Why is there electric field outside a battery?

[Joker93]

Correct me if I'm wrong,but a battery's electric field is like an electric field of a capacitor consisting of two plates.But we know that the electric field outside the two plates is zero.So why does an electric field form outside the battery and into the circuit?

 

[Drakkith]

Ask the same question about a capacitor. When you connect it to a circuit, why does the capacitor's field extend into the circuit and create voltage and current?

 

[Joker93]

@Drakkith so i assume it's for the same reasons!but why does that happen?I don't understand this phenomenon even when talking about capacitors.Because we learned that when you have a capacitor with two plates,when the plates are much larger than the distance between them,then the electric field inside is uniform while outside the capacitor is zero.So,yeah,WHY does the electric field go outside the capacitor(and battery) and into the circuit?Is it because what i said about the length of the plates is practically not infinite,so the electric field outside is only APPROXIMATELY zero?So,in real life it is just very very small?

 

[mr166]

In a capacitor the charge (power) is stored in the dielectric ( The matter between the 2 plates ). Electrons are moved from one side of the dielectric to the other while charging and are trapped in that position because the dielectric is an insulator. Batteries are more of a chemical reaction with the charging and discharging currents creating new chemical compounds within it.

 

[Joker93]

@mr166 so the chemical reaction causes electric field outside the battery(the emf)?
And except that,there is charge accumulation on each end of the battery also.So shouldn't this create an electric field that is different from the electric field from the chemicals of the battery?

 

[mr166]

I am not sure what you mean when you say electric field outside of a battery. There is a potential difference (voltage ) between the two battery terminals just as there is a potential difference between the two leads of a charged capacitor.

 

[Joker93]

I am not sure what you mean when you say electric field outside of a battery. There is a potential difference (voltage ) between the two battery terminals just as there is a potential difference between the two leads of a charged capacitor.

If you connect the end of the battery called A and B with a wire then you create two paths A to B.You have the path from inside the battery and you have the longer path through the circuit A-> B.The path from A to B through the circuit is what i call outside the battery and in a circuit!
So,you have charge accumulation on A and B,so just like the capacitor it should create an electric field through the second path that i meantioned.

 

[mr166]

Yes electrons will flow through the wire. This is know as an electrical current not a field. That is where my confusion started. In a capacitor this flow of electrons will end when both sides of the dielectric have an equal number of electrons. In a battery the electrons will stop flowing when the chemical reaction runs out of energy.

 

[Joker93]

Yes electrons will flow through the wire. This is know as an electrical current not a field. That is where my confusion started. In a capacitor this flow of electrons will end when both sides of the dielectric have an equal number of electrons. In a battery the electrons will stop flowing when the chemical reaction runs out of energy.

yes,i know what a current is.But those charges at the end of the battery must create an electric field inside the circuit in order to get the electron is moving through it.Otherwise,no force is applied on them,so there is no reason to have a current.My question is a little bit more complex than what you understood.

 

[mr166]

Well there is a unit of measurement assigned to that "field" and it is called voltage. In a capacitor of a given size adding more electrons to one side of the dielectric increases the voltage across the plates. In a battery the voltage is determined by the chemical reaction inside of it. Voltage is the equivalent of water pressure inside of a pipe.

 

[Joker93]

Well there is a unit of measurement assigned to that "field" and it is called voltage. In a capacitor of a given size adding more electrons to one side of the dielectric increases the voltage across the plates. In a battery the voltage is determined by the chemical reaction inside of it. Voltage is the equivalent of water pressure inside of a pipe.

i agree,BUT there is also accumulation of charge on the sides of a battery just the same way as at the plates of a capacitor

 

[mr166]

The voltage between the battery poles and capacitor leads does set up an electrical field.

 

[Joker93]

The voltage between the battery poles and capacitor leads does set up an electrical field.

yes,so why the voltage does not drop with respect to the distance from the battery?Because a field means a drop or rise in voltage!

 

[mr166]

Well if you are talking about current flowing through a wire or a load between both poles there is not any electric field only voltage changes at each point in the wire due to resistance. If you are talking about measuring the voltage change in the air gap between the poles I suppose that if you had the correct equipment you could measure different voltages in the field between the poles. Using one pole as the reference point the voltage should increase as you approached the other pole.

 

[Joker93]

Well if you are talking about current flowing through a wire or a load between both poles there is not any electric field only voltage changes at each point in the wire due to resistance. If you are talking about measuring the voltage change in the air gap between the poles I suppose that if you had the correct equipment you could measure different voltages in the field between the poles. Using one pole as the reference point the voltage should increase as you approached the other pole.

