What Happens to Electric Fields in a Circuit?

[Sahara]

Hello,

First of all, I'm not sure if this is in the right section, and I know it's probably a very basic question compared to all you levels of knowledge. So please if it isn't in the right section, let me know where I should post it, and if not thanks in advance for the help.

I'm trying to get an idea of what happens in a circuit in terms of electrons electric fields, etc. I know how to solve for basic things (voltage, current, etc) but I never really understood what exactly happens in the wires.

So this is what I think I understand in a simple resistor voltage source circuit. Please correct me if any of it is wrong!
The voltage source creates a potential difference across its terminals. So the conventional positive charges in the wire have a higher potential energy near the positive terminal and this allows them to do work. As they travel through the wires, they dissipate this energy through the resistances until they reach the same level of potential as the negative terminal of the battery. That's my general idea/theory of what happens.

Now moving on to the battery, there is an electric field across it that is caused by the charge inbalance. So point 1, 2 and 3 in the diagram have a different potential and moving a positive charge from those points would cause this charge to gain/lose potential energy. Is this correct?

This is where I get confused:
In the wire connected to the battery I'm trying to figure out if there is an electric field. If there were one, a charge that is at point 4 would reach a lower potential once it is at point 5 (this would be caused just by the fact that it is further away in the electric field (im ignoring the resistance in the wire)). But everytime I quantitatively analyse a circuit, the potential at those two points is the same. So the only explanation that i could think of is that there is no electric field in the wire so there is no potential drop between those two points. But this confused me even more cause if there isn't an electric field, what would cause the motion of electrons in the first place?

If anyone can enlighten me, it would be greatly appreciated!

 

[Adjuster]

Hello Sahara,

This is the sort of question which helps us realize how much we take for granted, without really understanding it! As an electrical engineer, I only have ideas about what happens in real-world circuits where wires always have some resistance.

Referring to your circuit, let's call the battery voltage V, the resistor resistance R, and the wiring resistance r. (The path you have labelled 4 to 5 will be part of r). The current I flowing in this circuit will be I = V/(R+r). As r tends towards zero, the current tends towards V/R. Note that however small r gets, there will always be just enough potential drop across it (hence a weak but sufficient field along the path) for this current to flow.

At this point, we need a real Physicist to help us understand what would happen if r actually dropped to zero. That's not just academic, I guess, as there are such things as superconductors. I look forward to hearing more about this: I just hope any explanations will be in relatively simple terms, preferably without recourse to vector calculus.

 

[dlgoff]

This is where I get confused:
In the wire connected to the battery I'm trying to figure out if there is an electric field. If there were one, a charge that is at point 4 would reach a lower potential once it is at point 5 (this would be caused just by the fact that it is further away in the electric field (im ignoring the resistance in the wire)). But everytime I quantitatively analyse a circuit, the potential at those two points is the same. So the only explanation that i could think of is that there is no electric field in the wire so there is no potential drop between those two points.

The circuit you are showing is like a capacitor connected to a battery. Once it is charged up, there is no current hence the potential at point 4 and 5 would be at the same value. Your drawing correctly shows the E-field normal to the plate surfaces. On the connecting wires, the E-field would be normal to it's surface, hence would produce a radial field. To actually calculate the field values, you would need to use Gauss' law. Here is a good visual analysis of how a field between http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elesht.html#c2" would be obtained.

 

[Mike_In_Plano]

The typical way of looking at the metal in the wire is referred to as the river of electrons theory. Essentially, metals freely share the electrons in their outer shells, and these electrons are constantly shifting about driven by vibrations between the all the various parts (nucleus, inner electrons, and outer electrons).

The electrons will start a net motion if an e-field is applied or if the electrons find a region of lower energy.

When electrons follow the e-field, this is referred to as drift. It's not normally a smooth process as electrons are subject to bumping into things and going all possible directions, but on average, they follow the e-field. This is the means by which the current travels through wires. It does include resistance (due to electrons crashing about) and it does require a voltage drop (to generate an e-field and motivate the electrons). Less current requires less of an e-field, hence the resistance tends to be fairly constant as long as the temperature, crystalline structure, etc of the wire are constant.

Alloyed and impure wire tends to have crystals with boundaries that interrupt the flow of the electrons. Thus, these materials have a higher resistance. Internal strain within these wires also has an effect, so a freshly drawn wire will typically have a higher resistance than one that has aged or been annealed.

Without an electric field present, we say that the electrons diffuse. If the electrons find an energy barrier, such as a different type of metal that has different energies in it's outer orbitals, than the electrons will diffuse across the barrier rather much a like a bunch of ping pong balls settling from a higher surface to a lower one.

This is the basis of the Seebeck effect. In thermocouples, the electrons diffuse across the energy barrier until enough voltage develops (from the electrons crossing) that the energy required to move across the barrier is equal to the energy required to move back. Thus, the height of the ping pong balls is equal on both sides.

Hope this Helps,

- Mike

 

[vk6kro]

If you have a thick copper wire (maybe 10 mm diameter), it would need to be conducting hundreds of amps to maintain a voltage of even 1 volt across it. Ohms Law always applies to conductors.
So, unless you have such a current flowing, you can say that copper wires are at the same potential on all parts of them.

An electron at 4 in your circuit will not move to 5.
An electron right near the resistor will move into the resistor because a different electron moved from the power source into the conductor.
It is a bit like dropping a brick into a full bucket of water. The water that sloshes over the sides is probably not the water that was displaced.I find the term "potential difference" adequate to describe this sort of circuit.
Neither the positive or negative terminal is superior to the other. It is just the difference between them that matters.If there is a complete circuit external to the power source, current will flow and the external voltage drops will add up to the voltage of the power source.

It is a simple and powerful concept.