# How can electrons flow all the way through the circuit?

In summary: Electric potential energy is stored in the electric field.When charges are separated, the electric field decreases in intensity along the path of separation because the potential energy of the charges is now divided.
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:

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?

JC2000 and etotheipi
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?

etotheipi and vanhees71
jbriggs444 said:
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.

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.

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.

DaveE
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.

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.

hutchphd
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.

sophiecentaur
@sophiecentaur , @DaveE Can you please help me in seeing how the field will be distorted? What would be the resultant shape of the field?

@sophiecentaur , @DaveE Can you please help me in seeing how the field will be distorted? What would be the resultant shape of the field?

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.

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.

valentin bogatu said:
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.

DaveE and sophiecentaur
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.

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.

sophiecentaur said:
That is very relevant at microwave frequencies.
True but not helpful because it is clear that the OP is thinking of the DC case.

anorlunda said:
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 equations.

anorlunda
And yes I believe in mathematics and will be okay if you could explain with the help of equations.
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.

anorlunda said:
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.

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.

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.

Last edited:
anorlunda said:
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.

## 1. How do electrons flow through a circuit?

Electrons flow through a circuit due to the presence of an electric field. When a voltage source, such as a battery, is connected to a circuit, it creates an electric field that pushes the electrons from the negative terminal of the source towards the positive terminal. This flow of electrons is what we call an electric current.

## 2. What is the role of conductors in allowing electrons to flow through a circuit?

Conductors, such as copper wires, have loosely bound electrons in their outermost energy level. When a voltage source is connected to a circuit, it creates an electric field that causes these electrons to move and flow through the conductor. The free movement of electrons in conductors allows for the flow of electricity through the circuit.

## 3. How can electrons flow through a circuit without getting used up?

Electrons are not consumed or used up as they flow through a circuit. This is because they are constantly being replenished by the voltage source, which provides the energy to move the electrons through the circuit. As long as the circuit is closed and the voltage source is connected, the flow of electrons will continue.

## 4. Can electrons flow in both directions through a circuit?

Yes, electrons can flow in both directions through a circuit. In a direct current (DC) circuit, the electrons flow in one direction, from the negative terminal of the source to the positive terminal. In an alternating current (AC) circuit, the direction of electron flow alternates back and forth as the current changes direction.

## 5. What factors affect the flow of electrons through a circuit?

The flow of electrons through a circuit can be affected by several factors, including the resistance of the circuit components, the voltage of the source, and the type of material the circuit is made of. The resistance of a material determines how easily electrons can flow through it, while the voltage of the source determines the strength of the electric field pushing the electrons. Additionally, the type of material can affect the speed at which electrons can move through the circuit.

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