Two line charges and a conducting cylinder are flying through space

[truth is life]

Homework Statement

Griffiths 3.36: Two straight wires having equal and opposite line charges are situated on either side of a conducting cylinder (all the wires and the cylinder are "long", so we can ignore edge effects--this is undergraduate electrostatics, after all!). The cylinder has no net charge and a radius R, the wires are an equal and opposite distance a > R apart from the cylinder's axis. Find the potential.

Homework Equations

Gauss' Law for the E-field (*much* easier than trying to calculate potential directly). Other than that, nothing really.

The Attempt at a Solution

I know this problem involves images; I'm just not sure where to place them (aside from inside the cylinder, obviously), and Griffiths is...less than helpful in figuring it out. If I did know where to put them, the problem would be easy, since all I'd need to do would be to put down the potential for a line charge and superposition the potentials from the two main charges and the images.

 

[diazona]

When I first started reading the title I was thinking "Two line charges and a conducting cylinder walk into a bar (magnet?)..."

Anyway, I don't remember ever doing this particular problem, but my first thought would be to go back to the application of the method of images to a spherical conductor, which I believe is discussed in some detail in Griffiths' book. See if you can apply a modified version of that to this problem.

 

[truth is life]

When I first started reading the title I was thinking "Two line charges and a conducting cylinder walk into a bar (magnet?)..."

It's what I was going for! Glad someone else thought it looked like a joke setup.

Anyway, I don't remember ever doing this particular problem, but my first thought would be to go back to the application of the method of images to a spherical conductor, which I believe is discussed in some detail in Griffiths' book. See if you can apply a modified version of that to this problem.

Thanks! I assumed that the image charges were on the same axis as the line charges (seemed reasonable), used the distance formula (between the image charge and r=0) he got for the spherical case, and got more or less the answer he got, the difference being which part of the fraction inside the natural log term you end up with is on top (he seems to have the negative part on top, while I have the positive part, which is rather mystifying). It still doesn't help me with the more general image case, though--Griffiths is really, really handwavy about how to do that--the best you get out of him is to *guess* where things are, which isn't much help...

 

[diazona]

That's pretty much the way I'd do it too. It definitely seems fair to assume that the images are in the same plane as the given line charges. The main difference from the spherical case is that instead of a point, you have a line, and instead of a sphere you have a cylinder. Or to think about it another way, if you take a slice through space perpendicular to the line charges (and perpendicular to the cylinder's axis), you have two points and a circle, and you're doing electrostatics in 2D instead of 3D (which is why you have natural logs in the formulas). If you're not sure whether your formulas are exactly correct, try coming up with some simple tests, like looking at the behavior as r goes to infinity or something like that.

And you're right to notice that the method of images does involve some guesswork. The fact is, it's not a generally applicable technique, in the sense that if you have a conductor of some arbitrary shape, there probably won't be any "nice" configuration of image charges you can use with it. There are only certain specific cases in which the method of images is really useful.

 

[truth is life]

That's pretty much the way I'd do it too. It definitely seems fair to assume that the images are in the same plane as the given line charges. The main difference from the spherical case is that instead of a point, you have a line, and instead of a sphere you have a cylinder. Or to think about it another way, if you take a slice through space perpendicular to the line charges (and perpendicular to the cylinder's axis), you have two points and a circle, and you're doing electrostatics in 2D instead of 3D (which is why you have natural logs in the formulas). If you're not sure whether your formulas are exactly correct, try coming up with some simple tests, like looking at the behavior as r goes to infinity or something like that.

Well, it gets the right answer so it must be right :) Anyways, I figured out the problem by plotting out the two functions; Griffith's version gives a negative potential close to the negative charge while my version gives a positive potential, which makes sense since a positive charge in a negative potential is in a lower energy state and that's what you would expect of a positive test charge coming closer to a negative charge. Still, I'm wondering why all the positive charge terms suddenly flipped sign.

And you're right to notice that the method of images does involve some guesswork. The fact is, it's not a generally applicable technique, in the sense that if you have a conductor of some arbitrary shape, there probably won't be any "nice" configuration of image charges you can use with it. There are only certain specific cases in which the method of images is really useful.

True, but Griffiths could at least give more than two examples and do a bit more discussion on how to *educatedly* guess. As I said, you are left pretty confused about how to go about doing it, since both examples have him essentially pulling the answer out of thin air and admitting as much.

 

[diazona]

True, but Griffiths could at least give more than two examples and do a bit more discussion on how to *educatedly* guess.

Actually, I don't know about that. Based on what I remember from the book, there really isn't much else to say about it. Whenever you use the method of images, it's just a matter of regurgitating one of the known examples: if it's a conducting sheet, you put the image charge at the same distance on the other side; if it's a sphere, you put it on the same radial line at R^2/r (IIRC); and so on (I forget if there are other cases).

The point I mean to make is that yes, it seems pretty arbitrary, but that's because it is arbitrary, and not because the book is neglecting to give you full information. (Again, based on what I remember, although I don't have my copy handy so I could be wrong)