Could someone explain to me what electricity really is?

[koliko987]

At this stage of my education there's a lot of things that we're just supposed to accept and roll with it. Last year electricity was pretty much Ohm's law and Kirchoff's rules, now I've done electric fields so I'd like to understand it in more detail. Let's consider a simple circuit with a battery, switch and a light bulb.
Does closing the switch generate a uniform electric field through the conductor? If so does this happen instantly? I mean if suddenly a planet appeared it would instantly have a gravitational field, does the same thing happen when we close the switch? Or is it just very fast and instant when the human eye is concerned?
Field strength is constant so force on the electrons should be constant so why do they move constantly at drift velocity and not accelerate?
I used to think of electrons carrying little packets of energy which they deliver to the bulb, the amount of which depends on the voltage. But what are these packets of energy really? Is it really kinetic energy that electrons gain from the battery which they deliver to the bulb. If so why don't they move more slowly once they deliver the energy?
Usually if we have a uniform electric field (such as between plates of a capacitor) the electric potential isn't constant, but depends on the distance at which charge is from the plates. But in a circuit if we follow the wire from the battery to the bulb the potential is.constant at that whole part of the wire. Why is this?
I guess the part I have the most problems is energy changes in a circuit.
If anyone can make this clear to me I'll be in your debt forever.

 

[ZapperZ]

First of all, let me ask you the obvious question. Are you familiar with the Drude Model?

If not, can you try to look it up first and see if this can answer many of your questions here. We can to recommend you some useful websites if you can't find any.

If you are familiar with it, can you point out the parts that you don't quite understand yet?

Zz.

 

[Hesch]

Does closing the switch generate a uniform electric field through the conductor?

No, the strength of the field depends on the resistance in the conductor. In your circuit nearly all the available field is located in the bulb, as nearly all of the resistance in the curcuit is located here. So in the conductor outside the bulb, the electric fieldstrength is almost 0. V = R*I, I is constant through the circuit.

If so does this happen instantly?

The electric field in vacuum will spread at a speed = speed of light ( 300000 km/s ). In a conductor the field will spread at a speed of typical 230000 km/s, due to inductance and capacitance in the conductor. The speed of electrons is typical 2 mm/s, depending on current and the crossarea of the conductor.

Field strength is constant so force on the electrons should be constant so why do they move constantly at drift velocity and not accelerate?

If you hold a ball in some height and you let it go, it will accelerate due to gravity, until air-resistance becomes as high as gravity force: F_gravity + F_air = 0. In a conductor you must substitute F_air by resistance force i the circuit, which is proportional to the speed of electrons. ( V = R*I, V is here "back-emf" due to resistance and current ).

I used to think of electrons carrying little packets of energy

No, electrons are not packets of energy. Of course an electron, floating through a conductor has some (ignorable) kinetic energy, but its speed and mass are very tiny. The energy is a result of the electric potential in the surroundings of the electron (where am I in this electric field ). So when the electron travels against the electric field, it will loose energy, which is changed into heat in the bulb (high resistance → high field-strengh → high loss of energy per mm ). The electron does not deliver its energy at arrival, but when passing a voltage drop.

 

[ZapperZ]

No, electrons are not packets of energy. Of course an electron, floating through a conductor has some (ignorable) kinetic energy, but its speed and mass are very tiny. The energy is a result of the electric potential on the surroundings of the electron (where am I in this electric field ). So when the electron travels against the electric field, it will loose energy, which is changed into heat in the bulb (high resistance → high field-strengh → high loss of energy per mm ). The electron does not deliver its energy at arrival, but when passing a voltage drop.

Sorry, but this is not correct.

The energy loss here is due to collisions of electrons with each other, with the lattice (or lattice phonons), and impurities/defects in the material. This is the origin of electrical resistance and how the Drude Model arrives at Ohm's Law. It has nothing to do with the electrons losing energy because it went through changing potentials. The changing potential/electric field simply drives the direction of electron transport. That alone does not cause resistivity.

Zz.

 

[Hesch]

The energy loss here is due to collisions of electrons with each other,

If you are able to keep one electron at a fixed position in vacuum and in some electric field at say 1000V electric potential, this electron will have an potential energy = 1000 eV.

Now you remove the electric field, and the potential energy of the electron will be 0 eV.

With what did that electron collide ?

 

[ZapperZ]

If you are able to keep one electron at a fixed position in vacuum and in some electric field at say 1000V electric potential, this electron will have an potential energy = 1000 eV.

Now you remove the electric field, and the potential energy of the electron will be 0 eV.

With what did that electron collide ?

None. But then the question is, what is causing the resistance?

