Battery Voltage: Potential & Chemical Reaction

[Physicslearner500039]

I have read that the potential V = E*d for a constant electric field E, so this is related to the battery voltage of some voltage say 12v etc. Because battery will produce voltage using chemical reaction. Above two are different concepts or related? Please advise.

 

[Dale]

They are the same

 

[vanhees71]

I would rather say that a battery a la Volta provides an electromotive force rather than a potential (difference). Here chemical energy is transformed into electric energy. The Wikipedia gives a nice introduction to the concepts:

https://en.wikipedia.org/wiki/Electromotive_force

 

[Physicslearner500039]

One related question i want to understand is since battery contains electrons inside it, the electric field lines exist in the air am i correct in this? In this case why do i need to connect a wire to the battery positive and negative terminals for the current to flow if i place the wire near the battery does the same thing happen? Please advise.

 

[vanhees71]

You need to connect the terminals in order to enable the electrons to flow since they are in a bound state within the battery. They cannot simply leave the battery into free air but need the conductor to quasi-freely move within it from one pole to the other:

https://en.wikipedia.org/wiki/Electrochemical_cell

 

[Physicslearner500039]

One last question i thought it is the electrons of the conductor (wire) which are moving to produce current. The movement is caused by the electric field of the battery.

 

[vanhees71]

Sure, the electrons move through the entire circuit, i.e., the conduction electrons in the wire move as well as those of the terminals. In the electrolyt the positively charged ions (the anions) move.

 

[Cutter Ketch]

One related question i want to understand is since battery contains electrons inside it, the electric field lines exist in the air am i correct in this? In this case why do i need to connect a wire to the battery positive and negative terminals for the current to flow if i place the wire near the battery does the same thing happen? Please advise.

I’ll probably get flamed for this, but I think the analogy of balls rolling down hill might help you understand the basic concepts. It isn’t perfect, and it breaks down if you go to far with it, but I think it can be helpful.

Think of the electrons as balls rolling down hill. The electric potential (voltage) is like the height of the hill. The higher the hill the more energy the charges gain going down hill and the more work they can do. The slope of the hill is the electric field.

Metal is like a box. If have an electric field (tip the box) the electrons are free to roll down hill till they hit the wall, but they can’t go any further. The walls of the box are not completely vertical nor infinitely high. You can tip the box up enough that the balls roll out of the box (“work function”) but it takes a very high angle (electric field). In most reasonable fields the balls can’t roll out of the box. If you place another box nearby the balls still can’t jump into the next box. In order for the balls to roll into the next box you must actually connect the boxes to get rid of the wall between them.

Continuing the analogy, the battery is like an elevator which lifts the balls up to the top of the hill. The voltage is the height the elevator can lift. From that height the balls can roll down hill through the circuit doing work. The number of balls rolling down at any moment is the current. More current can do more work, but the amount of work each individual charge can do is limited by the height of the hill (potential)

This analogy is good to get started, but it is just an analogy. Electricity has many important differences. Most importantly there isn’t actually a hill. The charges themselves create the forces and potentials due to their mutual repulsion (and attraction for the opposite sign, of course).

 

[rude man]

Ignoring internal resistance, a chemical battery has two E fields, $E_m$ and $E_s$. Emf = $\int \vec{E_m} \cdot \vec{dl}$ whereas voltage V = $\int \vec{E_s} \cdot \vec{dl}$. $E_m$ is the driving force pushing charge from - to + against the resistance offered by the $E_s$ field. The net E field in the battery is zero since the two fields exactly cancel each other: $E_m + E_s = 0$.

A voltmeter reads voltage, not emf: V = $\int \vec{E_s} \cdot \vec{dl}$.

A good analogy (offered by Yale prof. R. Shankar, cf. his excellent course on YouTube) is a ski lift. You have an ideal battery (no internal resistance) in series with a resistor. The lift pushes you (the charge) up the slope with $E_m$ against gravity ($E_s$), then at the top you slide down against friction until arriving at the bottom with zero speed, having dissipated your potential energy in the resistor as heat.

 

[sophiecentaur]

Continuing the analogy, the battery is like an elevator which lifts the balls up to the top of the hill. The voltage is the height the elevator can lift. From that height the balls can roll down hill through the circuit doing work.

