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Note

my questions are in form of explaining. its because I've scattered information (or assumptions) and it want to link them. that's probably why you can see me explaining to much?

Question part:

lets talk about electron speed vs "charge speed / current". so electron speed through a wire is actually slow? (drift of charge carriers)

as we know, \$1A= 1 \frac {Coulomb}{second} \$ is actually like an electron hitting one by another to get one electron out of the end of the wire (does it make sense?) i mean really how the charge is transferred.

knowing that \$1Coulomb \approx 6.241*10^{18} electrons\$

i only can imagine that \$ 6.241*10^{18} electrons \$ are getting out at the end [or entering the beginning] of the material (copper for example)

so at end of wire we are extracting this quantity of electrons per second? what happen to that end? does it get positive charge?

and that is what happens if we apply high current on small wire diameter, we are removing too many electrons from the cross section area, hence, it just cant be that positively charged so it act like insulator (resistance). (or the same applied to the beginning of the wire we are giving to much electrons that cant move)

oh, and how do you describe charging plate of a capacitor. i mean you have to actually remove the electrons from the plate, to get the two plated different charge (+ / -) and apply that electric field in the dielectric. And only then we will have the electrons fly away from the opposite plate as the electric force will apply to them?

so what is the electron speed ?! is it slow is it fast ?! (yes they are slow but .. you know, this raises above questions)

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closed as too broad by Chris Stratton, Elliot Alderson, Dwayne Reid, Finbarr, laptop2d Jan 8 at 19:50

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    \$\begingroup\$ Downvotes usually happen for poor questions or questions posted in the wrong stack, so they help us focus on questions with merit, so downvotes are helpful... \$\endgroup\$ – Solar Mike Jan 6 at 8:08
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    \$\begingroup\$ downvotes without feedback are helpful (for platform mechanism), but will not give feedback to the person who ask so he "correct his attitude" (which is also related to the platform right)? \$\endgroup\$ – Hasan alattar Jan 6 at 8:29
  • \$\begingroup\$ Well, if you read this (but I hope you have already) : electronics.stackexchange.com/help/dont-ask and this: electronics.stackexchange.com/help/how-to-ask you might avoid downvotes... \$\endgroup\$ – Solar Mike Jan 6 at 8:33
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    \$\begingroup\$ You have far to many different questions here. As for the speed: each electron is slow, but the neighbor bumping neighbor wave propagates at almost the speed of light. Plausible charge differences are minuscule compared to the number of electrons present. \$\endgroup\$ – Chris Stratton Jan 6 at 8:39
  • \$\begingroup\$ hello, Chris, my questions are related to the core of what you just said. they are probably physics and in atomic level but they are all related to electricity given that my background is in electricity. \$\endgroup\$ – Hasan alattar Jan 6 at 8:42
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so at end of wire we are extracting this quantity of electrons per second? what happen to that end? does it get positive charge?

No. In general there is no net change in charge. Any charge carrier that moves out of the end of the conductor is replaced by one entering at the other end. The net charge remains at zero.

... and that is what happens if we apply high current on small wire diameter, we are removing too many electrons from the cross section area, hence, it just cant be that positively charged so it act like insulator (resistance).

No. Current causes heating in the resistance of the material. The temperature stabilises at the point where power in equals heat out.

So what is the electron speed?

See Wikipedia's Drift velocity which should help. There is a worked example for a current of 1 A in a 2 mm diameter copper wire. The result is -

Therefore, in this wire the electrons are flowing at the rate of 23 μm/s. At 60 Hz alternating current, this means that within half a cycle the electrons drift less than 0.2 μm. In other words, electrons flowing across the contact point in a switch will never actually leave the switch.


And how do you describe charging plate of a capacitor. I mean you have to actually remove the electrons from the plate, to get the two plated different charge (+ / -) and apply that electric field in the dielectric. And only then we will have the electrons fly away from the opposite plate as the electric force will apply to them?

Application of the electric field causes the charges to flow. Wikipedia's article on Capacitance may help in this case.

