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Estimate the capacitance?

  1. Sep 1, 2005 #1
    Hi,

    My question: A homemade capacitor is assembled by placing two 9-in pie pans 5cm apart and connecting them to the opposite terminals of 9-V batter. Estimate the capacitance?

    I know that to get the capacitance, you can use C=Q/V or C=keA/d. Do I need to find the area of the two pie pans to be help me solve the problem or do I make a triangle and find the area of that.

    Thank You

    :smile:
     
  2. jcsd
  3. Sep 1, 2005 #2

    Astronuc

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    Staff: Mentor

    Since C=Q/V, unless one knows Q, then one will have to use C=keA/d.

    I think one can determine the area of a 9-in pie pan. Perhaps assume a circular pie-pan, i.e. 9-in dia.
     
  4. Sep 1, 2005 #3
    Hi,

    Work

    C=keA/d

    C=(8085 x 10^-12)(4.104 x 10^-2)/(5 x 10^-2)

    C= 7.26 x 10 ^-12

    Okay, I see

    Thank You
     
  5. Sep 2, 2005 #4
    Is C the charge? or Capacitance? Is it C = Q/ deltaV ?

    If you used pie pans, could they be filled with something to change the capacitance? like water?

    Is the capacitance value increased as the distance gets smaller between the plates?
     
  6. Sep 2, 2005 #5

    Astronuc

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    Staff: Mentor

    Brad, C is the Capacitance, Q is the charge.

    Putting a material between the conductive plates (some nonconducting (i.e. insulator) dielectric) will affect the capacitance.

    C=keA/d indicates that capacitance increases with increasing area and or decreasing distance.
     
  7. Sep 2, 2005 #6
    Okay, lets say you have two 5" square sheets of aluminum foil, and put regular paper in between, glue all three items together. The paper is just over 5" sq like 5.1" sq so the foil wont short out. That setup has some capacitance, C.

    Then you have two 5" square, well cubed blocks of aluminum, and put the same 5.1" paper between them. The distance is the same (assume no crushing of paper).

    Is the capacitance different? What "electrically" would be different?
     
  8. Sep 3, 2005 #7

    Astronuc

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    Staff: Mentor

    With more surface area (including the sides), the capacitance of the blocks should be greater than that of the thin foil - given the same square surfaces.

    For a given total charge - the surface charge density would be lower with more surface area.
     
  9. Sep 3, 2005 #8
    I can see how the surface charge would be spread out, more surface area means less density, because... (?) the electric charge is always conserved?

    Okay but is it all about just surface area? isnt there penetration into all of the atoms in a 5" cube of aluminum? It seems a sheet of foil would be sheets of atoms stacked to some thickness of a few mils or whatever thickness foil is, and I thought the charge filled the entire foil more or less evenly because Aluminum is a conductor.

    So if there was a 5" cube of Al, it seems the charge would also distribute evenly.

    But there is just the one surface where its facing the other cube, separated by the paper so because of..? the dipole effect? the charge would not be uniform in the cube, but it seems the capacitance must be affected by density of material, because more charge could build up in a cube than foil alone.

    would a 5" cube of foil, where the foil was shaped into cube form but empty on the inside, cause the same capacitance as a solid 5" cube of Aluminum?
     
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