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Capacitator problem

  1. Feb 16, 2007 #1
    1. The problem statement, all variables and given/known data
    You have aluminum foil and wax paper and you want to make a capacitator.
    a- describe how you would make a capacitator and and then estimate all appropriate dimensions in order to estimate the capacitance of your design, in Farads.
    b- Say you connect your capacitator to a 9volt battery, so that the potential difference between each side is 9 volts. (V) Find the amount of excess charge on each side of the capacitator.

    2. Relevant equations

    C= lQl/ delta V

    3. The attempt at a solution
    a- I dont even know how to approach this problem. I would assume that the aluminum can be used as the positive and the wax paper can be used as the negative. Maybe we can twist off strips of the aluminum foil and use them as conducting wires? We would need to measure the distance between the [+] and [-] sources. Im totally lost
    b- I dont know
  2. jcsd
  3. Feb 16, 2007 #2


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    Staff: Mentor

    Welcome to the PF, aquabum. The relevant equation that you listed is correct for part (b) of the problem statement. But first you need to calculate the capacitance that you can get by sandwiching an insulating layer (the wax paper) between two plates (using two pieces of the aluminum foil).

    There is an equation that is commonly used to calculate the capacitance C in terms of the dielectric properties of the insulator and the plate area A and the separation distance between the plates d. Are you familiar with that equation?

    EDIT -- BTW, I moved your post from Advanced Physics Homework to Intro Physics Homework. The Advanced Phyisics section is more for complex upper division homework problems.
  4. Feb 16, 2007 #3
    capacitance = epsilon naut x area / distance?
  5. Feb 16, 2007 #4


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    Staff: Mentor


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

    So now you need to figure out as much as you can about the epsilon and d (thickness) of wax paper, and decide what dimensions you want to make the Al foil plates. Then calculate the resulting capacitance that you can expect, being careful to use mks units and get your answer in Farads (it will be in more like uF, but that's still in "Farads").

    Quiz Question -- What happens if you stack several layers on top of each other? Al - Wax - Al - Wax - Al ? How can you connect the various Al layers to maximize the resulting capacitance?
  6. Feb 16, 2007 #5
    the Al layers need to be touching each other?
  7. Feb 16, 2007 #6
    The Al foil dimensions will be 2x2x2, therefore the area of each one will be 8cm^3.

    The distance or thickness of the wax paper will be 1cm.

    now, do I take area and divide it by distance?
  8. Feb 16, 2007 #7


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    Gold Member

    Just a note. Area is measured in squared units. Volume is measured in cubic units.
  9. Feb 16, 2007 #8
    right, sorry.

    So The Al foil dimensions will be 2x2, therefore the area of each one will be 4cm^2.

    The distance or thickness of the wax paper will be 1cm.

    now, capacitance = 4cm^2 / 1cm = 4 epsilon naut? Now I have to figure out what epsilon naut is. How could I do that?
  10. Feb 16, 2007 #9


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    Staff: Mentor

    The thickness of the wax paper is definitely not 1cm. You want to use the thinnest insulator possible in order to maximize the capacitance, right?

    And on the epsilon, you do not use [tex]\epsilon_0[/tex]. Why not? You could probably estimate the value of epsilon for wax paper from just looking up what a typical plastic's value is.
  11. Feb 16, 2007 #10
    ah i think i got it. I have to run to class now. Thanks for the help!
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