Why is the E-field inside a conductor zero?

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Homework Statement



Oh dear, veterens of PF who have tried to convince me that E = 0 inside conductors, I am having this problem again.

OKay, let's say that "charges do arrange themselves such that the E-field inside a conductor is always zero", then how can wires in circuits work? If the E-field inside the conductor is 0, how can the charges move? How can there be a current?


The Attempt at a Solution



Going in circles
 

Discussion

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The field inside a conducting wire, with an external field applied (end to end) is not zero.

The charges are in motion, and continue in motion, as though they were attempting to level out and nullify the field. But they can't ever accomplish that as long as the external EMF is maintained. New charges enter at one end, and leave by the other. This is why batteries "run down".
 
gneill said:
The field inside a conducting wire, with an external field applied (end to end) is not zero.

The charges are in motion, and continue in motion, as though they were attempting to level out and nullify the field. But they can't ever accomplish that as long as the external EMF is maintained. New charges enter at one end, and leave by the other. This is why batteries "run down".
Well, the field inside a conductor is practically 0. It is d/dx(IR) where R is the resistance of the wire. In other words the field is the current x resistance of the wire per unit length, the latter being a very small number measured in ohms/metre.

AM
 
Andrew Mason said:
Well, the field inside a conductor is practically 0. It is d/dx(IR) where R is the resistance of the wire. In other words the field is the current x resistance of the wire per unit length, the latter being a very small number measured in ohms/metre.

AM

I don't understand whaty you mean by current * resistance per unit length, you mean voltage per length? I don't understand what this has to do with 0 field inside the wire?
 
flyingpig said:
I don't understand whaty you mean by current * resistance per unit length, you mean voltage per length? I don't understand what this has to do with 0 field inside the wire?
Well, the units of potential are volts (Joules/coulomb). V=IR (Ohm's law). The units of resistance are Potential/current (eg. Volts/Amperes) or Ohms). The field, E, is equal to dV/dx. Since V = IR, E = d(IR)/dx. Since current is the same at all points in the conductor, this means that E = IdR/dx.

dR/dx is simply the resistance per unit length of the conductor which we could call [itex]\rho_L[/itex]. So:

[tex]\vec{E} = \vec{I}\rho_L[/tex]

AM
 
The units isn't giving me the intuition as to how charges flow if the E-field inside the conductor is 0? Do they flow outside?
 
The E field inside the conductor when current flows is NOT ZERO.

At least, it's not zero for DC current and low-ish frequency AC currents. If you're interested in some additional physics details when high frequency AC currents are involved, look up "skin effect". This may not be something you'll run across early on in your coursework.
 
flyingpig said:
The units isn't giving me the intuition as to how charges flow if the E-field inside the conductor is 0? Do they flow outside?
There is no field in an ideal conductor (ie zero resistance) in a circuit in which current flows. If there is no potential difference (V = IR = I x 0 = 0) there is no field. The field occurs in the resistance.

This is not difficult to understand. If there is no resistance, one does not need a "force" to keep current going through the conductor.

AM
 
Wait, if V = 0, how does that mean E = 0? But I am saying that it is the E-field that's driving the charges right? If there is no E-field they aren't going to move.
 
flyingpig said:
Wait, if V = 0, how does that mean E = 0? But I am saying that it is the E-field that's driving the charges right? If there is no E-field they aren't going to move.
dV = Edx If dx is not zero but dV is, what does that tell you about E?

Once moving, charges do not need a force to keep them moving if there is no resistance. So no field is needed to keep current flowing in a perfect (ideal) conductor. One is not accelerating the charges; merely keeping the current constant. The force is needed to push then through a resistance. So that is where the field occurs. If the electrons are pushed through a resistor there has to be a field and a potential difference from one end to the other.

AM
 
I still don't understand, especially when we are incooperating potential difference into this question. I can see it mathematically, but I have no intuition it for it whatsoever
 
Wait, just how could there be a 0 V? There must be a voltage drop if a current runs across the wire (with negligible resistance)
 
Also, inside a conductor, the e-field is 0 by definition because you apply an external field. But in many Gauss's law problems, there isn't an external field
 
flyingpig said:
Wait, just how could there be a 0 V? There must be a voltage drop if a current runs across the wire (with negligible resistance)
As the resistance goes to zero, the current created by a given voltage increases without bound--so essentially with zero resistance you could have any current regardless of voltage.

I think you have to keep in mind here that the "zero E-field inside a conductor" rule only works in equilibrium. If you have a conductor sitting in a region of zero electric field, then suddenly turn on an electric field, the electrons have to move to cancel out the electric field. With small resistance they can do that much more quickly (look up dielectric relaxation).
 
flyingpig said:
Also, inside a conductor, the e-field is 0 by definition because you apply an external field. But in many Gauss's law problems, there isn't an external field

Not sure what you mean by this. External field or not, there should be zero field inside a conductor. That idea does not come from Gauss's law; it just makes Gauss's law more useful
 
The electric field in a conductor is zero in electrostatic equilibrium. If a wire is carrying a current, it is not in electrostatic equilibrium. If the wire is hooked up to a voltage source (like a battery), that prevents it from ever reaching equilibrium.
 
diazona said:
The electric field in a conductor is zero in electrostatic equilibrium. If a wire is carrying a current, it is not in electrostatic equilibrium. If the wire is hooked up to a voltage source (like a battery), that prevents it from ever reaching equilibrium.
The point is that the E field is very small inside a real conductor with current flowing. E is proportional to the resistance/unit length of the conductor. If it is an ideal conductor with no resistance, the potential difference from one end of the conductor to the other with current flowing will be V = IR = 0. Since [itex]\Delta V = E\Delta x[/itex] the field E must be 0.

AM
 
Last edited by a moderator:
flyingpig said:
http://www.electron.rmutphysics.com...rs-Serway-Beichne 6edr-4/24 - Gauss's Law.pdf

Go to page 12 of the pdf, it derives why E-field is 0 inside a conductor

Now go to this site I found https://people.ok.ubc.ca/jbobowsk/phys102/phys102%20002%20Midterm%201%20solns.pdf

Go to the last page, there is no external field, how can there be a E-field of 0 inside the conductor?
I believe that diazona addresses the problem you are having with this concept.

Look at what Serway says in Chapter 27 (Section 3, I think).

In order for current to flow in a conductor having finite conductance, a non-zero E field is required within the conducting material itself. This is not an electrostatic situation.
 
Last edited by a moderator:
Then how am I suppose to know rather it is in electrostatic situation or not
 
No, but answer those Gauss's Law problem don't tell you they are static
 
No...but charges don't just become stationary just because we add a word called eletrcostatics, you need an external field to make them move so that they become 0
 
Please! I've been stuck on this concept for over 3 months now ...
 
Why E = 0 inside a conductor.

In my book, the way it was derived, it only happens when you apply an external E-field.

But in many of my HW problems concerning Gauss's law, this "external" field does not exist anywhere, yet they still claim E = 0 inside a conductor
 

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