Ohm's law, why current induces electric field and cause effect direction

In summary: Resistance is not a "mechanical" entity. It is an electrical phenomenon caused by the collisions of electrons.
  • #1
Fernsanz
57
0
Hi,

Lately I've been concerned about the Ohm's law [tex]J=\sigma E[/tex] and the physical interpretation of this law depending on what is considered the cause and what the effect.

More concretely, it is quite natural and intuitive for me the interpretation of this law in one direction: a electric field [tex]E[/tex] over a ohmic conductor will cause a current [tex]J[/tex]. No one would have problems working out that electric force acting upon electrons will cause them to move, i.e. setting up a current. We can certainly say that the electrical field is the cause in this case and the current is the effect, the result.

However, in the reverse direction, namely that a current [tex]J[/tex] injected somehow in a ohmic conductor will cause a electric field [tex]E[/tex], is by far more counterintuitive. Anyhow the current is generated previously by other means (for instance photoelectric or thermoionic effects) and injected into the conductor an electric field will arise. What is the explanation that acounts for that electric field caused just by movement of electrons in a conductor (which is nothing but a cristaline structure "covered" by a cloud of electrons)?

I would gladly read your explanations.

Thanks.
 
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  • #2
It is counter-intuitive indeed, but they are mutually inclusive, neither being the cause nor the effect.

I view it in the following way to see it better. By definition a constant voltage source is one that maintains a constant voltage despite varying load resistance. Let's say a current enters a conductor, the charge carriers in conductors are electrons. They have energy, but lose a portion of that energy when they collide with the lattice structure.

When an electron in the conduction band collides with an electron in the valence band, energy is transferred. So a current is a transfer of energy among electrons, where the actual motion/displacement of an electron is small, but the propagation of energy is near light speed.

The energy loss due to lattice collisions constitutes resistive loss. Thus the conduction electrons carry energy, a portion of which they lose due to resistance i.e. lattice collisions. By definition, the energy lost per unit charge is the voltage drop. Hence current entering a conductor incurs energy loss due to resistance/lattice collisions, resulting in a drop in voltage.

The converse also holds. A conductor with no current is instantly connected across a small voltage. The E field results in charge motion i.e. current.

Ohm's law is bi-lateral, where J & E are inclusive. Which one comes first is a moot question. You are thinking the right way.

Claude
 
  • #3
cabraham said:
The energy loss due to lattice collisions constitutes resistive loss. Thus the conduction electrons carry energy, a portion of which they lose due to resistance i.e. lattice collisions. By definition, the energy lost per unit charge is the voltage drop. Hence current entering a conductor incurs energy loss due to resistance/lattice collisions, resulting in a drop in voltage.
Claude

The energy lost per unit charge is not the voltage drop because that loss is caused by "mechanical" collision among electrons. Reasoning that way one could take an electron, throw it against a wall and conclude that, since the electron has lost all its energy after hitting the wall, there must be an electric field out of the wall and hence a voltage drop.

So the question is still in the air: why an electrical field appear just because electrons are moving? (Don't lose of sight the "electrical" nature of the force I am asking about which is the electrical field that appears in Ohm's low, not any other mechanical forces)
 
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  • #4
Fernsanz said:
The energy lost per unit charge is not the voltage drop because that loss is caused by "mechanical" collision among electrons. Reasoning that way one could take an electron, throw it against a wall and conclude that, since the electron has loss all his energy after hitting the wall, there must be an electric field out of the wall and hence a voltage drop.

So the question is still in the air: why an electrical field appear just because electrons are moving? (Don't lose of sight the "electrical" nature of the force I am asking about which is the electrical field that appears in Ohm's low, not any other mechanical forces)

Mechanical - electrical?! What is the difference? How do you think resistance takes place? If the electron can travel all the way through the conductor w/o colliding w/ the lattice, there is zero resistance, hence zero voltage drop. Superconductors achieve this property.

The "thermal energy" associated w/ an electron is given by 0.5*m*v^2 = k*T, where m is mass, v is velocity, k is Boltzmann's constant, & T is temperature (absolute).

If "resistance" is not a "mechanical" entity, then what is it? I'd like to know. Thanks in advance.

Claude
 
  • #5
cabraham said:
Mechanical - electrical?! What is the difference? How do you think resistance takes place? If the electron can travel all the way through the conductor w/o colliding w/ the lattice, there is zero resistance, hence zero voltage drop. Superconductors achieve this property.

The "thermal energy" associated w/ an electron is given by 0.5*m*v^2 = k*T, where m is mass, v is velocity, k is Boltzmann's constant, & T is temperature (absolute).

If "resistance" is not a "mechanical" entity, then what is it? I'd like to know. Thanks in advance.

Claude

If mechanical and electrical are the same for you good for you, but there is no explanation in your words on how an ELECTRIC field arises as the result of moving charges in that lattice. You have started talking about resistance, which I did not mention. Resistance is what steal energy from electrons. I'm not asking about where the electrons lose their energy, I'm asking about why an ELECTRIC field arises; an electrical field that would be measurable by measuring the force exherted on a test charge placed inside the conductor even if such test charge is not moving!

Please, I'm not expecting simple or naive answer; I expect a physical explanation on why moving charges in a ohmic conductor induce an ELECTRIC field, i.e., a field that would cause a test charge placed inside the conductor to move purely by electric force [tex]F=qE[/tex].
 
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  • #6
Fernsanz said:
If mechanical and electrical are the same for you good for you, but there is no explanation in your words on how an ELECTRICAL field arise as the result of moving charges in that lattice. You have started talking about resistance, which I did not mention. Resistance is what steal energy from electrons. I'm not asking about where the electrons lose their energy, I'm asking about why an ELECTRICAL field arise; an electrical field that would be measurable by measuring the force exherted on a test charge placed inside the conductor even if such test charge is not moving!

Please, I'm not expecting simple or naive answer; I expect a physical explanation on why moving charges in a ohmic conductor induce an ELECTRICAL field, i.e., a field that would cause a test charge placed inside the conductor to move purely by electrical force [tex]F=qE[/tex].

If moving charges collide into a capacitor, it induces an electrical field due to the collapse of the magnetic field. The collision of like-charged particles which stores potential energy in the electric field is what induces the electrical field in a wire. The potential energy results from the increase of the volume-averaged RMS value of the electric field. This is because the volume density of the energy in the electric field varies by (1/2)*(electric constant of permittivity)*E^2. The RMS value of E increases as like charges increase their proximity to one another. So the volume-averaged value of E^2 is greater in that case.
 
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  • #7
kmarinas86 said:
If moving charges collide into a capacitor, it induces an electrical field due to the collapse of the magnetic field. The collision of like-charged particles which stores potential energy in the electric field is what induces the electrical field in a wire. The potential energy results from the increase of the volume-averaged RMS value of the electric field. This is because the volume density of the energy in the electric field varies by (1/2)*(electric constant of permittivity)*E^2. The RMS value of E increases as like charges increase their proximity to one another. So the volume-averaged value of E^2 is greater in that case.

Thanks thanks thanks! That's the type of reasoning I was looking for. I don't see it clear though: if you get electrons closer to each other than they would be in equilibrium, any of those electron will feel a stronger field from its neighbours, but that's true in all direction! so the sum of stronger fields at any point would cancel out, even though it is true that the density of potential energy will be higher.

Or do you mean that it should be viewed more like a compresion-rarefaction process like a sound wave? That is, the first avalanche of electrons moving fast get really close to the front-end of standing-still electrons; then, because of proximity this standing still electrons will move forward getting closer to the next ones and so on. Is this a correct visualization?


Thanks again
 
  • #8
Fernsanz said:
Hi,

Lately I've been concerned about the Ohm's law [tex]J=\sigma E[/tex] and the physical interpretation of this law depending on what is considered the cause and what the effect.