This analogy should also include sections of (nearly) horizontal channel through which the balls run without losing potential energy. These sections correspond to IDEAL CONNECTING WIRES. The sloping sections only are the loads / resistances in the circuit.
Without this additional idea, the model is hard to understand. It also helps to justify why we don't tend to use Fields in circuit calculations because however the wires are routed and however 'wide' the load / resistance happens to be, the connection wires do not alter the PD where it counts although the Field between the wires can vary all over the circuit (as the wiring spaces vary).
I find that this is very often ignored in 'explanations' and that means that many students get confused and come to odd conclusions about just how necessary it is to consider the Fields around circuits, even for simple situations. It really isn't!

 

[vanhees71]

A voltmeter reads voltage, not emf: V = $\int \vec{E_s} \cdot \vec{dl}$.

Hm, on the other hand in general a voltmeter usually reads all kinds of EMF not only potentials! We had long discussions about this in connection with one of Lewin's lectures on youtube (at 38:45):



In DC of course you deal with magnetostatics and thus $\vec{\nabla} \times \vec{E}=0$ and thus you deal with voltages!

 

[sophiecentaur]

A voltmeter reads voltage, not emf: V = ∫→Es⋅→dl∫Es→⋅dl→ \int \vec{E_s} \cdot \vec{dl} .

You need to be careful when making generalisations about such things. The distinction is only relevant at the very fundamental level.
If you were sitting in a closed box with two terminals on the wall and if you had a 'good enough' voltmeter, how would you know what was producing the reading on it? Alternatively, I could ask you how you could 'know' that what you are reading is NOT an emf. A humble Potentiometer (metre wire and galvo') will take no / negligible current and a 'Voltmeter' can easily be made which does the same trick.

 

[vanhees71]

Well, you could change the wiring of your volt meter to check whether you have a potential field or not within your closed box!

 

[sophiecentaur]

Well, you could change the wiring of your volt meter to check whether you have a potential field or not within your closed box!

There are only two terminals in my box so what would you do about that? The only measurable potential field would be the V divided by the separation of the terminals.
You could introduce a variable known resistive load and that could allow you to extrapolate to the no-load situation (emf). But 'my' voltmeter would be clever enough to be doing that in any case.

 

[vanhees71]

In fact a volt meter measures general EMFs. If you cannot do anything than to connect it to two terminals you cannot decide, whether you have a potential field or a more general EMF in the example in Lewin's lecture in the youtube video quoted in #11. If you can move the connections of your voltmeter to the terminals you can at least check, whether the geometry of the loop changes the reading. If it changes you can be sure that you don't have a potential field.

 

[rude man]

Hm, on the other hand in general a voltmeter usually reads all kinds of EMF not only potentials! We had long discussions about this in connection with one of Lewin's lectures on youtube (at 38:45):

I know, depending on which discussion you refer to I was very much a part of it.

Anyway - a voltmeter reads volts, not emf's. It records emf's indirectly if and and only if there is an attendant electrostatic field present. A current driven purely by emf does not have a voltage. For example, if a ring of uniform resistance R is irradiated by a time-changing B field there is current = emf/R but a properly connected voltmeter reads zero between any two points along the ring since there is no electrostatic field present.

Did you watch Mr. Mabilde's critical YouTube video post from that time? He had the wrong explanation for Dr. Lewin's demonstration (100 and 900 ohm resistors etc.) results and came up with only an indirect way of measuring voltage along the conductors but he was right in criticizing Lewin's assertion that "Kirchhoff was wrong".

Lewin's explanation was badly misleading IMO. Had he mounted his voltmeter directly above the coil he would have read 0.4V as Mr. Mabilde did. That is the one and only correct voltage. All others are corrupted by including the voltmeter circuit (mainly the leads) partly or wholly within the B field. Still, paradoxically, the voltmeter itself does read a purely electrostatic potential.

I invite all to carefully read my recent Insight article on the subject. The situation is far more subtle than might be assumed. https://www.physicsforums.com/insights/a-new-interpretation-of-dr-walter-lewins-paradox/

 

[rude man]

In a closed box with twterminals on the wall and if you had a 'good enough' voltmeter, how would you know what was producing the reading on it?

You would know you're looking at an electrostatic potential.

I assume you saw Dr. Lewin's lecture with the 100 and 900 ohm resistors within a wire loop irradiated by a time-changing B field. If you mistakenly included the meter circuit (meter plus leads) within the loop as Dr. Lewin unfortunately and seemingly unwittingly did, you would still read a purely electrostatic potential - generated by the meter leads.. Had he lifted the leads directly above the coil he would have read 0.4V as Mr. Mabilde showed in his YouTube rebuttal video which I hope you also saw.

