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LC circuit, space between dielectric plates

  1. May 12, 2013 #1
    1. The problem statement, all variables and given/known data

    If the max E-field in the capacitor is E=1.2(103) N/C and the space between the plates is filled with a dielectric of constant 100,000, what is the separation between the plates?

    2. Relevant equations

    I know C=kεoA/d where k is the dielectric constant, A is the area of one of the plates, and d is the separation.

    3. The attempt at a solution

    So this problem came with multiple parts. Initially we are given the capacitor has capacitance 30μF, the inductor has inductance 0.7H, and the total energy is 44μJ. Its a simple closed loop with no battery. Parts a-e i found that the time for the energy to reach the inductor is 7.2ms, the charge on one side of the capacitor at t=0 is 51.4μC, the current at 7.2ms is 11.2mA, the inductor has length 4cm, radius 2.8cm, and 3010 turns, and the B-field in the inductor is 1.06mT.

    The question I'm asking about is part f....the next part g asks for the area of one of the plates.
    I just simply dont see how there is enough information to deduce either the space between the plates or the area of one. The equation above shows that that means there is two unknowns.
    Seeing as we are given the E-field strength, I thought we could maybe use

    E=σ/kεo,

    I solve for σ, and plugged in everything else, and the value I got out for σ was 1.06. The same value I got for the B-field strength...I'm just not really sure where to go with any of this, been stuck for a while.
     
  2. jcsd
  3. May 12, 2013 #2

    TSny

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    Homework Helper
    Gold Member

    Can you find the maximum potential difference Vmax between the plates of the capacitor? If so, then you can find the distance d between the plates using a relationship among Vmax, Emax, and d.
     
  4. May 12, 2013 #3
    Ahhhh thank you. I wasn't aware there was a way a could find the potential difference in the plates. Now I found the separation, and in turn the area. I appreciate the help!
     
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