More concretely, it is quite natural and intuitive for me the interpretation of this law in one direction: a electric field [tex]E[/tex] over a ohmic conductor will cause a current [tex]J[/tex]. No one would have problems working out that electric force acting upon electrons will cause them to move, i.e. setting up a current. We can certainly say that the electrical field is the cause in this case and the current is the effect, the result.

However, in the reverse direction, namely that a current [tex]J[/tex] injected somehow in a ohmic conductor will cause a electric field [tex]E[/tex], is by far more counterintuitive. Anyhow the current is generated previously by other means (for instance photoelectric or thermoionic effects) and injected into the conductor an electric field will arise. What is the explanation that acounts for that electric field caused just by movement of electrons in a conductor (which is nothing but a cristaline structure "covered" by a cloud of electrons)?

I would gladly read your explanations.

Thanks.

Maybe I'm missing something here. Can you tell me where in Ohm's law does it state that moving charges in a conductor will can an electric field to be induced, and that this electric field is the "E" in Ohm's law?

Note that I can shoot electrons in vacuum (we do that all the time in a particle accelerator) and after the accelerating structure, it can simply coast without any applied field. So you have current, but no applied field that cause this current, nor does it induced the SAME field in reverse. It will have other fields, such as magnetic field or time varying electric field, but this electric field is NOT the same electric field as referred to in Ohm's Law, i.e. constant axial field in the direction of motion.

Zz.
 
  • #9
ZapperZ said:
Maybe I'm missing something here. Can you tell me where in Ohm's law does it state that moving charges in a conductor will can an electric field to be induced, and that this electric field is the "E" in Ohm's law?

In an Ohmic conductor the Ohm's law says that [tex]J=\sigma E[/tex]. So, according to this constitutive relation whenever a [tex]J[/tex] exists in the conductor you have a electric field [tex]E=\frac{1}{\sigma}J[/tex]. Pretty obvious, isn't it?

Are you suggesting that we can have the current with or without electric field depending on how we generate that current? i.e., are you suggesting that we can't read the Ohm's law in the direction [tex]J[/tex] implies [tex]E[/tex] but we can read it in the direction [tex]E[/tex] implies [tex]J[/tex]? If you are claiming this then we should warn electrical engineers around the world because the relation V=IR is not valid when the current is the first cause (for example when the current is generated in a photodiode); and obviously this is not the case
 
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  • #10
Fernsanz said:
Thanks thanks thanks! That's the type of reasoning I was looking for. I don't see it clear though: if you get electrons closer to each other than they would be in equilibrium, any of those electron will feel a stronger field from its neighbours, but that's true in all direction! so the sum of stronger fields at any point would cancel out, even though it is true that the density of potential energy will be higher.

Or do you mean that it should be viewed more like a compresion-rarefaction process like a sound wave? That is, the first avalanche of electrons moving fast get really close to the front-end of standing-still electrons; then, because of proximity this standing still electrons will move forward getting closer to the next ones and so on. Is this a correct visualization?

Interestingly enough I have come across this post from the thread "Why current attain steady state": https://www.physicsforums.com/showpost.php?p=2778731&postcount=5
It is quite clarifying and supports my previous picture of sound wave to explain the propagation of the electric field.

This also explain that electric field induced when the current is the "first cause" is exactly the same as the one obtained from Ohm's law, because in any case the current has to be the dynamic solution of an existing electric field. So now it's clear for me: the Ohm's law is a true constitutive relation which relates two quantities no matter which one is the "first cause"

Thanks AJ Bentley
 
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  • #11
The answer to the OP question "how is it that if the current is first that it can produce an E field?" is that the injected charge carrriers provide the said E field.

Remember that electrons & holes being charged, possesses an E field of their own. Let's say we have a conductor, a good one but not a superconductor. Let's say it is connected across a current source suddenly. The holes & electrons enter the conductor. Upon arrival, they bang into the lattice resulting in heat dissipation known as resistance. Each charge carrier can approach a limited average velocity owing to collisions.

Hence a charge distribution will be effected. At the end where electrons enter, an accumulation of e- results in an E field since e- possesses such a field. Likewise with holes. h+ at the opposite end. If you prefer, a hole entering is equivalent to an e- exiting. So one end has an accumulation of h+, the other has e-. There is an E field whose lines start on a hole & end on an electron, as well as a potential. The potential V, is just the line integral of E over the path.

In a superconductor, SC, there is no resistance, hence no E field. When e- enter a SC, they incur no collisions, & their tendency is to diffuse apart from one another. When they reach the surface, they cannot continue outward since energy is needed. So a charge accumulation exists on the surface of a SC. Hence, in the absence of interior charge unbalance, there can be no E field on the interior of the SC.

Did this help?

Claude
 
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  • #12
cabraham said:
The answer to the OP question "how is it that if the current is first that it can produce an E field?" is that the injected charge carrriers provide the said E field.

I'm not searching an answer anymore, I have found out by which mechanism an E-field is induced. As I said in my last post is by acelerating charges producing a e-field wave down the conductor.

cabraham said:
Remember that electrons & holes being charged, possesses an E field of their own. Let's say we have a conductor, a good one but not a superconductor. Let's say it is connected across a current source suddenly. The holes & electrons enter the conductor. Upon arrival, they bang into the lattice resulting in heat dissipation known as resistance. Each charge carrier can approach a limited average velocity owing to collisions.

Hence a charge distribution will be effected. At the end where electrons enter, an accumulation of e- results in an E field since e- possesses such a field. Likewise with holes. h+ at the opposite end. If you prefer, a hole entering is equivalent to an e- exiting. So one end has an accumulation of h+, the other has e-. There is an E field whose lines start on a hole & end on an electron, as well as a potential. The potential V, is just the line integral of E over the path.

First, in a metalic conductor like, for instance, copper, there are no holes. So the reasoning has nothing to do with holes.

Second, and once again, the E-field is not produced as a result of charge accumulation. An accumulation of electrons radiates an E-field in all directions, not just in the direction of current, so an accumulation would accelerate electrons which are behind it and would slow down electrons which are before it, i.e. would spread electrons out of the accumulation in all directions. Furthermore you are seeing the current inside the conductor as a succession of charged zones which is not true since the conductor carrying a current is homogeneous

cabraham said:
In a superconductor, SC, there is no resistance, hence no E field. When e- enter a SC, they incur no collisions, & their tendency is to diffuse apart from one another. When they reach the surface, they cannot continue outward since energy is needed. So a charge accumulation exists on the surface of a SC. Hence, in the absence of interior charge unbalance, there can be no E field on the interior of the SC.

Again touching the resistance subject which is not related which my question. And, I insist a last time, you can not explain a directed current on the basis of charge accumulation. Charge acummulation does not push neighbour electrons in a preferred direction, so no net current can be generated (as much electrons will be pushed forward as backwards).

As I said I have found the answer I wrote in my previous post. Thank you anyway.
 
  • #13
cabraham said:
The answer to the OP question "how is it that if the current is first that it can produce an E field?" is that the injected charge carrriers provide the said E field.

I'm not searching an answer anymore, I have found out by which mechanism an E-field is induced. As I said in my last post is by acelerating charges producing a e-field wave down the conductor.

cabraham said:
Remember that electrons & holes being charged, possesses an E field of their own. Let's say we have a conductor, a good one but not a superconductor. Let's say it is connected across a current source suddenly. The holes & electrons enter the conductor. Upon arrival, they bang into the lattice resulting in heat dissipation known as resistance. Each charge carrier can approach a limited average velocity owing to collisions.

Hence a charge distribution will be effected. At the end where electrons enter, an accumulation of e- results in an E field since e- possesses such a field. Likewise with holes. h+ at the opposite end. If you prefer, a hole entering is equivalent to an e- exiting. So one end has an accumulation of h+, the other has e-. There is an E field whose lines start on a hole & end on an electron, as well as a potential. The potential V, is just the line integral of E over the path.

