# A Sample smaller than parallel plates in capacitance setup

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1. Jul 20, 2016

### SSChemist

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.

2. Jul 20, 2016

### Nidum

Why can't you just make some new plates the right size to match the samples ?

3. Jul 20, 2016

### SSChemist

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.

4. Jul 20, 2016

### Nidum

(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 .

5. Jul 20, 2016

### Nidum

6. Jul 22, 2016

### CWatters

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?