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Consider a neutral conductor with $n$ free carriers that are uniformly distributed inside the conductor when there is no electric field, $E_z = 0$. From the continuity equation, I can write down the charge drift-diffusion equation as

$ \frac{\partial n}{\partial t} = D \frac{\partial^2 n}{\partial z^2} - \mu E_z \frac{\partial n}{\partial z} $

where $D$ allows for diffusion, $\mu$ is the mobility, $E_z$ is an electric field applied, and only the the $z$-direction is focused on for simplicity. As I already said, initially $E_z = 0$ so the $n$ is uniform throughout the conductor.

Now the electric field is turned on. What happens? Well, intuitively one one would surmise that the free charges move against the electric field (assuming electrons here) which will leave the conductor positively charged on one end and negatively charged on the other. However, if I solve the continuity equation, I get $n$ to be uniform still -- nothing happens. You can see why this is, mathematically, from the above equation. The gradient of $n$ is initially 0 which sets $\partial n/\partial t = 0$ and nothing interesting comes of it. Compare this to the situation where you start with a charge gradient - i.e. a delta impulse at $z = 0$ - where the drift of the charge impulse can see in the time evolution.

Why is the math not matching my expectations?

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I think the problem is with your boundary conditions. Nowhere did you specify the metal is of finite size. If you exert a field on an infinite piece of metal, all electrons will drift in the direction of the field, but you won't see any change in the density of the electrons, since the electrons leaving any patch to the left are replaced by electrons coming from the right, and the right side will never "run out" of electrons since you're initially assuming an infinite number of electrons.

Put more simply: You should intuitively expect $n$ not to change, since an infinite uniform sea of electrons will remain an infinite uniform sea of electrons regardless of whether that sea is stationary or accelerating. The density doesn't bunch up anywhere, because there's no edge for it to bunch up on.

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  • $\begingroup$ I thought about that but when I put in hard boundaries, I think I got the same thing...I'll go back and check on that $\endgroup$ – user5419 Aug 24 '17 at 17:00

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