Electric Field Direction in Simple Battery Circuit

[quietrain]

lets say i have a simple battery circuit

1) why is the static electric field ,produced by the potential difference of battery, parallel to the wires? shouldn't it be pointing radially outwards? (for +ve)

2) is the electric field inside the wire or outside the wire? if its outside, then how does it drive the current?

thanks!

 

[RedX]

lets say i have a simple battery circuit

1) why is the static electric field ,produced by the potential difference of battery, parallel to the wires? shouldn't it be pointing radially outwards? (for +ve)

Why can't it be both? If there are no moving charges, then the charges on a conductor arrange themselves so that their electric field has no tangential component, for if there were a tangential component, then there would be movement, which is a contradiction. So the electric field would be normal (which in the case of a wire, is the same as radial). If the charges are moving, then there can be a tangential component to the field, and a radial component. In fact, if the wire has resistance, then there must be a tangential component. But nothing says there can't be a radial component: the reason the charges don't leave the wire is because they are attracted to the metal. Don't metal objects want as many electrons as they can get, even if they have a surplus?

2) is the electric field inside the wire or outside the wire? if its outside, then how does it drive the current?

thanks!

Again, why not both? As you imply, there must be an electric field on the inside. But there can be an electric field on the outside too. I don't see any reason why it can't be on the outside.

 

[yungman]

lets say i have a simple battery circuit

1) why is the static electric field ,produced by the potential difference of battery, parallel to the wires? shouldn't it be pointing radially outwards? (for +ve)
2) is the electric field inside the wire or outside the wire? if its outside, then how does it drive the current?

thanks!

I am not expert, I am here to learn also, this is my understanding:

In static condition, electric field only exist on the wire if the wire is not perfect conductor. If the wire is perfect conductor, there is no static electric field generated. Remember $$\vec J =\sigma \vec E \;\hbox { or }\; \vec E = \frac {\vec J}{\sigma}$$?

The smaller the $\sigma$, the smaller the field. And remember, in the wire, being a good conductor, there is NO charge accummulate in the wire, what goes in one end, come right out the other end. So don't look at it as if it is a static case where you have charge gethered and produce E.

Important note: Remember you specified is STATIC case now! If the current varys, then is a different story. The reason is even in static case, DC current do generate B circling around the wire, but in static case, B is decoupled from the E. In varying case, $$\vec E = -\frac {\partial \vec B}{\partial t}$$ and B and E are coupled together. This mean is AC case, the B around the wire varys and generate another E that is different from $$\vec E = \frac {\vec J}{\sigma}$$

 

[quietrain]

In static condition, electric field only exist on the wire if the wire is not perfect conductor. If the wire is perfect conductor, there is no static electric field generated. Remember $$\vec J =\sigma \vec E \;\hbox { or }\; \vec E = \frac {\vec J}{\sigma}$$?

The smaller the $\sigma$, the smaller the field.

from your equations, when the conductivity decreases, shouldn't that E field increase?

 

[quietrain]

Why can't it be both? If there are no moving charges, then the charges on a conductor arrange themselves so that their electric field has no tangential component, for if there were a tangential component, then there would be movement, which is a contradiction. So the electric field would be normal (which in the case of a wire, is the same as radial). If the charges are moving, then there can be a tangential component to the field, and a radial component. In fact, if the wire has resistance, then there must be a tangential component. But nothing says there can't be a radial component: the reason the charges don't leave the wire is because they are attracted to the metal. Don't metal objects want as many electrons as they can get, even if they have a surplus?



Again, why not both? As you imply, there must be an electric field on the inside. But there can be an electric field on the outside too. I don't see any reason why it can't be on the outside.

but from this thread

https://www.physicsforums.com/showthread.php?t=1656

the 3rd post mentioned that there is no radial E-field

 

[yungman]

from your equations, when the conductivity decreases, shouldn't that E field increase?

Yes. That's the reason for perfect conductor, there is no longitudinal E.

 

[quietrain]

Yes. That's the reason for perfect conductor, there is no longitudinal E.

what do you mean by longitudinal E? you mean E parallel to the wire or radial or perpendicular?

 

[yungman]

what do you mean by longitudinal E? you mean E parallel to the wire or radial or perpendicular?

Along the wire.

 

[quietrain]

Along the wire.

? but if there's no electric field along the wire, then how does charge move across the wire?

i am getting confused

am i right if i say,in a simple circuit, with battery , resistor and wire,

the E-field produced by the battery is along the wire? (inside and outside?)

 

[Phrak]

I'd be confused too. Mattson was completely wrong in his post about no radial electric field. But that post is nearly 8 years old.

A wire does not exist in isolation. For each point along the length, the wire has some potential relative to it's surroundings. In the case of a coaxial cable this is fairly well defined by the shield. For an unshielded wire, the potential difference between each point on the wire and the potential of the environment, which is usually not simply arranged, contributes to the electric field perpendicular to the wire.

In the scenario you are trying to imagine, you have a wire in isolation, far from any other charges, going from some positive potential to a negative potential. Even this is not a common arrangment where there is no nearby return path for charge.

 

[quietrain]

I'd be confused too. Mattson was completely wrong in his post about no radial electric field. But that post is nearly 8 years old.

