Sample smaller than parallel plates in capacitance setup

AI Thread Summary
The discussion centers on measuring the capacitance of samples smaller than the parallel plates in a capacitance setup, which complicates obtaining accurate dielectric constants. The main issue arises from the capacitance measurement including contributions from both the sample and air, making it difficult to isolate the sample's dielectric properties. The user has attempted to approximate the dielectric constant by treating the setup as two capacitors in parallel, but recognizes this method has limitations due to area discrepancies. Suggestions include using a probe-like setup for better sample fitting and a substitution method to measure capacitance with and without the sample to infer dielectric properties. Overall, accurate dielectric constant measurements require careful consideration of the area contributions in the capacitance calculations.
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Hi,

My setup for measuring parallel plate capacitance consists of an LCR meter and a parallel plate test fixture. The area of the parallel plates is much larger than the area of the samples I am measuring. From what I've seen in literature, and my physics classes, the sample area should be either equal or larger than that of the parallel plates. I haven't been able to find any information on capacitance measurements where the sample area is smaller than that of the plates. My goal is to obtain accurate values for dielectric constants of materials.

The problem: When measuring capacitance of a sample with smaller area than the area of the parallel plates, the obtained capacitance includes a contribution from both air and the sample.

The question: How does one "subtract" the air contribution? Is there a way to obtain accurate values for dielectric constant with the setup described above?

My attempt: The closest approximation of dielectric constants I was able to get was when I treat the entire setup (air + sample) as two capacitors connected in parallel; that is, C(total) = C(air) + C(sample). Using this approximation, most of the dielectric constants were within 10% error with a few being higher than 10%. However, I know that this is not correct because the area (A) in the parallel capacitance equation ( C = k e A / d) refers to the area of the parallel plates, not the sample. In my case, for C(sample), I am using the area of the sample itself.

Any thoughts would be helpful.
 
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Why can't you just make some new plates the right size to match the samples ?
 
I would do that but I wouldn't have a way to connect the leads to the newly made plates. Also, I have a variety of sample sizes, so one plate size/shape would not fit all. Keep in mind, the samples are tiny, areas of about 25 mm2 and the area of the plates on the test fixture is 3180 mm2.
 
(1) Something more like a probe than a plate as such . Copper rod . Machined end face is the plate . Reduce diameter of bar for a distance above plate then revert to full diameter for holding and connections .

(2) I have some memory of reading about a substitution method for the case of plates bigger in area than the dielectric separator . Can't give a reference but works something like testing with sample in place and then removing sample without altering plate separation and then inferring the dielectric properties from the difference of capacitance readings .
 
probe v2.png
 
Your approach of treating it as two capacitors in parallel should work. I don't follow why you think it doesn't. The area of C(sample) should be the area of the sample. The area of C(air) should be the area of the plates minus the sample. Is that what you were doing?
 
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