First, in a metalic conductor like, for instance, copper, there are no holes. So the reasoning has nothing to do with holes.

Second, and once again, the E-field is not produced as a result of charge accumulation. An accumulation of electrons radiates an E-field in all directions, not just in the direction of current, so an accumulation would accelerate electrons which are behind it and would slow down electrons which are before it, i.e. would spread electrons out of the accumulation in all directions. Furthermore you are seeing the current inside the conductor as a succession of charged zones which is not true since the conductor carrying a current is homogeneous

cabraham said:
In a superconductor, SC, there is no resistance, hence no E field. When e- enter a SC, they incur no collisions, & their tendency is to diffuse apart from one another. When they reach the surface, they cannot continue outward since energy is needed. So a charge accumulation exists on the surface of a SC. Hence, in the absence of interior charge unbalance, there can be no E field on the interior of the SC.

Again touching the resistance subject which is not related which my question. And, I insist a last time, you can not explain a directed current on the basis of charge accumulation. Charge acummulation does not push neighbour electrons in a preferred direction, so no net current can be generated (as much electrons will be pushed forward as backwards).

As I said I have found the answer I wrote in my previous post. Thank you anyway.
 
  • #14
Fernsanz said:
I'm not searching an answer anymore, I have found out by which mechanism an E-field is induced. As I said in my last post is by acelerating charges producing a e-field wave down the conductor.

First, in a metalic conductor like, for instance, copper, there are no holes. So the reasoning has nothing to do with holes.

Second, and once again, the E-field is not produced as a result of charge accumulation. An accumulation of electrons radiates an E-field in all directions, not just in the direction of current, so an accumulation would accelerate electrons which are behind it and would slow down electrons which are before it, i.e. would spread electrons out of the accumulation in all directions. Furthermore you are seeing the current inside the conductor as a succession of charged zones which is not true since the conductor carrying a current is homogeneous

Again touching the resistance subject which is not related which my question. And, I insist a last time, you can not explain a directed current on the basis of charge accumulation. Charge acummulation does not push neighbour electrons in a preferred direction, so no net current can be generated (as much electrons will be pushed forward as backwards).

As I said I have found the answer I wrote in my previous post. Thank you anyway.

I used "holes" in a loose sense. Also, peer-reviewed e/m fields texts insist that the charge distribution inside the conductor is indeed a source of E field. Where did you read otherwise? Of course a single e- radiates an E field in all directions. But a group of e- at one end, & a void of e- at the other end of a conductor has an E field directed from +ve to the -ve end. There is an E field thast extends outward as well, & it does slow down e- "before it".

Why would peer reviewed uni texts say that if it were not so? What are your references? I never expalined a directed current based on charge accumulation. The charge accumulation accounts for the E field inside the good conductor (not SC). Please don't put words in my mouth.

I try to help you by giving free advice which takes years to acquire, & you rudely rebuke me with straw man arguments, correcting me on issues I never raised. Just out of curiosity, how far along did you get regarding education. Are you a practicing scientist/EE? Just curious. You have a lot of sass & moxie with little chops to back it up. The problem with these forums is that everybody thinks they know more than everyone else. You exemplify that attitude.

Claude
 
  • #15
Fernsanz said:
In an Ohmic conductor the Ohm's law says that [tex]J=\sigma E[/tex]. So, according to this constitutive relation whenever a [tex]J[/tex] exists in the conductor you have a electric field [tex]E=\frac{1}{\sigma}J[/tex]. Pretty obvious, isn't it?

Are you suggesting that we can have the current with or without electric field depending on how we generate that current? i.e., are you suggesting that we can't read the Ohm's law in the direction [tex]J[/tex] implies [tex]E[/tex] but we can read it in the direction [tex]E[/tex] implies [tex]J[/tex]? If you are claiming this then we should warn electrical engineers around the world because the relation V=IR is not valid when the current is the first cause (for example when the current is generated in a photodiode); and obviously this is not the case

In physics, you cannot simply rearrange an equation and then reinterpret it and expect it to make sense. That's why physics isn't mathematics. There is such a thing as the dependent and independent variable! It is based on causality - A causes B, but B may not necessarily causes A.

I've just shown you an example where I could have a current without the use of an external electric field to generate that current. Want another example? Look at the current from a photoemission process. Look ma! No electric field!

I still want to see this experiment where you shoot electrons into a conductor, and this process, in turn, generates the corresponding E field that is similar to the E-field that causes the current in the reverse process. I shoot electrons into metals all the time (it's part of my job) - we call it either a beam stop, or a Faraday Cup. In none of these do you generate the identical E-field in the conductor that is described by Ohm's Law.

Zz.
 
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  • #16
ZapperZ said:
Maybe I'm missing something here. Can you tell me where in Ohm's law does it state that moving charges in a conductor will can an electric field to be induced, and that this electric field is the "E" in Ohm's law?

He can't. It cannot be done. Ohm's law considers no such things. The "E" in Ohm's law is not the "E" of the real world. It is fictitious. It does not even necessarily equal the projection of the "E" over a differential length of the conductor. That only works if the electric field inside the wire is uniform and if the electric field outside the wire approaches each wire element at the same angle. Only in very a contrived circumstance with a bizarrely shaped electric field could you meet those conditions in a curved wire element. If the real "E" varies over the length of the conductor, speaking of the "E" in Ohm's law is like talking about the tooth fairy.

ZapperZ said:
Note that I can shoot electrons in vacuum (we do that all the time in a particle accelerator) and after the accelerating structure, it can simply coast without any applied field. So you have current, but no applied field that cause this current, nor does it induced the SAME field in reverse. It will have other fields, such as magnetic field or time varying electric field, but this electric field is NOT the same electric field as referred to in Ohm's Law, i.e. constant axial field in the direction of motion.

And because an electric field (changing or not) is needed to generate a current, the "E" of Ohm's law is not the electric field.

ZapperZ said:
In physics, you cannot simply rearrange an equation and then reinterpret it and expect it to make sense. That's why physics isn't mathematics. There is such a thing as the dependent and independent variable! It is based on causality - A causes B, but B may not necessarily causes A.

Some people allege that a changing electric field does not "generate" a magnetic field and that it's just a correlation. Many of those same people allege the opposite to be the case as well.

Ohm's "law" is very quaint. For example, [tex]\mathbf{J}=\sigma\mathbf{E}[/tex] applies only for a uniform field in the direction of the current. What kind of "law" is that?

There is no "right-hand law" in physics. Left-handed materials exist. We have something called "the right-hand rule". Ohm's "law" should be called "Ohm's rule". I did a Google search, and it turns out many concur.

ZapperZ said:
I've just shown you an example where I could have a current without the use of an external electric field to generate that current.

Want to give an example of a situation without an electric field that can generate a current? You could try gravity, but you have to assume it is not fundamentally due to electromagnetism. You could also hook up a battery to a remote control. One could argue that it is external to the conductor, but it is part of the same circuit, so it still works because its field is "internal" to that circuit just as the current it produces will be.

ZapperZ said:
Want another example? Look at the current from a photoemission process. Look ma! No electric field!

The photon has an electric field. It is changing, but so what? It is still an electric field.

ZapperZ said:
I still want to see this experiment where you shoot electrons into a conductor, and this process, in turn, generates the corresponding E field that is similar to the E-field that causes the current in the reverse process. I shoot electrons into metals all the time (it's part of my job) - we call it either a beam stop, or a Faraday Cup. In none of these do you generate the identical E-field in the conductor that is described by Ohm's Law.

Zz.

You're right, it's not identical.
 
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  • #17
kmarinas86 said:
[tex]\mathbf{J}=\sigma\mathbf{E}[/tex] applies only for a uniform field in the direction of the current.

