Control transfer of static electricity

Sorry I would like to understand the following:
In a neon bulb as discribed in the worksheet static electricity, exercise9 functions as follows, we transfer enough static electricity(in fact electrons) through the 2 wires of the neon bulb and then these electrons are transfered basically to the atoms of neon in the bulb causing them to became ions(charged atoms) . These ions will gather on one electrode creating an excess on one side and then we will have the discharge in form of light to the other electrode. I think, if wrong please correct me.
But what I would like to understand is that on which basis the electrons will choose one side, since the two electrodes are the same and they have not come in contact to somehow loose some of their electrons.
Regards

I believe you can only hold one wire. The other has to be left untouched. Otherwise, there will be no voltage difference to ionize the neon gas an give that flash of light.

If both electrodes are at the same potential, there is no voltage difference between them to create ionization in the neon. The only possible discharge path would be through the glass envelope to an external conductor.

Hello,

Here are some links to pages in the EDUCYPEDIA.
On electrostatics.
http://www.educypedia.be/education/electrostatics.htm
On creating high voltages.
http://www.educypedia.be/electronics/electricityhighvoltage.htm
And electrostatic experiments.
http://www.educypedia.be/education/Electrostaticsexperiments.htm

I hope you can find a solution for your question over there.

Greetings,
Bertus

thank you for your fast replies
Regards

Something I find really strange in displacement of electrons from one component to another is that: Electrons have the same charge, so they should repel each other even when moving to the less charged bodies(with less electrons), but they dont , they keep moving in the same direction just to fill the gap and create the balance state between the two bodies. Is it that the attraction between Proton and electron is far bigger than the repulsion between electron and electron? because for me it is the only explanation I have found.
regards
Sorry I used to ask to much question to myself and try to figure out somehow answer myself. But I find always interesting to hear what the others think.

Something else: It is said that the electric current is the displacement in an orderly manner of electrons through a material (let assume a conductor). Let take as example a bar made of copper. To move its eletrons we need somehow energy I assume. This energy might come from a battery. But what really happens? Does the battery provides somekind of energy to the electrons in the copper bar for them to start moving or what? if yes how does it happens? how these electrons start moving, I really dont understand.
regards

Hello,

For understanding currentflow of electrons take a look at the attached PDF.
It comes from this page:
http://technology.niagarac.on.ca/courses/etec1120/notes.htm
This link comes from the EDUCYPEDIA again.
http://www.educypedia.be/electronics/generalelectronics.htm
The EDUCYPEDIA has a large amount of these pages.
http://www.educypedia.be/electronics/electronicaopening.htm

Greetings,
Bertus

Tanx Bertus, but it still not gives me satisfaction.
See the paper mention the fact that the goal is to move the electrons located in the valence shell of our material, again lets take the bar of copper.
It has one electron at the valence shell and by moving all these one electron(of each atom of copper) with rather small force (what is called voltage, ok let assume)we are able to make them flow through the copper bar and then create finally our current flow, ohhh great finally.
But that is not my focus, my focus is how this force is created, really how? Since the voltage source(battery) is just supposed theoritically to move electrons from one side (side with excess of electrons, the - pole) to the side lacking electrons(+ pole) just like in static electricity.
So where does the force come then from?
You know for me as soon as the voltage source starts pumping electrons through the copper bar(transfer of electrons from - pole to + pole), it will fill the valence shell of each atom of the copper bar, and as soon as all valence shells are full ,electrons will keep flowing through the copper bar since no place is more available. But this contradicts the process of current flow through a conductor, since the electrons of the valence shell should in fact be removed , not the valence shell filled.

Have you read the AAC eBook?

http://www.allaboutcircuits.com/vol_1/chpt_1/1.html

http://www.allaboutcircuits.com/vol_3/chpt_9/1.html

Tanx Bill but it still doesnt answer my Question.
Let me make it simpler. The Voltage it usually called the "pressure that help to move electrons". For me it is just an analogy. I'd like to know the physics behind.
How the electrons coming from the - pole of the battery pull out the electron of the copper atom valence shell. really how?

