How can electrons flow all the way through the circuit?

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Electric currents and the things within are generally explained through the help of intuitive water current examples, where potential difference is explained through the pressure difference and electric current is explained as the flow of water. But I like to think in terms of some driving force instead of pressure differences, beacuse it seems more intuitive to me that force causes the motion. So, I’m water current example, I won’t think in terms of “uphill” and “downhill” but in terms of “force acting per unit of fluid” and hence fluid will move in the direction of net force.

So, my question is how electric field is caused and distributed inside the wire such that electrons flow all the way from negative terminal to the positive terminal. See this diagram:

79B4F84F-D33A-4E5C-AD12-F76622D6A98C.jpeg


Blue plate is positively charged and yellow plate is negatively charged, therefore we have straight field lines (ignoring the boundary lines) going from positive plate to the negative plate. All the red segments are conducting wires.

I agree that field lines will push the electrons from C to A and similarly from B to F, but how free electrons of our conducting wire will move from F to E to D and to C. Because for a part of the segment CD and FE the electric field is perpendicular to the movable path of electrons, and for the other part there is no field. Segment ED lies completely out of the field. So, electrons will move all the way through the circuit? Who will push them?

I hope I made myself clear. Please do give your valuable explanation and knowledge.
 
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Answers and Replies

  • #2
jbriggs444
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I agree that field lines will push the electrons from C to A and similarly from B to F, but how free electrons of our conducting wire will move from F to E to D and to C. Because for a part of the segment CD and FE the electric field is perpendicular to the movable path of electrons, and for the other part there is no field. Segment ED lies completely out of the field. So, electrons will move all the way through the circuit? Who will push them?
If the electrons were pushed from B to F but not from F to E then electrons would pile up at F.

What happens due to electrostatic repulsion when electrons pile up at F?
 
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  • #3
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What happens due to electrostatic repulsion when electrons pile up at F?
Mutual repulsion. But they cannot go back to B, so only way left for them is to go up to E.
 
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jbriggs444
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Mutual repulsion. But they cannot go back to B, so only way left for them is to go up to E.
That mutual repulsion amounts to an electrical field. The equilibrium condition will be with a non-zero field everywhere that is just right to sustain the same current flow through every section of the wire.
 
  • #5
Meir Achuz
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When a wire is put in place, the E field within the wire follows the wire, whatever E was before the wire was put in place. E=\rho J. E causes j, but j can only flow along the wire, so E must also be along the wire.
 
  • #6
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j can only flow along the wire if the wire goes anywhere.
Suppose you put the wire EF in place but not DE.
The potential at B and F is initially more negative than the potential in free space such as E.
The presence of conducting wire EF, in absence of current and ohmic drop, would cause potentials to equalize between E and F (which was equal to B) bringing electric field to free space. The electric field along the wire is zero, and actual charge density is concentrated towards the E end. Indeed, the charge density is also not constant across the thickness of the wire, because the repulsion from the plate will cause electrons to concentrate on the side of the wire away from the plate.
That unless the electric field at the tip of the wire at E is strong enough to cause corona discharge in air or field emission from the wire itself.
 
  • #7
sophiecentaur
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See this diagram:

View attachment 263887

Blue plate is positively charged and yellow plate is negatively charged, therefore we have straight field lines (ignoring the boundary lines) going from positive plate to the negative plate. All the red segments are conducting wires.
This is the problem when you try to discuss Electric circuits in terms of Fields. If ACDEFB is a wire of uniform resistance then why would those field lines be "straight"? The shape of the field will not be straightforward. For uniform resistance wire, there will be field lines following the path of the wire but it will be different elsewhere.
That's the good thing about using Potential Energy unless you really really need to discuss fields.
 
  • #8
DaveE
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Another way of saying what everyone else said:
Your assumption that the e-field is uniform and directed from one plate to the other is incorrect once the conductor is introduced. The presence of the wire will cause the field to be distorted, dramatically so in or near the wire.
 
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@sophiecentaur , @DaveE Can you please help me in seeing how the field will be distorted? What would be the resultant shape of the field?
 
