# What is happening in those wires?

#### AnalogKid

Joined Aug 1, 2013
8,376
A voltmeter connected to two different points on wire A will read very close to zero potential difference the whole time; but the same meter connected from wire A to wire B will reveal a significant potential difference at the start of the discharge period. What is the physical cause of this potential difference?
The physical cause is the resistor and capacitor between the two wires. Two adjacent points on either wire A or B are separated by a very small resistance. But the two wires are separated by whatever the value of R is. Via Ohm's Law, the voltage drop across R can be calculated.
There must be something different about wires A and B during the period in question. What is it?
No, there mustn't. You are locked into an idea that is incorrect. If you can't get passed that, nothing here will help. For a simple DC circuit, most of the "water analogy" applies. A wire is a garden hose ***already full of water*** (electrons). When you open the faucet, more water (electrons) go into one end of the hose, and almost immediately water/electrons come out the other end of the hose. OK, a major flaw in the water analogy is that a complete circuit is not needed for flow, but I hope you see the concept here. So electrons are physically moving through the hose. There are two consequences, already pointed out: heating caused by resistance, and deflection caused by the electric and magnetic fields.

NOTE (a big note): the electrons in the wires in a DC circuit do not move at the speed of light, or anything near it. In the hose analogy, note that water starts coming out the far end of the hose almost immediately, even though the water entering the hose will take many seconds to reach the end. This is the difference between the way electromagnetic energy moves down a wire, and the way electrons move through it. Think of a ceiling light in a room, controlled by a wall switch, and powered by DC instead of AC. Flip the switch, light comes on too fast to perceive any delay. But electrons at the switch contact could take 30 minutes (not a typo) to reach the light bulb.

ak

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#### AnalogKid

Joined Aug 1, 2013
8,376
how do we account for the measurable potential differences in the wires throughout the hour in question?
The potential difference you are measuring is not in the wires. By your own statement, when both meter probes are on the same wire, the potential difference is too small to measure. When you move one of the probes to a different wire, *you moved one of the probes.* ***You are not measuring the same thing.*** It's not apples and oranges, it's apples and fish.

ak

#### AnalogKid

Joined Aug 1, 2013
8,376
And for the change of the water under pressure, I had always been taught the water is not compressible so I would think it does not change,.
Like all fluids, water is compressible. But not very much. And for the purposes of the water analogy, it is incompressible. Also, hoses do not expand under pressure, there is no hydrostatic friction, etc.

ak

#### ebeowulf17

Joined Aug 12, 2014
3,274

#### Gerry Rzeppa

Joined Jun 17, 2015
170
The potential difference you are measuring is not in the wires.
I understand why you might say that. I realize that the potential difference between the two wires is ultimately the result of there being more electrons on one of the capacitor's plates than the other. So it seems reasonable to imagine, when the circuit in question is first completed, that some of the excess electrons on the capacitor's bottom plate will move quickly to equalize the electron concentration in the entire lower leg of the circuit, and that a flow of electrons will after that proceed through the resistor, at a decreasing rate, to the electron-poor top leg of the circuit (and ultimately to the capacitor's top plate).

If this is what actually happens, the measured potential difference between the wires is due to the different concentrations of electrons in the wires during the discharge period.

If this is not what happens, please describe what does.

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#### AnalogKid

Joined Aug 1, 2013
8,376
If this is what actually happens, the measured potential difference between the wires is due to the different concentrations of electrons in the wires during the discharge period.
No, it is not.

The difference is due to the greater-than-zero ohm resistance between them.

Both wires have the same number of electrons (per copper atom) at all times. You keep bringing up a potential difference "in" a wire, and for all practical purposes (such as describing electricity to children) there is n.o.n.e. Again, you are not measuring a potential difference in a wire, you are measuring the difference between wires. "in" vs. "not in". Biiiiig difference.

ak

#### AnalogKid

Joined Aug 1, 2013
8,376
So it seems reasonable to imagine, when the circuit in question is first completed, that some of the excess electrons on the capacitor's bottom plate will move quickly to equalize the electron concentration in the entire lower leg of the circuit,
Not to me it doesn't. There is no equalizing of the lower leg. This is a failure in the water analogy, because everyone knows that when you turn on a spigot when the nozzle at the far end of the hose is closed, water rushes into the hose until the pressure in the hose equals the pressure in the house. Wires do not do that. The electron concentration in the wire does not change. Counter-intuitive? Yes. True? Yes.

