# B Why is the electric field in an ideal wire zero?

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1. Nov 23, 2017

### Elmer Correa

I’ve looked at the answers given to the previous times this question has been asked, but I still don’t seem to understand how this holds in the case of a closed circuit. Here’s an explanation given before:

“Think of the wire as a horizontal cylinder. If you apply an electric field pointing to the left, the electrons in the wire will move to the right, so that eventually they collect on the right side, and there is a deficit of electrons on the left. This distribution of charge (positive on the left, negative on the right) produces a field of its own, pointing to the right, which works against your applied field. This process will continue, until there is no net field left inside the conductor; the equilibrium is reached once there is no more field and thus the electrons experience no net force.”

This makes enough sense to me if we’re talking about a cylinder, but not a closed loop. Isn’t the whole point of an emf source in a circuit to prevent this sort of cancellation of fields? Instead of allowing electrons to clump up at the positive terminal of a battery, the battery “forces” the charges to the negative terminal to repeat the another cycle through the circuit, so how is it that the electric field in the conducting material of the wire has to necessarily be equal to zero? How would the electrons in the wire ever be able to redistribute themselves into an equilibrium?

I think this also may comes back to a misunderstanding I have about resistance. I’ve always thought of it as this sort of hand wavy property of a material that predicts the ratio of the potential difference through it to the current that runs through it. What part of this property actually allows an electric field to exist to establish a potential difference in a material?

2. Nov 23, 2017

### lomidrevo

The electric field in a conductor is zero in case of electrostatics (static charges). Once you attach a battery to a circuit, the charges starts to flow through the conductor creating the current, and of course there will be a non-zero electric field in the circuit (otherwise the charges wouldn't move)

EDIT: sorry, that is not truth for ideal wire, it can be zero even the current is flowing (no force is needed to maintain the current when there is no resistance)
The way I understand it: there is no voltage drop on ideal conductor without any resistance, ie. the potential difference is 0, therefore electric field must be zero.

Last edited: Nov 23, 2017
3. Nov 23, 2017

### Staff: Mentor

The zero voltage drop across a wire in a typical circuit is just a simplifying assumption. In reality there is some voltage drop, but if you have good wires then it is small enough that you just ignore it and pretend it is 0

4. Nov 23, 2017

### Elmer Correa

I still don’t see how that is exactly; what about the lack of resistance makes this voltage drop so close to zero? Wouldn’t the work done on the electrons moving in the circuit moving a certain distance in a given electric field be the same regardless of the presence of a resistor component in the circuit, or a given resistance per unit length in a wire?

5. Nov 23, 2017

### Staff: Mentor

Yes. But for a given current the electric field is smaller for lower resistance components. I.e. the electric field in a wire is usually not a “given” quantity.

6. Nov 23, 2017

### Elmer Correa

Ok, so resistance in a material affects the strength of electric fields in it directly then? Is there a reason for this that I may have missed while learning about Ohm’s Law or is it more complicated and safe to take it as a given until some more advanced course reveals why this is the case?

7. Nov 23, 2017

### Staff: Mentor

Well, sometimes teachers presenting Ohms law only give the circuit theory version $V=IR$ and neglect the Maxwell’s equations version $J=\sigma E$. In that version the connection is clear.

8. Nov 23, 2017

### Elmer Correa

That equation was presented, but my teacher described it in terms of a given electric field causing a certain current density, not the other way around which you described. If resistivity is the ratio of electric field over current density, and the reisitivity is zero, that simple relationship doesn’t really prove to me that the field has to be zero, just that the current in the material will approach infinity. How would the field be affected?

9. Nov 23, 2017

### Staff: Mentor

Hmm, that’s a pity. Such an equation should not be construed as a one way cause and effect relationship. Causes always preceed effects, and Ohm’s law does not imply any temporal ordering. Ohm’s law simply describes a relationship without any identification of cause or effect.

10. Nov 23, 2017

### Elmer Correa

I see. So in a simple circuit with a single resistor and battery, an electric field will only exist inside the resistor component? This seems pretty bizarre to me; is this aspect of Ohm’s Law grounded mostly in experimental evidence, or does theory actually exist that explains why resistance is necessary for the formation of an electric field?

11. Nov 23, 2017

### Staff: Mentor

And the battery.

Ohm’s law is not a universal law of nature. It is just an empirical relationship for certain materials. Many materials violate Ohm’s law, but where it holds it does not imply a specific cause-effect relationship.

12. Nov 23, 2017

### davenn

NO, that isn't what Dale said

it's small but not zero

so there is no bizzareness

13. Nov 24, 2017

### sophiecentaur

That's just part of the relationship and, as Dale says, it is not a 'one way' relationship. What are you doing with the Current in this thought experiment? Current, PD and Resistance are related by an equation (which is not necessarily Ohm's Law!!!) - that's all.

14. Nov 24, 2017

### Elmer Correa

What confused me was the part of the relationship that suggests that an electric field’s strength can change based on the properties of the material it passes through. It’s not particularly intuitive to me, but accept it I shall.

15. Nov 24, 2017

### sophiecentaur

You wouldn't be surprised if you replaced one resistor in a chain of resistors with a piece of R=0 wire and found that there was no PD across it, would you? Perhaps your problem is the thought experiment that you are using to come to your conclusion. In real life, parts of an experiment 'fight each other' and come to an amicable conclusion what they should all be doing. Volts settle down, tensions in springs equalise out, temperature gradients are formed. The 'Relationship R=V/I " will apply to the mutually agreed conditions.

16. Nov 24, 2017

### davenn

here's some reading material for you that should help you do more than "just accept", rather it should help you learn and understand which is the better path

1) Surface Charges on Conductors Carrying Steady Currents by B. R. Russell

2) The Electric Field Outside a Stationary Resistive Wire Carrying a Constant Current by A. K. T. Assis, W. A. Rodrigues Jr., and A. J. Mania.

3) Static Surface Charge on Current-Carrying Conductors by R. Cade and D. Soto Bello

Dave

17. Nov 24, 2017

### cabraham

Very well stated. I hope every person reads this & heeds it.

Claude

18. Nov 25, 2017

### DrZoidberg

That's only the case if there is more than one resistor in the circuit. Otherwise the field strength won't change if the value of the resistor is changed as long as it's physical dimensions stay the same.
This is easier to imagine if you think of water flowing through pipes. If there are two pipes and a pump connected in series, the pressure difference between the ends of one of the pipes can change if that pipe changes it's diameter or length. If there was only one pipe though it would stay constant as long as the pump outputs a constant pressure.

19. Nov 28, 2017

### FrankMlinar

Think of the "ideal" wire as a normal wire as the resistance goes to zero. That is, in the limit. Also assume the current (or current density) is constant, not the voltage. The electric field will get smaller and smaller, and in the limit, will tend towards zero.

20. Nov 30, 2017

### swampwiz

You're probably having a problem in discerning the level of abstraction. The case of a wire and a normal resistor will be such that the wire has such a smaller value of resistance that the value of the resistor can be presumed to be the value for the resistor and the wire in series. However, for the case of a short circuit, the wire *is* the resistor, so the resulting current will (theoretically) be very larger (what would end up happening is that the wire undergo a very large amount of heating from the resistance, causing the wire to melt and therefore open the circuit, or the battery will not be able to produce that level of current (which ends up being like the battery has some resistance).