E & M, E-potential and energy due to charged plate/ rod with infinite length

[maiad]

Homework Statement

An infinite sheet of charge that has a surface charge density of 25.9 nC/ m2 lies in the yz plane, passes through the origin, and is at a potential of 1.32 kV at the point y = 0, z = 0. A long wire having a linear charge density of 81.1 nC/ m lies parallel to the y-axis and intersects the x-axis at x = 3.15 m. Determine, the potential along the x-axis at x = 0.827 m between wire and sheet.

Homework Equations

The Attempt at a Solution

I actually don't even know where to start to begin with so some hints would be appreciated

 

[Mindscrape]

Have you studied the method of images? Basically it says that if you have a plane at a particular potential, and a charge above that plane you can create a virtual charge (an image charge) below that plane to give the plane the potential it needs. Start off with finding the potential for the images. I, or someone else (or if you catch it), will get to the charge density later.

 

[maiad]

I have only learned about gauss's law, property of charged conductors, and Electric potentials ;/ which mostly just consist of equations that we can use

 

[Mindscrape]

Well, method of images is just an extension of electric potentials. What if you treated the plane as a kind of charge mirror. You could create a potential that was
V=Vline1-Vline2+DCterm
where Vline1 is the potential from the first line charge, Vline2 is the potential of the second line charge, and the DC term is there because the problem wants to be annoying and not have the plane be grounded. The goal is for Vline1 and Vline2 (opposite in charge and location from Vline 1) to cancel their potentials at the (0,0) point. So, once you convince yourself of that geometry, work out what the potential is of 2 line charges (one of opposite charge) separated by 3.15m*2 would be.

Understand?

 

[maiad]

not quite

 

[Mindscrape]

Hmm, maybe try looking at the wikipedia page on method of images to start.
http://en.wikipedia.org/wiki/Method_of_image_charges

The first example on the wiki entry is a simple 1 charge and grounded plane problem. They make an opposite image charge to give the grounded plane a potential of 0 at z=0, where the plane is located. If you understand this problem, you can then imagine if we had a line of charges, then we need a line of images.

 

[maiad]

ok so i tried another approach
1. E=σ/2ε
2. V=σx/2ε (electric potential for the plane)
3.E=2κλ/(d-x)
4. V=-2λk ln(d-x) (electric potential for the rod)
x being the distance from the origin to the point between the two conductors

I found the potentials in reference to the point but... I'm not sure what i do with the initil potential given.