# Thread: Electrons - when they move from Atom to Atom - where do they end up?

1. Originally Posted by pfalcon
work in joules (J)
thus
current is proportional to work per volt-second. NOT energy per volt-second.
The change in energy equals the work and they have the same units. Does not change the fact that your units did not agree.

Originally Posted by pfalcon
J = kg*m*m/s/s
V = kg*m*m/C/s/s
V = kg*m*m/C/s/s = J / C
so
C = J / V
which is what I had in the prior post.

Originally Posted by pfalcon
Coulomb is the base unit for the magnitude of energy not joules. Joules and Volts are derived measurements of behaviour.
Coulomb is not an energy unit. Your units will not agree (see prior post).

2. Originally Posted by pfalcon
The interaction is typically the most dense in the metal because that's where the field center is located. But it doesn't have to interact inside the metal or at least the interaction can be balanced within the metal incurring the same result. Sort of like the beads on an abacus you don't have to push on the center to make it move down the wire but it's still constrained to the wire, the center still rests in the wire, and the field density is greatest in the wire.
Consider the simple, steady-state DC case with negligible resistance in the wire (approximately a zero axial electric field component). The net radial electric field is zero inside the wire because the wire has become polarized at equilibrium. This polarization produces an internal electric field that balances the applied electric field from the battery.

The remaining electric field is radially directed between the surfaces of the two conductors and is where the traveling energy resides.

With a current in a resistive wire, we never reach equilibrium in the axial direction because we have a charge pump (battery) that keeps the charges circulating (current).

3. Originally Posted by Besoeker
Not so. It's joules.
Originally Posted by mivey
Coulomb is not an energy unit. Your units will not agree (see prior post).
No. You won't find a definition of Joule that doesn't refer to it as work. Work is a limited characteristic of energy referenced to units of time and distance. If you neglect the static forces of a charge then the remaining forces are proportional to work. Which is exactly what most current equations do - they neglect the static forces. Coulomb is the base unit. Joule is the coulomb in motion.

CV = J

(Energy)*(Work potential per Unit of Energy) = Work

Meaning, as you have, when you only deal with Work equations if follows since energy is proportional to work those equations can treat energy and work as though they were identical.

4. Originally Posted by mivey
Consider the simple, steady-state DC case with negligible resistance in the wire (approximately a zero axial electric field component).
Negligible resistance in a wire is akin to a hockey puck on ice. It's axial component is independant of the resistance. Your equations are not going to work unless there is real resistance to create an equilibrium state. Otherwise any axial component will create acceleration.

Originally Posted by mivey
The net radial electric field is zero inside the wire because the wire has become polarized at equilibrium. This polarization produces an internal electric field that balances the applied electric field from the battery.
Polarized?!
This is more again to pushing an object across a rough surface where friction negates the acceleration from a force. The friction balances the force to create a net zero and therefore constant velocity. The forces in the wire reach an equilibrium. But forces in equilibrium are not the same as without energy. That the terms in equilibrium are crossed out when solving most of your equations is not the same as their not existing.

Originally Posted by mivey
The remaining electric field is radially directed between the surfaces of the two conductors and is where the traveling energy resides.
Which immediately fails the test of a capacitor since that requires only one conductor to create a current. There is no second conductor for the field to travel between. Think lightning bolt if you must. Leyden jar if you will. Electricity requires an energy source and an energy sink. It does NOT require two conductors. It does NOT require a return path. Those are artifacts of electrical generators and practical application of electricity. Important artifacts that you won't run your AC without - important for our usage of electrical energy not important to the transmission of electrical energy.

Originally Posted by mivey
With a current in a resistive wire, we never reach equilibrium in the axial direction because we have a charge pump (battery) that keeps the charges circulating (current).
We always reach equilibrium in a resistive wire. The voltage (potential for work) that pushes the current is balanced by the resistance (frictional work) to create a constant current (velocity). Otherwise the current would go exponential. Or are you proposing that Newton's second law doesn't apply?

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Originally Posted by pfalcon

We always reach equilibrium in a resistive wire. The voltage (potential for work) that pushes the current is balanced by the resistance (frictional work) to create a constant current (velocity).
I do not agree with current as velocity. Current is like the number of cars that pass a point on a highway per unit of time; it doesn't tell you how fast they are moving. If there are more lanes (lower resistance), they don't need to travel as fast. Or water in a hose; to get the same number of gallons per hour (current) through a smaller hose, the water must move faster.

6. Originally Posted by pfalcon
No. You won't find a definition of Joule that doesn't refer to it as work. Work is a limited characteristic of energy referenced to units of time and distance. If you neglect the static forces of a charge then the remaining forces are proportional to work. Which is exactly what most current equations do - they neglect the static forces. Coulomb is the base unit. Joule is the coulomb in motion.

