1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Homework Helper

    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

    User Avatar
    Homework Helper

    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

    User Avatar
    Homework Helper

    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!!!!!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Similar Discussions: Dielectric with a parallel-plate capacitor finding minimum area
Loading...