Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Mathematically, wouldn't the loss of electrons increase voltage?

  1. Sep 28, 2013 #1
    Dumb question, but something I don't get.

    I see this expressed in many basic examples of what voltage is. You have 1 object that has too many electrons and 1 object has too little. The potential difference between them is what is expressed as/in volts. If you connect a conductive wire between them, the electrons will flow from the object with the excess electrons (negative) to the object with lower amounts of electrons (positive). Here's my (lame) attempt at drawing it here =) (Forgive the extra "." only way I could get the spacing).

    ______________ ........................................._______________
    |.....................|.........................................|......................|
    |...6 electrons...|---------------------------->|...2 electrons.....|
    |_____________|.........................................|______________|

    There's a difference in electrons (assume coulombs of electrons. I wrote electrons to help visualize it in my drawing) of 4 coulombs creating voltage. 2 coulombs would need to flow from object 1 to 2 to equalize the electrons between them, bringing the difference to zero and eliminating the voltage.

    My question:

    If voltage = joules / coulombs, then as electrons flow from object 1 to object 2 and is losing electrons, wouldn't that increase the voltage in that mathematical equation? Your denominator is going down, so mathematically, wouldn't voltage increase?
     
    Last edited: Sep 28, 2013
  2. jcsd
  3. Sep 28, 2013 #2
    if you spend a work of 50 joules , this means you either spent the whole 50 joules on one electron and created a P.D. of 5o volts , or you spend them on two electrons and create a p.d. of 25 volts .

    but by the way its not about electrons flowing from where the number is high to where the number is low , its about overcoming the potential difference even if the number of electrons is the same( more electrons usually mean more potential difference but thats not always the case ) .
    in case of batteries , its not about equalizing the number of electrons , its about equalizing the potential difference across the poles of a battery , also i find it better not to think of the electric current in terms of electrons , think of it in terms of charge . electrons are weird little things !
     
  4. Sep 28, 2013 #3
    Think of it as a two-step process.

    1. The charge distribution creates an electric field, which is equivalent to voltage difference between the sides.

    2. In the presence of this electric field, the charges experience a force which makes them move from higher voltages to lower voltages.

    There is an interdependency between the charge positions and the electric field. One depends on the other, which depends on the other....


    Now voltage is DEFINED so that its downhill gradient points in the direction of the electric field (so 1 to 0 means the electric field points left to right). As the electrons move, the electric field that they create changes, and the voltage changes. Eventually, with 4 on each side, the voltage will be constant throughout the conductor. This is the steady state.

    Voltage can be thought of in terms of the amount of Joules one Coulomb loses when going from e.g. 1 V to 0 V, but this isn't particularly useful when the voltage depends on where the Coulomb of charge is! Your case here, where the electric field generators (the charges) are moving, is an electrodynamic problem, so you need to be more careful when using this formula. Most of the time the electrostatic approximation is fine, charges either move so fast you assume they're always in a stationary state (e.g. in perfect conductors), or they don't move at all (e.g. in perfect insulators).

    Your case is an electrodynamic problem because you're asking for the details of the electrons' movements as they equilibrate within a conductor. That means you can't assume the voltage distribution is constant.

    Logically, try to think of it this way:

    Charge distribution causes electric field/voltage
    Electric field/voltage causes forces to act on charges
    As these charges move in this field, they lose or gain energy.

    The energy they lose/gain is the voltage times their charge.

    So calculate the voltage, then the energy, not the other way around.
     
    Last edited: Sep 28, 2013
  5. Sep 28, 2013 #4
    Hmmm...
    1. How can you have potential difference without a difference in electrons?
    2. Better not to think of electric current in terms of electrons? I've just started this, but everything I've read/seen describes it as a flow of electrons. You saying it's not?

    Setting those aside for a moment, let me stick with the traditional definitions as they pertain to my question... Using my example, again, if there are 6 coulombs in object 1 and 2 coulombs in object 2, as electrons flow from object

    Assume joules = 100. As the difference decreases I would have expected the voltage to go down (assuming constant joules).

    Volts = joules / coulombs (shown as differences below)
    Volts = 100 / 4 = 25
    Volts = 100 / 3 = 33
    Volts = 100 / 2 = 50
    Volts = 100 / 1 = 100

    What am I doing wrong here?
     
  6. Sep 28, 2013 #5

    Dale

    Staff: Mentor

    Yes, it would, but that is only very indirectly related to the scenario you were talking about.

    In the equation V=E/Q, if you reduce Q but keep E constant then V will indeed increase. In other words, if you want to store the same amount energy in a smaller amount of charge then you need a larger amount of voltage.

    However, in the scenario you described above E is not constant. As the electron goes from one plate to the other it can do work. So both E and Q are going down. When both numerator and denomintor are going down the effect on the ratio is ambiguous and depends on which is going down more, which you have to determine from other factors.
     
  7. Sep 28, 2013 #6
    Ah. Ok. That answers it. I didn't realize joules would also naturally go down as well. Consider my question solved! =)

    Thanks!

    Though I still don't see how you have potential difference without difference in electrons or how current is not a flow of electrons?
     
  8. Sep 28, 2013 #7

    Dale

    Staff: Mentor

    Current could be a flow of electrons or protons or ions. Similarly, a potential difference could be due to a difference in protons or in ions rather than electrons. You also can get potentials due to changes in magnetic fields even in vacuum with no charges whatsoever.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Mathematically, wouldn't the loss of electrons increase voltage?
Loading...