Neither the field nor the current density need to be uniform. If the conductor is isotropic (conducts equally well in all directions), then at each point the current density and the electric field have the same direction, but possibly different directions at different points.

If the conductor is anisotropic, the current density and electric field can be in different directions at a given point. In this case you have to use a tensor for the conductivity.
 
  • #18
ZapperZ said:
In physics, you cannot simply rearrange an equation and then reinterpret it and expect it to make sense. That's why physics isn't mathematics. There is such a thing as the dependent and independent variable! It is based on causality - A causes B, but B may not necessarily causes A.

I think this is one of the best quotes I have ever seen on this forum. I encourage everybody reading this thread to really let ZapperZ's quote sink in. It is directly relevant to the topic at hand as well as to many of the misconceptions that commonly occur to students (both formal and informal) of physics.

Quite often, equations will be written in an "aesthetic" form. The left hand side (LHS) may sometimes be the cause, other times it may be the effect, yet other times the equation will simply be a mutual relation.

In the present case of Ohm's law,
[tex]J=\sigma E[/tex],
J is the effect and E is the cause. I suppose this makes perfect sense writing the equation this way because it places the dependent variable on the LHS and the independent variable on the RHS, i.e. it is in the from y(x)=mx.

Not to get off-topic, but I believe a few examples might be helpful to some...

Newton's second law: F=ma
F is the cause and a is the effect; forces cause bodies to accelerate, not vice versa. Why is it commonly written in the opposite form as Ohm's law? Easier to remember, looks nicer... who cares?

The most famous equation in all of physics:
[tex]E=mc^2[/tex]
This is simply, what I called above, a mutual relation. Rest energy is mass, mass is rest energy.
 
  • #19
cabraham said:
I used "holes" in a loose sense. Also, peer-reviewed e/m fields texts insist that the charge distribution inside the conductor is indeed a source of E field. Where did you read otherwise?

Of course that charge distribution is a source of E field, nobody is telling otherwise. But it can not be the source of net current in a preferred direction.

cabraham said:
Of course a single e- radiates an E field in all directions. But a group of e- at one end, & a void of e- at the other end of a conductor has an E field directed from +ve to the -ve end. There is an E field thast extends outward as well, & it does slow down e- "before it".

So, are you suggesting that the two ends of the wire act as a capacitor where electrons accumlate at one end and fault at the other end and that this situation is what causes the whole electric field?

cabraham said:
Why would peer reviewed uni texts say that if it were not so? What are your references? I never expalined a directed current based on charge accumulation. The charge accumulation accounts for the E field inside the good conductor (not SC). Please don't put words in my mouth.

Most engineering books give too much simplified views of physicis situations, they are not intended to give a deep insight of physics and so there is no reason to think they are totally right when one tries to go beyond "little balls called electrons" and consider "classical electrodynamics". So, raising the authority principle as a valid argument I think is not the type or reasoning a scientist should accept. Anyway, one of my references is Jackson's "Classical Electrodynamics"

It was no my intention to put words on your mouth.

cabraham said:
I try to help you by giving free advice which takes years to acquire, & you rudely rebuke me with straw man arguments, correcting me on issues I never raised. Just out of curiosity, how far along did you get regarding education. Are you a practicing scientist/EE? Just curious. You have a lot of sass & moxie with little chops to back it up.

I'm a physicist and telecomunication engineer (something like a mix of Electrical Engineering and Computer Engineering in Spain). So I think I have gone through all those years to acquire the knowledge. Yet I have questions no matter how many years I have studied; don't you?


cabraham said:
The problem with these forums is that everybody thinks they know more than everyone else. You exemplify that attitude.

I agree with both quotes. Literally.
 
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  • #20
ZapperZ said:
In physics, you cannot simply rearrange an equation and then reinterpret it and expect it to make sense. That's why physics isn't mathematics. There is such a thing as the dependent and independent variable! It is based on causality - A causes B, but B may not necessarily causes A..

If the equation is a constitutive relation you can rearrange the equation cause it express a relation of neccessity for that medium. For example in vacuum, [tex]B=\mu_0 H[/tex], so no matter which one do you consider the cause if you have one you can obtain the other with this constitutive relation

ZapperZ said:
I've just shown you an example where I could have a current without the use of an external electric field to generate that current. Want another example? Look at the current from a photoemission process. Look ma! No electric field!

It seems you don't read my posts. First, it was precisely me who mention the photoemission process as a differente way of obtaining current. But the big big mistake you are making is that none of your examples refer to an OHMIC medium. I stick to, and my question is about, ohmic media. So yes, I want just a single example where you have a current in a ohmic medium without E-field.

ZapperZ said:
I still want to see this experiment where you shoot electrons into a conductor, and this process, in turn, generates the corresponding E field that is similar to the E-field that causes the current in the reverse process.

Once again you haven't read my post where I provided the experiment (so simple and usual) you are demanding. I will repeat it: take a circuit composed of a photodiode (current source which shoot [tex]I[/tex] amps) conected to a resistor (the ohmic conductor). As you yourself have pointed, the generation of the current in the photodiode requires no electric field; however, once that this current cross the resistor you have an electric field (or a voltage [tex]V[/tex], as you prefer). So, from the current generated with no electric field you end up having a electric field. And I bet you wil apply Ohm's law V=IR to know the voltage across the resistor, right? So this is a clear proof that the Ohm's law is valid no matter which magnitude is the cause.

It is so funny the picture of an electrical engineer analyzing a circuit with a current source and thinking "wait wait, I have to take into account how this current is generated, because if it is being generated in a way that does not require an electric field and then pumped into my circuit then I can not apply Ohm's law V=IR in my circuit because the "first cause", "the driving force", is I and not V. I will have to find out myself the relation between V and I for the resistor in this case... I better change my job", hahaha. So ridiculous that should be enough for you to think twice what you are claiming.

cmos said:
In the present case of Ohm's law,
[tex]J=\sigma E[/tex],
J is the effect and E is the cause.

Wrong. It can be read the other way around also. Just read the reasoning above

cmos said:
Newton's second law: F=ma
F is the cause and a is the effect; forces cause bodies to accelerate, not vice versa. Why is it commonly written in the opposite form as Ohm's law? Easier to remember, looks nicer... who cares?

Again, no matter which one is the cause I can assure that if a mass m is moving with an acceleration a I can solve for F and the other way around.
 
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  • #21
cmos said:
I think this is one of the best quotes I have ever seen on this forum. I encourage everybody reading this thread to really let ZapperZ's quote sink in. It is directly relevant to the topic at hand as well as to many of the misconceptions that commonly occur to students (both formal and informal) of physics.

Quite often, equations will be written in an "aesthetic" form. The left hand side (LHS) may sometimes be the cause, other times it may be the effect, yet other times the equation will simply be a mutual relation.

In the present case of Ohm's law,
[tex]J=\sigma E[/tex],
J is the effect and E is the cause. I suppose this makes perfect sense writing the equation this way because it places the dependent variable on the LHS and the independent variable on the RHS, i.e. it is in the from y(x)=mx.

Not to get off-topic, but I believe a few examples might be helpful to some...

Newton's second law: F=ma
F is the cause and a is the effect; forces cause bodies to accelerate, not vice versa. Why is it commonly written in the opposite form as Ohm's law? Easier to remember, looks nicer... who cares?

The most famous equation in all of physics:
[tex]E=mc^2[/tex]
This is simply, what I called above, a mutual relation. Rest energy is mass, mass is rest energy.

Neither J nor E is the "cause" or "effect". Either one can take place first. An inductor energized, terminated in a perfect short has a current w/ zero voltage. A resistor is placed across the short. Then the short is opened via a switch, and the inductor de-energizes into the resistor.

The current pre-existed the voltage. When the SC was removed, the initial voltage across the resistor is determined by I*R. This is a clear example where I is the independent variable, & V is dependent.

