here turns out to be relatively simple.

### Exercise

EXAMPLE 5.6.1: ELECTRIC FIELD ASSOCIATED WITH AN INFINITE LINE CHARGE, USING GAUSS’ LAW.

Use Gauss’ Law to determine the electric field intensity due to an infinite line of charge along the

axis, having charge density

(units of C/m), as shown in Figure 5.6.1.

Gauss’ Law requires integration over a surface that encloses the charge. So, our first problem is to determine a suitable surface. A cylinder of radius

that is concentric with the

axis, as shown in Figure 5.6.1, is maximally symmetric with the charge distribution and so is likely to yield the simplest possible analysis. At first glance, it seems that we may have a problem since the charge extends to infinity in the

and

directions, so it’s not clear how to enclose all of the charge. Let’s suppress that concern for a moment and simply choose a cylinder of finite length

. In principle, we can solve the problem first for this cylinder of finite size, which contains only a fraction of the charge, and then later let

to capture the rest of the charge. (In fact, we’ll find when the time comes it will not be necessary to do that, but we shall prepare for it anyway.)

Here’s Gauss’ Law:

where

is the electric flux density

,

is a closed surface with outward-facing differential surface normal

, and

is the enclosed charge.

The first order of business is to constrain the form of

using a symmetry argument, as follows. Consider the field of a point charge

at the origin (Section 5.5):

We can “assemble” an infinite line of charge by adding particles in pairs. One pair is added at a time, with one particle on the

axis and the other on the

axis, with each located an equal distance from the origin. We continue to add particle pairs in this manner until the resulting charge extends continuously to infinity in both directions. The principle of superposition indicates that the resulting field will be the sum of the fields of the particles (Section 5.2). Thus, we see that

cannot have any component in the

direction because none of the fields of the constituent particles have a component in that direction. Similarly, we see that the magnitude of

cannot depend on

because none of the fields of the constituent particles depends on

and because the charge distribution is identical (“invariant”) with rotation in

. Also, note that for any choice of

the distribution of charge above and below that plane of constant

is identical; therefore,

cannot be a function of

and

cannot have any component in the

direction. Therefore, the direction of

must be radially outward; i.e., in the

direction, as follows:

Next, we observe that

on the right hand side of Equation 5.6.1 is equal to

. Thus, we obtain

The cylinder

consists of a flat top, curved side, and flat bottom. Expanding the above equation to reflect this, we obtain

Examination of the dot products indicates that the integrals associated with the top and bottom surfaces must be zero. In other words, the flux through the top and bottom is zero because

is perpendicular to these surfaces. We are left with

The side surface is an open cylinder of radius

, so

, a constant over this surface. Thus:

The remaining integral is simply the area of the side surface, which is

. Solving for

we obtain

Remarkably, we see

is independent of

, So the concern raised in the beginning of this solution – that we wouldn’t be able to enclose all of the charge – doesn’t matter.

Completing the solution, we note the result must be the same for any value of

(not just

), so

and since

:

This completes the solution. We have found that the electric field is directed radially away from the line charge, and decreases in magnitude in inverse proportion to distance from the line charge.

Suggestion: Check to ensure that this solution is dimensionally correct.

Ellingson, Steven W. (2018) Electromagnetics, Vol. 1. Blacksburg, VA: VT Publishing. https://doi.org/10.21061/electromagnetics-vol-1 CC BY-SA 4.0 Swipe left and right to change pages. Get the latest tools and tutorials, fresh from the toaster.

### What are you looking for? 