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When an electrical cord from, say, a lamp, is plugged into an AC wall socket, I'm aware that an electric field forms around the entire length of the cord and even before the lamp switch is flipped on.

1) So would the field around the length of the plugged-in lamp cord, prior to the lamp switch being turned on, be considered an electric field or a static electric field?

2) If it's a static electric field, then would the field, prior to the switch being flipped on, be considered DC even though the current to soon flow through the wire is AC?

3) What is specifically creating the field around the cord (whether static-electric or not) up through that point (before flipping the on switch)?

4) And the field around the length of the plugged-in lamp cord, prior to the switch being turned on, would be a near field, correct?

Thanks so much

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  • $\begingroup$ This website may help with some general understanding: who.int/peh-emf/about/WhatisEMF/en. $\endgroup$ – Ernie May 3 '15 at 15:25
  • $\begingroup$ What I'm still trying to find out, is, in the case of AC, are there static electric charges spaced around the surface of the wires as in the case of DC? Would you have static charges on the surface (before having a current) with those surface charges remaining in place while electrons are flowing through the created path? If so, are the static surface charges (on the wire) creating the electric field within the wire where AC is flowing or is it the charges within the flow that are creating the electric field within the wire. And is the electric field within the AC wire static before a current? $\endgroup$ – adam3033 May 3 '15 at 16:57
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You say the lamp is plugged into a AC outlet, but then talk of a "wall switch". Apparently you mean that this switch controls the power to the outlet, and that a switch on the lamp is kept on, or that the lamp has no switch. If so, you should clarify this as a switched AC outlet, since most aren't.

In the case of a switched AC outlet, the switch will be in series with the hot side. When off, no current flows, and both sides of the lamp cord will be at the neutral line potential, which should be connected to earth ground nearby, usually where the power feed enters the house and near the breaker panel.

Because both wires of the cord are essentially connected to earth ground, there will be no electric field between them and ground, or anything else connected to ground. There will be a weak electric field between these wires and other wires that are connected to the AC hot line, but this is true of all other conducting object in the house that is connected to ground, not just the wires of the lamp cord. Basically, for practicle purposes and common interpretation, there is "no field" around the lamp cord.

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  • $\begingroup$ Lets go with an example of a switch that's on a lamp but turned off (and plugged in) and the wall switch is already on. In this case, would you have an electric field or a static electric field around the length of the cord? $\endgroup$ – adam3033 May 3 '15 at 16:42
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    $\begingroup$ @adam: That's not what the question asked about. It clearly says prior to the wall switch being turned on. But, yes, there would be a electric field around the cord, although it would go to zero in the direction where the neutral wire is in front and the hot wire in back. $\endgroup$ – Olin Lathrop May 3 '15 at 23:21
  • $\begingroup$ @adam3033 Note that the question you asked in your comment was already answered explicitly in my answer... $\endgroup$ – Floris May 3 '15 at 23:57
  • $\begingroup$ OlinLathrop -- Thank you. But just to clarify, would it be a static electric field (DC field) or an electric field (AC field) that's around the cord before the lamp switch is turned on? (And sorry for the error by saying wall in the body text -- will edit that.) $\endgroup$ – adam3033 May 4 '15 at 1:56
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Depending on the location of the switch, the answer will change. A properly wired lamp would have no signal on the live (phase) wire, and therefore there would be no field. However, if you interrupt the neutral wire (or the switch is in the lamp, not the wall) then you will have a varying AC field because the voltage on the wire changes (and thus a small amount of charge will flow: the wire is a cylinder which has some finite capacitance w.r.t. infinity). And yes, since there is another (neutral) wire nearby, it may be a near-field (but again that depends a bit on the wiring).

A clarification:

If you have a "live" wire connected to the AC mains, its voltage will fluctuate from +155 V to - 155 V (for 110 V AC - like in the USA). The capacitance of 12 AWG wire is approximately 0.12 pF source. Therefore, charging the wire takes a (very) small amount of charge - about 19 pC for a meter of wire. When the wire goes from -155 to + 155 V in 1/120th of a second (60 Hz system) you have an average current of 4.5 nA (peak of 7 nA) for a 1 meter length of wire - and more if the wire is longer.

Note that this will generate a very tiny magnetic field (orders of magnitude smaller than when the lamp is lit and the current is on the order of amps) but the same electric field as if the lamp was turned on - because with the lamp on, (almost) the entire voltage drop will still be across the lamp-plus-switch, so the live wire going to the lamp will still swing through the full mains voltage.

So the answer to your question(s) are:

  • there will be a field; it will be an AC field;
  • with the lamp off, it will be only an electrical field;
  • with the lamp on, there will also be a magnetic field;
  • since there is a nearby neutral wire, the field will be localized (it will drop off faster than if there was no nearby neutral wire - the charge induces an opposite charge on the neutral).
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  • $\begingroup$ You mean you'd have a varying AC field around the length of the cord (around the outside of the cord) before turning on the switch that's on the lamp? Or would the AC field only be around the cord after the switch on the lamp is turned on? $\endgroup$ – adam3033 May 3 '15 at 16:36
  • $\begingroup$ I saw your comment where you mentioned that you had already answered my question. But I am wondering about before the switch is turned on. Your answer would imply that the lamp switch is already on as you mentioned charge flowing. I need to be clear on whether there is either a static electric field (DC field) or just a plain electric field (AC field) around the cord before the lamp switch is turned on (not wall switch -- wall switch was an error I made). $\endgroup$ – adam3033 May 4 '15 at 1:51
  • $\begingroup$ You really should change the question if it was not the question you meant to ask - that's why there is an "edit" function. If you are plugged into an AC socket and no lamp current is flowing, there will be a changing potential on the wire - and in fact that will support a very small amount of current flowing (because as I said, the wire has a small amount of capacitance). It is an AC field. $\endgroup$ – Floris May 4 '15 at 1:53
  • $\begingroup$ Floris -- Sorry about the edit that I meant to get to. And thanks for your response. So cord plugged in and lamp not turned on still allows for some current flow leading to a varying AC field around the cord. And you're saying that the moving charges that are a part of the small amount of flow end up producing the AC field that's around the cord? $\endgroup$ – adam3033 May 4 '15 at 2:27
  • $\begingroup$ Thanks for that detailed response and for the link. I should have said, though, it's the attraction and repulsion of the moving charges that are a part of the small amount of flow in the wire that end up producing the AC electric field that's around the cord. The movement of charges, alone, would create a magnetic field around the cord (from what I understand). But, the AC magnetic field would probably not occur until the lamp switch is turned on and it's a full flow within the wire in the cord. Please correct me if I'm wrong about any of this. $\endgroup$ – adam3033 May 4 '15 at 2:50
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The electrons are in random motion within the cord even when it is plugged and not switched on. The motion of the electrons in this is case is random i.e., there is no preferred direction of motion of electrons or vector sum of all the thermal velocities is zero. Each electron within this conductor acts like a point source of electric filed and these micro sourced electric field interact with each other and causes tremendous chaos (just like bees around a bee hive).

Now when you switch ON, the first half of AC cycle, increasing electric current increase the electric field in one particular direction and electron being negatively charged moves in opposite direction. When AC cycle changes, the electrons move in opposite direction .

In all the above cases electrons "drift" along the wire in one particular direction thereby generating electric current and electric field. This field remains within the conductor only and exerts a force of $F = qE$ on each electron. The static source will generate a static field but when when everything is moving then at each point in the conductor at any instant the net electric field is vector sum of all the fields.

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  • $\begingroup$ In your first paragraph, you say the motion of the electrons is random with no preferred direction of motion. But in the last paragraph, you state that in 'all the above cases' electrons "drift" along the wire in one particular direction. Can you please clarify. Thanks $\endgroup$ – adam3033 May 3 '15 at 16:30
  • $\begingroup$ And when you speak of the random motion of electrons before the cord is plugged in (in your first paragraph), would those electrons be considered static? They are the static source generating the static field? $\endgroup$ – adam3033 May 3 '15 at 17:56

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