By repulsion? If yes it will fill that valence shell again, since we have a continous flow coming from that pole. The logic still unclear for me.
Regards

Hello,

Here is an animation how electrons move in a piece of material:
http://www.kangwon.ac.kr/~sericc/sci_lab/physics/e_current/e_current.html
The text will be unreadable, but the pictures will tell the story.
It again comes from an EDUCYPEDIA page:
http://www.educypedia.be/education/physicsjavalaboelec.htm

Greetings,
Bertus

Sorry guys,
what I think is this(It is just my thinking): I think the electrons coming from the negative pole of the battery act on the electrons on the valence shell of the copper as follow:
let assume our bar of copper has just 2 atoms(just a supposition, there are of course far more), just for the sake of the argumentation.
These two atoms are neutral, meaning that they have only one electron in their valence shell. When the electron coming from the negative pole arrive at the first atom of the copper bar, it will expell the valence shell electron of this atom from its orbit (less nrg required since only one electron) and another electron will replace it of course, but the expelled electron will expell also the electron of the other atom(the electron at the valence shell ) and thus we will get our first electron out of the copper bar(this will happen so fast normally according to me that we will running this cyclic loop until we stop sending out electrons from the negative pole of the battery= off the battery) just like a Push-push.
regards

Diablo,

You seem to be confused about the mechanism of charge movement through a conductor. All the assistance you are getting from those trying help you, and the links they refer you to doesn't seem to do much good either, right? Now I am going to try. So let's first list all the things that are wrong about what you might think, and what has been suggested.

1) Current is charge flow per unit of time. Charge flows, current does not. It is wrong to say "current flow". Current flow means charge flow flow, which is redundant and ridiculous. Always say "charge flow" or "current exists".

2) Voltage is not pressure, although it is often represented as such, especially as a analog.

3) Current does not exist because of a "bunching" of charge carriers (electrons) at one end of the wire as compared to the other end of the wire.

Metals, especially those with only one weakly bound electron in the valance level, constitute a material with a sea of freely moving electrons. It doesn't take much energy to make them jump from one atom to another, displacing the valence electron of the atom they arrive at, and leaving a vacancy from where they departed for another electron to fill. So without any outside influence, the metal wire is full of charges freely moving around while staying uniformly distributed.

Next, what is voltage? It is not pressure or energy. It is the energy density of the charge. The MKS units of voltage are joules/coulomb, right? When a voltage is connected to a wire, it sets up a higher energy density of the charge at one end of the wire with respect to the other end. The energy for this is in the form of a electrostatic field. The energy density of the charge will try to even itself out by moving the carriers from the higher energy density to the lower energy density. Along the path to the other end of the wire, the carriers encounter collisions with the core atoms of the metal, collisions with other carriers, and other quantum impediments that we lump into resistance. These impediments take energy away from the charge carriers and dissipate it as heat. Therefore the charge carriers arrive at the other end of the wire with less energy density than they started out with. Sometimes it is called a "voltage drop". So charge movement is best explained as a energy transfer phenomenon, which it really is, instead of a pressure or force analog, which it is not.

The actual movement is sometimes described to the movement of marbles throught a hose. The actual movement of a marble is very slow, but the responce of a marble in to a marble out is very fast. Similarly the actual drift velocity of a electron in a metal is about the same as cold molasses flowing out of a container. But its response is the speed to set up the electrostatic field, which is the speed of light. Since there are an unimaginable number of charge carriers in a good conductor, a high charge flow or high current comes from quantity, not speed. You are right when you say the voltage source acts as a charge pump. For each charge carrier that leaves the voltage source, another enters it from the other terminal.

In conclusion, think of current/voltage as an energy exchange operation instead of a force or pressure concept.

Ratch

Remember also that at the same time charge carriers (electrons) are being pushed out from the negative pole of the battery, other charge carriers are being sucked into the positive pole of the battery.

Ratch your explanation seems logic, but @ some places It seems confusing. For example:
1- "When a voltage is connected to a wire, it sets up a higher charge density at one end of the wire with respect to the other end" . I think you should say "When a voltage is connected to a wire, it sets up a higher energy density(in fact the nrg per Coulomb) at one end of the wire with respect to the other end.