A wire does not exist in isolation. For each point along the length, the wire has some potential relative to it's surroundings. In the case of a coaxial cable this is fairly well defined by the shield. For an unshielded wire, the potential difference between each point on the wire and the potential of the environment, which is usually not simply arranged, contributes to the electric field perpendicular to the wire.

In the scenario you are trying to imagine, you have a wire in isolation, far from any other charges, going from some positive potential to a negative potential. Even this is not a common arrangment where there is no nearby return path for charge.

? wow its getting more confusing :(

ok my scenario is very simple

just a battery, resistor and wire forming a close circuit.

now i assume the battery produces an E-field that drives the current? since it has a potential difference?

so what is this E-field shape? is it along the wires parallel? is it radially outwards along the wire?

i mean, an E-field has to be present for charges to move right?

or am i wrong on that also? :(

 

[RedX]

Think about the physics. If the conductor is not perfect, then there are voltage drops all along the wire due to the wire's resistance. So a loop connecting two points along the wire should have some potential difference along the endpoints. In particular, there are two paths you can take between the endpoints: the path that is the wire, or a path that goes outside the wire. The integral of the electric field dotted with displacement should be the same along both paths. So there must be an electric field outside the wire. But there are no charges outside the wire so the field is divergence-less there (Gauss' law - any field line coming into a surface must exit the surface when there are no charges). So the field outside the wire had to have come from the field inside the wire, so there must be a radial component to the field at the wire.

So there are charges on the wire that create fields both radial from the wire and parallel to the wire. So does this mean the wire is not locally neutral? Yes. That's how electric fields are created in circuits. Anywhere there is large resistance there is only small flow of current. Therefore charges flowing from low resistance material to high resistance material pile up at the boundary between the low and high resistance materials. This pile up creates a strong electric field at areas of high resistance until the field is strong enough to push the current through the high resistance material at the same rate current is flowing into the high resistance material. Therefore a high resistance material has a greater voltage drop, since the electric field is higher, and the electric field is higher because at the ends of the high resistance material are a lot of charges caused from a pile-up.

 

[quietrain]

Think about the physics. If the conductor is not perfect, then there are voltage drops all along the wire due to the wire's resistance. So a loop connecting two points along the wire should have some potential difference along the endpoints. In particular, there are two paths you can take between the endpoints: the path that is the wire, or a path that goes outside the wire. The integral of the electric field dotted with displacement should be the same along both paths. So there must be an electric field outside the wire. But there are no charges outside the wire so the field is divergence-less there (Gauss' law - any field line coming into a surface must exit the surface when there are no charges). So the field outside the wire had to have come from the field inside the wire, so there must be a radial component to the field at the wire.

So there are charges on the wire that create fields both radial from the wire and parallel to the wire. So does this mean the wire is not locally neutral? Yes. That's how electric fields are created in circuits. Anywhere there is large resistance there is only small flow of current. Therefore charges flowing from low resistance material to high resistance material pile up at the boundary between the low and high resistance materials. This pile up creates a strong electric field at areas of high resistance until the field is strong enough to push the current through the high resistance material at the same rate current is flowing into the high resistance material. Therefore a high resistance material has a greater voltage drop, since the electric field is higher, and the electric field is higher because at the ends of the high resistance material are a lot of charges caused from a pile-up.

wow... i see...

for such a simple circuit , there is actually so much theory behind :X

i need to learn more. thanks everyone

 

[yungman]

lets say i have a simple battery circuit

1) why is the static electric field ,produced by the potential difference of battery, parallel to the wires? shouldn't it be pointing radially outwards? (for +ve)

2) is the electric field inside the wire or outside the wire? if its outside, then how does it drive the current?

thanks!

From your question, you ask why the E along the wire and not radiating outward, I think I tried to make things too simple by just not talked about the complicated interaction. In real world, things are not that simple. What I was trying to do is make it a very simple circuit. The longitudinal E is cause by the voltage drop along the wire due to the resistance of the wire( which is due to $\sigma$). In my posts, I just assume ideal with no interaction.

If you start looking at real world as simple as your battery that drive a resistor through the wire case, the battery is a dipole and you have to account for all the field lines and all, you never get to understand anything!

Point is there are multiple cause of E. Your question just ask one of the many E components.

From this and the other posts you make, my suggestion is reading the book over and over in ENGLISH ( don't keep thinking about the math). Also get another used EM book particular if you study physics, get an engineering electromagnetic textbook like Cheng or Ulaby that look at EM in another point of view. Every book explained a little different and from reading many books, many times, you'll get the feel.

This is the 4th time I gone through the few books, I started with Ulaby, then Cheng, now Griffiths. I tell you, every time I study over, I get new insight. The first time, I concentrate on the math ( the microscopic point of view.), get the equation out of the way. Then I start to read the books in English like a story book to find the intricate stuff, then read over and over! I don't mean to be offensive, from the question you ask, seems like you are mixing all the formulas together, it takes time to see the difference. I am still learning, that is the reason I take the risk to join in this discussion and everytime I join in like this, I learn, then go back and read more!

Good luck, it just take time. This is like a new language and it does not come natural.

 

[Phrak]

For your simple circuit, there are radial components of the electric field, but along the length of the wire is E=V/L, where L is the length of the wire and V is the voltage across the battery.