Why is there electric field outside a battery?

In summary: The difference is that in a capacitor,the charge accumulates on only one side of the dielectric. In a battery,the charge accumulates on both sides of the dielectric.
  • #36
sophiecentaur said:
So how do Fields help you in finding out what goes on?
Where do Ohm's Law and Kirchoff's Laws come into your model of things?

It was not my intention to discuss all the aspects of the E-field (... what goes on?...). In post#24 I have answered a question with two sentences.
Right otr wrong?
What do you mean with "model of things"? I really don`t know what you are speaking of. I did not describe any "model".
I simply have stated that there is an E-field within a resistive body which allows the movement of charges.
If you consider this as wrong - don`t hesitate to tell me.
As an engineer I am accustomed to clear statements (and I have problems with "...what goes on... and "...model of things..).
____________________________
"People go to a shop and buy a "potentiometer", not very often to measure things but to produce a variable voltage ratio."
I know how to use a potentiometer. However, I was trying to explain to you where the name comes from (it is not a "potential divider")
 
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  • #37
LvW said:
It was not my intention to discuss all the aspects of the E-field (... what goes on?...). In post#24 I have answered a question with two sentences.
Right otr wrong?
What do you mean with "model of things"? I really don`t know what you are speaking of. I did not describe any "model".
I simply have stated that there is an E-field within a resistive body which allows the movement of charges.
If you consider this as wrong - don`t hesitate to tell me.
As an engineer I am accustomed to clear statements (and I have problems with "...what goes on... and "...model of things..).
____________________________
"People go to a shop and buy a "potentiometer", not very often to measure things but to produce a variable voltage ratio."
I know how to use a potentiometer. However, I was trying to explain to you where the name comes from (it is not a "potential divider")

I am sure you know how to use a Potentiometer (as in school, with the galvo). But, go into an electronic component shop and ask for one and we both know what you will be given. The voltage division with two resistors of different materials is dictated by their resistance and not only by their length (proportional to 1/Field)
So you want my 'nuts and bolts' response. You are, in fact using a "model", which is what you are using to present your ideas about this topic. (Another language thing)
If you wanted to predict the current through an oddly shaped wire, would you do a line integral of Esds/Rs etc etc or would you use the Potential Drop across the wire, divided by its total resistance? We both know that the same answer would result with a straight or distorted wire (at DC, of course), whether you used the I=V/R or the long winded integral method.
And you still haven't commented on my post about the two Parallel Resistors with different shapes but the same R value.
Bottom line is that no one would argue against the fact that the microscopic level description can use Fields. I have read long, turgid articles which justify the Field approach to electrical circuits but it always seems that they don't actually achieve much; they are preaching to the converted (but not to users of the approach). The authors still use I=V/R, once they walk away from the word processor. It is true that the Energy approach to problems is nearly always more straightforward than trying to use Forces. Potential and Field are closely related (of course) but why choose the more complicated form of description?
 
  • #38
sophiecentaur said:
You are, in fact using a "model", which is what you are using to present your ideas about this topic.
I did nothing else than to state that - as a precondition for a current within a wire or a resistor - there must be an E-field. Right or wrong?
If you like to call this description a "model" - OK. Furthermore, I never have claimed that I have presented my "ideas" (How could I ?).

sophiecentaur said:
If you wanted to predict the current through an oddly shaped wire, ...
I never wanted to "predict" any current. Please, read again my short answer in post#24.

sophiecentaur said:
And you still haven't commented on my post about the two Parallel Resistors with different shapes but the same R value.
I don`t know what you are expecting. Did I classify your result ("Different fields but the same current") as wrong?
Of course, we have different fields.
So what? I repeat: I did not predict any current! Why do you suggest me more than I have written?
 
  • #39
This is getting pointless and it would be too tiresome to pick over all of your past posts in detail. You have suggested (you may say 'implied') that the field across a resistor determines the current through it. Many different fields will produce the same current through different resistors with the same nominal value. So how is your idea worth discussing? How does the field help you work out anything? Potential Difference is what counts. Do not agree?
Do you understand the accepted way that a potential divider works and how it involves Electrical Potential and not field?
You see, I have, in my mind, an innocent reader of your posts, who has no idea whether they are 'gospel' or just a bit of alternative speculation. There are many such visitors to PF. I look upon it as a disaster if one of them gets such a 'non-standard' view as yours and assumes that is how Electricity works. Can you substantiate your idea with any reference of substance? My ideas come straight out of any decent textbook so I do not need to justify them.
 
