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I was reading from several introductory E&M materials, and they all state that $$I=neA\vec{v}_d$$where $n$ is number of free charge carriers, $e$ is the elementary charge of electron, $A$ is cross-sectional area, and $\vec{v}_d$ is drift velocity.

To derive number of electrons that will pass through cross-section, we need to see that for some small time interval $dt$, segment of wire in gray area will pass through cross-section area $A$.That segment have volume of $dV=Adx=A\vec{v}_ddt$ From definition of current, we know that $$I=\frac{dQ}{dt}$$ Number of free electrons that will go through cross-section is therefore equal to the number of free electrons in the shaded area. Now, this conductor have some charge density $\rho_{free}$, and to get number of electrons we need to integrate this density over cross-section area $A$.

Segment of wire that will go through cross-section

This however is not what they do in materials i've read, and they instead assume that conductor have uniform charge density over cross-section. Under that assumption, equation they provide makes sense.

How strong this assumption is? I'm aware of the Drude model, but i'm not sure if this assumption is considered part of the Drude model. Is this assumption wrong?

I know there are 3 questions inside, but they are so related that asking in 3 separate posts will have 90% of duplicate text.

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  • $\begingroup$ I am not sure exactly what your problem is. $n$ is typically an charge carrier density, so the number of free charges $dN$ in your volume would be $dN = n dV = n A v_d dt$, corresponding to a charge $dQ = n e A v_d dt$ which gives the right answer. $\rho_\mathrm{free}$ is simply another way of writing the free charge density $n e$. $\endgroup$
    – QuantumApple
    Commented May 2, 2020 at 13:17
  • $\begingroup$ Imagine this. Cross-section area is made out of two different materials. Left half of cross section is made out of with first material that have $n=c_1$, and other half of cross section with material that have $n=c_2$. What is $dN$ equal to now? It's not $n*A$? In this case, it's easy to guess correct formula, but imagine more complicated setup with lot of impurities in the conductor that are not mixed in this easy symmetric way. $\endgroup$
    – Stanko Kovacevic
    Commented May 2, 2020 at 13:40
  • $\begingroup$ This formula is typically applied inside a single, homogeneous material. You are working with elementary volumes, which is why the charge density can be approximated to be constant, at least locally. $\endgroup$
    – QuantumApple
    Commented May 2, 2020 at 13:49
  • $\begingroup$ At a junction between two materials with carriers density $n_1$ and $n_2$, in a stationary regime, you must have $I_1 = n_1 e v_{d1} A = I_2 = n_2 e v_{d2} A$, so $n_1 \neq n_2$ but this is typically compensated by $v_{d1} \neq v_{d2}$. If $n_2 < n_1$ for instance, $I$ might generate a larger electric field $E_2$ compared to $E_1$, leading to $v_{d2} > v_{d1}$. But in the simplest case, you are looking at homogeneous materials so none of this really matters. Is it the "uniform charge density" that is bothering you? $\endgroup$
    – QuantumApple
    Commented May 2, 2020 at 13:50

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