  • #10
sophiecentaur
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@sophiecentaur , @DaveE Can you please help me in seeing how the field will be distorted? What would be the resultant shape of the field?
If you try to search for easy to read information about this you may find it's not as readily available as lots of other Electricity topics. People usually get away fine without worrying about it! I found this link that may help you. If you download the article (button near the bottom) you can see a treatment of E fields in the very simplest setup. Note that the model used is for perfectly conducting wires and a bulb filament is the only resistive element. The fields inside the filament will be tight and parallel to the resistance wire. Your idea is two steps along the road.

Your earlier diagram is less representative of real circuits as it has a long resistive wire and no explicit connecting wires. This, I think, is because you are trying to pursue the water-in-pipes analogy. I do not think much of the water analogy because it leads you into problems such as - where the source and load are not clearly identified in the context of electricity. Are the pipes the resistors too? I think one should leave the water model as soon as you possibly can because the analogy is too attractive and too limited.
 
  • #11
Imagine cars waiting at the red light. They make the electric current. The green light is the electric field that propagates through the row of the waiting cars. When the drivers react instantaneously all the cars in the row start immediately.
 
  • #12
anorlunda
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Imagine cars waiting at the red light. They make the electric current. The green light is the electric field that propagates through the row of the waiting cars. When all the drivers react instantaneously all the cars in the row start immediately.
That kind of visualization won't help you. The best way to understand fields is by mathematics. Are you able to study the math of fields and circuits?

In circuits, the objects that move each have their own field and the sum of all those fields affect the total field. It makes it too complicated to visualize in a simple picture for anything more than the simplest cases. Your circuit with bends is far from the simplest.
 
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  • #13
etotheipi
I'm a bit late to the party since @jbriggs444 has given the optimal answer, however I thought I might add this quite nice explanation from Griffiths concerning this exact question:
If the current were not the same all the way around (for instance, during the first split second after the switch is closed), then charge would be piling up somewhere, and—here’s the crucial point—the electric field of this accumulating charge is in such a direction as to even out the flow. Suppose, for instance, that the current into the bend is greater than the current out. Then charge piles up at the “knee,” and this produces a field aiming away from the kink. This field opposes the current flowing in (slowing it down) and promotes the current flowing out (speeding it up) until these currents are equal, at which point there is no further accumulation of charge, and equilibrium is established. It’s a beautiful system, automatically self-correcting to keep the current uniform, and it does it all so quickly that, in practice, you can safely assume the current is the same all around the circuit, even in systems that oscillate at radio frequencies.
 
  • #14
sophiecentaur
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Current and Volts will vary around any circuit of significant length. Although the effect is more striking for circuits involving transmission lines, the EM wave (for any change in input Volts) propagates at 2/3c or so which is about 200cm per ns. That is very relevant at microwave frequencies.
 
  • #16
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True but not helpful because it is clear that the OP is thinking of the DC case.
Yes, I’m thinking of DC case first. Actually, I’m not responding to replies because I’m reading the paper suggested by @sophiecentaur . And yes I believe in mathematics and will be okay if you could explain with the help of eqautions.
 
  • #17
anorlunda
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And yes I believe in mathematics and will be okay if you could explain with the help of eqautions.
Rather than starting with circuits, I think you should study electrostatic fields first, and what the field represents. Here's a very clear explanation.


If you are not familiar with the Coulomb Force, then start with lecture 1.
 
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  • #18
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Rather than starting with circuits, I think you should study electrostatic fields first, and what the field represents. Here's a very clear explanation.


If you are not familiar with the Coulomb Force, then start with lecture 1.
Thanks for suggestions, I know electrostatic fields and Coulombic force. My level is of Statics covered in Griffiths. I know and understand Faraday’s Law and I know some basic to intermediate level of Multivariable Calculus.
 
  • #19
etotheipi
This article by Bruce Sherwood and Ruth Chabay is quite nice, although still a little light on the equations. But it does deal with surface charges.
 
  • #20
anorlunda
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Can you please help me in seeing how the field will be distorted? What would be the resultant shape of the field?

From Wikipedia. It shows sample field distortions caused by conductors. It depends on the geometry of the field and the geometry of the conductors. The examples all show closed loops, but even open loop conductors distort the fields.


1591115445627.png
 
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  • #21
sophiecentaur
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True but not helpful because it is clear that the OP is thinking An of the DC case.
But the caveat was necessary imo. People can rush off with all sorts of ideas, left to themselves.
 

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