Method of Images: Infinite, charged wire above a grounded plate

In summary, the question asks for the potential and induced charge on an infinitely long wire with charge density \lambda, parallel to and a distance d above an infinite, grounded, conducting sheet. The solution involves constructing an infinitely long wire with opposite charge density and disregarding the sheet. The potential of an infinitely long charged wire can be calculated using the formula (-\lambda/2\pi)Ln(r/a), where r is the distance from the wire and a is an arbitrary reference point. The potential can also be rewritten as -(\lambda/4\pi)Ln(r^2/a^2) for easier calculation. There is a similar problem in Jackson's with two grounded planes meeting at right angles.
  • #1
SonOfOle
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Homework Statement


Given an infinitely long wire, with charge density [itex]\lambda[/itex], parallel to and a distance [itex]d[/itex] above an infinite, grounded, conducting sheet, what is the potential, [itex]V(x,y,z)[/itex] above the sheet?

What is the induced charge on the sheet?


Homework Equations




The Attempt at a Solution



This is a method if images problem. Construct a infinitely long wire with charge density [itex]-\lambda[/itex] a distance [itex]-d[/itex] below the plane of the sheet and then disregard the sheet (it creates the same potential, V=0, at the plane of the sheet).

However, I ran into a snag when I tried to find the potential of an infinitely long charged wire. Is the potential infinite?
 
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  • #2
No... if you have the Griffiths book please look at page 80. There is a very important discussion in there where students mostly pass reading it very fast!

Due to the argue in the last past of page 80, you can not choose the infinity as your potential reference point. you can choose it is a arbitrary point named a, for example so the potential of a infinite charged line will be:

(-lamda/ 2pi) Ln (r/a)

which r is the distance from the wire
 
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  • #3
aha! Thanks astrosona. That's something I've forgotten about.
 
  • #4
In fact, due to the properties of logarithms, you can rewrite this as

[tex]\phi = -\frac{\lambda}{4\pi} \ln \frac{r^2}{a^2}[/tex]

which might make some things simpler to calculate, because [itex]r^2 = x^2 + y^2[/itex], and you no longer have to bother with square roots.

Your problem is pretty simple, but in Jackson's there is a similar problem with two grounded planes, meeting at right angles, with the line charge located at (x_0, y_0). :P
 

1. What is the Method of Images?

The Method of Images is a mathematical technique used in electrostatics to solve boundary value problems involving conductors and charges. It involves creating a mirror image of a given charge or conductor to satisfy boundary conditions and simplify the problem.

2. How does the Method of Images work?

The Method of Images works by creating a virtual charge or conductor that is a mirror image of the original charge or conductor. The sum of the electric fields from the original and virtual charges satisfies the boundary conditions, allowing for the solution of the problem.

3. What is an example of using the Method of Images?

An example of using the Method of Images is solving for the electric field above a grounded plane caused by an infinite, charged wire. The image charge would be a mirror image of the wire located below the plane, with the same magnitude of charge but opposite sign.

4. What are the benefits of using the Method of Images?

The Method of Images allows for the simplification of complex electrostatic problems and provides a straightforward solution. It also provides insight into the behavior of electric fields near conductors and charges.

5. Are there any limitations to the Method of Images?

Yes, the Method of Images is limited to problems with simple geometries and boundary conditions. It also assumes an idealized, two-dimensional situation and does not account for the effects of other charges or conductors in the vicinity.

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