# Why don't high-voltage electrons in a metal wire escape?

• Mayan Fung
In summary: Yes, that's correct. The flow of electrons creates a negative charge on the wire, which in turn attracts positive charges (or repels negative charges) from the surroundings, creating a neutralizing effect. This allows for a continuous flow of electrons without any buildup of charge on the wire. However, if the wire is connected to a voltage source, the potential difference will be maintained and the flow of electrons will continue. This is described by Ohm's Law, which states that the current (I) is directly proportional to the voltage (V) and inversely proportional to the resistance (R) of the wire, expressed as I=V/R.
Very picturesque but my physical appreciation is a bit overwhelmed with mental images of the flora and fauna in this mountainside canal.

A good toy model for an electron in a metal is a finite potential pot. For simplicity I make a 1D model:
$$V(x)=-V_0 \Theta(0<x<L), \quad V_0>0.$$
The quantum mechanical time-independent Schrödinger equation (determining the energy eigenstates) can be solved analytically.

Now you can consider two plates forming a capacitor and putting a constant voltage on it. The voltage is between the two plates. In order to get an electron out of the (negatively charged) plate you'd need a voltage difference of some eV over a region of the order of a Bohr radius for an electron bound with its nucleus in the cathode. To get this you'll need a huge voltage between the plates. Even with air between the plates, which you can ionize at sufficiently large voltages between the plates, you need quite high voltages to do so and thus getting a conducting medium between the plates due to the partially ionized air (depending on the conditions like humidity, if I remember right, you need several kV voltage to trigger such a "gas discharge").

In contradistinction to that the photoelectric effect is not with a static voltage between plates but by shining an electromagnetic wave at the plate. In a bit hand-waving way you can argue that you only have to absorb one photon from the light source by one bound electron in the metal, and that's why with a sufficiently high frequency (for metals like zinc in the UV range is already sufficient) you can kick out the electron also with light of very low intensity, and this happens within a few nanoseconds after switching on the UV light, which shows that you don't need to wait to accumulate enough energy at the one bound photon located in a sphere of a about one Bohr radius around its nucleus (which would take pretty long compared to the correct quantum description).

Chan Pok Fung said:
Combining the above thoughts, I propose two possibilities where both contradicts to reality.
You seem to think that electrons are little packets of energy that are delivered to the load. That's false. You are jumping all over the place, from thermionic emission to the power grid.

Your strategy for learning about electricity is a bad one. You should go back and learn about simple circuits, about electrostatics, electrodynamics, and Maxwell's Equations, and Poynting's Theorem. Forget that you ever heard the word electron until all that is done. Believe me, electrons first is the most difficult possible way to learn about electricity.

Most of all, you can't get an education by asking questions on an Internet forum. This forum helps students who are studying the traditional way.

sophiecentaur and etotheipi
Well, I think in classical electrodynamics to think about charges and currents a continuum-mechanical picture is of great help. You get very far in establishing the constitutive relations using this classical continuum-mechanical picture of charges and currents. The real problem is only a point-particle picture, which is only in an approximate sense consistent and you need pretty advanced math to establish it, leading to the Landau-Lifshitz approximation of the LAD equation, which is the best you can do in this classical point-particle picture.

etotheipi
Chan Pok Fung said:
So the electrons only gain the energy very gently, but they accumulate energy throughout the journey in the wire.
Here's an easy 'mechanical' argument that tells you the Kinetic Energy of electron motion is actually irrelevant. Consider a light but strong bicycle chain. The cyclist could be supplying a couple of hundred Watts of power to the wheels, pedalling once a second. The chain could be, say 0.5kg mass and any part of it would be traveling at say 0.1m/s over the sprocket. The total KE of the chain will be mv2/2=0.025J. So the KE of the chain represents 0.025/200 = 0.125% of the Energy supplied every second the wheels. The remaining 99.875% of the work is being transferred by the tension of the chain and its distance traveled (force times distance) and not the KE of the chain.

Bear in mind that the total mass of the 'mobile' electrons in a wire is 1/(1800X60) *of the mass of the positive copper ion cores and that they are drifting at a mm/s. What is transferring the Power along the wire is the fields 'pushing' the electrons along - the KE of the electrons is much much less of a proportion of the energy transferred than for the bicycle chain links.

*electron mass is about 1/1800 of the mass of a proton and the atomic mass of copper is about 60

kith, vanhees71, anorlunda and 1 other person
sophiecentaur said:
Here's an easy 'mechanical' argument that tells you the Kinetic Energy of electron motion is actually irrelevant. Consider a light but strong bicycle chain. The cyclist could be supplying a couple of hundred Watts of power to the wheels, pedalling once a second. The chain could be, say 0.5kg mass and any part of it would be traveling at say 0.1m/s over the sprocket. The total KE of the chain will be mv2/2=0.025J. So the KE of the chain represents 0.025/200 = 0.125% of the Energy supplied every second the wheels. The remaining 99.875% of the work is being transferred by the tension of the chain and its distance traveled (force times distance) and not the KE of the chain.

Bear in mind that the total mass of the 'mobile' electrons in a wire is 1/(1800X60) *of the mass of the positive copper ion cores and that they are drifting at a mm/s. What is transferring the Power along the wire is the fields 'pushing' the electrons along - the KE of the electrons is much much less of a proportion of the energy transferred than for the bicycle chain links.

*electron mass is about 1/1800 of the mass of a proton and the atomic mass of copper is about 60

This is a good one, thanks! I have long been mistakenly understanding the power transmission mechanism of electricity.

anorlunda and sophiecentaur
Chan Pok Fung said:
This is a good one, thanks! I have long been mistakenly understanding the power transmission mechanism of electricity.
Along with millions of other people. Glad it helped.

There's a very clear and detailed discussion of this fact (one should also note that the electrons in a household wire move with ridiculously small velocities of the order of ##\text{mm}/\text{s}##) in

Sommerfeld, Lectures on Theoretical physics vol. 3

for the simple case of a DC current.

sophiecentaur