6
$\begingroup$

I have been thinking about ways of teaching electronics and I'm wondering if the following is true...

For starters, when we talk about voltage as energy per unit charge, is this energy manifest simply as the kinetic energy of the electron?

So, all other things being equal, will the electrons coming out of a 4 volt battery will have twice the velocity of electrons coming out of a 1 volt battery because of ${1 \over 2}mv^2$?

This would seem to imply that the voltage drop across different parts of the circuit is essentially a comparison of the electron speeds at those points.

Furthermore doesn't this imply that the electrons are nearly halted by the time they reach the positive terminal of the battery? This makes intuitive sense to me, it seems pedagogically sound, and it provides an explanation for how electrons "know" to give up their energy across the circuit -- electrostatic repulsion communicates later resistances to earlier parts of the circuit.

How much of the above is correct? The one things that makes me suspicious is that, in this picture, electron density at the positive terminal of the battery is very high, and repulsive forces would prohibit this. On the other hand, if the Joules per Coulomb are not coming from kinetic energy, then what? The field? But wouldn't that just dissipate as light?

Something is wrong but I don't know what.

Improve this question
$\endgroup$
2
  • $\begingroup$ Related: physics.stackexchange.com/q/17741/2451 and links therein. $\endgroup$
    – Qmechanic ♦
    Jul 11 '14 at 20:12
  • $\begingroup$ Suggestion to the question (v2): Ask for completeness how the drift-velocity of the electrons depends on the applied voltage. $\endgroup$
    – Qmechanic ♦
    Jul 11 '14 at 20:20
10
$\begingroup$

If there was nothing in the way then an electron leaving the anode of a 4 volt battery would have a kinetic energy of 4 electronvolts by the time it reached the cathode.

However the mean free path of electrons in metal wires is exceedingly short so electrons never build up anything like this velocity. The end result is that the electron velocities are randomised but with an small average velocity from the anode to the cathode called the drift velocity. To give you a feel for the difference the electron velocity corresponding to 4eV is about a million m/sec but drift velocities are typically around 1 m/sec.

To make an analogy - suppose you are standing in a pleasant breeze of one m/sec. According to the Maxwell-Boltzmann distribution the velocity of air molecules at room temperature is 300 to 400 m/sec but their directions are random and cancel out. You're left with a net motion of the air of 1 m/sec which is analogous to the drift velocity of electrons in a circuit.

The best way to teach electronics to beginners is to use the hydraulic analogy. Even though it might seem clunky at first glance the hydraulic analogy is remarkably useful. You can even model circuit elements like capacitors.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Like the "pleasant breeze" analogy. $\endgroup$
    – Floris
    Jul 11 '14 at 17:12
  • 1
    $\begingroup$ The drift velocity is typically around a few centimeters/hour. $\endgroup$
    – Physiks lover
    Jul 11 '14 at 20:58
5
$\begingroup$

For starters, when we talk about voltage as energy per unit charge, is this energy manifest simply as the kinetic energy of the electron?

Voltage is potential energy per unit charge. An analogy: voltage is to charge as altitude (as on the surface of Earth) is to mass.

So if you lift a 1kg rock off the ground by a height of 1m, you've added some gravitational potential to that rock. Does it manifest as kinetic energy?

Well, if you drop it in a friction-less environment, maybe. But any number of other things could happen. The space within the 1m where the object falls could be filled of some viscous goo. In that case, most of the gravitational potential energy is converted to heat by way of friction. Or, we might design an apparatus to turn a generator as the mass falls, converting its gravitational potential energy into some kind of electrical energy. Maybe we charge a battery, and it becomes chemical energy.

And of course, at the bottom of the fall it probably hits the ground, transferring momentum to the Earth and sending off energy as sound, among other things. It doesn't stay as kinetic energy for long, it most real situations.

And so it is with electricity, too. You probably won't find anything analogous to a frictionless place to "drop" electric charge. If the current goes through a resistor, then it is converted to heat. If it goes through a motor, then it's converted to mechanical energy.

