Identical Wires with different voltage--does electron behavior differ?

[LouisL]

Homework Statement:: I am studying on my own, so I don't have a specific homework statement, but want to make sure I am thinking about things correctly.

What I am wondering is if you have equivalent wires, let's say both are made of copper, and one wire has three times the voltage of the other wire, I would assume the higher voltage wire has electrons that are moving faster (greater Amperage), with higher translational, rotational, and vibrational energy than the electrons in the wire with lower voltage and the speed of these faster electrons causes them to have a higher Kinetic Energy than the lower voltage wires? Am I correct?
Relevant Equations:: I=VR and KE=1/2Mv^2.

I=VR, so V=I/R and 3V=3I/R. In terms of calculating the Kinetic Energy of a section of each wire, you would use KE= 1/2 M V^2 . Would velocity in this equation equal the amps in the wire OR perhaps the average drift velocity of electrons in the wire or something entirely different? Am I thinking about this correctly? Thanks.

 

[kuruman]

Electrons in a metal do not behave like rocks in free fall. Read this about drift velocity. To answer your question, more current means higher drift velocity. However, when electrons move in a conductor through a potential difference ΔV, the loss of potential energy is not a gain of kinetic energy. The drift velocity and hence the average electron kinetic energy remain the same. The loss of potential energy goes into heat generation (Ohmic loss). Thus the higher the voltage in a given wire, the higher the current which means the higher the rate $I^2R$ at which heat is generated.

 

[Vanadium 50]

but want to make sure I am thinking about things correctly.

What I am wondering is if you have equivalent wires, let's say both are made of copper, and one wire has three times the voltage of the other wire

You are not thinking about this correctly. Voltage is measured with respect to some stated potential. "Three times the voltage" doesn't make sense, since the only thing that is physical is voltage difference.

 

[kuruman]

You are not thinking about this correctly. Voltage is measured with respect to some stated potential. "Three times the voltage" doesn't make sense, since the only thing that is physical is voltage difference.

Isn't "voltage" synonymous to "electric potential difference"? Electric potential is what needs a reference. I understand "three times the voltage" to mean "three times the potential difference".

That is made clear here: https://en.wikipedia.org/wiki/Voltage.

 

[davenn]

I=VR, so V=I/R and 3V=3I/R.

this is not correct which kinda messes up your suppositions

Am surprised the other 2 guys didnt pick up on that

I=VR No, I = V/R etc for the rest of your statement
cheers
Dave

 

[berkeman]

Voltage is measured with respect to some stated potential. "Three times the voltage" doesn't make sense, since the only thing that is physical is voltage difference.

Most likely the OP is thinking in terms of circuits, where the reference potential is "ground". So the voltage difference is just the voltage with respect to ground.

 

[kuruman]

Most likely the OP is thinking in terms of circuits, where the reference potential is "ground". So the voltage difference is just the voltage with respect to ground.

That was my thinking too. One wire is connected to a 10 V battery and the other to a 30 V battery.

 

[berkeman]

Vaporize those wires!

 

[davenn]

since this is homework and you have laid it out as such,
I have asked for it to be moved to the homework section @berkeman
And hence also, I will not give direct answers

OR perhaps the average drift velocity of electrons in the wire or something entirely different?

Yes it is ... google it and come back with an answer for drift velocity

I would assume the higher voltage wire has electrons that are moving faster (greater Amperage)

This is not a correct assumption either

do a google search on the definition of an Ampere
come back and state what you find out. You will discover it has little to do with drift velocity

 

[berkeman]

Seems like a fairly general question so far, so we'll leave it here for now. Thanks Dave.

 

[Vanadium 50]

Most likely the OP is thinking in terms of circuits, where the reference potential is "ground". So the voltage difference is just the voltage with respect to ground.

I'm not disputing that. But he wanted to know if he is thinking about it correctly, and he's not. (Well, technically, he's not writing about it correctly) Whether this will bite him later or not depends on where he wants to go.

 

[anorlunda]

The way the OP is worded, it could mean two isolated open-circuit segments of wire with no current but different potentials. In those cases, there is no motion of the electrons due to potential. None. Zip. Compare those electrons with them motion of books on different levels of shelves in the library. Different gravitational potential, but no motion.

But if the question means potential of wires in a circuit, that gives me the excuse to repost one of my favorite examples.

 

[kuruman]

since this is homework and you have laid it out as such,
I have asked for it to be moved to the homework section @berkeman
And hence also, I will not give direct answers

Truth be told. This thread was initially posted in the Introductory Physics Homework forum. I reported it as not belonging there and asked that it be moved here to Classical Physics and it was. I did so based on OP's statement that this is not related to homework assigned by an external agent.

 

[davenn]

Truth be told.

... you didnt catch the errors in his post tho

 

[LouisL]

Thanks everyone for your answers. I obviously have a lot of studying to do. I will revisit this post after I get a better grip on the subject.

 

[etotheipi]

You are not thinking about this correctly. Voltage is measured with respect to some stated potential. "Three times the voltage" doesn't make sense, since the only thing that is physical is voltage difference.

