The above is pretty much all there is to the question. By "can you," I mean will it be properly effective / are there any tips? If not, why? It seems that this ought to work, but I can't find info online.
Thanks.
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\$\begingroup\$ you mean as plates? Or as dielectric? \$\endgroup\$– Marcus MüllerCommented Jul 16, 2020 at 22:29
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\$\begingroup\$ @MarcusMüller I mean can you use it as the dielectric, between my two copper plates. \$\endgroup\$– EEEEslCommented Jul 16, 2020 at 22:30
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1 Answer
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Sure; the capacity of a plate capacitor (Wikipedia is your friend!) proportional to three things:
- The area of the opposing plates (the larger, the better)
- the inverse of their distance (the closer, the better)
- the dielectric constant of the dielectric (the higher, the better)
Your isolating will probably have dielectric constant higher than that of air.
What it also might have might be a higher or lower isolation strength (i.e. at which voltage there will be arcing through it).
Also, there might be higher losses than in an air-filled capacitor when using the capacitor at high frequencies.
answered Jul 16, 2020 at 22:34
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\$\begingroup\$ Thanks! I'll mark as answered when it lets me! \$\endgroup\$– EEEEslCommented Jul 16, 2020 at 22:36
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\$\begingroup\$ The dielectric constant ought to be around 6-7. \$\endgroup\$– EEEEslCommented Jul 16, 2020 at 22:40
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\$\begingroup\$ so that is the factor you get over an air gap of the same thickness. \$\endgroup\$ Commented Jul 16, 2020 at 22:41
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\$\begingroup\$ Why would the loses be greater? \$\endgroup\$– EEEEslCommented Jul 16, 2020 at 22:44
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\$\begingroup\$ Dielectric losses can have a number of reasons, but in essence: if there's a few charges in a not-perfect isolator, they will be "pulled around" by the changing electric field caused by AC over your capacitor. And if these moving charges are met with resistance, that will convert energy from the electric field to heat. That effect happens on mulitple scales, from polar molecules oscillating around their own center to actual movement of the isolator. I don't know your rubber well enough to quantize that! However, the way you ask this: irrelevant at frequencies you care about, probably. \$\endgroup\$ Commented Jul 16, 2020 at 22:49