Gauss' Law - difficulty understanding this example

In summary, the charge on the inner wire attracts an opposite charge to the inner surface of the cylinder, neutralizing the electric field over the transition into the metal cylinder. This explains why the charge per unit length for the inner surface is -λ.
  • #1
NewtonianAlch
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Homework Statement



[PLAIN]http://img851.imageshack.us/img851/5741/86853377.jpg


The Attempt at a Solution



For the long straight wire, the charge per unit length λ is already given, this is distributed on the surface of the wire. The electric field is zero for anything less than r1, because there is no enclosed charge inside the metal.

In the solutions booklet it says that the charge per unit length for the inner surface of the cylinder is -λ so that the internal charge adds to zero for r2 < r < r3.

What do they mean by that, and how did they get -λ ?

I understand that for r greater than r2, but less than r3, the same principle applies as it did earlier, the charge is on the surface, and no enclosed charge inside the metal, so therefore the electric field is zero inside. But why is -λ the charge per unit length for the inner surface of the cylinder?
 
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  • #2
The charge on the inner wire attracts an opposite charge to the inner surface of the cylinder. This charge is what neutralizes the electric field over the transition into the metal cylinder. Otherwise the electric field due to the charge on the inner wire would extend through the metal cylinder, and we know that this doesn't happen!
 
  • #3
Ah yes, that makes sense now that you put it that way! Of course...
 

1. What is Gauss' Law and what does it explain?

Gauss' Law is a fundamental law in electromagnetism that relates the electric flux through a closed surface to the charge enclosed by that surface. It explains the relationship between electric charges and the resulting electric fields.

2. How is Gauss' Law expressed mathematically?

Gauss' Law is expressed as ΦE = qenc / ε0, where ΦE is the electric flux, qenc is the enclosed charge, and ε0 is the permittivity of free space.

3. Can you give an example of Gauss' Law in action?

An example of Gauss' Law in action is the electric field created by a point charge. The electric flux through a spherical surface surrounding the point charge is directly proportional to the charge enclosed by that surface.

4. What is a closed surface and why is it used in Gauss' Law?

A closed surface is a surface that completely encloses a given region in space. It is used in Gauss' Law because it allows us to calculate the electric flux passing through the surface and relate it to the charge enclosed by that surface.

5. How is Gauss' Law related to Coulomb's Law?

Gauss' Law and Coulomb's Law are closely related, with Gauss' Law being a more general form of Coulomb's Law. While Coulomb's Law relates the force between two point charges, Gauss' Law relates the electric field at a point to the charge enclosed by a surface surrounding that point.

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