I think that you confuse some things.I do not know at which level your physics course is,but you certainly got some thing wrong here.Electric field is a force field to be precise.If you put a test charge in an electric field,coulomb force will be acted on it and it will cause it to move.Charge movement means current(if you have multiple charges in the electric field going in the same direction-just like in a circuit).And the electric field is just the gradient of potential.Which means the rate of change of potential in respect to the 3 dimensions of space.So,having a non-zero electric field means having a rate of change of voltage.And in a circuit we do have electric field but we do not have a rate of change in potential in the "wire" sections(where there is no battery or resistors,only the plain wire-check the image above).We have change of potential only inside the battery and inside resistors and capacitors.I am only asking:why isn't there a change in potential EVERYWHERE along the wire due to the constant electric field inside the circuit?

 

[mr166]

I may have missed the point here but can you have an electric field in a conductor where the electrons are free to move?

 

[Joker93]

I may have missed the point here but can you have an electric field in a conductor where the electrons are free to move?

yes,of course.

 

[Drakkith]

So,yeah,WHY does the electric field go outside the capacitor(and battery) and into the circuit?Is it because what i said about the length of the plates is practically not infinite,so the electric field outside is only APPROXIMATELY zero?So,in real life it is just very very small?

My guess is that yes, the field outside is only approximately zero.

 

[Joker93]

thank you!

 

[pranav p v]

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Correct me if I'm wrong,but a battery's electric field is like an electric field of a capacitor consisting of two plates.But we know that the electric field outside the two plates is zero.So why does an electric field form outside the battery and into the circuit?

i have the same doubt...did you get it?? i read from a book that there exist an electric field due to chemical reaction which is non conservative ,which is responsible to make the line integral not zero..but i don't know how to visualize it ..if u got he answer please post here

 

[cnh1995]

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i have the same doubt...did you get it?? i read from a book that there exist an electric field due to chemical reaction which is non conservative ,which is responsible to make the line integral not zero..but i don't know how to visualise it ..if u got he answer please post here

This image from post #15 is helpful. Line integral of E.dl outside the battery is equal to the line integral of E.dl inside the battery. So when you traverse a complete loop, you can see ∫_{closed loop}E.dl=0.

 

[pranav p v]

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?

 

[sophiecentaur]

Well there is a unit of measurement assigned to that "field" and it is called voltage.

This statement accounts for many of the confusions about this topic. Voltage and Field are not the same. Field is the gradient of the Voltage. Take a 1.5V battery and bring the leads close together so that they are 1mm apart. The Field between the ends is 1.5kV per metre! You achieved this massive field simply by choosing to put two parts of the circuit that close.
Volts are to do with Energy. Field is to do with Force on a charge. You must not mix them up.

 

[LvW]

Yes - of course, between two poles of opposite polarity there is an E-field. A soon as there is a conducting material between both poles the field is mainly forced "into" this conductor. And it is the field which causes the movement of charges called "current".

 

[sophiecentaur]

Yes - of course, between two poles of opposite polarity there is an E-field. A soon as there is a conducting material between both poles the field is maily forced "into" this conductor. And it is the field which causes the movement of charges called "current".

That's ok as far as it goes but simply saying that there is a 'Field' across the battery is not enough to describe the situation. Take two resistors, wired in parallel to a 1.5V battery. They both have a resistance of 1Ω, only one is 1cm long and the other is 10cm long. We know that I = V/R so they will both pass 1.5A. But the Field, that you insist is the important bit, is 150V/m and, across the other, it is 15V/m. Different fields but the same current. We use Potential Difference to describe the situation in circuits because it is the Relevant Quantity. Field can be anything you like, depending on the circuit layout and the dimensions of the components.

 

[LvW]

But the Field, that you insist is the important bit, is 150V/m and, across the other, it is 15V/m. Different fields but the same current.

Yes - of course. Different field strength and different voltages. Both are directly connected.
Another example: A large (idealized) potentiometer. The E-fields between both ends and the pick-up contact in between is diretly proportional to the distance between the points of measurment - and the same applies, of course, to the voltages.

 

[sophiecentaur]

Yes - of course. Different field strength and different voltages. Both are directly connected.

How can they be different voltages? They are both connected in parallel to a 1.5V battery. Potential and current the same but Fields are different. This is basic stuff and your ideas have to fit in with what you can measure, surely.

 

[cabraham]

Yes - of course, between two poles of opposite polarity there is an E-field. A soon as there is a conducting material between both poles the field is mainly forced "into" this conductor. And it is the field which causes the movement of charges called "current".[/QUOTE]

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. Currents cannot be created by an electric field unless said E field loses energy in the process. A capacitor is a prime example. If a resistor is connected in parallel with a charged cap, with a series switch open, then that switch is closed, a current takes place. The cap E field gives rise to this transient current. But the energy in said E field decreases as does the current, after several time constants the field cannot support current because it has expended its energy.

