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.
Best Answer
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}\:\text{mol}^{-1}\$ (standardized in 2019.)
The number of atoms in \$1\:\text{cc}\$ of copper is then \$\frac{N_A\,\cdot\, \rho_{_\text{Cu}}}{m_{_\text{Cu}}}\approx 8.49\times 10^{22}\,\frac{\text{atoms}}{\text{cc}}\$. Assuming each atom of copper is able to donate one conduction band electron, then you'd expect the same number of conduction band electrons per cc. So we'd estimate \$8.49\times 10^{22}\,\frac{\text{electrons}}{\text{cc}}\$ in the conduction band.
There is a slight modification to this made by Fermi-Dirac statistics. Based upon the Fermi energy of copper, which is \$E_{F_{_\text{Cu}}}=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:
$$\left[\frac{8\pi\,\sqrt{2\,m_e^3}}{h^3}\right]\cdot\bigg[\frac23\sqrt{E_{F_{_\text{Cu}}}^{\:\:3}}\bigg]\approx 8.411\times 10^{22}\,\frac{\text{electrons}}{\text{cc}}$$
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.
Normally, in copper we might expect the orbital shells to fill up in the usual order of energy: 1s, then 2s, then 2p, then 3s, then 3p, then 4s, and then back to 3d. (The 4s is lower in energy than 3d.) Simplistically following that approach, we'd write: \$1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^9\$. That gets all 29 electrons placed. But it's wrong. There's a special stability that occurs when shell levels are completely filled up and in this case the 3d (which holds 5 pairs, or a total of 10 electrons) prefers to fill up completely. (It sums out to a lower average energy using 3d this way, than leaving 3d slightly shy of being full.) So the result is that copper is actually: \$1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^1\$. It is this \$4s^1\$ electron that copper donates to the conduction band.
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.)