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Capacitance of parrallel plates decrease

  1. Feb 13, 2005 #1

    Alkatran

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    Small question: The capacitance of parrallel plates decreases as you increase the distance between them, why?

    I know the voltage increases, because the electric field is constant, but how does this stop more charge from entering the plate? I mean our voltage relative to some far-off charge is excessively high and yet this doesn't change our capacitance by much?

    Just a conceptual thing. I can do the work, I just need to know WHY it works!
     
  2. jcsd
  3. Feb 13, 2005 #2
    i may be wrong

    but, is this simply because the electric force between the two plates is an inverse square law, [tex] F_e = \frac {kq_1q_2}{d^2}[/tex]
    where the q's would be the charges of the plates, k being the constant, 9.0e9 and d being the distacne between them. So as the distance increases between the plates the static electric force bewteen them decreases

    by the way that is Coulomb's Law

    anyways I could have interpreted your question wrong
     
  4. Feb 13, 2005 #3
    my previous post may have been wrong, I wasn't exactly sure of the relationship between electrostatic force, using Coulomb's law, and Capacitance, however, I was eager to figure it out and googled a site that may help, http://dept.physics.upenn.edu/courses/gladney/phys151/lectures/lecture_feb_03_2003.shtml
    they show the relationship of Capacitance and distance as:
    [tex] C=\frac{Q}{V}=\frac{Q}{Q_d/e_oA}=\frac{e_o A}{d}[/tex]
     
    Last edited: Feb 13, 2005
  5. Feb 13, 2005 #4

    Doc Al

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    By definition, capacitance is C = Q/V. Normally, charge can only enter or leave the plates if there is an available path. (Like if the capacitor is connected to a battery.)

    The electric field is constant only if the charge is constant: If you disconnect the battery before increasing the plate separation. The charge per volt will decrease.

    On the other hand, if you keep the battery connected as you increase the separation the charge will change, since the field decreases. The charge per volt will again decrease.
     
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