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Homework Help: Finding capacitence, used C = (EoA)/d, but didn't work!

  1. Oct 9, 2005 #1
    This seems like a plug n' chug problem but i missed it somehow..
    A parallel-plate air-filled capacitor having area 38 cm2 and plate spacing 2.0 mm is charged to a potential difference of 450 V. Find the following values.

    (a) the capacitance
    wrong check mark pF
    (b) the magnitude of the charge on each plate
    nC
    (c) the stored energy
    ┬ÁJ
    (d) the electric field between the plates
    V/m
    (e) the energy density between the plates
    J/m3

    So i used C = (EoA)/d
    A = .38
    V = 450
    d = .002m
    C = [(8.85e-12)(.38)]/.002
    C = 1.68e-9, but the want it in pF, p is e-12, so i moved the decimal over to amek it 1680pF, but it was still wrong, did i mess up converting it from e-9 to e-12?
     
  2. jcsd
  3. Oct 9, 2005 #2

    Physics Monkey

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    38 cm^2 = .0038 m^2
     
  4. Oct 9, 2005 #3
    hm...1 meter = 100 centimeters
    38cm/100cm = .38m
     
  5. Oct 9, 2005 #4

    lightgrav

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    Area is NOT measured in "cm", but in cm x cm .
    as in 10cm x 3.8 cm ... Convert both of them!
     
  6. Oct 9, 2005 #5
    I know area is m^2, not m, or cm^2, not cm...
    The problem already gives me the Area, so its already cm^2:
    38 cm^2, so i want to conver that into meters, so .38 m^2. 2.0mm is the distance, i don't square that also.
     
  7. Oct 9, 2005 #6

    Physics Monkey

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    [tex] 38 \, cm^2 = 38 \, cm^2 * \frac{1 \, m}{100 \, cm}* \frac{1 \, m}{100 \, cm} = 38*10^{-4} \, m^2 = .0038 \, m^2 [/tex]
     
  8. Oct 9, 2005 #7
    ohhhh my bad, sorry i'm slow hah, thanks guys!
     
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