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  • #40
sophiecentaur said:
You have suggested (you may say 'implied') that the field across a resistor determines the current through it.
No - that`s not true. The opposite is true (see my post#24 and #38).
sophiecentaur said:
So how is your idea worth discussing? How does the field help you work out anything?
I don`t know which "idea" you are referring to. And I do not intend to "work out anything".
The OP`s question was about "the electrical field" - and I gave an answer (post#24). That`s all.
sophiecentaur said:
Do you understand the accepted way that a potential divider works and how it involves Electrical Potential and not field?
I am not your student. To me, there is no need to discuss with you "how a potentiometer works".
sophiecentaur said:
You see, I have, in my mind, an innocent reader of your posts, who has no idea whether they are 'gospel' or just a bit of alternative speculation. There are many such visitors to PF. I look upon it as a disaster if one of them gets such a 'non-standard' view as yours and assumes that is how Electricity works. .
My "non-standard view" is described in my first post#24. (E-field enables charge movement within a resistor).
It was not my intention to describe "how electricity works". I have answered a question regarding the E-field. That`s all.
In contrast to my "alternative speculation", it would be interesting to hear about YOUR answer to the question "which force causes the charges to move within a resistor"?
______________________

For my opinion, your replies are not free from personal "side blows".
I do not like such "discussions".
Hence, I suggest to stop here.
Regards
LvW
 
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  • #41
OK
 
  • #42
LvW said:
Quote Claude: Begging to differ, but the "field" inside the battery does NOT "cause" the current inside the battery. The charges inside the battery are propelled by the reduction/oxidation chemical reaction, or "redox" for short

Is there anybody who has claimed this?

It is incorrect to think that the E field "causes" the current.
Correct. But it is true to say that an E-field allows a (load) current.
Ok, let's explore that. I'm not sure what you mean by "E field 'allows' current." Of course circuit theory avoids fields & transmission line (t-line) explanations, focusing solely on Kirchoff's 2 laws, Ohm's law, & conservation of energy & charge. But since fields are part of the OP question, we can discuss this issue with fields & related topics like t-lines.

A battery has a redox reaction taking place which drives ions towards the terminals. Positive ions are forced towards the positive terminal & likewise for negative. The redox action is doing work, the stored chemical energy is decreased while the electric potential energy is increased. The E field is built up by the internal battery current which is generated by redox action. Thus redox generates current, this current then generates an E field. But when a pair of wires with a load resistance at the other end & a switch at the battery end is connected across the battery, what happens?

When the switch is closed, the E field across the battery terminals is felt by the free electron's in the metal wires. Immediately electrons in the positive lead move towards the positive terminal, due to Lorentz force, F=qE, or Coulomb force if you prefer, since the buildup of positive charges on the battery + terminal attracts electrons. Likewise electrons get repelled from the negative battery terminal. What determines the current flow in the wires (t-line) is the characteristic impedance, Zo, of the t-line. What is important to know is that energy propagates from the battery towards the load at a speed below light, say it is c/2 for this example. When an electron vacates its parent atom's shell, it leaves behind a hole, which ionizes said atom positively. Now this positive ion will attract an electron from its neighboring atom, which leaves behind a hole as well, becomes ionized positive, which attracts an adjoining atom's free electron, etc. The electrons move very tiny distances at very tiny speed, but the energy propagates at a significant fraction of light speed.

The current & the voltage are both moving towards the load in unison. The t-line is an ordinary pair of wires with uniform spacing, having a purely resistive Zo. But what happens when the wave reaches the load resistor? The charges, electrons in this case, are already in the conduction band & energy is already being transported. These electrons continue into & through the resistor load. The resistor is a crystal lattice structure. Electrons crash into lattice ions, & some electrons drop in energy level from conduction band down to valence band. Energy was lost since valence is a lower energy state than conduction (valence more negative than conduction). To conserve energy, a photon is emitted per Planck's law E=hf, in the form of infrared emission, which we feel as heat. A resistor carrying current feels warm for this reason.

