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Find the maximum potential difference that can be applied to a capacitor

  1. Oct 8, 2009 #1
    1. The problem statement, all variables and given/known data
    Determine a) the capacitance and b) the maximum potential difference that can be applied to a Teflon-filled parallel-plate capacitor having a plate area of 1.75 cm^2 and plate separation of .04 mm.
    2. Relevant equations

    A=.0175
    d=.04e-3
    k=2.1
    C=k(epsilon-not)(A/d)
    Vmax=E*d
    Qmax=C*(delta)Vmax=C(Emax*d)

    3. The attempt at a solution

    C=2.1*8.85e-12*(.0175/.04e-3)=8.13e-9 F

    If Vmax=E*d, how do I solve for E to find Vmax?

    Not really sure where to go with this...
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 8, 2009 #2
    Don't you need to know the dielectric strength of Teflon?
     
  4. Oct 9, 2009 #3
    That would be "k"-->k=2.1

    I guess I need to know how to determine what the electric field is in between the plates, as a function of the dialectric.
     
  5. Oct 9, 2009 #4

    Redbelly98

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    Staff Emeritus
    Science Advisor
    Homework Helper

    k is the dielectric constant. Dielectric strength is the maximum electric field that a material can withstand.

    Chances are you were provided with dielectric strength values -- probably in your textbook. Check the relevant section, or try looking in the book's index.
     
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