2- "Therefore the charge carriers arrive at the other end of the wire with less energy density than they started out with". Does it mean that the electrons located at the valence shell are not displaced? Because through this definition it looks like it is just the variation of the nrg state of the electrons coming from the voltage source that is considered. But this could contradict the fact that connecting a battery to a wire will provide the required nrg to move the electron(s) located at the valence shell of their atom not reducing the nrg state of the entry electrons toward the exit.

But your definition make really sense. In fact it means if we increase voltage , we just increase the nrg state of electrons coming out of the - negative pole of it..

regards

Diablo,

Ratch your explanation seems logic, but @ some places It seems confusing. For example:
1- "When a voltage is connected to a wire, it sets up a higher charge density at one end of the wire with respect to the other end" . I think you should say "When a voltage is connected to a wire, it sets up a higher energy density(in fact the nrg per Coulomb) at one end of the wire with respect to the other end.

Click to expand...

You are correct, I should have said "higher energy density". I will modify my post to reflect that fact.

What is "nrg"? http://www.acronymfinder.com/NRG.html . When writing, an obscure acronym should be defined at least once, and the acronym itself written in capital letters.

2- "Therefore the charge carriers arrive at the other end of the wire with less energy density than they started out with". Does it mean that the electrons located at the valence shell are not displaced? Because through this definition it looks like it is just the variation of the nrg state of the electrons coming from the voltage source that is considered. But this could contradict the fact that connecting a battery to a wire will provide the required nrg to move the electron(s) located at the valence shell of their atom not reducing the nrg state of the entry electrons toward the exit.

Click to expand...

The valence electrons are constantly being displaced and replaced because they are so loosely attached to the core atoms. It is like a long lake with a feeder stream at one end and a outlet at the other end. The water that comes in is not the same at the water that goes out. Not until a relatively long time has passed. But the amount of water in the lake balances itself very fast provided the outlet is open.

Until I find out what "nrg" means, I don't know what your question is.

Ratch

Try "energy". Pronounce each letter in "nrg" individually.

This is why we urge all posters to use standard English.

Thanks Beenthere, I would have never figured it out on my own.

Diablo,

2- "Therefore the charge carriers arrive at the other end of the wire with less energy density than they started out with". Does it mean that the electrons located at the valence shell are not displaced? Because through this definition it looks like it is just the variation of the nrg state of the electrons coming from the voltage source that is considered. But this could contradict the fact that connecting a battery to a wire will provide the required nrg to move the electron(s) located at the valence shell of their atom not reducing the nrg state of the entry electrons toward the exit.

Click to expand...

I don't know how you come to the above conclusion. The electrons simply jumping from one atom to another atom next to it do not use or release a total net energy. It all balances out. But when the electrons are moved from on end of the wire to the other end, it takes and dissipates energy in the form of heat, because there is resistance in the wire caused by collisions and other subatomic effects in moving the electrons through the wire. Think of a vertical windshield of a car in a rain. When stationary, no energy is needed to move the windshield. When moving, there are many collisions with the raindrops which use up the kinetic energy of the car, and the energy has to be replaced by the car engine.

Ratch

you hit at 80 % the point.
But I think the electron(s) located at the valence shell need additonal energy to jump from one atom to another one even when they are loosely.

And I think this energy is provided to them by radiation from the higher energy state electron replacing him in its orbital (since they can not come in contact due to repulsion).
In fact for any electron to move from his energy state to a higher or lower energy state, it has to gain or lost energy, and by going through this change of state it is expelled from its orbital. Thus for electron(s) located at the valence shell they will just leave the atom since they are located at the last shell. Well, it is what think.
But what you said make really sense and I enjoyed the exchange.
Sorry for the abbreviation, the remark has been taken into account.
Thanks to you guys

regards

Another question: I read somewhere this: "The electrical resistivity of a metallic conductor decreases gradually as the temperature is lowered". But I dont understand this.
What I thought was this: By decreasing the temperature of a conductor one is able to bring the loosely electron(s) at the valence shell to a more stable state, therefore one will need a higher energy to move those electrons, consequently these electrons will resist more, and the resistance property(electric resistivity) will increase.
Please I would like to know what is wrong in my logic, because it lead to a totally contradictory result.
regards