Of course, capacitors tend to behave as voltage sources, the counterpart of the inductor, which behaves like a cuurent source. A charged cap has a V, w/ zero I. A switch is closed, & V is connected across a resistor. In this case, I = V/R. The initial current is determined by the voltage & resistance, V being the independent variable.

It works both ways. Which is dependent & independent variable can be either. But real world power sources like generators & batteries are constructed optimized & operated in the constant voltage mode. Thus the mindset that "V is independent, I is dependent variable" has taken root. It is nothing more than a prejudice having no basis in science.

Ohm's law can be written 3 ways, neither more correct than the other. Which form works best is application dependent.

Force & acceleration are mutually dependent. Physicists have repeated this since I can remember. In modern physics, F = m*a is discarded in favor of

F = dP/dt. Force & acceleration are not cause/effect related. For either to be a cause, it must exist independently of the other. Neither can do that. Ask any physicist.

Claude
 
  • #22
Fernsanz said:
Of course that charge distribution is a source of E field, nobody is telling otherwise. But it can not be the source of net current in a preferred direction.



So, are you suggesting that the two ends of the wire act as a capacitor where electrons accumlate at one end and fault at the other end and that this situation is what causes the whole electric field?



Most engineering books give too much simplified views of physicis situations, they are not intended to give a deep insight of physics and so there is no reason to think they are totally right when one tries to go beyond "little balls called electrons" and consider "classical electrodynamics". So, raising the authority principle as a valid argument I think is not the type or reasoning a scientist should accept. Anyway, one of my references is Jackson's "Classical Electrodynamics"

It was no my intention to put words on your mouth.



I'm a physicist and telecomunication engineer (something like a mix of Electrical Engineering and Computer Engineering in Spain). So I think I have gone through all those years to acquire the knowledge. Yet I have questions no matter how many years I have studied; don't you?




I agree with both quotes. Literally.

The accumulated charge only was brought up to account for the small E field in the conductor. No, this charge accumulation is not "driving" the network current. You asked if current entering a conductor can give rise to an E field, & why so. I answered. The current entering the conductor is part of a larger network driven by an independent power source, i.e. generator, battery, photodiode, etc. The energy conversion in the prime mover produces the current & associated E field.

The charge density in the conductor accounts for the internal E field, but does not motivate the network current.

As far as my peer-reviewed textbooks not being sophisticated enough, what texts do you suggest I read? Even solid state physicists refer to resistance as the result of collisions. This is their position in conductors as well as semiconductors. How do you account for heating loss due to resistance if collisions are not the right answer? Many would love to know.

I realize that undergrad EE/physics texts cannot go as deep into the sub-atomic world as needed, but they do not preach heresy. They gloss over things & give simplified models, but they don't outright propagate untruths. Again, if resistance is not mechanical, owing to collisions, please enlighten us as to what it really is.

You claim that I exemplify the "know it all attitude", but I rely on the work of countless others for my claims. The person with the know-it-all attitude is the one who defies peer-reviewed texts & journals, stating "here's how it really is", w/ no proof to back it up. Just show me references where resistance is not mechanical, heat is not the result of collisions, & I will be grateful, as will many others.

Nothing personal at all. You just don't seem comfortable with my explanation, but I researched this topic carefully to assure that I do not post heresy. Please clarify. Thanks in advance.

Claude
 
  • #23
Fernsanz said:
If the equation is a constitutive relation you can rearrange the equation cause it express a relation of neccessity for that medium. For example in vacuum, [tex]B=\mu_0 H[/tex], so no matter which one do you consider the cause if you have one you can obtain the other with this constitutive relation

...

Again, no matter which one is the cause I can assure that if a mass m is moving with an acceleration a I can solve for F and the other way around.

Fernsanz, you're both completely missing the point and hitting it right on the spot. Of course, if you can mathematically manipulate any equation to solve for whatever quantity you want, then it must be true. But this doesn't change the physical interpretation of which is cause, which is effect, and which is neither.
 
  • #24
cabraham said:
Neither J nor E is the "cause" or "effect". Either one can take place first. An inductor energized, terminated in a perfect short has a current w/ zero voltage. A resistor is placed across the short. Then the short is opened via a switch, and the inductor de-energizes into the resistor.

The setup you speak of requires that the inductor is connected to a DC voltage source. The "termination by a perfect short" implies that the inductor is placed across the source. The voltage source drives the current through the inductor.

cabraham said:
The current pre-existed the voltage. When the SC was removed, the initial voltage across the resistor is determined by I*R. This is a clear example where I is the independent variable, & V is dependent.

See above.

cabraham said:
It works both ways. Which is dependent & independent variable can be either. But real world power sources like generators & batteries are constructed optimized & operated in the constant voltage mode. Thus the mindset that "V is independent, I is dependent variable" has taken root. It is nothing more than a prejudice having no basis in science.

How then would you make a constant current source? The only way to do it is to do it the way actual current sources are made, by putting a feedback mechanism into a variable voltage source.

cabraham said:
Ohm's law can be written 3 ways, neither more correct than the other. Which form works best is application dependent.

Yes, equations can be inverted to solve for whatever you want. That is what makes mathematics so useful, especially for investigating physical phenomena. But the physical interpretations do no change with how you manipulate an equation.


At the risk of going off topic...
cabraham said:
Force & acceleration are mutually dependent.
Yes, since forces cause bodies to accelerate.

cabraham said:
In modern physics, F = m*a is discarded in favor of
F = dP/dt.

If by modern you mean since 1687. The quantity m*a is simply a special case of dp/dt; the latter being the most general case. So in general, forces cause bodies to change their momentum (Newton's 1st and 2nd Laws).

cabraham said:
Force & acceleration are not cause/effect related. For either to be a cause, it must exist independently of the other. Neither can do that.

You're logic is completely flawed, if force and acceleration were independent of each other, then neither could be a cause of the other.

cabraham said:
Ask any physicist.

I should give you that advice, but it seems you would rather argue with them with a closed-mind and erroneous claims.
 
  • #25
jtbell said:
Neither the field nor the current density need to be uniform. If the conductor is isotropic (conducts equally well in all directions), then at each point the current density and the electric field have the same direction, but possibly different directions at different points.

If the conductor is anisotropic, the current density and electric field can be in different directions at a given point. In this case you have to use a tensor for the conductivity.

Which E are you talking about?

Read what ZapperZ said to Fernsanz:

ZapperZ said:
Can you tell me where in Ohm's law does it state that moving charges in a conductor will can an electric field to be induced, and that this electric field is the "E" in Ohm's law?

Zz.

ZapperZ's implied answer is "no". You can read the rest of what said. He was testing "Fernsanz".

So what is the answer to my question?

Signed,
Kmarinas86
 
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  • #26
cmos said:
cabraham said:
Force & acceleration are mutually dependent. Physicists have repeated this since I can remember. In modern physics, F = m*a is discarded in favor of F = dP/dt. Force & acceleration are not cause/effect related. For either to be a cause, it must exist independently of the other. Neither can do that.

You're logic is completely flawed, if force and acceleration were independent of each other, then neither could be a cause of the other.

Of FORCE and ACCELERATION:
1. Claude Abraham says they are mutually dependent.
2. Claude Abraham says they are not cause/effect related.
3. Claude Abraham says they cannot exist independently.

IMPLICATIONS:
1. Claude Abraham does not believe acceleration can be observed without force.
2. Claude Abraham does not believe force can be observed without acceleration.
3. Claude Abraham believes that no acceleration means no force.

ONLY ONE PROBLEM FOR CLAUDE ABRAHAM:
1. This only works if you equate "force" as the "vector sum of all forces" on a given object.

