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Dielectric with a parallel-plate capacitor finding minimum area

  1. Apr 23, 2008 #1
    The dielectric to be used in a parallel-plate capacitor has a dielectric constant of 3.70 and a dielectric strength of 1.90×10^7 V/m . The capacitor is to have a capacitance of 1.05×10^−9 F and must be able to withstand a maximum potential difference of 5600 V.

    What is the minimum area the plates of the capacitor may have? use 8.85*10^-12 for permittivity of free space.

    A=______ m^2.

    Well i know that the capacitance equals permittivity times area times dialectric constant divided by distance between plates. The distance i got by dividing the voltage by the dialectric strength. Is that correct?

    the final answer i got was 9.47*10^11 but it was wrong. Help please!
     
  2. jcsd
  3. Apr 23, 2008 #2

    alphysicist

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    Hi sonrie,

    That's quite large for the area of a capacitor! I think you might have just made a calculation error. What numbers did you use for both of your calculations?
     
    Last edited: Apr 23, 2008
  4. Apr 23, 2008 #3
    The max intensity of field can be 1.9 x 10^7. Now what is the field inside a capacitor?
    Relate the intensity to the potential difference.And use the capacitance formula.
     
  5. Apr 23, 2008 #4
    For the distance i got 2.95*10^10
    so the equation looks like
    1.05*10^-9= 8.85*10^-12 *Area/ 2.95*10^10 = 3.50*10^-12 which is different than the one i posted before but its still wrong! Help please!
     
  6. Apr 23, 2008 #5

    alphysicist

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    Hi sonrie,

    When you get a distance like 2.95 x 10^10 something must be wrong. That is a huge distance! What did you do to find that number?

    The equation that you then used next to solve for A is missing the dielectric constant. For a parallel plate capacitor with a dielectric, we have:

    [tex]
    C = \kappa \epsilon_0 \frac{A}{d}
    [/tex]
     
  7. Apr 24, 2008 #6
    To find the distance if did the following 5600/1.90*10^7 = 2.94*10^7
     
  8. Apr 24, 2008 #7
    sorry its 2.94 *10^10
     
  9. Apr 24, 2008 #8
    With the formula that you provided by equation will look like this:

    1.05*10^-9= 8.85*10^-12 *3.70* A / 2.94*10^10 so i just solve for A? then my final answer would be final answer was 9.47*10^-11 still wrong
     
  10. Apr 24, 2008 #9

    alphysicist

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    sonrie,

    That distance is huge! The radius of the earth is only about 6 x 10^6 meters or so.

    It looks like you just entered it into the calculator wrong. I get:

    5600 / ( 1.9 x 10^7 ) = 0.000295
     
  11. Apr 25, 2008 #10
    your Right! so my equation will be 1.05*10^-9= 8.85*10^-12*3.70*A/2.95*10^-4 . I finally got it RIGHT!
    Thank You So Much! Have a GREAT DAY!!!!!
     
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