Hello all. I wasn't exactly sure of the best phrasing for title, so I hope I've attracted the right folks to answer this question. It has to do with the physical description of circuits, and the electric fields within them.

The question is: What is the electric field in an ideal wire, in a circuit, for a wire that is connected to ground?
Here's what I mean: Imagine a circuit with a battery and a resistor. The battery is 1.5V, the resistor is a carbon resistor, nothing special, with value R. We have the (+) terminal of the battery, a wire (called wire A), then the lead of the resistor. The other lead of the resistor is connected to ground, and that wire is called wire B. The (-) terminal of the battery is also ground. Here it is, with electric fields shown:

Note, the electric field within the resistor is not shown. The red field, and the green field, simply distinguish between the field outside of the conductor (red) and the field within the conductor (axial field, green). For sake of argument, all wires and leads are 'superconductors', and the battery is ideal. The electric field of interest in this question is the green electric field.

The first assumption is that there is no voltage drop across the grounded wire from the resistor lead to the (-) terminal of the battery i.e. it is a net with 0V at all points. But this is about the physical description, so the distance and organization matters for the circuit. The second assumption is that a current is flowing through the circuit. The third assumption is that a non-zero electric field must exist in order for a current to flow in a conductor.

Now, the fourth assumption, is that the definition of electric potential is the negative of the line integral along the path from a reference (0V) to the point of interest. Take the reference as the negative side of the battery, and the point of interest is the closest point to the resistor lead on the grounded wire, wire B.

The proposed question comes from the fact that you have current, and therefore an assumed electric field, along the wire's axis, but in order for the electric potential to be 0V, you must have either a) a zero-length path, or b) a zero electric field. The third option would be an electric field that is conservative, i.e. a net electric field of zero that still changes. There seems to be, in all cases, something of a contradiction. If the entirety of the voltage gradient is dropped across the resistor, and there is no voltage dropped across the ground wire, why does current flow?

Now, some solutions to the question that I have considered are: perhaps there is a mechanical aspect, because of course charge is moving through the wire, and if current was not flowing through the ground wire into the battery, then you would start to build up charge and of course that doesn't happen. When I picture a voltage, I imagine naturally the entire circuit at once so considering only the ground wire has been quite confusing for me, as you can probably tell. What resolution is there to my problem?

Any help is greatly appreciated. Thanks in advance.
Sam

 

https://www.tu-braunschweig.de/Medien-DB/ifdn-physik/ajp000782.pdf

https://www.tu-braunschweig.de/inde...oken=90bbd237cd4ef87cf2941a99eb3c3d610de3d0d3

EDIT: update link

 

Hello all. I wasn't exactly sure of the best phrasing for title, so I hope I've attracted the right folks to answer this question. It has to do with the physical description of circuits, and the electric fields within them.

The question is: What is the electric field in an ideal wire, in a circuit, for a wire that is connected to ground?
Here's what I mean: Imagine a circuit with a battery and a resistor. The battery is 1.5V, the resistor is a carbon resistor, nothing special, with value R. We have the (+) terminal of the battery, a wire (called wire A), then the lead of the resistor. The other lead of the resistor is connected to ground, and that wire is called wire B. The (-) terminal of the battery is also ground. Here it is, with electric fields shown:
View attachment 115704
Note, the electric field within the resistor is not shown. The red field, and the green field, simply distinguish between the field outside of the conductor (red) and the field within the conductor (axial field, green). For sake of argument, all wires and leads are 'superconductors', and the battery is ideal. The electric field of interest in this question is the green electric field.

The first assumption is that there is no voltage drop across the grounded wire from the resistor lead to the (-) terminal of the battery i.e. it is a net with 0V at all points. But this is about the physical description, so the distance and organization matters for the circuit. The second assumption is that a current is flowing through the circuit. The third assumption is that a non-zero electric field must exist in order for a current to flow in a conductor.

