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

AI Thread Summary
The discussion focuses on solving a problem involving an infinitely long charged wire above a grounded conducting sheet using the method of images. To find the potential above the sheet, an image wire with charge density -λ is placed a distance -d below the sheet, allowing the potential at the sheet to be set to zero. The potential of the infinite charged wire is clarified as not being infinite; instead, it is expressed as (-λ/2π) ln(r/a), where 'r' is the distance from the wire and 'a' is an arbitrary reference point. This formulation simplifies calculations by avoiding square roots through the use of logarithmic properties. The conversation also references similar problems in advanced texts, indicating the complexity of the topic.
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Homework Statement


Given an infinitely long wire, with charge density \lambda, parallel to and a distance d above an infinite, grounded, conducting sheet, what is the potential, V(x,y,z) 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 -\lambda a distance -d 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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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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aha! Thanks astrosona. That's something I've forgotten about.
 
In fact, due to the properties of logarithms, you can rewrite this as

\phi = -\frac{\lambda}{4\pi} \ln \frac{r^2}{a^2}

which might make some things simpler to calculate, because r^2 = x^2 + y^2, 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
 
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