CV = J

(Energy)*(Work potential per Unit of Energy) = Work

Meaning, as you have, when you only deal with Work equations if follows since energy is proportional to work those equations can treat energy and work as though they were identical.
Can you provide some reference material? I'm all for exploring new things, but if it is outside standard accepted electrical theory I would like some background material for a frame of reference.

Thanks.

7. Originally Posted by pfalcon
Negligible resistance in a wire is akin to a hockey puck on ice. It's axial component is independant of the resistance.
The axial component of the electric field is not independent of the resistance. It is directly related.

Originally Posted by pfalcon
Your equations are not going to work unless there is real resistance to create an equilibrium state. Otherwise any axial component will create acceleration.
We've covered that. If there is an axial component, there is resistance and thus a voltage drop along the conductor. Basic circuit theory stuff.

Originally Posted by pfalcon
Polarized?!
Yes. That is what happens when you apply an electric field across a metal conductor. Basic physics.

Originally Posted by pfalcon
This is more again to pushing an object across a rough surface where friction negates the acceleration from a force. The friction balances the force to create a net zero and therefore constant velocity. The forces in the wire reach an equilibrium. But forces in equilibrium are not the same as without energy. That the terms in equilibrium are crossed out when solving most of your equations is not the same as their not existing.
The net electric field is zero. The field from the polarized metal offsets the applied field from the battery. Basic physics again.

Originally Posted by pfalcon
Which immediately fails the test of a capacitor since that requires only one conductor to create a current. There is no second conductor for the field to travel between. Think lightning bolt if you must. Leyden jar if you will. Electricity requires an energy source and an energy sink. It does NOT require two conductors. It does NOT require a return path. Those are artifacts of electrical generators and practical application of electricity. Important artifacts that you won't run your AC without - important for our usage of electrical energy not important to the transmission of electrical energy.
Electrical energy transmission actually does not require any conductor...not even one. I've already covered that.

A capacitor requires the presence of both an electric field and magnetic field to store or release energy. Besides, please recall that we were discussing conductors and the energy traveling to the load.

Originally Posted by pfalcon
We always reach equilibrium in a resistive wire.
No we do not. More basic physics. The current in a wire keeps the metal from reaching equilibrium.

Originally Posted by pfalcon
The voltage (potential for work) that pushes the current is balanced by the resistance (frictional work) to create a constant current (velocity). Otherwise the current would go exponential. Or are you proposing that Newton's second law doesn't apply?
I said nothing of the sort. One of my points, and the point made in the paper I referenced, was that the axial component of the electric field pushed the current along against the resistance of the wire. The component of the Poynting Vector that is directed radially into the wire equals the wire losses. That was covered in detail in the paper.
Last edited by mivey; 06-25-12 at 08:56 PM. Reason: spelling

8. Originally Posted by ggunn
I do not agree with current as velocity. Current is like the number of cars that pass a point on a highway per unit of time; it doesn't tell you how fast they are moving. If there are more lanes (lower resistance), they don't need to travel as fast. Or water in a hose; to get the same number of gallons per hour (current) through a smaller hose, the water must move faster.
Yes, ggunn, it's closer to a flow rate than a velocity. Adjust the discussion accordingly.

9. Originally Posted by mivey
Can you provide some reference material? I'm all for exploring new things, but if it is outside standard accepted electrical theory I would like some background material for a frame of reference.

Thanks.
Originally Posted by NIST
http://physics.nist.gov/cuu/Units/units.html
energy, work, quantity of heat joule J N·m m2·kg·s-2
electric charge, quantity of electricity coulomb C - s·A
According to NIST which I assume is a sufficiently reliable source for you, the base SI units for the Joule are Newton-Meters. That's force applied over a distance which is exactly the definition of Work aka Work energy, Energy of work.

When people use work equations they typically stop using the word work as it becomes inconvenient and repetitious. In work equations other forms of energy are cancelled out. Balanced work energy is typically cancelled out. That doesn't mean they cease to exist. It just means they have no impact on that specific equation. Net zero NOT equal zero.

10. Originally Posted by pfalcon
No. You won't find a definition of Joule that doesn't refer to it as work. Work is a limited characteristic of energy referenced to units of time and distance. If you neglect the static forces of a charge then the remaining forces are proportional to work. Which is exactly what most current equations do - they neglect the static forces. Coulomb is the base unit. Joule is the coulomb in motion.

CV = J

(Energy)*(Work potential per Unit of Energy) = Work

Meaning, as you have, when you only deal with Work equations if follows since energy is proportional to work those equations can treat energy and work as though they were identical.
Hey ho...........
A coulomb is a measure of electric charge.
Not energy.

A joule is energy.
One volt times one amp times one second.

It's an impracticably small unit of energy for billing purposes so you pay for electrical energy in kWh as a rule.

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