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Parrallel Plate Capacitor Model Question

  1. Jul 7, 2012 #1
    Hello, i am new to the forums here so forgive me if I have mistakenly started a matching thread that already exists.

    So i have a task for building a parallel plate capacitor for two aluminum plates connected to a battery on opposite terminals. The target capacitance is 2000 microfarads. with a distance of 0.062 inches between the plates. Oh yea, the units of length are inches.

    Since the units for the general equations for capacitance are in either CGS or MKS units, this leaves me confused. So, i have a few questions before completing the experiment.

    Questions:
    1) Do the parallel plates have to be identical; meaning that they have to be of the same material?
    2) For the dimensions of the plates, do they have to be absolutely the same? meaning that can one be smaller or larger than the other?
    3) When finding the area of the plates, are the thicknesses of the plates taken into account. (I know Area is L * W, and adding the thickness would mean Volume; however, since voltage is flowing through the plates, i wonder about it.)
    4) Since the units of length that i am obligated to use are in inches would that interfere with the equations that are given for Capacitance such as:

    C = εrε0[itex]\frac{A}{d}[/itex]

    C = εr[itex]\frac{A}{4*Pi*d}[/itex]


    5) One final question, for the equation would be why is the 4*pi applied in the second equation or is it used to omit ε0 constant?

    Please and thank you in advance for anyone's time.
     
  2. jcsd
  3. Jul 10, 2012 #2

    berkeman

    User Avatar

    Staff: Mentor

    Welcome to the PF.

    The plates do not have to be identical, but you will get a capacitance closest to that first equation if the plates are the same area and are spaced close together compared to their 2-D size.

    You do not take the thickness of the plates into account. The first equation is correct -- I'm not sure where the 2nd equation came from, but at least when using MKS units, the first equation is used.

    And just because you are given the spacing of the plates as 0.062", that doesn't mean that you can't just convert that into meters and do the rest of the calculations in MKS units. After all, Farads are MKS units. :smile:
     
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