# Why is there electric field outside a battery?

Tags:
1. Mar 17, 2015

### 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?

2. Mar 17, 2015

### Staff: Mentor

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?

3. Mar 18, 2015

### Joker93

@Drakkith so i assume it's for the same reasons!but why does that happen?I dont 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?

4. Mar 18, 2015

### 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.

5. Mar 18, 2015

### 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?

6. Mar 18, 2015

### 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.

7. Mar 18, 2015

### Joker93

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.

8. Mar 18, 2015

### 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.

9. Mar 18, 2015

### Joker93

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.

10. Mar 18, 2015

### 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.

11. Mar 18, 2015

### Joker93

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

12. Mar 18, 2015

### mr166

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

13. Mar 18, 2015

### Joker93

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!

14. Mar 18, 2015

### 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.

15. Mar 18, 2015

### Joker93

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?

16. Mar 18, 2015

### 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?

17. Mar 18, 2015

### Joker93

yes,of course.

18. Mar 18, 2015

### Staff: Mentor

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

19. Mar 18, 2015

### Joker93

thank you!

20. Oct 26, 2016

### pranav p v

\
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 plz post here

21. Oct 26, 2016

### cnh1995

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 loopE.dl=0.

22. Oct 27, 2016

### 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???

23. Oct 27, 2016

### sophiecentaur

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.

24. Oct 28, 2016

### 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".

Last edited: Oct 28, 2016
25. Oct 28, 2016

### sophiecentaur

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.