Line Image of a Charged Cylinder

[Bobbo Snap]

Homework Statement

A long conducting cylinder bearing a charge [itex] \lambda [/itex] per unit length is oriented parallel to a grounded conducting plane of infinite extent. The axis of the cylinder is at distance [itex] x_0 [/itex] from the plane, and the radius of the cylinder is [itex] a [/itex]. Find the location of the line image, and find also the constant [itex] M [/itex] (which determines the potential of the cylinder) in terms of [itex] a [/itex] and [itex] x_0 [/itex].

The Attempt at a Solution

My instructor is giving us problems from a different book than we are using in class and there are some gaps in the material covered. That said, we haven't covered anything but point images (i.e., no "line images"). It seems to me that the location of the image would be at a distance [itex] -x_0 [/itex] on the other side of the plane by symmetry. Is this correct? And I have no idea how to find [itex] M [/itex]. Does anyone have any suggestions?

 

[Bobbo Snap]

Anyone? Am I correct in thinking that the image is on the other side of the plane? Any hints on how to to find [itex] M [/itex]?

 

[TSny]

The line image is on the other side, but not at a distance ##x_0##. The problem is rather involved, so you have some work cut out for you. You might start with finding some references on the method of images.

http://www.ece.mcmaster.ca/faculty/nikolova/4FJ4_downloads/tutorials/T02_CircuitParametersTLs.pdf I found that contains some discussion of the type of geometry in this problem. In fact, if you can follow the material starting on page 8, I believe you should be able work out the solution.

 

[Bobbo Snap]

Thanks, that does look like the geometry of this problem. I think the ratio [itex]\rho_1/\rho_2[/itex] is the constant [itex]M[/itex] it's asking for but I'm not sure how the potential values were found. I will study it some.

 

[paranormal]

I would suggest approaching this problem by first reviewing the concept of line images and their properties. Line images are created when a charged object is placed near a conducting plane, resulting in a symmetrical arrangement of charges on both sides of the plane. The image charge is located at a distance equal to the object's distance from the plane, but on the opposite side. In this case, the line image would be located at a distance -x_0 from the plane.

To find the value of M, we can use the method of images to create an equivalent system with a known solution. We can imagine a point charge placed at a distance x_0 from the plane, with a magnitude of charge equal to the total charge per unit length of the cylinder (λ). This point charge will create the same electric field as the cylinder, and we can use this to determine the value of M.

Using the formula for potential due to a point charge, we can calculate the potential at the surface of the cylinder (r = a) as V = kλ/a. We can then equate this potential to the potential of the equivalent point charge system, which is V = kQ/x_0, where Q is the charge of the point charge and x_0 is the distance from the plane. Solving for Q, we get Q = λa. This is the charge of the equivalent point charge.

Now, we can use the formula for potential due to a point charge to calculate the potential at any point along the axis of the cylinder. The potential at a distance x from the center of the cylinder is given by V = kQ/(x_0 + x). We can equate this potential to the potential of the cylinder system, which is V = M/x. Solving for M, we get M = kQx_0/(x_0 + x). Plugging in the value of Q, we get M = kλax_0/(x_0 + x).

In summary, the line image is located at a distance -x_0 from the plane, and the constant M is given by M = kλax_0/(x_0 + x). I would also recommend discussing this problem with your instructor or classmates to gain a better understanding of line images and their properties.