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Parallel plate capacitor problem

  1. Jun 30, 2009 #1
    A parallel plate capacitor is formed using a material whose dielectric constant is 3,00 and whose dielectric strength is 2x10^{8} V/m. The capacitance is 0,25x10^{-6}F and the capacitor must withstand a potential difference maximum of 4000V. Find the minimum area of the plates of the capacitor

    I have

    C = (E. * A * k)/d

    How to calculate d ?
     
  2. jcsd
  3. Jun 30, 2009 #2

    LowlyPion

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    Re: Capacitor

    How will the dielectric strength constrain the distance separation?

    Then solve for A.
     
  4. Jun 30, 2009 #3
    Re: Capacitor

    the dielectric strength (2x10^8 V/m) is the separation of the plates?
     
  5. Jun 30, 2009 #4

    LowlyPion

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    Re: Capacitor

    No. That's a property of the material if it was 1 meter thick.

    How thick does it need to be to yield a 4000 V rating?
     
  6. Jun 30, 2009 #5
    Re: Capacitor


    V = (q*d)/(E.*A*k)

    It is related to d and A again
     
  7. Jun 30, 2009 #6
    Re: Capacitor

    d = 4000 / 2x10^8 = 2x10^{-5}

    A = (C*d)/(E.*k) = 1,88x10^{-7}

    It is ?
     
  8. Jun 30, 2009 #7

    LowlyPion

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    Re: Capacitor

    Like I said, find d.

    For this problem minimum d is the thickness required to provide a 4000 V separation between the plates. That means that with a dielectric strength of 2*108 you need a thickness of 4000/2*108 2*10-5 m.

    Then apply that to Area calculation.

    Edit: I see you have in your next post. So yes. That looks like the right way to do it.
     
  9. Jun 30, 2009 #8
    Re: Capacitor

    Thank you very much!
     
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