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Percentage of energy in a dielectric placed between a parallel plate conductor

  1. Apr 21, 2009 #1
    1. The problem statement, all variables and given/known data

    A dielectric of dielectric constant 3 fills three fourth of the space between the plates of a parallel plate capacitor. What percentage of the energy is stored in the dielectric?

    E(r) = 3, distance between plates = d, thickness (t) = (3/4) * d, C=capacitance of the capacitor, q= charge of the capacitor, V= potential difference, A =area of plates, E(0) = permittivity of free space

    2. Relevant equations

    E = C(V^2)/2,
    C = E(0)A/( (d-t) + t/E(r) ),
    C= q/V

    3. The attempt at a solution

    I solved this (sort of) as follows:
    The capacitance C' if there was no dielectric = E(0)A/ d
    The capacitance C with the dielectric = 2 E(0)/ d [ after giving t = 3/4 d and simplifying ]
    The energy E' in the absence of dielectric = C'(V^2)/2
    The energy E in the presence of the dielectric = C(V^2)/2 = 2E'
    percentage of energy in dielectric = (2E' - E')/2E' * 100 = 50 %

    I don't know if this approach is correct. Please help.
     
  2. jcsd
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