-1
$\begingroup$

I was reading Chapter 29 on Halliday-Resnick-Krane regarding this but couldn't understand these things.

  1. Suppose you have a conductor in an electric field. In normal conditions the field inside the conductor is zero, and charges deposit on the surface of the conductor. But then why does the electrons move in a wire (say in my house's electrical wiring) ? It's written "suppose there were a mechanism to remove electrons from the top of the slab, carry them around an external path, and re-inject them at bottom of the slab" - yes, it's clear if you do that there would be electron flow, but how does the mechanism to remove electrons and inject them back work ?

  2. Why does electrical effects seems to occur immediately while the drift speed is very slow ? (There's the garden hose analogy, but I don't understand it clearly)

  3. Although current flow is not dependent on the surface through which you're measuring it (i.e it's same irrespective of you take the surface to be slanted w.r.t to the direction the electrons are flowing or directly perpendicular to the direction the electrons are flowing), does't current density changes on which surface you're measuring it ? (i.e high when measured w.r.t a surface perpendicular to the flow of electrons but low w.r.t a surface slanted to the flow of electrons ?) Then how is the current density in a hollow sphere or conical frustum or objects like that defined ?

  4. (Related with the above question) How do you define resistance for a object which is not like a rod (say antipodal points in a hollow sphere or two ends of a conical frustum) ? The formula $R = \frac{V}{I}$ (or $R = \frac{\rho L}{A}$) don't work for the things I mentioned, because although $V$ is constant, $I$ is not constant through various slices of the objects right ?

  5. When electrons flow through a conductor (like an electric wire), what's the electric field inside the wire ?

$\endgroup$

closed as too broad by AccidentalFourierTransform, Jon Custer, Kyle Kanos, Sebastian Riese, sammy gerbil Jul 1 '18 at 9:45

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.

  • $\begingroup$ It is preferred if you can post separate questions instead of combining your questions into one. That way, it helps the people answering your question and also others hunting for at least one of your questions. Thanks! $\endgroup$ – Kyle Kanos Jun 29 '18 at 10:13
  • $\begingroup$ Many of your questions have already been asked and answered on this site. Please make a search before posting questions. $\endgroup$ – sammy gerbil Jul 1 '18 at 9:47
1
$\begingroup$

Your questions are entirely in the classical domain, there can be no confusion about this.

1) This is essentially what a car battery does. In it a chemical potential separates charges. A chemical reaction occurs if the terminals are connected to to a resistor.

2) The speed is determined by the speed of electromagnetic excitations inside the material. In a metal the plasma oscillation transfers electromagnetic energy. In a dielectric optical phonons are involved.

3) Indeed the current density per surface unit depends on the orientation of the surface , in a trivial manner.

4) You can always define the resistance between two points contacting a conductor of any shape as R=V/I.

5) The electric field is $\vec E = \vec J / \sigma$ where $J$ is the current density and $\sigma$ is the material conductivity.

$\endgroup$
  • $\begingroup$ For 2) can you please explain a bit (I don't understand "plasma osciallation" or "optical phonon" or such big words) ? Also for 4), how do you measure the $I$ ? Through which surface ? $\endgroup$ – cdt Jun 28 '18 at 19:21
  • $\begingroup$ You are asking quite a few questions so I have to be concise. The plasma oscillation of a metal has all electrons moving at the same time in the same direction, see en.wikipedia.org/wiki/Plasma_oscillation. An optical phonon is a lattice wave in which the ions and the electrons move in opposite directions, so that it can couple to light. $\endgroup$ – my2cts Jun 28 '18 at 21:55
  • $\begingroup$ @AlexKChen. Regarding point 2., think of 10 billard balls placed in a row next to each other. You hit the first one and immediately see the last one move. The push propagates very fast. Similarly, the electric repulsion force between electrons propagates insanely fast. A "push" at the beginning immediately is felt at the other end, even though no electron has moved very far yet. That is why you see the light turn on right away when flipping the switch - the signal is propagated very fast. But the electron at the plug is not the one flowing through the light bulb at that moment. $\endgroup$ – Steeven Jun 28 '18 at 22:44
  • $\begingroup$ The speed by which a force travels through a billiard ball is the speed of sound in the material. $\endgroup$ – my2cts Jun 28 '18 at 23:35
0
$\begingroup$

You are confusing two frameworks, the classical electromagnetic one, when talking of conductors and fields, and the quantum one when talking of electrons.

The classical electric field emerges from the quantum mechanical level. Going to the level of electrons and nuclei and lattices in conductors sees the details which have to build up to zero macroscopic electric field in the conductor, in macroscopic units, because the transition from classical to quantum mechanical is mathematically smooth. In a conductor you can add bodily electrons and subtract electrons from the surface only, that's all. The lattice inside remains tied up with the quantum mechanical forces keeping electrons around nuclei and atoms in their places.

For 1 and 2. At the quantum level electromagnetic forces move with the velocity of light, the atoms transfer the electrical impulse to each other , which goes with the velocity of light. The electrons participating in this impulse have to move step wise, electrons absorbed at the source of the potential difference will leave a positive hole which will attract an electron from the atom behind, and the transfer will take place with velocities according to the medium and the forces, much less than the velocity of light. They jump from atom to atom, and that gives the drift velocity. The impulse itself, because it goes with the velocity of light is practically instantaneous.(think of a stadium wave, the signal itself can travel much faster than the people who are waving their hands would be able to run)

I have answered already 6, it is zero, because atoms and molecules are neutral and stay neutral. (within an atom it is another story).

Maybe somebody else will answer 3 4 and 5.

$\endgroup$
  • $\begingroup$ I know only Druke model (since that's what is written in the book to explain Ohm's law which is in the same chapter), but there it's written electrons are like free moving gas particles inside the atom and collides with the lattice particles. What's moving "step wise" and "positive hole"? And this is probably a silly question but for 6) If the field is zero inside the wire then why does the electrons move inside the wire without any net force acting on them ? $\endgroup$ – cdt Jun 28 '18 at 15:17
  • $\begingroup$ It is quantum mechanical transitions in solids, i.e solutions with coulomb potentials but classical fields and forces are an apporximation. Positive hole means an atom is missing an electron, and has positive charge, and can attract an electron from the the atom behind it in the wire, which will neutralize and move the positive backwards. No classical field because all this happens at atomic distances, overall everything inside is neutral so cannot build a macroscopic field $\endgroup$ – anna v Jun 28 '18 at 18:53
  • $\begingroup$ this is the type of model hyperphysics.phy-astr.gsu.edu/hbase/electric/miccur.html you are discussing, but overall the number of positive charges equals the number of negative within the conductor, except that the negatives move, i.e. are subtracted on one side and added on the other. $\endgroup$ – anna v Jun 28 '18 at 18:57

Not the answer you're looking for? Browse other questions tagged or ask your own question.