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I have something of a theoretical question. In any given electrical/electronic circuit there seems to be an electric field developed inside (and outside the conductors?) to push the electrons in the circuit.

My question is, how does this electric field has so many different values in different components, I.e different potential values across different components!

How does this electric field propagate and decide which values to take, therefore develop the potential difference?

If there is no potential difference across some elements, perfect conductor, then there's no electric field so what really pushes the electrons? How is process energyless? I know that J = sigma*E, but my question is how this E gets to be decided?

Is current built up instantaneously in the circuit or it needs time? If it needs time, as I suppose it does, thus means that electrons which are moving will collide with the resting electrons right?

I am an electronic and computer engineering major so these topics are hardly discussed.

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  • \$\begingroup\$ Part of your question is answered here: Is there no electric field inside a conductor? . It would probably make it more likely to get good answers if you edit out that part of the question and focus on the parts that aren't answered elsewhere. \$\endgroup\$ – The Photon Dec 11 '18 at 6:23
  • \$\begingroup\$ Also related: Why are wires in simple circuits approximated as equipotentials?. \$\endgroup\$ – The Photon Dec 11 '18 at 6:26
  • \$\begingroup\$ "Matter & Interactions," 3rd edition or later, Chabay and Sherwood (chapter 19, or so.) \$\endgroup\$ – jonk Dec 11 '18 at 7:09
  • \$\begingroup\$ Neither of the links answers my question. The first link doesn't discuss with the details I need and the other answer is utterly irrelevant. The second link talks about approximation. I understand the approximation, and I'm not talking about it. \$\endgroup\$ – Raafat Abualazm Dec 11 '18 at 9:39
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    \$\begingroup\$ For a non-ideal conductor, Ohm's law holds. For an ideal conductor Ohm's law also holds, but you have the indeterminate form: \$ I =\frac{0}{0} \$, for \$ I=\frac{V}{R} \$, \$\endgroup\$ – Chu Dec 11 '18 at 14:11
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Lets assume you have a voltage source. Let it be for ex. 10VDC and an ideal one. It has no other circuit connected than its output wires. The voltage source keeps a constant voltage between its output wires. The electric field lives wildly when you move the wires. If the wires are say 15cm apart each other, the average E-field strength between the wires is 10V/15cm ie. about 0,67V/m. That's because voltage sources keep the voltage, the field adapts to the geometry of affected materials.

Idealizing a source of electricity by thinking it as voltage source is often a good approximation because we get electricity from processes which give to free electrons certain energy. If the process happens to arm free electrons with energy = 10 eV it shows out as voltage = 10V. In batteries that energy is released by chemical processes where the electron structure of the materials change or as we say "a chemical reaction occurs"

If you connect your voltage source to a circuit the field pulls free electrons until a new balance is found and the electric field again gets its final form. Hopefully you know "the influence" ie. how electrons move in a conductor until the generated unbalance of the charge exactly compensates the external field. That doesn't happen immediately because changes in electric field propagate in the space only about 30 centimeters/nanosecond. That's the speed of light. During the transient an electromagnetic wave propagates through the circuit. It's in the space between wires, very little of it is in the metal.

There can be also slower settling processes than the electromagnetic wave which reflects forth and back and attenuates gradually due losses in resistances and radiation out of the circuit. Capacitors for ex. which get discharged through big resistors can have very slowly increasing voltages.

The speed of the electrons in the metal is very slow. There's no need to keep high speed because there are so much free electrons in metals that all reasonable currents are possible with less than 1mm/s drifting. Colliding electrons do not cause the current to come out from a wire, it's the electric field. Generally electrons do not collide, they are so small and they push each other away if someone comes too close. You should know that materials are mainly empty space between atomic particles.

All phenomenas in circuits are finally caused by fields and their interaction with materials. Voltage and current are only handy ways to wipe the fields out of our minds and keep the false illusion of simplicity (=it's all in wires) alive. It works to some degree, as we know, but for ex. microwave circuits cannot be understood with Ohms and Kirchoffs laws.

