If you have a circuit of a battery, a Resistance and a light connected with a wire by example: in the middle the wire is cutted in two pieces. if you connect each piece of wire with a metal object and put those twee metal objects together so the current can flow to the resistance and the light.
How is it possible that the electrons can flow from the wire to one of the metal objects into the other metal object and again in the wire? I do understand how electrons move in metals and wires, but i don't know how the transaction of electrons between those components goes. Is it really that simple that electrons just can flow from one object into another of is the (kinetic) energy passed through when an electron clashes with the edge and that energie is passed through to the other object?
[Physics] tranfser of electrons between two metals and a wire
electricity
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First off, you should be aware that your question is in a field fraught with inconsistent terminology. I posted a little guide on my website:
http://sjbyrnes.com/fermiterminology.html
In particular watch out for the words "voltage" (either difference in electric potential, or difference in electrochemical potential), and "chemical potential" (which might mean either electrochemical potential or internal chemical potential. I think you're using it inconsistently even within your question.)
Anyway, the reason electrochemical potential (a.k.a. fermi level) is always constant in equilibrium is because electrochemical potential is the total free energy of the system per electron. If the electrochemical potential is higher at point A than point B, then the system can reduce its free energy by having an electron travel from A to B. Since all systems tend to minimize their free energy, this process keeps happening until eventually the electrochemical potential is flat (equilibrium).
The same cannot be said of electric potential (a.k.a. galvani potential). Just because there is an electric field pointing from A towards B, does NOT mean that an electron has lower free energy at A than B. Maybe A and B are different materials, with different workfunctions, and different electronic wave functions. Maybe moving from B to A would decrease the entropy of electrons in the system (something like this happens in a p-n junction). To make a long story short, the macroscopic electric field is just one of many influences on electron motion. The electrochemical potential is the final result from adding together all these influences.
For the other question, the contact potential between two metals has no "threshold voltage", etc. It actually has no effects that you can observe at all, unless you're willing to spend hundreds of thousands of dollars on fancy scientific equipment (e.g. kelvin probe) Try it at home! Pinch a copper wire with steel alligator clips. You'll see it looks like a normal single conductor, if you have good contact.
The cause of your confusion is probably this unfortunate terminology situation. The word "voltage" can mean either "the thing you measure with a voltmeter" (which is actually electrochemical potential difference) or it can mean "the integral of electric (E) field" (a.k.a. galvani potential). After you have connected the copper and the steel, there is immediately an equilibrium, where a voltmeter will measure 0 volts, and the electrochemical potential difference is 0, even though there is a nonzero galvani potential, let's say 0.1V. What that means is that 0.1V of electric field is pulling electrons one way, and it's exactly cancelled by 0.1V of other influences pulling electrons the opposite way. When you apply a voltage, the forces become out of balance by exactly that amount. Current flows according to the electrochemical potential difference, which is the sum total of all influences on the electron motion.
Suppose we take some object, for example a conducting sphere, and start with it at the same electrical potential as its surroundings. Now we add one electron the sphere, and because the sphere now has a higher negative charge that its surroundings the potential of the sphere will be slightly lower (i.e. more negative) than its surroundings. Adding the electron has created a potential difference. Now add a second electron and the potential difference gets bigger. Add more and more electrons and the potential difference keeps getting bigger.
A side note: this is basically how a Van de Graaff generator works. Charge is added to the sphere at the top by mechnical means, and this can create a big enough potential difference to generate some impressive sparks.
Anyhow, if we transfer a charge $Q$ to our metal sphere the resulting voltage change is given by:
$$ V = \frac{Q}{C} \tag{1} $$
where the constant $C$ is called the capacitance of the sphere (strictly speaking it's the self-capacitance). For spheres we can work equation for the out the capacitance fairly easily, and in fact it's given by:
$$ C = 4\pi\varepsilon_0 r $$
where $r$ is the radius of the sphere. For objects with different shapes the equation for the capacitance will be different, but the key point is that any object of any size and shape has some capacitance and we can use equation (1) to work how what voltage difference is created when we add charge to it.