SEVERAL PROBLEMS FOR YOU:
1. I quote the following from you: "You're logic is completely flawed, if force and acceleration were independent of each other, then neither could be a cause of the other."
2. "You're logic" is wrong.
3. Claude Abraham said, "Force & acceleration are mutually dependent."
4. Claude Abraham said, "Force & acceleration are not cause/effect related. For either to be a cause, it must exist independently of the other."
5. Claude Abraham did not say, "Force and acceleration are independent of each other."
6. You did not prove that mutual dependence of events implies causation.
7. You did not understand that Claude Abraham used the term "mutual dependence" to make a point about mutual existence (i.e. one does not exist without the other), not causality.
8. You say, "If force and acceleration were independent of each other, then neither could be a cause of the other." You imply that this proves his logic is wrong. However, if (A), then (B), then Claude Abraham says of this quote (A="force and acceleration were independent of each other"=false) (B="neither could be a cause of the other"=null). The word "could" in (B) is a direct reference to "force and acceleration...independent of each other" in (A). Per points (3.) and (5.), Claude Abraham must find your statement irrelevant to his arguments. (B) is (null) rather than false because in order to evaluate it requires knowledge of (A). This says to Claude Abraham: (B) requires (A) to be meaningful. If (A) is false, then (B) can only be known if (A), which is false, is known. That's why his position on (B) is "null". In consideration of all points above as what they really mean, his conclusion does not violate your proposition.
9. You were corrected by 23 year-old business student who does not major in science or liberal arts.

Signed,
Kmarinas86
 
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  • #27
cabraham said:
Neither J nor E is the "cause" or "effect". Either one can take place first. An inductor energized, terminated in a perfect short has a current w/ zero voltage. A resistor is placed across the short. Then the short is opened via a switch, and the inductor de-energizes into the resistor.

The current pre-existed the voltage. When the SC was removed, the initial voltage across the resistor is determined by I*R. This is a clear example where I is the independent variable, & V is dependent.

Of course, capacitors tend to behave as voltage sources, the counterpart of the inductor, which behaves like a cuurent source. A charged cap has a V, w/ zero I. A switch is closed, & V is connected across a resistor. In this case, I = V/R. The initial current is determined by the voltage & resistance, V being the independent variable.

It works both ways. Which is dependent & independent variable can be either. But real world power sources like generators & batteries are constructed optimized & operated in the constant voltage mode. Thus the mindset that "V is independent, I is dependent variable" has taken root. It is nothing more than a prejudice having no basis in science.

Ohm's law can be written 3 ways, neither more correct than the other. Which form works best is application dependent.

Force & acceleration are mutually dependent. Physicists have repeated this since I can remember. In modern physics, F = m*a is discarded in favor of

F = dP/dt. Force & acceleration are not cause/effect related. For either to be a cause, it must exist independently of the other. Neither can do that. Ask any physicist.

Claude

This is a rather strange post because it appears to be self-inconsistent.

First of all, as a condensed matter physicist, I could derive Ohm's Law using statistical mechanics using nothing more than the model for free electron gas. With NO applied E-field, you get no net drift velocity in the electron gas, thus, no net current. It is only UPON the action of such a field do you get, within the Drude model, such a current, resulting in the ever-familiar Ohm's law. Please see Chapter 1 in Ashcroft and Mermin's Solid State text.

Now, all you have to do to falsify this is to show that there is some spontaneous current that does not depend on the external field, and that upon the generation of such a current, the SAME E-field as above is the result. If you are successful, I can use it as our new accelerating mechanism at our particle accelerator.

Till then, all I have is E-field causes the current in Ohm's Law.

Zz.
 
  • #28
Fernsanz said:
If the equation is a constitutive relation you can rearrange the equation cause it express a relation of neccessity for that medium. For example in vacuum, [tex]B=\mu_0 H[/tex], so no matter which one do you consider the cause if you have one you can obtain the other with this constitutive relation

Again, you are considering only the MATHEMATICAL aspect of the equation. Mathematically, you can rearrange it any way you like. But it doesn't mean that you can switch around the interpretation. The "interpretation" part, which gives it meaning, depends on the PHYSICS.

It seems you don't read my posts. First, it was precisely me who mention the photoemission process as a differente way of obtaining current. But the big big mistake you are making is that none of your examples refer to an OHMIC medium. I stick to, and my question is about, ohmic media. So yes, I want just a single example where you have a current in a ohmic medium without E-field.
Once again you haven't read my post where I provided the experiment (so simple and usual) you are demanding. I will repeat it: take a circuit composed of a photodiode (current source which shoot [tex]I[/tex] amps) conected to a resistor (the ohmic conductor). As you yourself have pointed, the generation of the current in the photodiode requires no electric field; however, once that this current cross the resistor you have an electric field (or a voltage [tex]V[/tex], as you prefer). So, from the current generated with no electric field you end up having a electric field. And I bet you wil apply Ohm's law V=IR to know the voltage across the resistor, right? So this is a clear proof that the Ohm's law is valid no matter which magnitude is the cause.

It is so funny the picture of an electrical engineer analyzing a circuit with a current source and thinking "wait wait, I have to take into account how this current is generated, because if it is being generated in a way that does not require an electric field and then pumped into my circuit then I can not apply Ohm's law V=IR in my circuit because the "first cause", "the driving force", is I and not V. I will have to find out myself the relation between V and I for the resistor in this case... I better change my job", hahaha. So ridiculous that should be enough for you to think twice what you are claiming.

I believe cmos has responded to your scenario. Furthermore, the current generated in the photodiode DOES require a field. It is why you need a PN junction inside one of those things. And you should look up why one needs such a junction (read about the depletion zone). The phenomena of a photodiode is NOT the same as a typical photoemission process.

The microscopic aspect of Ohm's Law came out of the Drude Model. Could you please start with that and show me where you can get a current that, in turn, causes the identical E-field to be generated? The relevant reference will be any Solid State text, such as Ashcroft and Mermin or Kittel.

Zz.
 
  • #29
cabraham said:
Neither J nor E is the "cause" or "effect". Either one can take place first.

Did the chicken cause an egg and an egg cause a chicken? What about the chicken's diet and the diet of its young? What about their requirements to exist?

The cause of J is the sum and derivatives of all externally-originated fields extending through that point. The cause of E is the sum and derivatives of all externally-originated fields extending through that point.

The cause of an E may be an external E and/or an external B.

The cause of an B may be an external E and/or an external B.

And so forth...

Observers beware,
Causes are everywhere,
But their effects are local,
So put on your bifocals.

cabraham said:
An inductor energized, terminated in a perfect short has a current w/ zero voltage. A resistor is placed across the short.

Inductive voltage drop = Inductance * change of current / change in time

cabraham said:
Then the short is opened via a switch, and the inductor de-energizes into the resistor.

Origin of I (after short):
Inductive voltage drop < 0 volts

cabraham said:
The current pre-existed the voltage.

Origin of current (before short):
Inductive voltage drop > 0 volts

cabraham said:
When the SC was removed, the initial voltage across the resistor is determined by I*R. This is a clear example where I is the independent variable, & V is dependent.

Origin of current (before short):
Inductive voltage drop > 0 volts

Origin of I (after short):
Inductive voltage drop < 0 volts

V = Inductive voltage drop

cabraham said:
Of course, capacitors tend to behave as voltage sources,

So does shorting an inductor.

Inductors and capacitors are storage devices. They can dissipate energy in the form of electricity, which means they are capable of producing voltage and current (as well as V/m and A/m fields).

cabraham said:
the counterpart of the inductor, which behaves like a cuurent source. A charged cap has a V, w/ zero I. A switch is closed, & V is connected across a resistor. In this case, I = V/R. The initial current is determined by the voltage & resistance, V being the independent variable.

It works both ways. Which is dependent & independent variable can be either.

Well of course they can.

cabraham said:
But real world power sources like generators & batteries are constructed optimized & operated in the constant voltage mode. Thus the mindset that "V is independent, I is dependent variable" has taken root. It is nothing more than a prejudice having no basis in science.