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  • \$\begingroup\$ Can you clarify what you said? I have a hard time wrapping my head around what you said \$\endgroup\$ – Raafat Abualazm Dec 11 '18 at 14:51
  • \$\begingroup\$ @ Raafat You have entered a learning-experience. Go explore each of the words, each of the phrases, you find confusing. Flip thru some more books. Talk with people. You are learning, and that takes time. Let your brain have time to explore. Be patient. \$\endgroup\$ – analogsystemsrf Dec 11 '18 at 15:38
  • \$\begingroup\$ Yes, I have entered a learning experience, but all I am saying that some things said up there isn't clear to me. My apologies! \$\endgroup\$ – Raafat Abualazm Dec 12 '18 at 6:10
  • \$\begingroup\$ I understand somewhat you say now. I guess. You gave me a wonderful insight. Much thanks. \$\endgroup\$ – Raafat Abualazm Dec 12 '18 at 6:20
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Copper

Copper's density at \$20\,^\circ\$C temperature is \$\rho_\text{Cu}\approx 8.96\,\frac{\text{g}}{\text{cc}}\$. It's atomic mass is \$m_\text{Cu}\approx 63.546\,\text{u}\$. Avogadro's constant is \$N_A=6.02214076\times 10^{23}\:\frac{1 }{mol}\$ (due to become the standard value in 2019.) You can compute from this the number of atoms in \$1\:\text{cc}\$ of copper as \$\frac{N_A\,\cdot\, \rho_\text{Cu}}{m_\text{Cu}}\approx 8.49\times 10^{22}\,\frac{\text{atoms}}{\text{cc}}\$. If each atom were able to donate one conduction band electron, then you'd expect that many conduction band electrons per cc, as well. So we'd estimate \$8.49\times 10^{22}\,\frac{\text{electrons}}{\text{cc}}\$.

There is a slight modification to this made by Fermi-Dirac statistics. Based upon the Fermi energy of copper, which is \$7.00\:\text{eV}\$, an integration is performed over the product of the electron state density term and the Fermi-Dirac distribution term. The result is \$\left[\frac{\pi\,8\sqrt{2\,m_e^3}}{h^3}\right]\cdot\bigg[\frac23\sqrt{E_{F_\text{Cu}}^3}\bigg]\approx 8.411\times 10^{22}\,\frac{\text{electrons}}{\text{cc}}\$.

Note that this is slightly lower than the more simplistic assumption. But not so far apart that the simplistic assumption isn't a practical one. (Don't forget the above is for pure copper, which isn't actually used in most copper wire you will find. But at least there is some consistency and we can say that 99.1% of the copper atoms donate a conduction band electron.)

Multiplying the electron charges by the charge of an electron shows that there is about \$1.3476\times 10^{4}\:\frac{\text{Coulomb}}{\text{cc}}\$ in pure copper.

Drift velocity and mobility

From the above, it's not too difficult to work out about how fast these electrons are moving when a current flows. Suppose you have a copper wire of 20-gauge; the diameter is 32 mils. This works out to a cross-section of \$\approx .51887\: \text{mm}^2\$.

Now you can compute the drift velocity for \$1\:\text{A}\$ easily: \$\frac{1\:\text{A}}{1.3476\times 10^{4}\:\frac{\text{Coulomb}}{\text{cc}}\:\cdot\: 0.51887\: \text{mm}^2}\approx 143 \:\frac{\mu m}{s}\$. That's not a large velocity.

With a \$10\:\text{V}\$ supply and a requirement of \$10\:\Omega\$ (a 20-gauge wire that is \$300.3\:\text{m}\$ long) to get \$1\:\text{A}\$, the electric field intensity must be only about \$33.3\:\frac{\text{mV}}{\text{m}}\$. Since conduction band electron mobility is the drift velocity divided by the electric field intensity, you can easily work out that the copper mobility figure is \$\mu_\text{Cu}\approx 4.3\times 10^{-3}\:\frac{\text{m}^2}{\text{V}\cdot\text{s}}\$.