The point of all this is that the ends of your battery also have a capacitance. Suppose the chemical reaction in the battery transfers a charge $Q$:
The if the capacitance of the anode is $C_-$ and the cathode $C_+$, the voltage change due to transfer the charge will be give by equation (1) as shown on the diagram. The total voltage will be:
$$ V = V_+ - V_- = \frac{2Q}{C_b} $$
where I've assumed the capacitances of both ends are the same and I've used $C_b$ for them both. What actually happens in a battery is that the reaction runs and transfers charge until the voltage $V$ builds up to the battery voltage. At that point no more charge can be transferred and the reaction stops.
This probably all seems a bit long winded, but having gone gone through all this we can answer your questions really easily:
- when you attach a long wire to the ends of the battery you will increase the capacitance. The battery voltage is constant, and we can rearrange equation (1) to give:
$$ Q_T = \tfrac{1}{2} C_T V \tag{2} $$
where $C_T$ is the new bigger total capacitance with the wire attached. So when you increase the capacitance, $C_T$, you increase the charge $Q_T$. That means the reaction in the battery will restart and transfer more electrons until the total charge rises to $C_T$.
- the electrons spread out across the ends of the battery and the wire. If we call the capacitance of just the wire $C_w$ and the capacitance of the end of the battery $C_b$, then the total capacitance is $C_T = C_w + C_b $. If we put this into equation (2) we get:
$$\begin{align} Q_T &= \tfrac{1}{2} (C_w + C_b) V \\ &= \tfrac{1}{2} C_wV + \tfrac{1}{2} C_bV \\ &= Q_w + Q_b \tag{3} \end{align}$$
where $Q_w = \tfrac{1}{2} C_wV$ is the charge on the wire and $Q = \tfrac{1}{2} C_bV$ is the charge on the battery. So if you disconnect the wire it keeps a charge $Q_w$. The size of the charge is given by $Q_w = \tfrac{1}{2} C_wV$.
- The potential difference between the wires is just the battery voltage $V$. If you connected them together with some suitable voltmeter, that meter would show a voltage of $V$.
In practice the capacitance of a piece of wire is going to be very small, and the charge you'd build up on the wires will be tiny. Nevertheless, the capacitance will be greater than zero so there will be some charge.
Best Answer
Metals consist of small crystals; within each crystal exists the "free electrons" which are shared by all of the atoms in the crystal lattice. The number of free electrons per atom depends upon the details of the atoms, but is most often 1 or 2. The free electrons are visualized as "the electron sea" in the Drude model, devised ~1900, and is semi-classical. Introductory condensed matter texts often start with this model. The situation is slightly more complicated with alloys, but the same ideas hold.
In the electron sea the electrons are electrically shielded from each other by (a) the net positive charges of the ion cores and (b) the uncorrelated motions, essentially random, of the free electrons, which are described using the kinetic theory of gasses.
The crystal boundaries serve to impede the free flow of these electrons from one small crystal to the next, and also serve as scattering sites which continually randomize the motions. The velocities of the free electrons are quite large. When an external electric field is applied, it appears as a net "drift velocity" in the electron sea. This is the current in that piece of metal.
When you bring two clean pieces of metal together all of the above is still true, but there is an additional restriction: each crystal has an effective "crystal voltage" on its interior, and for crystals of the same type it should be the same. But when different metals are joined, the difference in the crystal voltage causes a voltage drop when going in one direction, and an increase in the other. This voltage difference is known as the Seebeck effect, discovered in 1821. Since the internal voltages change slightly with temperature, it is possible to measure temperature change electrically; this is the physical basis for the thermocouple.
So adding additional metal increases the total resistance of the circuit, depending upon the resistivity of the additional metal, its dimensions, and other properties.
The current is the net flow of electrons; each individual electron barely moves, but the effects are passed down the line. With alternating currents for every move forward, there is a corresponding move backward -- hence no net motion from the electron drift at all.