So my point is this. Electrons in the resistor are not moving immediately due to a pre-existing E field, but rather, when the electrons arrive at the resistor, there is an E field at the resistor terminals, but the electrons in the resistor bulk material do not immediately feel the E field presence. Interior electrons are surrounded by & bonded to neighboring atoms. Interior electrons are much closer to neighboring atoms than to the charges arriving at the terminals. The terminal electrons of the resistor get attracted or repulsed by the arriving electrons on the t-line & the electron-hole generation & propagation takes place in the resistor. A layer of charge builds up on each end of the resistor, & its E field opposes that of the battery E field. Current decreases & approaches steady state value of Vbatt/Rload.

If the resistor value is larger than the t-line value of impedance, the current will decrease, as the charge accumulation opposes the battery E field. If load resistor is smaller than Zo, charge accumulation is much smaller & the battery E field incurs less opposition, overall current is greater than the large resistance case. Anyway, the current inside the resistor does not require a pre-existing E field to commence. As soon as the electrons arrive, the resistor current has begun. The E field at the interior of the resistor is smaller than that at the terminals since the crystal lattice has many charges with their own E field that neutralize the external E field.

In a nutshell, that is my understanding based on reference texts by Artley, Kittel, Sze & Ng, Muller & Kamens, etc. I will clarify if needed, feel free to ask questions. BR.

Claude :-)
 
  • #43
cabraham said:
I'm not sure what you mean by "E field 'allows' current."

Here is what I mean (excerpt from wikipedia):

J=sigma*E

where J is the current density at a given location in a resistive material, E is the electric field at that location, and σ (Sigma) is a material-dependent parameter called the conductivity. This reformulation of Ohm's law is due to Gustav Kirchhoff.

(It happens not often that I quote wikipedia - however, in this case it seems reasonable).
LvW
 
  • #44
pranav p v said:
consider a parallel plate capacitor, in that uniform field there inside but no electric field outside right?then how charges in the conductor moves when it connected between the plates(outside)??if conductor connected inside the capacitor ,charges definitely moves bcz field is present there...but outside?:frown:
Before the connection of the plates it was a uniform surface charge density into the wires and the total electric field is zero at all points inside the wires. After the connection the surface charge density varies along the wire and create a net electric field inside the wire. This field creates the current.
 
  • #45
sarecon said:
Before the connection of the plates it was a uniform surface charge density into the wires and the total electric field is zero at all points inside the wires. After the connection the surface charge density varies along the wire and create a net electric field inside the wire. This field creates the current.
Hello @sarecon .
:welcome:
You are replying to a thread that is over 7 years old and has not been replied to in nearly 6 years.
 
  • #46
sarecon said:
Before the connection of the plates it was a uniform surface charge density into the wires and the total electric field is zero at all points inside the wires. After the connection the surface charge density varies along the wire and create a net electric field inside the wire. This field creates the current.
Thanks for the answer.Can you elaborate your answer?After the connection,how does surface charge density vary?(There exist already established electric field, whose value is zero outside the capacitor )
 
  • #47
When I first started doing electronic engineering decades ago, a guru told me not to complicate things by studying electric fields, just remember one formula, which is Ohm's Law. Of course, I don't completely agree with him.
 
  • #48
alan123hk said:
When I first started doing electronic engineering decades ago, a guru told me not to complicate things by studying electric fields, just remember one formula, which is Ohm's Law. Of course, I don't completely agree with him.
But there might be people here who can answer this.
 
  • #49
cabraham said:
When an electron vacates its parent atom's shell, it leaves behind a hole, which ionizes said atom positively.
I think this is a very naive model of conduct ion in metals. Individual 'valence' electrons in the metal are not associated with any one particular metal atom. They are attracted (and repelled) by all nearby atoms (which are actually Positively charged Ion Cores). Very little energy is required to move an electron through the bulk metal. This structure also accounts for the high strength and ductility of metals; it's referred to as Metallic Bonding (individual bonds don't 'let go'). This is usually not dealt with in elementary Science teaching. Strange, when you think that metals are very significant components in constructed items.

Imo, @alan123hk 's guru had the right idea, at least when dealing with most of EM. When you have to deal with the difference between 'resistors' of different dimensions and shapes (Long, short, coiled etc.) with the same resistance values, it's a big step and I think you can only approach it once your basics are fully there. My old (Third edition) Kittel on solid state Physics discusses it at a reasonably understandable level.
 
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