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Coulombic charge of a battery pole

  1. Nov 19, 2007 #1
    Occasionally I need to undrstand electric fields and I usually find an equation that allows me to solve my problem without understanding the solution. Dangerous but.... Anyway I've encountered another field problem and now I'd like to understand the solution to the problem.

    Put in general terms I'd like to understand what the charge in Coulombs might be on the positive post of a 12 V battery that is fully charged but in which there is no external connection (other than air) between the positive and negative posts. Further, if the post can be treated as a point source is it reasonable to use Coulombs Law to calculate the electric field at a given distance (say 10 cm) directly above the post.

    Intuitively it seems the post should bear a coulombic charge and thus act as a point source but I'm having trouble proving to myself that my intuition is in agreement with fact.
     
  2. jcsd
  3. Nov 23, 2007 #2

    rl.bhat

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    Homework Helper

    There is no accumulation of charges on the the positive or negative posts. The electrolyte inside the battery contain positive and negative ions and they are equal in number. The posts are neutral when the circuit is open.
     
  4. Nov 23, 2007 #3
    Most sites say that there is an electric field associated with the posts of the battery. It is clear that the field 10 cm above a post is proportional to Q (charge on the post in Coulombs divided by square of the distance away from the post. What I can't arrive at is the relationship between the voltage at the pole and the Coulombic charge (Q) at the pole.

    Thanks for your opinion.
     
  5. Nov 23, 2007 #4

    rbj

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    oh, there's a little bit, but very, very, very little.

    think of the two posts as "plates" in a capacitor spaced about by about 25 cm and charged to 12 volts.
     
  6. Nov 23, 2007 #5
    That's right but I need to know what that little number is.
     
  7. Nov 24, 2007 #6

    rbj

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    oh, okay. this is not very precise, but will get you in the ballpark.

    for a parallel plate capacitor, where the spacing of the plates is small compared to the diameter of the plates, the capacitance is:

    [tex] C = \frac{\epsilon_0 A}{d} [/tex]

    where C is capacitance, A is the plate area, d is the spacing.

    now, the problem is that the posts are not exactly parallel plates, but you might want to approximate the plate area with the vertical cross-section area. the bigger problem is that the spacing is much larger than the sqrt of the "plate" area, but hell, it's just an approximation to an order of magnitude (i hope).

    the other formula you need is the amount of charge on the plates of a capacitor, if the plates are charged to a known voltage:

    [tex] Q = C V [/tex]

    where Q is the charge (you want), C is the capacitance, and V is the voltage.

    so now you have to get your ruler out, measure how big the posts are and how widely they are separated.
     
  8. Nov 24, 2007 #7
    Thanks, I can see that this is an approximation but that is what I need at the moment. From the last equation it apears that if I know the capacitance in Farads and the voltage in volts I can directly calculate the charge (Q), right? One other thing, in the first equation epsilon subzero is the dielectric constant, correct? And if I had a conventional battery it would be the value for air?

    I'll work on this problem some more on Sunday and get back to you if I encounter any additional problem. I do appreciate your help. Walt
     
    Last edited: Nov 24, 2007
  9. Nov 24, 2007 #8

    rbj

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    yup, and given those units, the charge would be in coulombs.

    learn to use LaTeX, okay? you can quote any of us to see how we do it. i'll answer your question when you do that.

    not if the conventional battery is sitting around in a tank of oil or something like that. so the dependency is not regarding the conventionalness of the battery, but what material is separating the posts (which you are calling the "plates" of a capacitor).
     
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