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  • \$\begingroup\$ so you are saying if we have 1 amp, we will have large amount of electrons entering and exiting the wire? or you charge will only propagate through wave and electron speed (drift velocity) is not related to the current? \$\endgroup\$ – Hasan alattar Jan 6 at 9:22
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    \$\begingroup\$ (1) If you have 1 A you will have 1 C entering and leaving the wire every second. (2) As shown in the Wikipedia article, the drift velocity, \$ u = \frac {I}{nAq} \$ so it is proportional to the current but inversely proportional to the cross-sectional area. (This would be the same as water current in a pipe.) (3) Note that your questions and comments lose credibility due to lack of capitalisation.and punctuation. May I encourage you to write properly in standard English. \$\endgroup\$ – Transistor Jan 6 at 9:32
  • \$\begingroup\$ The calculation (drift velocity) makes sense. thanks. \$\endgroup\$ – Hasan alattar Jan 6 at 13:05
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In many ways, electron flow is much like that of water. Voltage is pressure, current is flow rate, conductors are pipes. When you pump 1 liter of water into a (filled) pipe, 1 liter of water gets pushed out at the other end. If you seal the other end, not allowing water to exit the pipe, you cannot push more water into the pipe because water is almost incompressible, just like the electrons in a conductor. You also cannot pump water out of the pipe without allowing water to be sucked in at the other end.

And, one molecule of water pumped into the pipe will push a different molecule out at the other end. How long it takes until a specific molecule has travelled through the pipe depends on the volume (length and diameter) of the pipe and the flow rate (volume per time unit). For electrons it's exactly the same.

The reason in both cases is that there's a repelling force between the molecules (or electrons) which travels much faster (speed of sound or speed of light) than the particle itself.

Electrical resistance directly corresponds to the resistance encountered by flowing water. A pipe (conductor) of smaller diameter poses more resistance than one of bigger diameter. Hence you need more pressure (voltage) to get the same flow rate (current) through the smaller pipe (conductor).

In this water model, a capacitor's analog is a piece of pipe with a rubber membrane in the middle. You can push water into the pipe and against the membrane. The higher the pressure is, the more water can be pumped into the pipe before the membrane's tension balances out the force. As the membrane deforms from the pressure at one side, it passes that motion on to the other side pushing the molecules on that side out of the pipe at the far end. When you release the pressure built up previously, the tension in the membrane will relax, moving the water back in the opposite direction. Notice that, just as in a capacitor, the membrane prevents particles to travel from one side to the other, yet the force gets passed on as long as water flows, i.e. as long as the capacitor is still charging.

Electrically (not magnetically), even self-inductance has its analog: The volume of water in the pipe will have a certain moment of inertia while it is flowing, storing some energy. If you instantly close a valve, that inertia will cause a bigger pressure before and a negative pressure behind the valve, even some oscillation, for some time until the water comes to rest.

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  • \$\begingroup\$ So what you are saying is that for 1Ampere, there are \$6.241*10^{18}\$ electrons entering the input, and same amount is exiting at the end? \$\endgroup\$ – Hasan alattar Jan 6 at 12:27
  • \$\begingroup\$ That is correct. \$\endgroup\$ – pjc50 Jan 6 at 12:33
  • \$\begingroup\$ Yes, for 1 Ampère, 1 Coulomb of electrons enters the conductor per second, and the same number of electrons exits at the other end. This also reflect's in one of Kirchoff's law: Every single electron entering a conductor must cause exactly one electron to exit, and vice versa. \$\endgroup\$ – JimmyB Jan 6 at 12:35
  • \$\begingroup\$ after learning about mol and copper density. it makes sense after i made calculation number of atoms is huge. thanks. \$\endgroup\$ – Hasan alattar Jan 6 at 13:05
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    \$\begingroup\$ Yes, it's the huge number of electrons in a conductor ("volume of the pipe") that makes the electrons travel very slowly for even big currents. And, possibly unintuitively, we need many, many times more flowing water molecules to do something useful than electrons. That's why the water we use from the tap or in a river travels much faster than electrons in conductors of much smaller diameter. \$\endgroup\$ – JimmyB Jan 6 at 13:14

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