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

Simple Battery Misunderstanding

  1. Jun 14, 2017 #1
    If a small voltage - say 1V, is applied to the negative terminal of a 1.5V alkaline battery (although I don't think the type of battery matters?) will this create a current between the 1V and the negative terminal until the charges are somehow balanced? If not, what will this situation do to the battery?I know you are able to connect batteries in series to achieve a larger voltage than a single battery, but I'm not exactly sure what occurs inside the batteries to make this occur.

    I've learnt quite a bit about electronics during my degree but the details of batteries haven't been explained at all and this problem above has always confused me.

    Thanks!
     
  2. jcsd
  3. Jun 14, 2017 #2

    davenn

    User Avatar
    Science Advisor
    Gold Member

  4. Jun 14, 2017 #3

    jim hardy

    User Avatar
    Science Advisor
    Gold Member
    2016 Award

    Then you are accustomed to "ideal voltage source" and "ideal current source".

    Until you've familiarized yourself with the chemistry inside a battery, just think of it as an ideal voltage source and some resistance in series, with a battery label wrapped around it.

    The phrasing of your question implies that you don't quite grasp the concept of "voltage" aka "potential difference".
    We get lots of questions on that subject. Here's the most recent: https://www.physicsforums.com/threads/physical-basis-of-voltage.917327/

    See also this not too terribly old one
    https://www.physicsforums.com/threads/what-does-more-less-voltage-mean.852532/#post-5347090
     
  5. Jun 14, 2017 #4

    sophiecentaur

    User Avatar
    Science Advisor
    Gold Member

    Firstly, Voltage must be applied across two points. Nothing can 'happen' to the battery unless the other terminal is involved in some way - otherwise the mean potential of the two terminals will change but nothing else. But how much, will depend what is actually connected to the other terminal of your 'voltage source'.
    Objects cannot affect their Electrical Potential by pulling themselves up by their own bootstraps. Try a diagram of what you are asking about. You stand to get a better answer. :smile:
     
  6. Jun 14, 2017 #5

    CWatters

    User Avatar
    Science Advisor
    Homework Helper

    I'm not disagreeing with or contradicting other replies but possibly yes depending on how you model the set up...

    Consider this...

    1) The 1.5V battery starts off with it's negative terminal in contact with a metal plate. We will call this plate "Earth".
    2) The battery is then raised up on insulators so that there is a gap between the negative terminal of the battery and the Earth plate.
    3) This gap will form a crude capacitor.
    4) The 1V source is inserted between the Earth plate and the battery negative terminal (eg in parallel with the capacitance detailed in 3)
    5) Applying a voltage to the capacitor causes it to become charged to that voltage.
    6) Charging a capacitor requires a flow of electrons.

    We can estimate the charge that will flow for something like an AA battery or coin cell in the above model..

    The equation for the charge on a capacitor is Q = CV

    So the amount of charge that flows depends on the capacitance. For something like a small button cell or AA the capacitance will be small. The general equation for a flat plate capacitor is..

    C = kε0A/d

    For air kε0 is about 1.26 x 10-6 H/m (Error see follow up posts below)
    Lets say the diameter of the cell is 0.01m (=1cm) and d the air gap is the same.
    Then the capacitance would be around...

    C = 1.26 * 10-6 * Pi * 0.0052/0.01
    = 9.9 * 10-9
    call it 10nF (Aside: that's a bit larger than I expected but I can't see if/where I made a mistake)

    To charge the stray capacitor to 1V the charge required would be

    Q = C*V = 10*10-9 Coulombs

    I think we can even calculate how many electrons will flow?...

    The charge on an electron is about 1.6 × 10-19 Coulombs so..

    No = 10*10-9 / 1.6 × 10-19

    = 62 * 109 electrons.

    Your mileage may vary.
     
    Last edited: Jun 14, 2017
  7. Jun 14, 2017 #6

    CWatters

    User Avatar
    Science Advisor
    Homework Helper

    Darn it. Google gave me permeability rather than permittivity!!

    That 1.26 x 10-6 H/m should be 8.9 x 10-12 C2.N-1.m-2

    I knew the units look odd.
     
  8. Jun 14, 2017 #7

    CWatters

    User Avatar
    Science Advisor
    Homework Helper

    So the capacitance should be

    C = 8.9 * 10-12 * Pi * 0.0052/0.01
    = 7 * 10-14
    = 0.07 * 10-12
    = 0.07pF

    Much closer to what I expected.

    That makes the charge 7 * 10-14 Coulombs

    and the number of electrons

    = 7 * 10-14 / 1.6 × 10-19
    = 437,500
     
  9. Jun 14, 2017 #8
    Ah thanks, I am still to read the links but I'm guessing an applied voltage like this shouldn't affect the battery in any way then. Thankyou!
     
  10. Jun 14, 2017 #9
  11. Jun 15, 2017 #10

    CWatters

    User Avatar
    Science Advisor
    Homework Helper

    If the battery isn't connected to anything that's correct.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted