Electric field or static electric field around a plugged-in lamp cord (when lamp is not turned on)? 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
 A: 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).

A: 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.
A: 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.
