A simple electric model: A line of charge

One situation that we sometimes encounter is a string of unbalanced charges in a row. As a simplified model of this, we can look at a straight-line string of charge that has infinitely small charges uniformly distributed along a line. Although this doesn't sound very realistic, we'll see that it's not too bad if you are not too close to the line (when you would see the individual charges) and not too close to one of the ends. A variety of diagrams can help us see what's going on.

A set of charges along a straight line

We can use one of the ubiquitous simulation programs available on the web to look at what the electric field for a string of point charges along a straight line really looks like. (The Physics Classroom has a nice electric field simulator.) We've put 6 identical positive charges along an approximate straight line. The program has put the electric field vector due to these 6 charges down at every point on a grid. Each vector gives the direction of the field and, by its intensity (darkness of the vector), the strength of the field.

It's a little hard to see how the field is changing from the darkness of the arrows. Another way to see it is from coloring the arrows. In the figure below, the red arrows represent stronger field with the intensity decreasing as the color goes through yellow, green and blue.

We can see that close to the charges, the field varies both in magnitude and direction pretty wildly. But as you get even a little bit away it settles down and is smooth. Anywhere along the middle of the line the field points straight away from the line and perpendicular to it. You can see the "edge effect" changing the direction of the field away from that as you get towards the edge. And not that as you get farther from the line, the edge effect works its way in towards the center.

A model of a uniform infinite line of charge: linear charge density

In order to get a simple model, let's imagine that we could make our charges as tiny as we wanted. We could then describe our charge as a linear charge density: an amount of charge per unit length.

We'll ignore the fact that the charges are actually discrete and just assume that we can treat it as smooth. This is like treating water as having a density of 1 g/cm³. Really, it depends on exactly how many molecules of water you have included. But as long as we have lots of molecules in even the smallest volume we allow ourselves to imagine, we're OK talking about a density. Same thing here, but we are going to ignore the with of the individual charges and treat them as if they are an ideal (geometric) line. Then if we have a charge of $Q$ spread out along a line of length $L$, we would have a charge density, $λ = Q/L$.

Below we show four lines of different lengths that have the same linear charge density.

The first has a length $L$ and a charge $Q$ so it has a linear charge density, $λ = Q/L$. The second has a length $2L$ and a charge $2Q$ so it has a charge density, $λ = 2Q/2L$. The third has a length $3L$ and a charge $3Q$ so it has a charge density, $λ = 3Q/3L$.

The fourth line is meant to go on forever in both directions — our infinite line model. It therefore has both an infinite length and an infinite charge, but if each piece of the line is just like our first one we can still say the line charge has a linear charge density of λ, even if we can't say what its total charge or length is.

What does the field near an infinite line charge look like?

From the picture above with the colored vectors we can imagine what the electric field near an infinite (very long) line charge looks like. In a plane containing the line of charge, the vectors are perpendicular to the line and always point away. It's a bit difficult to imagine what this means in 3D, but we can get a good idea by rotating the picture around the line. It's sort of like a cross between a snake and a hedgehog. At every point around the snake there is a spine pointing out and away. Something like the picture at the right. The red cylinder is the line charge. The E field at various points around the line are shown. (The other cylinders are equipotential surfaces.)

Why does it look that way?

Let's take a look at how the field produced by the line charge adds up from the little bits of charge the line is made up of. Take a look at the figure below.

We are considering the field at the little yellow circle in the middle of the diagram. Now we break up the line into little segments of length $dx$. The one right beneath the yellow circle is colored red. By Coulomb's law it produces an E field contribution at the yellow circle corresponding to the red arrow — pointing up. Now a useful observation is that for every bit of charge on the left side of the line — say the green one — there is a corresponding one on the other side of the center, an equal amount away. These two produce green contributions pointing away from themselves.

We note three things:

1. As we get further away from the center (say from red to green to purple), the contribution gets smaller since the distance of the charge from our observation point gets larger.
2. Note that for the paired contributions that are not at the center, the horizontal components of the two contributions are in opposite directions and so they cancel.
3. As we get further away from the center (say from green to purple) the individual vectors tip out more. This means that more of their magnitude comes from their horizontal part.

The result is that as you get further away from the center, the contributions to the E field at our yellow observation point matter less and less. It really is only the part of the line that is pretty close to the point we are considering that matters. (This is why we can get away with pretending that a finite line of charge is "approximately infinite.")

How does it depend on stuff?

Can we quantify the dependence? We can actually get a long way just reasoning with the dimensional structure of the parameters we have to work with.

As with most dimensional analysis, we can only get the functional dependence of the result on the parameters. Fortunately, that's often the most important part of what the equation is telling us. (To get the number out in front, we actually have to do the integral, adding up all the contributions explicitly.)

Let's suppose we have an infinite line charge with charge density $λ$ (Coulombs/meter). If we are a distance of d from the line, how strong do we expect the field to be?

Well, we know that what we are doing is adding up contributions to the E field. The E field from a point charge looks like

$$E= \frac{k_Cq}{r^2}$$

the Coulomb constant, times a charge, divided by a length squared. Our result from adding a lot of these up will always have the same structure dimensionally. There's always a $k_C$ and it's messy dimensionally so let's make our dimensional analysis easier and factor it out: we'll just look at the dimension of $E/k_C$. This is just a charge over a distance squared, or, in dimensional notation:

$$\bigg[\frac{E}{k_C}\bigg] = \bigg[\frac{q}{r^2}\bigg] = \frac{\mathrm{Q}}{\mathrm{L}^2}$$

That is, $E/k_C$ has dimensions of charge divided by length squared. So for a line charge we'll have to have this form as well, since it's just adding up terms like this.

But for an infinite line charge we aren't given a charge to work with. The charge is infinite! We therefore have to work with $λ$. This is a charge per unit length so it has dimensions $\mathrm{Q/L}$. We only have one length to work with — the distance from the line, $d$.

This tells us that the only combination we can make with the correct dimensions from this parameter set is $λ/d$. Therefore we must conclude that $E$ from the line charge is proportional to $k_C λ/d$ just by dimensional analysis alone.

Doing the integral shows that there is actually a factor of 2, so near a line charge the E field is given by

$$\frac{E}{k_C} \propto \frac{\lambda}{d} \quad \rightarrow \quad E = \frac{2k_C\lambda}{d}$$

Basically, we know an E field looks like a charge divided by two lengths (dimensionally). Since our charge already comes with one length, we only have room for one more.

Joe Redish 2/14/12