There are devices that launch electrons in essentially lossless conditions (vacuum tubes, CRTs), but most such devices have something at the end into which the electrons smack to do something else (in the case of the CRT, produce visible light). In these cases, the kinetic energy of the electrons immediately prior to impact would be proportional to the voltage through which they have fallen, but I don't suspect this is especially insightful for teaching electronics generally. If anything, it demonstrates that even if the current is constant in a series circuit, the speed of the electrons is not necessarily so.

To know what becomes of the electrical potential energy, you need to know what happened to the charge as it fell.

Improve this answer
$\endgroup$
2
  • $\begingroup$ A vacuum tube is an almost perfect "frictionless place to 'drop' electric charge". When the electrons hit the anode they are going very fast - and turn into heat (emitting bremsstrahlung). $\endgroup$
    – Floris
    Jul 11 '14 at 17:25
  • 1
    $\begingroup$ @Floris yes, good point. I've expanded the answer a bit. $\endgroup$
    – Phil Frost
    Jul 11 '14 at 17:32
2
$\begingroup$

Do not equate potential with kinetic energy. How fast electrons flow in a conductor has very little to do with their potential. You need to consider the current and the charge carrier density for that. Depending on the material you can have a few fast electrons or many more slower ones. In semiconductors the carrier velocity will be higher - which is why the Hall effect shows up more, for example.

Electrons (charge carriers) lose energy as they are scattered by the material they travel through. This is experienced as resistance. How much current flows is a function of voltage and resistance - so yes when you double the voltage and everything else is the same then average drift velocity of charge carriers doubles (same number of carriers traveling twice as fast). But this velocity is surprisingly low for a good conductor - I recommend you try calculating it to persuade yourself.

Improve this answer
$\endgroup$
1
2
$\begingroup$

For starters, when we talk about voltage as energy per unit charge, is this energy manifest simply as the kinetic energy of the electron?

No, not at all.

Recall that, in electric circuits especially, voltage is not measured at a point (in general, the potential at a point is not physically meaningful - only the difference in potential of two points is physically meaningful).

We speak of the voltage across (or between) two terminals or nodes in a circuit.

For example, we don't say that the positive terminal of a battery has a voltage of 1.5V, we say that the voltage across (between) the positive and negative terminals of the battery is 1.5V.

Now, as you point out, the volt is joules per coulomb. How to interpret this?

In the case of the battery above, the (ideal) battery does 1 joule of work moving 1 coulomb of charge from the negative terminal (lower potential) to the positive (higher potential) terminal of the battery.

If the battery is connected to a resistor, the energy gained by the coulomb of charge is converted to heat in the resistor.

The rate at which the work is done is the power and is given by the product of the voltage across and the current (rate of flow of electric charge) through.

Note that kinetic energy has not been mentioned. The battery (ideally) does the same amount of work 'pumping' 1 coulomb of charge from negative to positive regardless of the rate at which this done.

Furthermore doesn't this imply that the electrons are nearly halted by the time they reach the positive terminal of the battery?

Imagine, for example, a 1.5 volt battery connected across a length of uniform conductor with a total resistance of 1 ohm. The current through the conductor is, by Ohm's law, 1.5 amps and, since the conductor is uniform, the amount of charge flowing through any cross section anywhere along the conductor is 1.5 coulombs per second.

The electrons near the positive end of the battery do have less energy than the electrons near the negative end of the battery but that is due to their position in the electric field, i.e., their difference in potential energy.

Improve this answer
$\endgroup$
0
$\begingroup$

This question was given in Moroccan Olympiads:

They said: If we have the same circuit: In the first case the electric current measured is $I_1=1A$ and the second case we have $I_2=2A$. The speed of electrons will change ? give a mathematical explanation.

I'm going to use simple physics: (High school level)

We know that : $I=\frac{Q}{\delta t}$

And the speed is given by : $v=\frac{\delta d}{\delta t}$

So we can write the speed : $v=\frac{I.\delta d}{Q}$

Know we observe that the speed can be changed, the factor here is the electric current.

I see that it's the same as your question, I used what they gave at the final solution.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.