First, let's set some ground rules so this doesn't descent into pedantic borscht.

All perfect components: wires have zero resistance; capacitors have zero leakage current, ESR, and ESL; resistors have zero ESL, switches have zero resistance and zero bounce, and everything is rated to handle the various currents without damage.

Connect battery to capacitor through switch. Cap charges up to battery voltage.

Open switch. Capacitor will sit there fully charged forever.

Connect resistor to capacitor through switch. Voltage across the resistor changes instantly and electrons start moving.

Energy in capacitor is dissipated in resistor as heat. After five time constants, voltage across capacitor has decreased 99%.

That's it. For every electron that goes in one end of either wire or the resistor, another one comes out the other end.

Note that the above description does not go into electromagnetic wave propagation theory.

ak

#### SamR

Joined Mar 19, 2019
2,008
The analogy I was given many years ago was a tube filled with marbles. At the SAME time you push a marble into the tube one MUST come out of the other end no matter how long the tube is.

#### shortbus

Joined Sep 30, 2009
7,597
We need a sign here, one that says, "DON'T FEED THE TROLLS"

#### nsaspook

Joined Aug 27, 2009
7,179
We need a sign here, one that says, "DON'T FEED THE TROLLS"
This is a continuation of a many years old trolling conversation with @Gerry Rzeppa over many forums. IMO he has no intention of listening to anyone.

#### BobTPH

Joined Jun 5, 2013
2,345
Let me see if I understand what you're saying. Let's say I have two similar books, A and B, sitting on a table. When I lift book A above the table it is now at a different gravitational potential than book B, but it remains physically unchanged. Likewise, when I connect wires A and B to opposite ends of the charged capacitor, wire A is now at a different electrical potential than wire B, but it remains physically unchanged. The field of which you speak is therefore, at least in some respects, the electrical analog of gravity. Yes?
Exactly! Electric potential and gravitational are analogous.

Make that book a pan of water. Connect a U-shaped set of pipes going across, down, and across again at the height of the table. There is no difference between the two cross pipes or the water in them, but the water keeps flowing until the pan is empty.

Bob

#### Gerry Rzeppa

Joined Jun 17, 2015
170
...Connect resistor to capacitor through switch. Voltage across the resistor changes instantly and electrons start moving....
You say, "Voltage across the resistor changes instantly," which suggests that the electrical potential has instantly become different at each end of the resistor. Yet you insist that there is nothing different at the atomic level at those points. So from whence this potential difference?

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#### Gerry Rzeppa

Joined Jun 17, 2015
170
Exactly! Electric potential and gravitational are analogous.
Your answers seem quite different from the others. Unfortunately, I don't yet fully understand them.

Make that book a pan of water. Connect a U-shaped set of pipes going across, down, and across again at the height of the table. There is no difference between the two cross pipes or the water in them, but the water keeps flowing until the pan is empty.
As I understand it (correct me if I'm wrong), the water flows because the gravitational field is "everywhere" in and around the system in question -- which is why objects in different locations can have different potentials with no internal change in the objects themselves. Are you saying that the electrical field (generated by the difference in electron concentration on the capacitor's two plates) likewise surrounds the entire circuit in my original post? If so, will the potential across the resistor change if I "fold" the circuit to move the resistor closer to the capacitor (as the water will flow faster/slower if I move the pans to different relative positions within the gravitational field)? If I replace the resistor with an incandescent bulb, will the brightness of the lamp vary as I move the bulb around in this electrical field?

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#### crutschow

Joined Mar 14, 2008
24,709
I think the marble analogy of SamR is appropriate for a simplistic view of what is happening.
The marbles in the tube are basically incompressible and so are the electrons in the wire.
The wire with the greater negative voltage will have the electrons pushing against each other with a greater "pressure" carried by the electric field, which is transferred from one end of the wire to the other, but the number of electrons in the wire changes very little.

So physically there is no significant change in a wire carrying a potential as compared to one at zero potential.

#### ebeowulf17

Joined Aug 12, 2014
3,274
Not to me it doesn't. There is no equalizing of the lower leg. This is a failure in the water analogy, because everyone knows that when you turn on a spigot when the nozzle at the far end of the hose is closed, water rushes into the hose until the pressure in the hose equals the pressure in the house. Wires do not do that. The electron concentration in the wire does not change. Counter-intuitive? Yes. True? Yes.

