Question about how resistance works

[Tusike]

Homework Statement

Consider a simple circuit that looks like a rectangle (the cables), with 4.5 V attached on the bottom. The electrons are traveling from - to +, let's say I1 = 10A, which means they have an average velocity v1. Now put a 1kOhm resistance near the + side of the battery. I2 = 4.5mA anywhere in the circuit. My question is, how the hell does the electron that starts from the - end of the battery know that it shouldn't be traveling as fast? Isn't the E-field the same?

Homework Equations

None that I can think of.

The Attempt at a Solution

It's ready a theoretical question so I don't have an attempt. I know that they have to travel slower, because if not, they would clutter together at the resistance, and there's some law that doesn't allow it, I forget which (one of Maxwell's). I've always seen the potentials of a circuit represented by water falling down, so I tried applying the same here. Without the resistance = water falling in a tube, with a pump bringing it back up. With the resistance = near the bottom, make the tube really narrow, then after a few feet back to normal again. Applying what happens in the circuit in this situation, the water should start falling much slower than before; This can only be if the pump is pumping it slower. So does this mean that in the circuit, the battery's voltage is pumping the electrons slower or something like that?
Thanks for any help!

 

[Tusike]

So what I'm imagining now is that when there is a resistance in the circuit, less electrons start from the - end, but still at the same speed as without the resistance. This results in the lower I2 observed. However, inside the resistance, aren't they supposed to travel slower? And they would clot up exactly (more electrons) so that the I inside the resistance equals the same as in the rest of the circuit. But as I mentioned before, clotting up is against a law, so I still don't understand.

 

[hikaru1221]

It's very unclear and usually wrong to say "less electrons" or "more electrons". You don't even need to apply Maxwell's equations here. You only need the microscopic Ohm's law:

\frac{1}{\rho}\vec{E}= \vec{J}= -ne\vec{v}

where \rho is resistivity, \vec{J} is the current density, n is the number of electron per volume, and \vec{v} is the drift velocity of electron.

We have the current: I=\int _{A} \vec{J}d\vec{A} = JA where A is the cross-sectional area (assume that \vec{J} is the same at every point on A). Since I is the charge passing through the cross-section in 1 second, and the total charge must be conserved, then in series, I = const, right? Therefore:

\frac{1}{\rho}\vec{E}A = -ne\vec{v}A = const

So the E-field is not the same everywhere in general, as it depends on cross-sectional area and resistivity. Nothing can be concluded here, as the wire may get high resistance but be big and vice versa.

Similarly, for the drift speed, we have: nvA=const so it depends on the properties of the wire, i.e. the number of free electron per volume n (one property about the material) and the cross-sectional area (one property about the size). Again, nothing can be concluded if the only thing we know is the resistance of the wire. But we understand that electrons at different places may get different drift speeds, so the power supply doesn't try to pump slower or faster - it only supplies with the E-field in the wire. The definition of power supply is a potential-difference source after all, right? The rest, i.e. how electrons at one point move, is up to the E-field at that point and the properties of the wire at that point. The electron doesn't foresee anything.

So the question is, how the hell the electron knows where and when to slow down or speed up? It doesn't. The E-field tells it to do so. From the first equation, you can see that the drift speed is proportional to the E-field. An interesting model which accounts for this is the Drude model. Find it on Wikipedia

 

[Tusike]

Thanks that's all I needed to know!