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

\(\kappa=\epsilon^{\prime}+j\epsilon^{\prime\prime}\)

\(\kappa=\epsilon^{\prime}+j\sigma/\omega\)

where \(\kappa\) is the dielectric constant, \(\sigma\) is the conductivity and \(\omega\) is the frequency.
but I'm not sure how I would find the real part \(\epsilon^{\prime}\) in order to get \(\kappa\)
I've also got the equations

\(C=Q/V\)

\(C=\kappa\epsilon_{0}A/d\)

where \(\epsilon_{0}\) is the vacuum permittivity, \(A\) is the area of the plate and \(d\) is the distance between the plates.
Then for conductivity,

\(J=\sigma E\)

\(G=\sigma A/d\)

where \(J\) is the current density and \(G\) is the conductance
and so relating the two

\(C=\kappa\epsilon_{0}G/\sigma\)

but that still leaves me with two unknowns, \(C\) and \(\kappa\)
I feel like there's a really simple, really obvious solution that I'm just not seeing...
Can anybody help me out?

 

I'd ask your advisor where they would recommend you look to find a relationship between conductance and dielectric constant. Perhaps turn it around and ask him, given the dielectric constant of free space, is he claiming you can find the conductance of free space?

If your problem is that you don't have a meter that measures capacitance, then spend the $10 to $20 to get one that does!

What values of capacitance are you expecting to see? That will determine whether a cheap meter will do what you need.

 

My advisor for this project initially suggested that I find the conductance (or conductivity?) in order to calculate the dielectric constant...

I think there's a good chance he misspoke or you misunderstood what he said. A suggested, you should talk to him.

I'm no expert but I think the dielectric constant of a material is distinct from its conductivity and there is no fixed relationship between these two physical properties. It's like trying to measure melting point by looking at the color. You can't get there from here. I'll be glad to learn if I'm way off.