True that.

cabraham said:
Ohm's law can be written 3 ways, neither more correct than the other. Which form works best is application dependent.

No kidding, isn't that what they teach in high school?

I have not looked yet, but I would not be surprised if anyone here disagrees with this.

Someone might correct you by saying there are more than 3 ways to write Ohm's law *gasp - omgish!*. It utterly pointless to argue with people who respond like that except to embarrass them by replying with common sense.

Next...

cabraham said:
Force & acceleration are mutually dependent. Physicists have repeated this since I can remember. In modern physics, F = m*a is discarded in favor of F = dP/dt.

True enough.

cabraham said:
Force & acceleration are not cause/effect related.

If you consider the time delay of propagation of fields at the speed of light and their dissipation with distance, this is not true at all.

Does the Sun's light cause the Earth to be exposed to radiation (relatively non-lethal doses at that)? Of course. Does space weather (not limited to solar weather) transfer force to Earth weather? Of course! Is the reverse true? Many orders of magnitude less!

cabraham said:
For either to be a cause, it must exist independently of the other. Neither can do that. Ask any physicist.

You mean:

For one to be a cause at the exclusion of the other, it must exist independently of the other.

A force is "applied" at the same point as the acceleration of a thing it causes. Does this mean they are "dependent"?

kmarinas86 (beginning of this post) said:
Observers beware,
Causes are everywhere,
But their effects are local,
So put on your bifocals.

cabraham said:
Claude
 
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  • #30
ZapperZ said:
This is a rather strange post because it appears to be self-inconsistent.

First of all, as a condensed matter physicist, I could derive Ohm's Law using statistical mechanics using nothing more than the model for free electron gas. With NO applied E-field, you get no net drift velocity in the electron gas, thus, no net current. It is only UPON the action of such a field do you get, within the Drude model, such a current, resulting in the ever-familiar Ohm's law. Please see Chapter 1 in Ashcroft and Mermin's Solid State text.

Inductors and capacitors are both sources of electrical power.

Disagreement exists when one alleges that inductors are without voltage and capacitors are without current.

Without voltage, what is a V/m field?

Without current, what is an A/m field?

Without both, how do you obtain electrical power?

What does one mean when saying "power without E-field". They could mean discharging an inductor. But an inductor can have a voltage when the current is changing! It is called an inductive voltage drop (or a gain if we are talking about coils in an electrical generator). A charged particle moving inertially dissipates no power. Therefore specific kinetic energy of a particle must change for power to be a relevant factor.

ZapperZ said:
Now, all you have to do to falsify this is to show that there is some spontaneous current that does not depend on the external field, and that upon the generation of such a current, the SAME E-field as above is the result. If you are successful, I can use it as our new accelerating mechanism at our particle accelerator.

Till then, all I have is E-field causes the current in Ohm's Law.

Zz.

What is an external field? It is any field external to what you got. If you are measuring the field at a point in space, external fields are fields which exist outside that point. How many points do you think space has? If you are measuring the field in a conductor, then your "external" field can include the battery connected across that conductor. Your external field can be "internal" in your circuit and "external" to the part of the circuit you are measuring.
 
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  • #31
ZapperZ said:
Furthermore, the current generated in the photodiode DOES require a field. It is why you need a PN junction inside one of those things. And you should look up why one needs such a junction (read about the depletion zone). The phenomena of a photodiode is NOT the same as a typical photoemission process.

You have reached a point where you say absolute no senses: the deplection zone is what causes the E-field that generates the current? Oh my God, that's too much. The electric field in the deplection zone opposses to the current! The deplection zone is only a result of equilibrium between diffussion and conduction, it has nothing to do with a net current. Have you ever seen a diode giving current on its own? lol.

So, according to you there is no way of generating a current without an electric field? It is funny how you are getting inconsistent because some post ago you named photoemission as a method to generate current without E-field. As for photoelectric effect, I would like to see how you try to explain it in terms of electrical field, lol.

Hence, being obvious that we can generate a current without electrical field (for instance by mean of photoemission or photoelectric effect, thermoionic effect, SC, etc, and if you claim otherwise it would be so ridiculous that I would leave the thread), take one of this methods and pump that current into a resistor. Every single people on Earth will use the Ohm's law V=IR to obtain the voltage across that resistor.

ZapperZ said:
The microscopic aspect of Ohm's Law came out of the Drude Model. Could you please start with that and show me where you can get a current that, in turn, causes the identical E-field to be generated? The relevant reference will be any Solid State text, such as Ashcroft and Mermin or Kittel.

Could you please start with telling me if there is any case in which you wouldn't use the Ohm's law V=IR when a current cross a resistor?

Could you please start with telling me if a person analyzing a circuit which has a current source has to ask about the nature of the current in order to know if it he/she can apply the Ohm's law V=IR?

Could you please start with telling me why my solar panels give a voltage across my house lamps?

Up to now cabraham has been the only person who has taken for granted what every electrical engineer knows (even by pure experience): the Ohm's law is always applicable no matter which is the cause. From there he has tried to provide an explanation on why a current implies an E-field. He and me can agree or disagree in the explanation but he is the only one trying to answer the OP question. The rest of you are more worried trying to pretend that you have answer for everything and clear ideas and, obviously, is not the case. Anyway, as I have said, I found the explanation in another thread in this forum, and that thread support my ideas.
 
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  • #32
Fernsanz said:
You have reached a point where you say absolute no senses: the deplection zone is what causes the E-field that generates the current? Oh my God, that's too much. The electric field in the deplection zone opposses to the current! The deplection zone is only a result of equilibrium between diffussion and conduction, it has nothing to do with a net current. Have you ever seen a diode giving current on its own? lol.

No, I didn't say anything that you said here. I was pointing out that such a device is NOT devoid of any E-field, contrary to what you were claiming! Read what you wrote and read the particular point that I was responding to.

So, according to you there is no way of generating a current without an electric field? It is funny how you are getting inconsistent because some post ago you named photoemission as a method to generate current without E-field. As for photoelectric effect, I would like to see how you try to explain it in terms of electrical field, lol.

Now it is my turn to tell you that you haven't been reading my post.

Note that I emphasized that the E-field that I was referring to is the identical E-field as described within Ohm's Law, NOT any other E-field. I already described many situations where I can have current without any external E-field. But these situations are not the same as the Ohmic E-field that is in question.

Could you please start with telling me if there is any case in which you wouldn't use the Ohm's law V=IR when a current cross a resistor?

Yes, when the resistor becomes a superconductor, or when the resistor is a semiconductor.

Could you please start with telling me if a person analyzing a circuit which has a current source has to ask about the nature of the current in order to know if it he/she can apply the Ohm's law V=IR?

No. And why is this relevant? The nature of a current source doesn't negate the fact that a component must also have an applied potential across its terminals. I work with constant current power supply all the time. There's nothing here to indicate that these things only pump current into a circuit, and the E-field comes later.

Up to now cabraham has been the only person who has taken for granted what every electrical engineer knows (even by pure experience): the Ohm's law is always appliable no matter which is the cause. From there he has tried to provide an explanation on why a current implies an E-field. He and me can agree or disagree in the explanation but he is the only one trying to answer the OP question. The rest of you are more worried trying to pretend that you have answer for everything and clear ideas and, obviously, is not the case. Anyway, as I have said, I found the explanation in another thread in this forum, and that thread support my ideas.

Ohm's law is valid under the majority of cases. This is because the material that we use for most circuits are metals that obey the Drude model under the majority of the situation that they are used, i.e. the APPROXIMATION that the conduction electrons can be modeled as free electron gas. This explicitly described in Chapter 1 of Ashcroft and Mermin. However, go a bit further and look at Chapter 3, and you'll see a whole chapter titled "Failure of the Free Electron Model"! There are materials and situations in which the Drude Model, i.e. the foundation for Ohm's Law, no longer works! This is not my opinion, it is a FACT!