Now, the fourth assumption, is that the definition of electric potential is the negative of the line integral along the path from a reference (0V) to the point of interest. Take the reference as the negative side of the battery, and the point of interest is the closest point to the resistor lead on the grounded wire, wire B.
View attachment 115706
The proposed question comes from the fact that you have current, and therefore an assumed electric field, along the wire's axis, but in order for the electric potential to be 0V, you must have either a) a zero-length path, or b) a zero electric field. The third option would be an electric field that is conservative, i.e. a net electric field of zero that still changes. There seems to be, in all cases, something of a contradiction. If the entirety of the voltage gradient is dropped across the resistor, and there is no voltage dropped across the ground wire, why does current flow?

Now, some solutions to the question that I have considered are: perhaps there is a mechanical aspect, because of course charge is moving through the wire, and if current was not flowing through the ground wire into the battery, then you would start to build up charge and of course that doesn't happen. When I picture a voltage, I imagine naturally the entire circuit at once so considering only the ground wire has been quite confusing for me, as you can probably tell. What resolution is there to my problem?

Any help is greatly appreciated. Thanks in advance.
Sam

Is the drawing wrong? Does the magnetic field form around the wire and expand outward?

 

There is nothing to make the "ground" wire anything special. It is at 0 V only because we chose to call the voltage at that node 0 V.

Since you have specified that the wires are ideal (and note that there is a big difference between an ideal conductor and a superconductor, so let's just stick with ideal conductors), then there is no electric field within the conductor. The can be seen from a couple of perspectives. If there were a field, then the charges within the conductor would be accelerating continuously and, since it is an ideal conductor, there would be no mechanism to hold them to a steady-state velocity. Second, if there is an axial component to the electric field within the conductor, there by definition there would be a voltage drop along the conductor, but again it is an ideal conductor and hence no voltage drop along it.

 

@hp1729 The drawing does not show the magnetic field, it shows two components of the electric field (axial and normal). @WBahn The question is not about whether or not the wire drops a voltage, which it cannot, or whether or not an electric field exists within the conductor, which it does, but rather what mechanism exists which would cause a current to flow in the wire, despite no voltage drop. As @nsaspook 's article pointed out, that mechanism is surface charges and interface charges on the conductor, which cause an electric field within the conductor. The arguments you present are helpful in pushing understanding, but it seems that charge flow in a conductor would not accelerate under constant electric field because of the path that charges take throughout the conductor. I cannot back this up mathematically at the moment, and for the case of superconductors, I am not completely sure about the conduction mechanism. The point remains that the electrons are not in free space, and that the constant force = constant acceleration argument is a likely an over-simplification.

In the article posted by @nsaspook , A semiquantitative treatment of surface charges in DC circuits, Muller explains that there are several misconceptions brought about by how physics is taught to students, such as "how can a conductor, which cannot have an electric field, cause current to flow with an electric field?" These questions are usually a result of teaching subjects in a simplified manner, which is fine, but without telling the students it's a simplified case, or a special case. That is not fine, and it causes much confusion, such as this. Another interesting paper is Understanding Electricity and Circuits: What the Text Books Don't Tell You by Ian M. Sefton of the University of Sydney. Sefton focuses more on how energy is transferred in DC circuits, after raising some objections to the conventional way electric circuits are taught.

Honestly, I can't believe how difficult it has been to understand this mechanism. I mean, I've worked with circuits for years, and thought I understood conduction pretty well. But no, I guess I really didn't!
Sam

 

@hp1729 The drawing does not show the magnetic field, it shows two components of the electric field (axial and normal). @WBahn The question is not about whether or not the wire drops a voltage, which it cannot, or whether or not an electric field exists within the conductor, which it does, but rather what mechanism exists which would cause a current to flow in the wire, despite no voltage drop. As @nsaspook 's article pointed out, that mechanism is surface charges and interface charges on the conductor, which cause an electric field within the conductor. The arguments you present are helpful in pushing understanding, but it seems that charge flow in a conductor would not accelerate under constant electric field because of the path that charges take throughout the conductor. I cannot back this up mathematically at the moment, and for the case of superconductors, I am not completely sure about the conduction mechanism. The point remains that the electrons are not in free space, and that the constant force = constant acceleration argument is a likely an over-simplification.

Look at the definition of voltage -- it is the line integral of the electric field. If an electric field exists in the conductor having a axial component, then there IS a voltage drop along the conductor. There HAS to be because that is what DEFINES a voltage drop! So if you want to insist on a perfect conductor that has no voltage drop across it, then you can't have an axial component of an electric field within that conductor. Period.

