Electric field inside a conductor

In summary, this conversation discusses the concept of potential difference and current in a simple circuit with ideal components. The voltage between two points is defined as the path integral of electric field, and the current is equal everywhere in the wire. The contradiction arises when assuming zero potential difference inside the wire, but the wire has a finite resistance. It is explained that in an ideal wire, electrons are not subject to any force and move steadily with a little push.
  • #1
anhnha
181
1
I need help to understand what is going on in this simple circuit. Consider all voltage source, resistor and wire are ideal. And also assume that the wire has the same cross-section area along its length (A = const (m^2)).
attachment.php?attachmentid=59350&stc=1&d=1370610004.jpg


The voltage between two points A and B is defined as path integral from A to B of electric field.
[tex]V_{AB} = \int ^B_A \overrightarrow {E}.\overrightarrow {dl}[/tex]

The current is equal to each other everywhere in the wire and the wire has the same cross-section area => the current density is also the same everywhere.
Acoording to Ohm's law:
[tex]\overrightarrow {J}=\sigma \overrightarrow {E}[/tex]

Because J ≠ 0, σ ≠ 0 => E ≠ 0 (1) everywhere in the wire (also very point from A to B)
But I also know that: [tex]V_{AB} = 0 [/tex]
[tex]V_{AB} = E\int ^B_A \overrightarrow {r0}.\overrightarrow {dl} = 0 [/tex]
If I choose the path is the straight line from A to B => integral from A to B of dl ≠ 0
[tex]\int ^B_A \overrightarrow {r0}.\overrightarrow {dl} ≠ 0 [/tex]=> E = 0 (2)
(1) and (2) can't exist simultaneously. Can you point out what I am wrong?
 

Attachments

  • Simple circuit.JPG
    Simple circuit.JPG
    4.8 KB · Views: 674
Physics news on Phys.org
  • #2
I think that when you say

[tex]V_{AB} = 0 [/tex]

You make the assumption that the resistance of the conductor is zero. This is only justified when the conductivity of the conductor is infinity. As J is finite then the value of E should tend to zero making the equations consistent and removing the contradiction.
 
  • Like
Likes 1 person
  • #3
Thanks, but the product ∞.0 is undefined.
J ≠ 0, σ = ∞ then can we say E = 0?
 
  • #4
anhnha said:
Thanks, but the product ∞.0 is undefined.
J ≠ 0, σ = ∞ then can we say E = 0?

Are you familiar with limits? The product isn't undefined it tends to J. We can say that E tends to zero but σ.E tends to J. The contradictions are from mixing ideal conditions like "potential difference inside a wire is zero!" and real conditions with wire having some finite resistance.
 
  • #5
I know the limit but not sure.
J = σE
J is a constant larger than zero.
Is there any rule to say that when σ approaches to infinity (∞) E has to approach to zero?
 
  • #6
If ## \sigma \rightarrow \infty ##, then what about ## E = \frac {J} {\sigma} ##?
 
  • Like
Likes 1 person
  • #7
anhnha said:
I know the limit but not sure.
J = σE
J is a constant larger than zero.
Is there any rule to say that when σ approaches to infinity (∞) E has to approach to zero?

Yes because (tending to infinity)*(something other than tending to zero) is always infinity it isn't any finite number.

On a side note your question raises some interesting questions like "in an ideal wire with no resistance there is no potential difference between two points on the wire so why do the electrons move?" I am not too sure about the answer except to say that they are moving towards the positive terminal of the battery.
 
  • Like
Likes 1 person
  • #8
voko said:
If ## \sigma \rightarrow \infty ##, then what about ## E = \frac {J} {\sigma} ##?

Ah, that is simple. I thought about it as J = σE and it makes more difficult!
 
  • #9
On a side note your question raises some interesting questions like "in an ideal wire with no resistance there is no potential difference between two points on the wire so why do the electrons move?" I am not too sure about the answer except to say that they are moving towards the positive terminal of the battery.
yes, that is also what I am confused. I asked this question in another thread but still not understand.
Hope someone can help.
 
  • #10
consciousness said:
On a side note your question raises some interesting questions like "in an ideal wire with no resistance there is no potential difference between two points on the wire so why do the electrons move?" I am not too sure about the answer except to say that they are moving towards the positive terminal of the battery.

In an ideal wire, a.k.a. a superconductor, electrons are not subject to any force, so they just keep going steadily. Simple inertial motion. All it takes is just a little push.
 
  • Like
Likes 1 person
  • #11
In an ideal wire, a.k.a. a superconductor, electrons are not subject to any force, so they just keep going steadily. Simple inertial motion. All it takes is just a little push.
Well, it makes sense.:smile:
 
  • #12
voko said:
In an ideal wire, a.k.a. a superconductor, electrons are not subject to any force, so they just keep going steadily. Simple inertial motion. All it takes is just a little push.

Cant say no to that! :smile:
 

1. What is the purpose of an electric field inside a conductor?

The purpose of an electric field inside a conductor is to maintain equilibrium and allow for the flow of electricity. The electric field inside a conductor helps to distribute the electric charge evenly and ensures that the conductor remains neutral.

2. How is the electric field inside a conductor different from that of an insulator?

The electric field inside a conductor is zero, while the electric field inside an insulator is non-zero. This is because the free electrons in a conductor can easily move and redistribute themselves, while the electrons in an insulator are tightly bound and cannot move as freely.

3. What factors affect the strength of the electric field inside a conductor?

The strength of the electric field inside a conductor is affected by the amount of charge present, the shape and size of the conductor, and the material it is made of. It is also affected by the presence of other nearby conductors or insulators.

4. Can the electric field inside a conductor change?

The electric field inside a conductor can change if the amount of charge inside the conductor changes. If the charge distribution inside the conductor is altered, the electric field will also change to maintain equilibrium.

5. How does the electric field inside a conductor relate to the surface charge density?

The electric field inside a conductor is directly proportional to the surface charge density. This means that as the surface charge density increases, the electric field inside the conductor also increases. This relationship is known as Gauss's law.

Similar threads

  • Introductory Physics Homework Help
Replies
1
Views
696
Replies
2
Views
170
  • Introductory Physics Homework Help
Replies
7
Views
898
  • Introductory Physics Homework Help
Replies
6
Views
898
  • Introductory Physics Homework Help
Replies
17
Views
285
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
7K
  • Introductory Physics Homework Help
Replies
13
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
26
Views
425
Back
Top