What does the E field look like in a real circuit?

In summary, the E field lines inside the wires point in the same direction as the current does. However, if the resistance of the wires is not small compared to the resistance of the resistor, the field lines are not orthogonal to the wires any more.
  • #1
DoobleD
259
20
When looking at the E field in a circuit, we usually find something like the images below (from here).

fig5.gif


voltagee.gif
This is with an idealized circuit, with idealized (0 resistance) wires linked by a resistor. I wonder what the same pictures would look like with real (non 0 resistance) wires?

I think it would be the same, except that now there would also be a E field inside the wires. Is this correct ?

However the pictures above do not show any E field inside the resistor, that seems strange to me.

I'm ignoring any transient state.
 
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  • #2
DoobleD said:
I wonder what the same pictures would look like with real (non 0 resistance) wires?
For typical setups: very similar. If the resistance of the wires is not small compared to the resistance of the resistor, the field lines are not orthogonal to the wires any more.

The magnetic field gets weaker everywhere if the overall resistance in the circuit increases, but its geometry does not change.
DoobleD said:
However the pictures above do not show any E field inside the resistor, that seems strange to me.
Probably just a drawing issue. You also don't see field lines in the power source.
 
  • #3
Thank you.

mfb said:
If the resistance of the wires is not small compared to the resistance of the resistor, the field lines are not orthogonal to the wires any more.

Why is that?

Also, can you confirm that the E field lines inside the (non ideal) wires points in the same direction as the current does? Due to some buildup of charges during transient state from what I read.
 
  • #4
DoobleD said:
Why is that?
You get a component along the wire from the electric field along the wire.
DoobleD said:
Also, can you confirm that the E field lines inside the (non ideal) wires points in the same direction as the current does?
It does not, there is still the orthogonal component.
 
  • #5
mfb said:
It does not, there is still the orthogonal component.
Are you talking about something besides an Ohmic conductor?
 
  • #7
mfb said:
You get a component along the wire from the electric field along the wire.

Is this component along the wire inside or outside the wire or both? And where does that component comes from?

Are you saying that an E field inside the wire tends to be parallel to it, while an E field outside tends to be orthogonal to it, so the addition of both make a net E field not quite orthogonal nor parallel, but kind of "tilted" ?
 
  • #8
mfb said:
No.
Well, if it is Ohmic then by definition ##J=\sigma E##, so I would say that the E field lines do point in the same direction as the current.
 
  • #9
In the interior of the cable, right.
Not outside (=where the sketch has been drawn), but I misread the question.
 
  • #10
DaleSpam said:
Well, if it is Ohmic then by definition ##J=\sigma E##, so I would say that the E field lines do point in the same direction as the current.

mfb said:
In the interior of the cable, right.
Not outside (=where the sketch has been drawn), but I misread the question.

Thanks. Why then is the outside field not orthogonal anymore? Is it somehow "influenced" by the straight field inside the wires?
 
  • #11
mfb said:
In the interior of the cable, right.
Not outside (=where the sketch has been drawn), but I misread the question.
OK, that makes sense.

@DoobleD inside the wire the current t is in the same direction as the E field. Outside the wire there is no current.
 
  • #12
DoobleD said:
Thanks. Why then is the outside field not orthogonal anymore? Is it somehow "influenced" by the straight field inside the wires?
What do you mean by "not orthogonal"? I am not sure what you are asking.
 
  • #13
DaleSpam said:
What do you mean by "not orthogonal"? I am not sure what you are asking.

Well I'm reusing the term of mfb. I understand it as "not perpendicular". That would mean that the E field just outside the wires has a non zero component along the wire.

I might have misunderstood mfb answer.

I understand that there is no current outside of the wires. My concern is about the E field everywhere. We all agree the E field inside the wires is in the same direction as the current. What about the E field outside the wires?
 
  • #15
DaleSpam said:
You might like this reference. It goes into considerable detail about the field lines inside and outside the wire.

http://depa.fquim.unam.mx/amyd/arch...ia_a_otros_elementos_de_un_circuito_20867.pdf

Thank you, that is exactly what I was looking for!

Reading it, as well as some other papers referenced on it, provided a lot of answers, not only regarding the E field behaviour but also energy flow and surface charges on wires. Actually you already gave me the link but I needed to learn about magnetic fields and the Poynting vector. Very nice. :)
 
  • #16
Yea, it is a really good paper, but definitely assumes some background knowledge to begin with.
 

Related to What does the E field look like in a real circuit?

1. What is an E field in a real circuit?

The E field, also known as the electric field, is a physical quantity that describes the strength and direction of the electric force experienced by a charged particle in a given point in space. In a real circuit, it is the result of the movement of charges through the circuit's components.

2. How is the E field created in a circuit?

The E field in a circuit is created by the presence of a potential difference, or voltage, between two points in the circuit. This potential difference causes charges to move from the higher potential point to the lower potential point, creating an electric current and an accompanying E field.

3. What factors affect the strength of the E field in a real circuit?

The strength of the E field in a real circuit is affected by the amount of charge present, the distance between the charges, and the material properties of the circuit's components. Additionally, the type of circuit (series or parallel) and the presence of any resistors can also affect the strength of the E field.

4. How is the E field measured in a real circuit?

The E field in a real circuit can be measured using a device called an electric field meter. This device measures the strength and direction of the E field at a specific point in the circuit. The unit of measurement for the E field is volts per meter (V/m).

5. How does the E field affect the behavior of a circuit?

The E field plays a crucial role in the behavior of a circuit. It is responsible for the movement of charges, which allows for the flow of electric current. The E field also determines the direction of the current flow and can affect the resistance of the circuit's components. In practical terms, a stronger E field can result in a higher current flow and potentially cause malfunctions or damage to the circuit.

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