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Using a simple lattice model of conduction, where electrons are accelerated by an electric field, and are slowed down by bumping into the lattice, you get the following equation for current density:

$\vec{J}_n=nqμ_n\vec{E}$

Let's imagine an ideal DC voltage source connected with perfectly conducting wires to a resistor. Just from intuition, as the electrons reach the lattice of the resistor, you'd think that there'd be a pileup of electrons, since they don't have as much mobility in the resistor (almost like a traffic jam at tight roads). Do electrons or other charge carries collect at the end of resistors? If they do, is this what creates a voltage drop across resistors (or equivalently, an electric field across a resistor)? This idea of a collection of charge seems to imply a capacitance to the resistor. Do real resistors display any in-built capacitance?

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Do real resistors display any in-built capacitance?

Yes, and series inductance too. In fact, physical resistors have self-resonance frequency and a Q. At "high-enough" frequencies, these non-ideal contributions must be taken into account.

See, for example, this lecture: The Hidden Schematic

enter image description here

If I recall correctly, there is a charge density gradient through the length of the resistor. I remember seeing a nice graphic of this but I haven't found it online as yet. I will post as soon as I do.

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  • $\begingroup$ Based again on the lattice model, resistance is proportional to length, and voltage drop is proportional to resistance, so the voltage drop is proportional to length. Poisson's equation relating voltage and charge density would then tell us the charge density is 0 throughout the resistor (only built up charge at the ends?). Is that consistent with what you remember of the charge density gradient? $\endgroup$ – David Nov 1 '13 at 2:24

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