This is better stated that electric potential is measured with respect to some specified reference potential, and that the only thing that is physical is potential difference, i.e. voltage.

Whilst those familiar with the subject can afford to be less strict with terminology, for a learner I think it's important to be clear about the idea of potential being a scalar field that is a function of position in the space, and voltage being a line integral of a conservative electric field between two points or equivalently a difference in the electric potentials at those two points.

 

[kuruman]

... you didnt catch the errors in his post tho

Are you referring to the three fundamental E&M equations?

Spoiler: What are they?

$$V=IR$$ $$I=\frac{V}{R}$$ $$R=\frac{V}{I}$$

 

[Delta2]

Homework Statement:: I am studying on my own, so I don't have a specific homework statement, but want to make sure I am thinking about things correctly.

What I am wondering is if you have equivalent wires, let's say both are made of copper, and one wire has three times the voltage of the other wire, I would assume the higher voltage wire has electrons that are moving faster (greater Amperage), with higher translational, rotational, and vibrational energy than the electrons in the wire with lower voltage and the speed of these faster electrons causes them to have a higher Kinetic Energy than the lower voltage wires? Am I correct?
Relevant Equations:: I=VR and KE=1/2Mv^2.

I=VR, so V=I/R and 3V=3I/R. In terms of calculating the Kinetic Energy of a section of each wire, you would use KE= 1/2 M V^2 . Would velocity in this equation equal the amps in the wire OR perhaps the average drift velocity of electrons in the wire or something entirely different? Am I thinking about this correctly? Thanks.

First of all I assume that by "Voltage" you mean the potential of a point as it is defined in electrostatics or as it is used in circuit theory relative to a ground point where we take the potential to be zero at ground.

Second the current in Amperes $I$and the average drift velocity $v$ in m/s are related by the equation $$I=nSve$$ where $e$ is the charge of electron, $n$ is a constant that depends on the material of the conductor (it is its average density in free electrons) and $S$ is the cross section area of the conductor. As you can see this equation doesn't involve the potential at the point where $I$ or $v$ are measured, so the potential does not relate to drift velocity, at least not in an explicit way.

We can use this equation in conjunction with your setup . Let's say that the current is $I_H=n_HS_Hv_He$ at the wire of high potential and $I_L=n_LS_Lv_Le$ at the wire of low potential. if the two wires are connected in series, then it will be $I_H=I_L$ and furthermore if they are from the same material then $n_H=n_L$ hence we can infer that $$S_Hv_H=S_Lv_L\Rightarrow \frac{v_H}{v_L}=\frac{S_L}{S_H}$$ that is that the drift velocity is inversely proportional to the cross sectional area. So the wire with the smaller cross sectional area has the bigger drift velocity and not the wire with the higher potential.

 

[davenn]

Are you referring to the three fundamental E&M equations?

yeah, he got them completely wrong, which, as I said, sort of messes up the rest of his assumptions

 

[vanhees71]

For me the discussion of the (naive) Drude model of conductivity was a revelation to understand DC (and later also AC) circuits. So here it is:

In a metal a part of the electrons are pretty free to move within the lattice of positively charged ions making up the metal, i.e., they are not bound to the specific atoms within the lattice. These are called conduction electrons. However, the electrons are not completely free to move, because they are scattered on the lattice due to "defects" (i.e., the deviation from an ideal crystal) as well as due to lattice vibrations (thermal motion).

Forgetting about all these microscopic details you can describe the motion of a conduction electron within the medium by a equation of motion taking into account the friction by a friction constant $m \gamma$, i.e.,
$$m \dot{\vec{v}}+m \gamma \vec{v}=-e \vec{E}.$$
Here I neglected the magnetic part of the Lorentz force (though in principle it's important to include them, if it comes to a really accurate description of the self-consistent Hall effect, but that can be safely neglected for all purposes of usual "household currents" as we shall see below).

If you have now a time-independent voltage (potential difference) across the wire you can assume that $\vec{E}=\text{const}$ pointing along the wire. If you wait some (very short) time after switching on the circuit, the electrons will be accelerated to a velocity such that the friction force balances the electric force, and then $\vec{v}=\text{const}$. Then you get
$$\vec{v}=-\frac{e}{m \gamma} \vec{E}.$$

Now there is a certain density $n$ of conduction electrons in the metal ($n$ is the number of conduction electrons per unit volume), and the current density thus is
$$\vec{j}=-e n \vec{v}=\frac{e^2 n}{m \gamma} \vec{E}.$$
The proportionality constant is nothing else than the electric conductivity
$$\sigma=\frac{e^2 n}{m \gamma}.$$
If you take typical densities of $n$ in a wire (e.g., estimating it by assuming that there's one conduction electron per atom) for a given current $I=A |\vec{j}|$, where $A$ is the cross sectional area of the wire, you come to a drift velocity $v \simeq 1 \text{mm}/s$, and this proves that it's safe to neglect the magnetic force on the conduction electron in the very beginning as far as the magnetic field within the wire is concerned that is made by this current itself, i.e., you can usually neglect the self-consistent Hall effect).