Derive the Electric Field Outside an Ideal Conductor

AI Thread Summary
To derive the electric field just outside an ideal conductor, start with the electric field expression for an infinite sheet of uniform surface charge density, E=σ/(2ε). The superposition principle can be applied, considering two parallel ideal wires with opposite currents, where the electric fields E1 and E2 from each wire are equal in magnitude but opposite in direction, resulting in E=E1+E2. The crucial point is that the electric field inside the conductor is zero, requiring all internal electric fields to sum to zero. Consequently, the electric field just outside the conductor is E=σ/ε. This approach highlights the relationship between surface charge density and the electric field in conductive materials.
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Homework Statement


Begin from the expression of the electric field outside an infinite sheet of uniform surface charge density, E=\frac{σ}{2\epsilon}. Derive the electric field just outside an ideal conductor: E=\frac{σ}{\epsilon}. Do NOT use Gauss's law.


Homework Equations


Not sure.


The Attempt at a Solution


Any tips on how I can solve this problem?

NOTE:
I removed my attempt at a solution, but I did try.
 
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Hi, I think you just use superposition principle in electric field. Imagining that there are two ideal wires in parallel with the current flowing in the opposite direction. Then the electric field between two wires is calculated as follows:
E = E1 + E2
where E1, E2 is the electric field of each ideal wire at the said point.
They have the same value but opposite direction.
E1 = E2 = σ/2ε
I think it is a bit like the electric field between two plates of capacitor.
attachment.php?attachmentid=60140&stc=1&d=1373349736.jpg

It is my opinion, maybe I am wrong.
 

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Anhnha's advice should put you on the right track; it's not exactly the same as a parallel plate capacitor, but the idea is very similar.

The key bit of knowledge for this particular problem is that the [static] electric field within the conducting material itself is zero. So you need to make the vector sum of all electric fields inside the conductor add to zero (some electric fields may be originating from charges located somewhere else, such as the opposite side of the conductor or whatnot [it doesn't really matter where the other charges are] -- whatever the case, all electric fields must all sum to zero inside the conductor). By doing so, what then is the electric field just outside the conductor? :wink:
 
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