When we derive Ohm's Law using the Drude Model, we assume at one point of time that $E=V/L$, when is fact, $E=dV/dL$, unless $E$ is constant, in which case the assumption $E=V/L$ is true. But I don't understand why the electric field in a conductor must be constant as current flows. Is there a convincing explanation that is perhaps related to the way atoms behave and orient themselves?
Also, if the assumption $V=E\cdot L$ makes sense, I can understand why Ohm's Law should work for a homogeneous electric circuit. However, I don't understand why it should work for a heterogeneous circuit – perhaps one with two different resistors connected in series. And please don't use the traffic jam analogy. Surely there must a more theoretical way to explain this (using Classical Physics).
Best Answer
An easy way to prove Ohm's law for electric fields that aren't constant is to first assume that the electric field is approximately constant over short lengths, just like $E=dV/dL$ suggests. Using that, you can derive Ohm's law for short lengths of material, $dV=IdR$. We'll assume that "current in = current out", which is true at steady-state. This allows us to integrate this equation (since current is a constant relative to both dV and dR), and you get regular Ohm's law $V=IR$. This is equivalent to saying that small resistances combine in series to form a net resistance for a material, for which Ohm's law also holds. This is regardless of how complex the geometry is that makes up the resistor.