- #1
SSChemist
- 2
- 0
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.
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.