Again, I invite all doubters to carefully read my recent Insight article on the subject. The situation is far more subtle than might be assumed. And perhaps dire admonitions of caution could be held in abeyance until then.

https://www.physicsforums.com/insights/a-new-interpretation-of-dr-walter-lewins-paradox/

 

[vanhees71]

I know, depending on which discussion you refer to I was very much a part of it.

Anyway - a voltmeter reads volts, not emf's. It records emf's indirectly if and and only if there is an attendant electrostatic field present. A current driven purely by emf does not have a voltage. For example, if a ring of uniform resistance R is irradiated by a time-changing B field there is current = emf/R but a properly connected voltmeter reads zero between any two points along the ring since there is no electrostatic field present.

Did you watch Mr. Mabilde's critical YouTube video post from that time? He had the wrong explanation for Dr. Lewin's demonstration (100 and 900 ohm resistors etc.) results and came up with only an indirect way of measuring voltage along the conductors but he was right in criticizing Lewin's assertion that "Kirchhoff was wrong".

Lewin's explanation was badly misleading IMO. Had he mounted his voltmeter directly above the coil he would have read 0.4V as Mr. Mabilde did. That is the one and only correct voltage. All others are corrupted by including the voltmeter circuit (mainly the leads) partly or wholly within the B field. Still, paradoxically, the voltmeter itself does read a purely electrostatic potential.

I invite all to carefully read my recent Insight article on the subject. The situation is far more subtle than might be assumed. https://www.physicsforums.com/insights/a-new-interpretation-of-dr-walter-lewins-paradox/

But that's the point! There is no "unique voltage" in this case, but the reading of the voltmeter depends on how it is connected to the circuit, including the coil with a time-varying magnetic flux. That's why a voltmeter reads general EMFs not only voltages (potential differences). It's hard to conceive to construct such a pure voltage-measuring device.

 

[rude man]

But that's the point! There is no "unique voltage" in this case, but the reading of the voltmeter depends on how it is connected to the circuit, including the coil with a time-varying magnetic flux. That's why a voltmeter reads general EMFs not only voltages (potential differences). It's hard to conceive to construct such a pure voltage-measuring device.

If you'll study my Insight blog ...

 

[Dale]

To all. Please note that this recent discussion is not directly relevant to @Physicslearner500039 original question. He just wanted to know if the voltage in a chemical reaction is also associated with an E field.

 

[rude man]

To all. Please note that this recent discussion is not directly relevant to @Physicslearner500039 original question. He just wanted to know if the voltage in a chemical reaction is also associated with an E field.

I think my comments could be relevant but are probably too advanced for an introductory physics student.

Perhaps it could be transferred to the Classical Physics forum if there continues to be interest in the subject.

 

[Dale]

@rude man no worries, I just didn’t want to cause any confusion for the OP or any lurkers that might think that the simple answer to the OP’s specific question is the subject of any controversy.

 

[rude man]

@rude man no worries, I just didn’t want to cause any confusion for the OP or any lurkers that might think that the simple answer to the OP’s specific question is the subject of any controversy.

OK. besides, I just noticed the thread was already in that forum!

 

[zoki85]

Actually voltmeter doesn't measure voltage but in most of cases uses/measures tiny amount of current and its' effects and converts it into voltage reading.
There are execeptions like electrostatic voltmeter and such of course

 

[vanhees71]

If you'll study my Insight blog ...

I already stumble over the very beginning of this Insight article. There's one electromagnetic field with electric components $\vec{E}$ and magnetic components $\vec{B}$ and not "two electric fields". In the following I simply write "electric and magnetic field" for simplicity.

The electromagnetic field obeys Maxwell's equations, particularly Faraday's Law (in SI units)
$$\vec{\nabla} \times \vec{E}=-\partial_t \vec{B},$$
i.e., if there is a time-varying magnetic field the electric field is not a potential field, and the reading of the volt meter depends on how it is connected to the circuit.

"Kirchhoff's" 1st rule is nothing than the integral version of Faraday's Law quoted above. Since nothing moves here, we have
$$\int_{\partial A} \mathrm{d} \vec{r} \cdot \vec{E}=-\dot{\Phi}_{\vec{B}},$$
where $\Phi$ is the magnetic flux of the area $A$ with $\partial A$ the boundary.