(As an aside, when the drift velocity reaches approximately the speed of sound in copper, about \$18.3\:\frac{\text{m}}{\text{s}}\$, the electrons generate significant phonons interacting with the solid state atomic lattice and providing yet another source of energy loss into heat beyond that due to scattering effects in the Drude model.)

Feedback and Current

The sea of mobile electrons in copper don't interact with each other. Their mutual repulsion ensure that they stay as far from each other as is possible within the conductor. Their mutual repulsion is on average canceled out by the attraction of the positive atomic cores, so by themselves they can't push each other through or off the copper wire. They just sit there as far apart as possible. (The Drude model treats them like a "gas cloud.")

Any excess charges will appear on the surface of the copper wire because the situation at the surface of the wire isn't exactly the same as it is in the center of the wire and excess charges at the center would repel nearby electrons towards the surface.

If you apply a battery across the copper wire, there will be an excess of electrons injected at the negative end and removed at the positive end. There will be surface charge build-ups at each end (more negative at one end; more positive at the other end) which will act to accelerate the interior electrons. The effect of switching on the battery happens at about the speed of light, rapidly distributing a charge gradient along the wire surface.

Feedback in the process rapidly equalizes the situation. A simple example might help in understanding the feedback process. Suppose you bend a wire while a circuit is operating. The electrons in motion don't know (at first) to "take the bend." Instead, they collide and start to pile up at the bend, which now starts to act upon oncoming electrons causing them to start taking the turn better. The electrons will continue to pile up until there is just enough of them to cause the existing current to "take the bend" just right.

If you want to read more about this process, see "W. G. V. Rosser's "Magnitudes of surface charge distributions associated with electric current flow," American Journal of Physics, 38 (1970), pp 265-266. If you want to watch a video where they demonstate this using pith balls and a high voltage source, you can watch this video.

The number of excess electrons involved on the surface is quite small. The calculations are a little more involved than I want to belabor here, but for the figure of \$33.3\:\frac{\text{mV}}{\text{m}}\$ indicated above in the 20-gauge wire example the number of electrons in the first centimeter's wire surface might be on the order of perhaps 1000 electrons or so. Completely negligible compared to the number of electrons in the copper found in that first centimeter (more than \$4\times 10^{20}\$.)

Summary

Like most things, there are models and then there are more models. (There are models "all the way down" until you reach the Schroedinger wave equation.) But hopefully, this provides a little bit of insight. (If not, my apologies.)

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  • \$\begingroup\$ Umm, so the electric that accelerates is built up by the excess electrons not the battery directly? I mean it is not internal to the battery itself? \$\endgroup\$ – Raafat Abualazm Dec 12 '18 at 6:13
  • \$\begingroup\$ Another thing, how materials decide which values of Electric field to be present in them? Why Electric field in 10K resistor is more than 10 Ohm resistor? I don't get this point. \$\endgroup\$ – Raafat Abualazm Dec 12 '18 at 6:16
  • \$\begingroup\$ But the explanation is great, I understood something now. \$\endgroup\$ – Raafat Abualazm Dec 12 '18 at 6:16
  • \$\begingroup\$ @RaafatAbualazm The internal chemistry in a battery is rather complex and, in some cases, even includes "tunneling effects." But a simple view of a battery is like a capacitor, except that the charge on the capacitor is being continually refreshed (replenished) by a conveyor belt of sorts transporting electrons from the (+) to the (-) terminal. These electrons are acted on both by the Coulomb forces of the charges at the "plates" and by the non-Coulomb chemical processes doing the transport. The first opposes the second, and the second must exceed the first, for the battery to work. \$\endgroup\$ – jonk Dec 12 '18 at 6:53
  • \$\begingroup\$ @RaafatAbualazm A resistor has far, far fewer conduction band electrons per cc. Look back at the equation and you can see that this is a divisor for the computation of drift velocity. So the drift velocity must be much, much higher. To achieve a higher drift velocity there needs to be a more intense electric field. If you think closely about 1000 identical \$10\:\text{k}\Omega\$ resistors in parallel with each other to make \$10\:\Omega\$, I think you will have an answer to your question on resistors. \$\endgroup\$ – jonk Dec 12 '18 at 7:06

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