# Relationship Between Conductance and the Dielectric Constant

1. Jun 5, 2013

### AnnaLinnea

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
$$\kappa=\epsilon'+j\epsilon''$$
$$\kappa=\epsilon'+j\sigma/\omega$$
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 $\epsilon'$in order to get $\kappa$
I've also got the equations
$$C=Q/V$$
$$C=\kappa\epsilon_{0}\frac{A}{d}$$ 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,
$$J=\sigma E$$
$$G=\sigma\frac{A}{d}$$
where $J$ is the current density and $G$ is the conductance
and so relating the two
$$C=\kappa\epsilon_{0}\frac{G}{\sigma}$$
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?

2. Jun 5, 2013

### the_emi_guy

Welcome

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

Is you AC source variable in frequency?