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Getting weird formula for Capacitance

  1. Dec 20, 2011 #1
    Okay, so I know
    [tex]C=Q/ΔV[/tex]

    And ΔV is the sum of the electric fields multiplied by the distance between the charges, so if the first charge has a charge of Q and the other has -Q with R distance between, the electric potential/voltage is:
    [tex]ΔV = [(kQ/r^2) + (kQ/r^2)]R[/tex]
    so
    [tex]ΔV = 2kQ/R[/tex].

    And [tex]C = Q/ΔV[/tex]
    so...
    [tex]C = Q/(kQ/R)[/tex]
    and...
    [tex]C = R/2k[/tex]
    and...
    [tex]C = R/2[1/(4Πε_0)][/tex]
    and
    [tex]C = 2ΠRε_0[/tex]

    What...?
    It would make some semblance of sense if R were inversely proportional to capacitance, but it isnt...
     
    Last edited: Dec 20, 2011
  2. jcsd
  3. Dec 20, 2011 #2

    Doc Al

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    Staff: Mentor

    Where ΔV is the voltage between two conductors.

    That's only true if the field is constant.

    Not sure what you're doing here with two point charges.

    In any case: In the expression for the field from a point charge, r is the distance from the charge. So r for one charge is different than the r for the other. Also, the field isn't constant, so you can't just multiply by the distance R.
     
  4. Dec 20, 2011 #3
    Well, I was just adding up the electric fields and doing V = EΔd...

    But I'm guessing that's not possible because the electric field isn't constant...?

    Okay, please forget about everything I typed up there, I guess it was a complete waste of time.


    So how DO you change the voltage in a capacitor? Wouldn't it be to just increase the charge?
    Actually, how do you FIND the voltage in a capacitor? Is it V = EΔd, because the electric field is constant in a capacitor?
     
  5. Dec 20, 2011 #4

    Doc Al

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    Staff: Mentor

    You charge a capacitor by hooking it up to voltage source (a battery, perhaps). The higher the voltage, the greater the charge stored on each conductor.
    For a parallel plate capacitor, the field is constant. So you could use that method.
     
  6. Dec 20, 2011 #5
    So could you substitute for voltage in the capacitance equation?

    [tex]C = q/EΔd[/tex]

    Then [tex]E_T[/tex] would be:
    [tex]E_T = kq/r^2 + kq/r^2[/tex]
    because
    [tex]E = kq/r^2[/tex]
    and both electric fields are going the same direction.

    And then I'd arrive at the same thing I did in my original post...

    [tex]E_T[/tex] = net electric field
     
  7. Dec 20, 2011 #6

    Doc Al

    User Avatar

    Staff: Mentor

    That's the field for a point charge. Nothing to do with the constant field found between the plates of a parallel plate capacitor.
     
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