I work with particle accelerators. We accelerate using such a scheme. Yet, I don't measure "resistivity". If I do this in a copper wire and apply the same amount of potential difference, I get quite a lot of resistance that it will heat up and melt the copper wire. So what's the difference? They both have the same potential difference. What is the origin of the energy loss that is causing the resistance?

Apply your analogy of the terminal velocity. Do you think you'll have such a resistance if you do this in vacuum? What is the ball colliding with to generate the resistance? Is the existence of the gravitational potential ALONE the cause of this resistive energy loss?

Zz.

 

[Delta2]

Usually if we have a uniform electric field (such as between plates of a capacitor) the electric potential isn't constant, but depends on the distance at which charge is from the plates. But in a circuit if we follow the wire from the battery to the bulb the potential is.constant at that whole part of the wire. Why is this?

You are correct with one exception. If the electric field is zero inside the connecting wire then it is trivially uniform (so everywhere inside the wire it is the same and zero) but the potential is constant because constant potential is needed for the electric field to be zero.

In real world, the connecting wire will have some small resistance, so a small non-zero uniform electric field will exist inside the wire in order to keep the current alive, and hence there will be some small potential difference between the start and end of the connecting wire.

 

[Hesch]

What is the origin of the energy loss that is causing the resistance?

The origin is loss of potential energy. This energy must be converted to some other energy. Σ energies in a closed system = 0. This other energy could be thermal energy. I don't think, that electrons can "collide" in a "normal" conductor, regarding the speeds of the electrons. I think their electron fields will push them apart before a collision takes place. If you charged some capacitor to the extreme, you could "compress" electrons on the surface of some plate, so that the probability of a electron-collision would be high. But I have never heard about capacitor-plates beeing heated up. You could charge such a plate to 1 TV, and it would still be cold as ice.

Do you think you'll have such a resistance if you do this in vacuum?

I don't think a "resistance" in vacuum can be defined. But there can be an electric field in vacuum, and when an electron is traveling through this field, it will still loose potential energy, and will change it into kinetic energy. This energy will in turn be converted to other energies ( e.g. X-rays ) arriving to the anode. But a bulb is not a particle accelerator, and not a X-ray machine.

So all in all I don't believe in mechanical collisions in a resistor, mechanical friction, etc., but mayby that when electrons close up, and their electric fields are disturbed, a back-emf and heat may be created ? ( I'm not a high-energy-atomic-physicist ).

 

[ZapperZ]

The origin is loss of potential energy. This energy must be converted to some other energy. Σ energies in a closed system = 0. This other energy could be thermal energy. I don't think, that electrons can "collide" in a "normal" conductor, regarding the speeds of the electrons. I think their electron fields will push them apart before a collision takes place. If you charged some capacitor to the extreme, you could "compress" electrons on the surface of some plate, so that the probability of a electron-collision would be high. But I have never heard about capacitor-plates beeing heated up. You could charge such a plate to 1 TV, and it would still be cold as ice.
I don't think a "resistance" in vacuum can be defined. But there can be an electric field in vacuum, and when an electron is traveling through this field, it will still loose potential energy, and will change it into kinetic energy. This energy will in turn be converted to other energies ( e.g. X-rays ) arriving to the anode. But a bulb is not a particle accelerator, and not a X-ray machine.

So all in all I don't believe in mechanical collisions in a resistor, mechanical friction, etc., but mayby that when electron closes up, and their electric fields are disturbed, a back-emf and heat may be created ? ( I'm not a high-energy-atomic-physicist ).

Maybe the advice I gave to the OP about looking into the Drude Model should apply to you as well. Whether you believe it or not, it is how Ohm's law was derived.

Secondly, you should also read up on charge transport in solids. There are numerous experiments that show the effects of electron-electron scattering, electron-phonon scattering, and electron-impurity scattering in metals that effect the mean-free path, lifetime, and consequently, the resistivity of metals.

Example: http://arxiv.org/abs/cond-mat/9904449

Edit: BTW, take note that if I have a superconductor, there will be no potential drop across the superconducting wire. I've created a short between the two ends. So clearly, the potential drop is due to the existence of the resistivity, i.e. that is the cause!Zz.

 

[Dale]

Does closing the switch generate a uniform electric field through the conductor? If so does this happen instantly?

No, it happens at the speed of light in the material.

I used to think of electrons carrying little packets of energy which they deliver to the bulb, the amount of which depends on the voltage. But what are these packets of energy really? Is it really kinetic energy that electrons gain from the battery which they deliver to the bulb. If so why don't they move more slowly once they deliver the energy?

In a typical circuit the energy is transported through the fields outside of the wire, not through the electrons inside the wire. http://en.wikipedia.org/wiki/Poynting_vector

 

[rumborak]

Oh wow, I was totally unaware of this fact. That is very cool (and somewhat bizarre).