The notion that you HAVE to have an electric field to SUSTAIN a current flow is simply not true. Look at a superconductor. A superconducting magnet placed in persistence mode will have a current flowing in it for months with only a barely perceptible drop (due to parasitics) with no electric field to push the electrons along.

 

I made the same observation in my original post, that an electric field cannot exist without a potential difference. This is the reason for the majority of my confusion. But we had different conclusions - I believed that there could be a piece missing from this general model (as I'd learned it), or that the voltage drop being looked at is not the only one that exists in that local area, and that some assumption may be causing the issue.

I don't mean to bother anyone by asking these questions, it is simply that until the whole thing is explained properly, I might still have objections that, while not valid in an objective sense, will continue to pester me. I shouldn't have assumed that the addition of the surface charges will finish the model, so I'm sorry about that.

Doing further research, and realizing the question I'm really asking is: Is there an electric field in a current-carrying ideal conductor? The answer, proven as you stated, is no, there is no electric field in an ideal conductor. What confused me is the starting conditions, i.e. how the current begins flowing. But that was not relevant to the question. I understand now that no electric field is required to sustain a current in an ideal conductor, which was really the key element! Thanks WBahn.

Now I do have questions about the starting conditions i.e. "turning on" the circuit. Transient conditions tend to be weird, so I don't know if I should change threads to discuss how current starts flowing in an ideal conductor that is in a voltage-resistor circuit? In any case, how does current start flowing in an ideal conductor?
Sam

Note: One response I found was that the potential "travels" back and forth, but this is not a very clear to me.

 

Look at the definition of voltage -- it is the line integral of the electric field. If an electric field exists in the conductor having a axial component, then there IS a voltage drop along the conductor. There HAS to be because that is what DEFINES a voltage drop! So if you want to insist on a perfect conductor that has no voltage drop across it, then you can't have an axial component of an electric field within that conductor. Period.

The notion that you HAVE to have an electric field to SUSTAIN a current flow is simply not true. Look at a superconductor. A superconducting magnet placed in persistence mode will have a current flowing in it for months with only a barely perceptible drop (due to parasitics) with no electric field to push the electrons along.

What does a "ground wire" gave to do with it? Which point is "ground'? Any point can be used as a reference, can't it?
Is there any "field" that travels as your arrows suggest? Isn't it the electromagnetic field moving outward, perpendicular to the wire (or crossing the wire), that pushes the electric current.

 

Look at the definition of voltage -- it is the line integral of the electric field. If an electric field exists in the conductor having a axial component, then there IS a voltage drop along the conductor. There HAS to be because that is what DEFINES a voltage drop! So if you want to insist on a perfect conductor that has no voltage drop across it, then you can't have an axial component of an electric field within that conductor. Period.

The notion that you HAVE to have an electric field to SUSTAIN a current flow is simply not true. Look at a superconductor. A superconducting magnet placed in persistence mode will have a current flowing in it for months with only a barely perceptible drop (due to parasitics) with no electric field to push the electrons along.

What does a "ground wire" gave to do with it? Which point is "ground'? Any point can be used as a reference, can't it?
Is there any "field" that travels as your arrows suggest? Isn't it the electromagnetic field moving outward, perpendicular to the wire (or crossing the wire), that pushes the electric current.

Are you sure you quoted the right post? Did I not point out that calling a wire "ground" has no bearing since it is just an arbitrary reference. Nor did I draw any arrows.

 

Are you sure you quoted the right post? Did I not point out that calling a wire "ground" has no bearing since it is just an arbitrary reference. Nor did I draw any arrows.

Sorry, I will rephrased the comment.
Do the arrows in the picture accurately rep[resent any force?

 

Sorry, I will rephrased the comment.
Do the arrows in the picture accurately rep[resent any force?

The green arrow does not exist if the wires are ideal conductors. In an ideal conductor the charges can "glide" through the conductor without requiring an impressed electric field to keep them moving along. In a physical conductor with finite resistance this is not the case and then there IS a (generally quite small) electric field in the conductor to keep the charges moving -- along with a small voltage drop along the conductor as a result.

