# E field in a real circuit

1. Oct 4, 2015

### DoobleD

When looking at the E field in a circuit, we usually find something like the images below (from here).

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.

2. Oct 4, 2015

### Staff: Mentor

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.
Probably just a drawing issue. You also don't see field lines in the power source.

3. Oct 4, 2015

### DoobleD

Thank you.

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. Oct 4, 2015

### Staff: Mentor

You get a component along the wire from the electric field along the wire.
It does not, there is still the orthogonal component.

5. Oct 4, 2015

### Staff: Mentor

Are you talking about something besides an Ohmic conductor?

6. Oct 5, 2015

No.

7. Oct 5, 2015

### DoobleD

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. Oct 5, 2015

### Staff: Mentor

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. Oct 5, 2015

### Staff: Mentor

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

10. Oct 5, 2015

### DoobleD

Thanks. Why then is the outside field not orthogonal anymore? Is it somehow "influenced" by the straight field inside the wires?

11. Oct 5, 2015

### Staff: Mentor

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. Oct 5, 2015

### Staff: Mentor

What do you mean by "not orthogonal"? I am not sure what you are asking.

13. Oct 5, 2015

### DoobleD

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?

14. Oct 5, 2015

### Staff: Mentor

15. Oct 7, 2015

### DoobleD

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. Oct 7, 2015

### Staff: Mentor

Yea, it is a really good paper, but definitely assumes some background knowledge to begin with.