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Relationship Between Conductance and the Dielectric Constant

  1. Jun 5, 2013 #1
    Hello all,
    I'm new to the forum so I hope I'm posting this in the correct section.
    I'm currently conducting an experiment in which I'm examining how the dielectric constant of a binary mixture changes with concentration and temperature using a parallel plate capacitor for measurements. The only problem, however is that the multimeter I'm using doesn't measure capacitance. Here's what I can measure:
    • DC volts
    • AC volts
    • DC current
    • AC current
    • Resistance
    • Temperature
    • Frequency

    My advisor for this project initially suggested that I find the conductance (or conductivity?) in order to calculate the dielectric constant, however I've searched and searched but I can't seem to find a way to relate the two.
    Here's what I got so far.
    I know that conductivity can be used to calculate the imaginary part of the dielectric constant
    [tex]\kappa=\epsilon'+j\epsilon''[/tex]
    [tex]\kappa=\epsilon'+j\sigma/\omega[/tex]
    where [itex]\kappa[/itex] is the dielectric constant, [itex]\sigma[/itex] is the conductivity and [itex]\omega[/itex] is the frequency.
    but I'm not sure how I would find the real part [itex]\epsilon'[/itex]in order to get [itex]\kappa[/itex]
    I've also got the equations
    [tex]C=Q/V[/tex]
    [tex]C=\kappa\epsilon_{0}\frac{A}{d}[/tex] where [itex]\epsilon_{0}[/itex] is the vacuum permittivity, [itex]A[/itex] is the area of the plate and [itex]d[/itex] is the distance between the plates.
    Then for conductivity,
    [tex]J=\sigma E[/tex]
    [tex]G=\sigma\frac{A}{d}[/tex]
    where [itex]J[/itex] is the current density and [itex]G[/itex] is the conductance
    and so relating the two
    [tex]C=\kappa\epsilon_{0}\frac{G}{\sigma}[/tex]
    but that still leaves me with two unknowns, [itex]C[/itex] and [itex]\kappa[/itex]
    I feel like there's a really simple, really obvious solution that I'm just not seeing...
    Can anybody help me out?
     
  2. jcsd
  3. Jun 5, 2013 #2
    Welcome

    Given AC volts and AC current, you can get impedance. From that you can get capacitance.

    Is you AC source variable in frequency?
     
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