**Required math: calculus, algebra**

**Required physics: electrostatics**

Reference: Griffiths, David J. (2007) Introduction to Electrodynamics, 3rd Edition; Prentice Hall – Problem 3.36.

Another example of the method of images. We are given two infinite charged wires, one with linear charge density and the other with . These wires lie in the plane along the lines . Between them is an infinite conducting cylinder of radius , with its axis on the axis. The cylinder carries no net charge. We are to find the potential everywhere outside the cylinder.

The solution to this problem relies on an earlier example in which we worked out the potential due to two charged wires on their own. We saw there that the equipotential surfaces for two charged wires were circular cylinders, so we should be able to set up an image configuration in which the surface of the cylinder can be replaced by image wires inside the cylinder.

In the charged wires problem, we had two wires at and found that the potential could be written as

We also found that for a constant potential , if we define the value we could get expressions for the radius and axis of the equipotential cylinders:

A useful relation between these two is

To apply the method of images and the above solution, we need to create one image for each wire. Consider first the wire on the left, at . We want to create an image of this wire at , where this location is inside the cylinder. We can map this image problem onto the two-wire problem by noting that the midpoint between the wire and its image is at . If we define , then the potential due to this wire-image configuration is

On the other side, the wire at has an image at . Defining , we get for the potential of this pair:

The only remaining problem is to find . We can use 4 for this. Consider the first wire-image pair. The distance is the distance from the midpoint between the wire and image to the axis of the cylinder, which is . The quantity in this relation is half the distance between the wire and its image, so in this case must be replaced by . We therefore have

Since potentials obey the superposition principle, we can just add the two potentials from the two image-wire pairs to get the total potential:

To convert this to cylindrical coordinates, we can use and :

Note that if , then the argument of the logarithm becomes 1, so on the surface of the cylinder. This happens even though the potential of each wire-image pair separately is *not* zero at . In fact, for the first pair, we have

and for the second pair, we get

Also, if or , . This corresponds to the plane, which is the plane of symmetry in the problem.

AndrewThank you for this. However, I don’t understand how you, using equation 4, came to replace a by ((a-b)/2)^2. Would you please explain?

AndrewAlso, why consider mid-points?

gwrowePost authorIn the original problem (two wires at , no cylinder), the wires are separated by a distance of , with being the distance from each wire to the midpoint. In the image problem, we have one wire at and the image wire at , so the distance from each wire to the midpoint is (in both cases). Therefore, to map the original solution onto the image problem, we replace by .