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In a simple circuit are there any charges accumulated at two ends of resistor? There is an conservative and thus electrostatic electric field and potential difference across the resistor. It seems it is only possible if positive and negative charges are accumulated at two ends to provide the electric field. Can these kind of extra charges be detected in a real resistor? How?

Same questions for inductor in the case of changing current in inductors. Thank you.

Kelvin S
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1 Answers1

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Microscopically, in a resistor electrons are scattered by lattice impurities and defects (e.g. surfaces), other electrons and phonons.

An approximate description of the process: In the classical approximation of the Drude model the scattering processes are reduced to a scattering time $\tau$, in between the scattering processes the electrons move balistically (that is freely accelerated by the electrical field).

So the actual process is highly statistical an non-trivial and the voltage drop has to be accounted to the electrical field losing energy due to the acceleration of the electrons and thus being reduced. (Actually the flow of energy of the electromagnetic field (Ponyting vector) around a resistor is inwards, when you consider the whole setup it will flow out of the source (where electrical and magnetic fields will have reversed directions compared to each other as the current has the same direction everywhere, but the electrical field direction is against the current in the source) into the wires.) So you can see, the situation is stationary but not static, there always is an energy flow going on. The dynamical equilibrium between this energy inflow and the energy losses of the electrons to the resistor cause the electric field to drop without charges accumulating. (And on the microscopic scale the process is not even stationary!)

In the case of the inductor the answer is simple and a clear no. The relevant Maxwell equations is:

$$ \nabla \times \vec E = -\frac{1}{c^2} \partial_t \vec B $$

This says, there is another way of generating electrical fields than accumulating charges. When the current through the wire changes the generated magnetic field changes, thus inducing an electrical field with closed field lines (that is, without sources and sinks). So actually the field involved here is not even conservative (just it's projection to the effectively one dimensional wire is).

So in a way the description of these processes by an electrical potential is cheating and only possible because the wires are effectively one dimensional (and roughly every one dimensional function has an antiderivative). In the entire 3d-space an electrical potential alone cannot account for the observed fields (one also will have to consider the vector potential there).

Sebastian Riese
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  • Thanks, but in a physics lecture (https://www.youtube.com/watch?v=EYYNRubHIno&list=PL5694B25F659D95B0&index=11) see 1:00:12, the lecturer explain things in terms of accumulated charge. That' what bothers me. What's happening there? – Kelvin S May 10 '15 at 04:35
  • In the part I watched (the next 2 minutes after the point you referenced), the lecturer just explains the analogy to the capacitor (where charge accumulates), without implying that charge accumulates in an inductor. – Sebastian Riese May 10 '15 at 11:23
  • Sorry, it should be from 41:00 to 45:00 he explained that charges are build up and conservative field was set up. If there is no charges accumulated, how can the E field be build up? – Kelvin S May 10 '15 at 14:41
  • In 43:00 to 43:30 he said there are always some charges to keep the E field. Anything wrong? – Kelvin S May 10 '15 at 14:54
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    Certainly there are always some charges accumulated (even a straight run of wire has a capacitance), but if you have a battery, there will be no charges like on a plate capacitor as voltage source! The electrical field is kept up by a dynamic equilibrium between the electrons in the material of the resistor and the EM field. The usual simple relations are certainly good approximations, but arise out of the complex microscopic behaviour only when you average over small space-time-volumes. I will try to add a more detailed passage on this later on. – Sebastian Riese May 10 '15 at 15:16
  • Thanks for reply. I have read from another source mentioning the charge accumulation: University Physics with Modern Physics 13th Edition in page 996 here it says "Then a negligibly small electric field is required to make charge move ... there have to be accumulations of charge on the terminals of the inductor and the surfaces of its conductors to produce this field." You can read it online in Google.

    I haven't heard of anything about charge accumulation before but now there are more than one source mention that. Therefore I make this post.

     – Kelvin S May 11 '15 at 02:24
  • Well I guess one way to think about it (although I am not sure a helpful way), is to consider the (idealized infinitesimal) capacitance of the circuit, then an infinitesimal accumulation of charges will be enough to cause a large voltage $U$ (as $C = Q/U \Rightarrow U = Q/C$). – Sebastian Riese May 11 '15 at 08:28