For the analysis I redraw your circuit and use the physical quantitities only. I've never seen the formalism used in the Insight article, and I don't understand it. Assuming that the resistances of the voltmeters are very large, $R_V \rightarrow \infty$ the analysis is very simple, as you see in the attached pdf.

 

[rude man]

"Kirchhoff's" 1st rule is nothing than the integral version of Faraday's Law quoted above.

Kirchhoff's law pertains only to electric fields beginning and ending in charges, what I call $E_s$. This is the crucial point so ubiquitously misunderstood by so many. Kirchhoff talked about "voltages", not emf's. Voltage is the line integral of $E_s$. A voltmeter reads voltages only, not emf's. If a voltmeter reads a voltage between points in a resistance carrying only an $E_m$-generated current then the voltmeter is being fooled by its own circuit. This is the crux of Prof. Lewin's misdirected lecture where the voltmeter circuit forms a part of the Faraday loop..

I've never seen the formalism used in the Insight article, and I don't understand it.

This is the problem. I have invited you and many other doubters to get hold of a copy of Stanford Professor Emeritus H. H. Skilling's excellent text, Fundamentals of Electric Waves, which is probably out of print but surely still available from ebay et al. There you can learn the difference between $E_s$ and $E_m$ fields.(I believe this text has already produced an awakening in other PF members with whom I've communicated in the past so you wouldn't be the first.)

I regret I cannot open yur pdf article for some unknown reason. I don't have Acrobat in my W10 pc since it costs an outrageous $13/month wheras my XP copy cost me a one-off $20. Sorry.

 

[Mister T]

I regret I cannot open yur pdf article for some unknown reason. I don't have Acrobat in my W10 pc since it costs an outrageous $13/month wheras my XP copy cost me a one-off $20. Sorry.

You don't need Acrobat, just the free Adobe Acrobat Reader.

 

[rude man]

You don't need Acrobat, just the free Adobe Acrobat Reader.

@MisterT, thanks but his pdf file was announced as "corrupted".

In general, not having acrobat I have found that I can still read pdf files using whatever W10 uses as the default program for pdf files. (I'm guessing Edge but I'm not sure). So it's doubtful that another pdf reader would open the corrupted file, assuming it really is corrupted.

But anyway, thanks for making the suggestion.

 

[vanhees71]

Kirchhoff's law pertains only to electric fields beginning and ending in charges, what I call $E_s$. This is the crucial point so ubiquitously misunderstood by so many. Kirchhoff talked about "voltages", not emf's. Voltage is the line integral of $E_s$. A voltmeter reads voltages only, not emf's. If a voltmeter reads a voltage between points in a resistance carrying only an $E_m$-generated current then the voltmeter is being fooled by its own circuit. This is the crux of Prof. Lewin's misdirected lecture where the voltmeter circuit forms a part of the Faraday loop.. This is the problem. I have invited you and many other doubters to get hold of a copy of Stanford Professor Emeritus H. H. Skilling's excellent text, Fundamentals of Electric Waves, which is probably out of print but surely still available from ebay et al. There you can learn the difference between $E_s$ and $E_m$ fields.(I believe this text has already produced an awakening in other PF members with whom I've communicated in the past so you wouldn't be the first.)

I regret I cannot open yur pdf article for some unknown reason. I don't have Acrobat in my W10 pc since it costs an outrageous $13/month wheras my XP copy cost me a one-off $20. Sorry.

No! Kirchhoff's Law is the integral version of Faraday's Law. Otherwise you cannot deal with precisely questions as we discuss here. Also you couldn't deal with AC circuit theory at all, at least not if significant inductances (coils) are involved, and also not with transformers, which are at least somewhat important also for practical purposes ;-)) I don't care about the history in this case.

The complete 1st Kirchhoff law, including moving parts reads (in SI units)
$$\mathrm{d}_t \int_A \mathrm{d}^2 \vec{f} \cdot \vec{B}=\dot{\Phi}_{\vec{B},A} = -\int_{\partial A} \mathrm{d} \vec{r} \cdot (\vec{E}+\vec{v} \times \vec{B}).$$
The 2nd Kirchhoff law is nothing else than charge conservation, i.e., that the sum over all currents at any node of the circuit is 0.

Also there's nothing wrong with Lewin's lecture. As I've just seen, he has also a supplementary pdf file with the correct analysis linked at youtube. I don't know, how to deal with texts other than with pdf. There are tons of free pdf readers on the marked, even for M$ (there's even a free acrobat reader).