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Capacitance (farads) am I doing this right?

  1. Feb 5, 2009 #1
    1. To make a parallel plate capacitor, you have available two flat plates of aluminum (area = 190 cm2), a sheet of paper (thickness = 0.10 mm, κ = 3.5), a sheet of glass (thickness = 2.0 mm, κ. = 7.0), and a slab of paraffin (thickness = 10.0 mm, κ = 2.0). Find each's capacitance; then enter the largest capacitance possible, of the three, in nF and the smallest in pF.



    2. (capacitance) = ((epsilon not)x(Area)x(k)) / (d)

    **d, or distance between the capacitors, would be the thickness of the paper/glass/paraffin right?**



    3. my answers are:

    5.885E-7 farads for paper
    5.885E-8 farads for glass
    3.363E-9 farads for paraffin

    -smallest = 3363 pF (paraffin)
    -largest = 588.5 nF (paper)

    when i enter the values on my webassign it says im wrong... are the values correct and just converted wrong or am i totally off [i did converted all the given values to meters before plugging them in to the equation]
     
  2. jcsd
  3. Feb 5, 2009 #2
    This is not what I get. I get different exponents. Can you show one example of how you calculated the capacitance?
     
  4. Feb 5, 2009 #3
    in example, for paper i did

    (Eo) (A) (k) (d)
    (8.85E-12)x(1.9)x(3.5)... then divided by .0001 (thickness)
     
  5. Feb 5, 2009 #4
    the equation i though to use was C = ( (Eo)x(A)x(k) ) / (d)
     
  6. Feb 5, 2009 #5
    The error lies in the conversion of the area; 190 cm[tex]^{2}\neq[/tex] 1.9 m[tex]^{2}[/tex].
     
  7. Feb 5, 2009 #6
    oh wow... i was in a rush and forgot it was area... hha thanks alot
     
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