First, let's set some ground rules so this doesn't descent into pedantic borscht.

All perfect components: wires have zero resistance; capacitors have zero leakage current, ESR, and ESL; resistors have zero ESL, switches have zero resistance and zero bounce, and everything is rated to handle the various currents without damage.

Connect battery to capacitor through switch. Cap charges up to battery voltage.

Open switch. Capacitor will sit there fully charged forever.

Connect resistor to capacitor through switch. Voltage across the resistor changes instantly and electrons start moving.

Energy in capacitor is dissipated in resistor as heat. After five time constants, voltage across capacitor has decreased 99%.

That's it. For every electron that goes in one end of either wire or the resistor, another one comes out the other end.

Note that the above description does not go into electromagnetic wave propagation theory.

ak
The analogy I was given many years ago was a tube filled with marbles. At the SAME time you push a marble into the tube one MUST come out of the other end no matter how long the tube is.
Both of these posts were very helpful from my point of view. Thanks guys!

That said, it raises an interesting question for me: A capacitor holds a charge by having plates charged with different electron densities, right? So, what is it about a capacitor that allows it to exist with different electron densities, while you say a theoretically perfect wire will not? Is it anything to do with materials, etc? Or is it simply the fact that it's a pair of surfaces at a set distance from one another?

The same theoretically perfect wire that enforces one electron in -> one electron out behavior in isolation would exhibit parasitic capacitance if it were in close enough proximity to another similar wire right? So does that mean that the capacitance of those wires has to do with their respective charges balancing each other?

#### Danko

Joined Nov 22, 2017
916
Charge is contained not in wire, but in medium around wire surface.
Therefore capacitance depends on surface area and dielectric constant of medium.
Physical properties of wire are not affected by surrounding charge.

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#### Gerry Rzeppa

Joined Jun 17, 2015
170
The analogy I was given many years ago was a tube filled with marbles. At the SAME time you push a marble into the tube one MUST come out of the other end no matter how long the tube is.
I think the marble analogy of SamR is appropriate for a simplistic view of what is happening. The marbles in the tube are basically incompressible and so are the electrons in the wire. The wire with the greater negative voltage will have the electrons pushing against each other with a greater "pressure" carried by the electric field, which is transferred from one end of the wire to the other, but the number of electrons in the wire changes very little.
So, what is it about a capacitor that allows it to exist with different electron densities, while you say a theoretically perfect wire will not?
We are referring to the movement of electrons in metallic conductors. Electric current in a conductor is the effect of free electrons which you can imagine as a cloud.
Thanks, folks, for taking my question seriously. But the "marble" and "cloud" analogies bring different pictures to my mind. The reference to a cloud of free electrons, in fact, reminded me of many books I've read about vacuum-tube amplifiers that invariably describe a cloud of electrons traveling from cathode to plate inside a triode (controlled by the grid). One of these books goes on to say:

So the "cloud" analogy is sanctioned in the literature and makes a lot of intuitive sense to me -- any "electron rich" area (in a capacitor, a vacuum tube, or a wire) will exhibit a negative electrical potential relative to any "electron poor" area in that circuit.

The "marble" analogy, on the other hand, which is also common in the literature, does a better job of explaining how an electrical potential applied to a circuit at one point is instantly transferred to other points in that circuit. But it does not as readily allow for electron-rich and electron-poor areas in a circuit, and does not provide as good a picture of "surge currents" that occur when a circuit is first activated (and which can be so damaging to various components on the business end of an electron "pile up").

Just thinking out loud, now: Is the reality a combination of the two? Clouds on the plates (in the tubes and in the capacitors), but marbles in the wires?

#### CharlesWMcDonald

Joined May 16, 2019
233
Exactly! Electric potential and gravitational are analogous.
No, they are not. You forget the proton. Electrons and protons are polar, they have a negative charge on the electron and positive on the proton. Gravitational forces exists between anything with mass and attract each other, there is no opposite repulsive force. You may attempt to use gravitational force on a water column to explain basic electronics but the analogy is flawed. It can't explain everything about potential and current flow.

#### CharlesWMcDonald

Joined May 16, 2019
233
Charge is contained not in wire, but in medium around wire surface.
Therefore capacitance depends on surface area and dielectric constant of medium.
Physical properties of wire are not affected by surrounding charge.