The red arrows also have problems because they are drawn using an implied assumption that the top wire has a net positive charge and the bottom wire has a net negative charge when both wires are essentially neutral.

 

The red arrows also have problems because they are drawn using an implied assumption that the top wire has a net positive charge and the bottom wire has a net negative charge when both wires are essentially neutral.

Is this not true for the ideal conductor case, or for non-ideal conductors as well? Such a diagram appears in two of the publications I've read (Understanding Electricity and Circuits by Sefton, and A semiquantitative treatment of charge in DC circuits by Muller, both listed above).

From Sefton



From Muller

 

Is this not true for the ideal conductor case, or for non-ideal conductors as well? Such a diagram appears in two of the publications I've read (Understanding Electricity and Circuits by Sefton, and A semiquantitative treatment of charge in DC circuits by Muller, both listed above).
View attachment 115769
From Sefton


View attachment 115770
From Muller

So now there is the Strong Force, Weak Force, Gravity, Magnetic field and Electric field???
What am I missing? That "Electric Field" is the result of the EM field? A result, not a cause? Not a force?
Is my attitude obsolete?

 

Hello Sam. I'm not real sure exactly what your asking, but I'll give it a shot.

Please ignore the static field between the conductors. Do you know what a gaussian surface is?

Using your circuit with NORMAL conductors......apply a gaussian positive surface around the circuit. Consider the negative terminal as zero positive electric field.

The largest radius of this gaussian surface will be at the positive terminal and the left hand side of the top conductor. As you come across the top conductor the radius of the surface will taper slightly to the top of the resistor. This slight taper in the top conductor surface is due to conductor resistance. The surface down across the resistor looks like a cone. The radius starts large at the top and tapers sharply at the bottom of the resistor. This taper is caused by the power supply voltage drop across the resistor. The bottom small cone at end of resistor, then tapers to zero at the negative terminal across the bottom conductor.

This surface represents how the positive electric power supply field is expended around the circuit. I hope you can picture that in your mind. There is a mirror negative field around the circuit also. Just reference the top terminal.

Don't let the static fields between the conductors fool you. The field that is doing the work, is thru the circuit.....not across it. The difference between the radii of gaussian surface is what powers circuit. The electric field goes around the circuit, not across it.

IF you replace the conductors with super conductors.......the taper will only happen at start up.......Then become a cylinder with no taper. It will still have a field. No taper will appear after start up with a super conductor.

Did I thoroughly confuse you?

And don't forget, the current itself.....generates an electric field.

 

Hello Sam. I'm not real sure exactly what your asking, but I'll give it a shot.

Please ignore the static field between the conductors. Do you know what a gaussian surface is?

Using your circuit with NORMAL conductors......apply a gaussian positive surface around the circuit. Consider the negative terminal as zero positive electric field.

The largest radius of this gaussian surface will be at the positive terminal and the left hand side of the top conductor. As you come across the top conductor the radius of the surface will taper slightly to the top of the resistor. This slight taper in the top conductor surface is due to conductor resistance. The surface down across the resistor looks like a cone. The radius starts large at the top and tapers sharply at the bottom of the resistor. This taper is caused by the power supply voltage drop across the resistor. The bottom small cone at end of resistor, then tapers to zero at the negative terminal across the bottom conductor.

This surface represents how the positive electric power supply field is expended around the circuit. I hope you can picture that in your mind. There is a mirror negative field around the circuit also. Just reference the top terminal.

Don't let the static fields between the conductors fool you. The field that is doing the work, is thru the circuit.....not across it. The difference between the radii of gaussian surface is what powers circuit. The electric field goes around the circuit, not across it.

IF you replace the conductors with super conductors.......the taper will only happen at start up.......Then become a cylinder with no taper. It will still have a field. No taper will appear after start up with a super conductor.

Did I thoroughly confuse you?

And don't forget, the current itself.....generates an electric field.

"Left" and "top" from who's view?

 

Did some one change the definitions again? Left means left and top means top.

Edit: does your screen show a mirror image?

 

Did some one change the definitions again? Left means left and top means top.

Edit: does your screen show a mirror image?

Ah, yes. I'll just turn my screen around.

 

I was referring to the circuit in post #1, gentlemen. Did I get something turned around? Sometimes I don't know where I'm pointed.