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I have difficulty understanding the picture below. I can understand the green magnetic fields. But why is there an outside, red, electric field? Isn't the electric field guided entirely through the wire? Are there arguments or measurements showing that there is an electric field as depicted? The blue vector is the Poynting vector, associated with energy flux.

I can understand an electric field around the battery if no current runs. But doesn't this disappear when the current runs?

enter image description here

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4 Answers 4

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Isn't the electric field guided entirely through the wire?

The wires are (very nearly) perfect conductors, and therefore the electric field within the wires is (very nearly) zero.

why is there an outside, red, electric field?

Because there is no field inside the wires, the wires are (very nearly) equipotential volumes. And you know there is a potential difference between the upper wire and the lower wire, of (let's say) 1.5 V, because one is connected to the anode of the battery and the other is connected to the cathode.

Therefore you know that if you take a point a on the upper wire and a point b on the lower wire and calculate (based on the definition of electrostatic potential difference)

$$V_{ab} = -\int_b^a\mathbf{E}\cdot d\mathbf{\ell}$$ you will get 1.5 V.

That tells you that the electric field in the region between the wires must be non-zero, and must point from the upper wire to the lower wire.

I can understand an electric field around the battery if no current runs. But doesn't this disappear when the current runs?

No. The electric field is there if there is a potential difference between the wires. It doesn't depend on whether current is running or not. (The magnetic fields, of course, do depend on the current)

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  • $\begingroup$ Nice answer. Still, I can't see any difference between the wire up and the wire down. They both are wires carrying current. Why should they have different potentials if they are made up of the same? I can imagine two different potentials at two sides of R or the battery. Or are there more electrons in the lower wire? $\endgroup$ Feb 5 at 19:13
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    $\begingroup$ @Felicia they have different potentials because there's a battery connected between them. Inside the battery there's a chemical reaction going on that deposits excess electrons on the anode and removes electrons from the cathode. As other answers point out, the wires will be slightly charged because of this (which also must be true in order for electric field to terminate on the wires). $\endgroup$
    – The Photon
    Feb 5 at 19:18
  • $\begingroup$ @ThePhoton the statement you told above "The wires are (very nearly)perfect conductors, and therefore the electric field within the wires is (very nearly)zero",in this are you talking about the condition when there is no potential difference across the ends of the wire Or we can say an external electric field is applied perpendicular to the cross-section of the wire and due to this EF the charges gets accumulated at the ends of the conductor and with this we get the field opposite to the external field by which the net E.F at some point inside the conductor is either zero or may be nearly? $\endgroup$ Feb 5 at 23:06
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The wires have a bit of self capacitance. This self capacitance means that when they are at a given voltage then they gain a corresponding amount of net charge. Having that net charge leads to an E field in the space around the wires, as drawn.

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  • $\begingroup$ Ain't the potential at V and R the same? $\endgroup$ Feb 5 at 16:11
  • $\begingroup$ @Felicia yes….. $\endgroup$
    – Dale
    Feb 5 at 16:35
  • $\begingroup$ But the electrons are still pulled along? $\endgroup$ Feb 5 at 16:40
  • $\begingroup$ @Felicia are you asking about the field inside the wire or outside? The electrons are inside the wire, but the arrows you are asking about in the question are outside. The fields are completely different inside and outside. There is a discontinuity in the E field at the surface of the wire due to the charges on the surface of the wire. $\endgroup$
    – Dale
    Feb 5 at 17:27
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    $\begingroup$ @Felicia if you want to find out the field inside the wire, calculate the resistance of the wire and see how much voltage drop there is along the length of the wire, then divide that by the length of the wire. You will find (unless the wire is very thin or made of a non-typical material, or the load resistor value is very low) that the voltage drop along the wire and the electric field within the wire are very small, compared to the other voltages and fields involved in the problem. $\endgroup$
    – The Photon
    Feb 5 at 19:39
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I can understand an electric field around the battery if no current runs. But doesn't this disappear when the current runs?

No.

The difference is the electric field in the conductor exerts a force on the free electrons creating current, whereas outside the conductor current cannot flow because of the lack of mobile charges. But the electric field still exists.

Hope this helps.

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$$\mathbf E = -\frac{dV}{dx}$$ Inside (because there is an internal resistance) and around the battery, it points from the highest (+) to the lowest potential (-) as showed in the picture. In the resistor, for the same definition, it has the same direction of the current.

Imagine a small positive test charge outside and close to the resistor. It is repelled by the upper side and attracted by the lower side. The same type of field configuration is also present inside the components, but is not restricted only to them.

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