A battery E field can "cause" current only short term, the redox reaction keeps replenishing said internal battery E field. If a battery delivers 1 amp to an external load, the redox reaction must deliver 1 amp inside the battery to replenish the E field energy. Long & short of it:

Inside battery - redox creates current. Current generates E field. Redox ---> current ---> E field.
Outside battery - E field ---> current.

The current which creates the E field is the same value as the load current created by E field. But the internal & external current are the same per Kirchoff.
It is incorrect to think that the E field "causes" the current. They are mutually inclusive. No causality in general.

Claude

 

[LvW]

How can they be different voltages? They are both connected in parallel to a 1.5V battery. Potential and current the same but Fields are different. This is basic stuff and your ideas have to fit in with what you can measure, surely.

Sorry, I did not read your answer very carefully. For my (first) understanding you spoke about two resistive bodies in series (because of the same current).
Therefore, my second example with the potentiometer. OK?

 

[LvW]

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.

 

[sophiecentaur]

Therefore, my second example with the potentiometer. OK?

So you have shot your argument in the foot, there. You refer to a "Potentiometer" (A Potential Divider) and not a 'Field Divider'. Field is a redundant concept in most aspects of Circuit theory - except inside some devices.

 

[LvW]

So you have shot your argument in the foot, there. You refer to a "Potentiometer" (A Potential Divider) and not a 'Field Divider'. Field is a redundant concept in most aspects of Circuit theory - except inside some devices.

I must admit that I do not understand your reply (.."redundant concept"...?). Let me explain:
Imagine a resistive body of 10cm length and a pick-up (wiper) between both ends.
If a current goes through this device we can measure (a) a voltage Vo between both ends and (b) two voltages V1 and V2 betwen the pick-up and the corresponding ends (Vo=V1+V2).
And these voltages are directly connected to the E-field (to be measured in V/m) within the body.

(By the way: I don`t think that the name "potentiometer" comes from "potential divider" - the potential of a single point is not "divided". Instead this term originates from its use as a measuring device ( ...meter): Two potentials are compared with each other using a galvanometer: One potential is unknown and the other one is made equal (zero potential difference) using such a tunable device called "potentiometer").

 

[sophiecentaur]

Let me explain:

By that, you must mean that you still believe you are right and the way Physics approaches 'Electricity' is wrong.
You sort of apologised, earlier on, for having misread my earlier post because you read "parallel" as "series". Before going any further, I suggest you re-read that post and include the term "parallel". The message in that post is that circuit behaviour does not relate to distances.
No one has claimed that there are no Electric Fields at work in circuits. The point is that Fields involve Spatial relationships but circuits largely behave independently of their layout. This is because they are Energy Driven and not Force driven. Can you produce one instance / reference where a simple circuit is analysed in terms of Fields, rather than Volts? Perhaps you could take a lesson from that.

 

[LvW]

Sophiecentaur - with all respect, I don`t know how to interpret your reply. Perhaps (probably ?) because of my limited knowledge of your language.
May I kindly ask you to tell me where I am wrong in my last contribution? I must admit, I don`t know what you are arguing against.

Reminder: In post#16 there was the question:
"...can you have an electric field in a conductor where the electrons are free to move?"
And my answer in post#24 was:
"Yes - of course, between two poles of opposite polarity there is an E-field. A soon as there is a conducting material between both poles the field is mainly forced "into" this conductor. And it is the field which causes the movement of charges called "current".

So - where is the problem?

 

[sophiecentaur]

Sophiecentaur - with all respect, I don`t know how to interpret your reply. Perhaps (probably ?) because of my limited knowledge of your language.
May I kindly ask you to tell me where I am wrong in my last contribution? I must admit, I don`t know what you are arguing against.

I guess you still have not re-read my past post. If you had, then you would get a clue about what I find wrong with that post.
RE "pot / potentiometer / Potential Divider" I suggest you read around (google the terms and find how they are used). People go to a shop and buy a "potentiometer", not very often to measure things but to produce a variable voltage ratio. It's a typical bit of Use of (Engineering) English, which, I know, can be confusing.
You choose a uniform wire as your example. Would you expect to get different results if the top section was coiled in a loop, with input and output very near each other and the bottom section stretched straight? But the field between the top and middle contacts could be 1000V/m and the field between middle and bottom contacts could be 0.5V/m. The Potential Ratio is what is relevant.
Also, if you take a long, thin resistor and a short, fat resistor with equal resistances, in series, the fields would not be the same but the Potential drops would be equal.
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?