I still don't understand why we are debating if Ohm's Law works or not. I don't ever recall in my original post of contesting the validity of Ohm's Law. I did contest the validity of your interpretation, i.e. that any kind of interpretation is valid based on mathematics. Since you claim that you can reverse the "order" of the cause-and-effect in this, I asked for you to go back to the http://people.seas.harvard.edu/~jones/es154/lectures/lecture_2/drude_model/drude_model_cc/drude_model_cc.html" [Broken], and show where, at the microscopic level, that you can get the electric field (the identical field in Ohm's Law) as a consequence of the drift velocity of the electrons. To me, this is the only way to show a convincing argument of what you are claiming.

Zz.
 
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  • #33
kmarinas86 said:
What does one mean when saying "power without E-field". They could mean discharging an inductor. But an inductor can have a voltage when the current is changing! It is called an inductive voltage drop (or a gain if we are talking about coils in an electrical generator). A charged particle moving inertially dissipates no power. Therefore specific kinetic energy of a particle must change for power to be a relevant factor.

That is certainly true. An inductor can have its own potential difference.

What is an external field? It is any field external to what you got. If you are measuring the field at a point in space, external fields are fields which exist outside that point. How many points do you think space has? If you are measuring the field in a conductor, then your "external" field can include the battery connected across that conductor. Your external field can be "internal" in your circuit and "external" to the part of the circuit you are measuring.

I don't know if I mentioned the phrase "external field". Since I was talking within the context of the Drude model, then I certainly was using the E-field within that scenario. It certainly doesn't mean external to the conductor or the circuit.

Zz.
 
  • #34
It has become clear to me that you are unable to understand the essence, neither the statement of the problem.

FACTS:
  1. A current can be generated without electric field
  2. A current crossing a resistor (ohmic resistor) will induce a voltage, i.e., an E-field

EXPERIMENT:
  1. Take a current I generated without electric field (possible by fact 1) and connect it to a resistor (ohmic resistor)
  2. Measure the voltage V across the resistor (it will happen by fact 2)

CONCLUSIONS:
  1. A current generated without electric field will induce an electric field i.e. a voltage acrross the resistor
  2. The value of that voltage satisfies the Ohm's law: V=IR

QUESTION:
Where does that elecric field arise from? By which mechanism?

The question is clear and direct, so answer it. When I told you "when don't you apply Ohm's law" your answer was "in superconductivity or semiconductors". What's wrong with you? didn't I say enough times that I'm talking about ohmic conductors? Are you avoiding by all means facing the problem?

Provide an answer for that question which is the only question in this thread. Any other subjects are only brought about by you and has no relevance to the question.

That's all, clear and simple.
 
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  • #35
cmos said:
The setup you speak of requires that the inductor is connected to a DC voltage source. The "termination by a perfect short" implies that the inductor is placed across the source. The voltage source drives the current through the inductor.



See above.



How then would you make a constant current source? The only way to do it is to do it the way actual current sources are made, by putting a feedback mechanism into a variable voltage source.



Yes, equations can be inverted to solve for whatever you want. That is what makes mathematics so useful, especially for investigating physical phenomena. But the physical interpretations do no change with how you manipulate an equation.


At the risk of going off topic...

Yes, since forces cause bodies to accelerate.



If by modern you mean since 1687. The quantity m*a is simply a special case of dp/dt; the latter being the most general case. So in general, forces cause bodies to change their momentum (Newton's 1st and 2nd Laws).



You're logic is completely flawed, if force and acceleration were independent of each other, then neither could be a cause of the other.



I should give you that advice, but it seems you would rather argue with them with a closed-mind and erroneous claims.

I don't believe what I'm reading. The inductor terminated in a perfect short has 0 volts across it. Not a voltage source, but a dead short. Please research this. It is elementary. Ask a professor. Inductors terminated in a SC short retain their current indefinitely. When the path is switched so that said inductor is terminated in a resistance, the current at the instant of transition remains as it was, then decays. The initial voltage across said resistor is I*R, then it decays. In other words I determines V.

How would I make a constant current source? The same way the power company makes a constant voltage source, except that I drive the turbine at constant torque. To obtain constant voltage, the power company uses constant speed drive. It's pretty easy. They use CVS because it offers lower losses than CCS. Also, CVS provides constant frequency, useful for synchronous machines.

A photodiode terminated in a short or virtual short (op amp inputs) outputs a constant current source behavior proportional to incident light intensity. This is a good example of a CCS. The feedback in the op amp loop does not control the constant current, the illumination does. The feedback forces the 2 op amp inputs to remain at virtual zero potential difference. The light controls the magnitude of the current.

Also, batteries can be constructed for CCS operation, but CVS works better for common cell materials (carbon-zinc, alkaline, NiMH, Li ion, etc.). But research on nuclear cells indicates that they lend themselves to CCS operation. Interestingly, not only are nucells CCS devices, but AC as well. A nucell outputs a constant ac current.

Seriously, this notion that voltage is the cause of current, force causes acceleration, & that CCS devices do not exist, is just a groundless prejudice. A generator can be operated for CVS or CCS output. Inductors terminated in a superconductor hold constant current indefinitely. If the SC has a resistance across it, & the SC is switched out, the resistor will dissipate the inductor energy. The initial voltage is forced by I*R. I determines V in this case, vice-versa for a cap.

Not too hard.

Claude
 
<h2>1. What is Ohm's Law?</h2><p>Ohm's Law is a fundamental law in physics that describes the relationship between voltage, current, and resistance in an electrical circuit. It states that the current through a conductor is directly proportional to the voltage across it, and inversely proportional to the resistance of the conductor.</p><h2>2. How is Ohm's Law used in practical applications?</h2><p>Ohm's Law is used to calculate the voltage, current, or resistance in a circuit. It is also used to design and troubleshoot electrical circuits, as well as to determine the power consumption of electrical devices.</p><h2>3. Why does current induce an electric field?</h2><p>Current induces an electric field because moving charges create a disturbance in the electric field around them. This disturbance causes the electric field to propagate and affect other charges in the vicinity, resulting in the flow of current.</p><h2>4. How does current cause an effect in a specific direction?</h2><p>The direction of the electric field induced by current is determined by the direction of the current flow. The electric field will always point in the direction that the positive charges would move, which is opposite to the direction of the current flow.</p><h2>5. What is the cause and effect relationship between current and electric field?</h2><p>The cause and effect relationship between current and electric field is that current induces an electric field, and the electric field affects the movement of charges in the circuit, which in turn affects the current. This relationship is described by Ohm's Law, which states that the current is directly proportional to the electric field and inversely proportional to the resistance.</p>

1. What is Ohm's Law?

Ohm's Law is a fundamental law in physics that describes the relationship between voltage, current, and resistance in an electrical circuit. It states that the current through a conductor is directly proportional to the voltage across it, and inversely proportional to the resistance of the conductor.

2. How is Ohm's Law used in practical applications?

Ohm's Law is used to calculate the voltage, current, or resistance in a circuit. It is also used to design and troubleshoot electrical circuits, as well as to determine the power consumption of electrical devices.

3. Why does current induce an electric field?

Current induces an electric field because moving charges create a disturbance in the electric field around them. This disturbance causes the electric field to propagate and affect other charges in the vicinity, resulting in the flow of current.

4. How does current cause an effect in a specific direction?

The direction of the electric field induced by current is determined by the direction of the current flow. The electric field will always point in the direction that the positive charges would move, which is opposite to the direction of the current flow.

5. What is the cause and effect relationship between current and electric field?

The cause and effect relationship between current and electric field is that current induces an electric field, and the electric field affects the movement of charges in the circuit, which in turn affects the current. This relationship is described by Ohm's Law, which states that the current is directly proportional to the electric field and inversely proportional to the resistance.

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