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Capacitance Problem

  1. May 4, 2008 #1


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    [SOLVED] Capacitance Problem

    1. The problem statement, all variables and given/known data

    A parallel-plate air capacitor is made by using two plates 16 cm square, spaced 4.7 mm apart. It is connected to a 12-Volt battery.

    What is the capacitance?

    2. Relevant equations

    [tex]|C = \epsilon_r\epsilon_0\frac{A}{d}[/tex]

    3. The attempt at a solution


    [tex]\epsilon_r[/tex] = 1.00059
    [tex]\epsilon_r = 8.85*10^{-12}[/tex]
    A = 0.0016m
    d = 0.0047m,

    And inserting into the formula, I get:

    [tex]|C = (1.00059)(8.85*10^{-12})frac{0.0016}{0.0047}[/tex]

    Which is 3.01*10^-12, which is apparently incorrect.

    Any uides where I have have gone wrong?

  2. jcsd
  3. May 4, 2008 #2


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    Staff: Mentor

    Check your cm/mm to m conversion.
  4. May 4, 2008 #3


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    4.7mm = 0.0047m,
    16 cm squared = 0.0016 square meters

    so it can't be the mm/cm converions?

  5. May 4, 2008 #4


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    Staff: Mentor

    Sorry, missed that squared thingy, looked at 16 cm -> 0.0016 m and it didn't look right... That's what happen when you don't pay attention to units.

    OTOH - why do you think 3 pF is wrong?
  6. May 4, 2008 #5


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    I entered the equation into Mastering Physics, and it just said it was wrong?

  7. May 4, 2008 #6


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    Staff: Mentor

    No idea how these things work in physics, in the chemistry I would tell watch significant digits...
  8. May 4, 2008 #7


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    In MP, you can enter the formula and it calculates it for you, so I entered:


    And MP Calculated it to be 3.01*10^-12, I have also had 3.02 and 3.00 (Both *10^-12), and none of these answers are accepted!!!?

    Any other ideas?

  9. May 5, 2008 #8


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    Any ideas will be greatlt appreciated since there are at least three more parts related to this question, and the next one is:

    What is the charge on each plate?

    To which I need the capcitance to use the formula:

    [tex]C = \frac{Q}{V}[/tex]

    So without the correct capacitance, I can't finish the questions!

    So, Any help will be very much greatly appreciated,

  10. May 5, 2008 #9
    What is Mastering Physics,some software for technical computing?

    And how exactly does it say that the result is wrong?

    Where did you get that value of [tex]\varepsilon_r=1.00059[/tex],in text of the problem?
  11. May 5, 2008 #10


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    Mastering Physics is online Coursework Software - it basically is online homework:

    The Value I gotfrom the corresponding text book, Young and Freedman, 12th Edition, which accomapnies the Mastering Physics

    All Mastering Physics Says is

    "Try Again;"

    So far I've tried 3.00, 3.01, 3.02, all x10^-12, none are correct?

  12. May 5, 2008 #11
    Sorry,I didn`t saw that it is an air capacitor.So maybe there is an error in program,althought probably not,but...
  13. May 5, 2008 #12
    Hey TFM,

    I have the same assignment and I was having the same problem. It seems it is an error in the language used by MP. The area isn't 16cm[tex]^{2}[/tex], it's 0.0256m[tex]^{2}[/tex]. The 16cm refers to the side length. Just to save you from entering the wrong answer too many times my value is (using your formulae);

    4.8*10^-11 F
  14. May 5, 2008 #13


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    Thaks for the info, Vuldoraq, Silly Mastering Physics :rolleyes:

    One small query further on,

    "The battery is disconnected and then the plates are pulled apart to a separation of 9.4 mm.

    What is the charge on each plate in this case?"

    I have already calculated the Capacitance, but wasn't sure how to work this one out, Q=CV doesnt work with 12V for this one.

  15. May 5, 2008 #14
    Your welcome.

    With the battery disconnected, and no new circuit made, the capacitor is unable to discharge. Therefore it retains the charge previously given to it.
  16. May 5, 2008 #15


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    Indeed it is, Thanks

    Thanks to all those who helped,

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