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Potential and electron distribution in circuits and cells

  1. Jun 13, 2008 #1
    could anyone shed some light on the following...

    the language might not be too precise here, but i'm trying to get a feel for the subject

    in a battery one terminal (metal) has an excess on electrons which have come from the other +ve terminal (metal), this causes a difference in potential energy between the electrons in the two terminals

    i'm thinking in the -ve terminal the electrons are more concentrated and therefore there will be a large repulsive force between them relative to the other terminal

    this imbalance of electrons is maintained by the chemical reactions within the cell

    if a metal wire is connected between the two terminals the area of more concentrated electrons exerts a greater force than the area of less concentrated electrons on the electrons in the wire and so a flow of electrons occur due to the difference in force

    what's on my mind is the distrubution of electrons throughout the wire and terminals

    if the wire contains a bulb is the following scenario correct..

    the electrons 'before' the bulb will have be equally concentrated throughout the terminal and wire and this concentration will be greater than the concentration in the wire and terminal on the other side of the bulb (where the electrons are also equally concentrated throughout the wire after the bulb and the +ve terminal)

    it is this difference in electron concentration which results in a force when moves the electrons through the bulb so do work

    and in the bulb filament is there essentially a gradient of electron concentration across both ends of the filament? I haven't been able to find any info on this

    i've focussed on electron concentration/distribution in my mind rather than potential since isn't it the distribution of electrons which results in the potential (difference)?

    if it is the case that a potential difference across a component occurs due to the difference in electron distribution across it then i have another question....

    if there are two van der graff domes, one charged and one not with a certain PD between them and then another dome is attached to the charged one, i'm imagining that the electrons will distribute across attached dome (doing work at the time perhaps and heating)
    and therefore reducing the potential between the charged attached domes and the other non charged dome

    so if a wire is attached to one side of a cell and the pd is measured should that result in a drop in pd? my argument would be that the excess electron at the -ve terminal of the cell will distribute themselves throughout the terminal-metal wire system resulting in a lower concentration of electrons and therefore if there is less force between them the potential across the tip of the wire and the + terminal will drop?

    so would there be a very small current flow the moment a wire is attached to a cell at one end only

    and would the decrease of the electron distribution at the -ve terminal result in the 'chemical reaction' depositing more electrons at the metal terminal of the cell?

    but then these would have to come from the +ve metal terminal?

    would an very long wire indeed attached to one end of a cell result in a 'dead' cell?

    if you've managed to read through this and make sense of my questions/thoughts and could shed some light on the matter, i've very much appreciate a reply, it's on my mind at the moment!

    any relevant links etc.. also much appreciated.

    thanks, richard.
  2. jcsd
  3. Jun 13, 2008 #2


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    Staff: Mentor

    Too much to address all at once, but I did want to clarify something from the beginning of your post. Think of the battery as a pump -- pumping electrons from the + terminal to the - terminal. One visualization that I use sometimes is to think of the electrons being like ball bearings, and you have a screw mechanism or some other kind of pump that lifts the bearings up as a linear string to some height. The bearings gain potential energy (gravitational in this analogy) as they are lifted up, which they can then give up as they fall from the top back to ground. If you close a door at the top of the lifting pump, the pump stops running as the bearings stop falling out of the top (this is like the situation when there is no external connection to the battery terminals). When you open the door (connect an external circuit), the bearings can then flow out of the pump (electrons flow out of the battery), and they give up all their PE as they fall to the ground (the electrons lose the PE they were given by the battery as they move through the external circuit from the - terminal back to the + terminal).

    Does that help with the battery basics part?
  4. Jun 14, 2008 #3
    thanks for your reply, i've heard the gravtiational analogy before

    the thing on my mind is, if the charge/electrons have more potential energy before a component than after (resulting in a pd) then in what form do they have that energy?

    to return to the gravitational energy analogy, then an object which is higher is able to fall since the gravitational force can do work on it and it willl then end up with a lower gravitational potential energy

    if charges/electrons have more potential energy of one side of a bulb say than another then is the force responsible for this difference the electrostatic force?

    if that's the case, then for the potential of electrons to be higher on one side of the bulb compared to the other wouldn't they have to be more concentrated there, if they are closer that at the other side that would result in a larger electrostatic repulsion force on one side and then it would be this force which would do the work in the bulb

    so wouldn't there be a charge concentration gradient wherever there was a pd?
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