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C is the capacitance
l is the length of the wires
a is the wire radius
d is the distance between the wire centers
ε is the permittivity of the media the wires are surrounded by.
The capacitance in air takes the 'relative' permittivity of free space to be 1.
The capacitance of 2 wires in parallel and using a dielectric is not simple.
If the wires are in a screened pair arrangement, then this defines the electric field.
If the two wires are part of an unscreened cable, then the dielectric will be part air and part plastic. Because of the higher voltage gradient of the electric field close to the conductor surface, then the plastic insulation will have a major effect even though the amount of insulation is not very much.
The relative dielectric constant for polyethylene and PTFE and polypropylene is close to 2.3.
For PVC and other odd materials, the dielectric constant can be MUCH higher due to the use in some of these materials of fillers and plasticisers. These materials are often used in order to improve the mechanical, or fire retardent or oil resistant properties.
So, do you have a specific cable in mind?
hope this helps.
I am actually working on changes in capacitance in different fluid dielectrics such as water. Assume two basic wires of known length and diameter and distance from each other. In aire they are X1, water, X2, etc.
What are you actually trying to do?
What you say is clear but I cant get what you are wanting to know.
Is it the conductor dielectric you are trying to calculate, or the dielectric constant of the medium in which the wires are immersed?
The following online calculator calculates the capacitance using the dielectric constant (relative permittivity) as input, to within almost 3 significant digits of match with the formula in post #2.
Curtis, of you run the numbers on this online calc vs the long hand from the equation on wiki, they are different values! I am curious if there is an issue with one on the using the radius or diameter? Since I cannot see how the online calc is worked, I cannot be certain. Just trying to verify this? Any other place we can gather or verify the equation? Hard to believe someone would go through the effort to make an online calc and not work it correctly but I guess that is why I like to run the numbers to verify.
Hi, fastline. I put the equation in post #2 into a spreadsheet and I get the same result as the online calculator, within reasonable precision. The post#2 equation uses the radius of the wire, the online calculator uses the diameter of the wire. The only thing that is not obvious is that the absolute permittivity entered into the equation in post #2 is not the same as the dielectric constant value entered in the online calculator. The absolute permittivity relates to the dielectric constant as follows:
ε = εr x ε0
where ε is the absolute permittivity in Farads/mm (F/mm) εr is the dielectric constant (relative permittivity, unitless) ε0 is 8.8541878176E−15 F/mm (the permittivity of a vacuum)
Of course, as always, check your decimal points and units conversions carefully.
using the formula in #2 is interesting but fraught with difficulty. With a pair of insulated wires, the 'effective permittivity' is a mixture of air and dielectric, and also the effective 'earth radius' is located at infinity.
So, how curtis is dealing with these uncertainties is not clear.
In post #4, you said you were working in a confined space so the basic formula will be incorrect because the effective screen diameter is not at infinity.
In post#5 I asked about the actual arrangement of your situation, because it is possible that a single coaxial arrangement will be more suitable. I cannot understand why, for the information you require, you need to have a pair of conductors. If you wish to measure capacitance of the pair arrangement, you will need a balanced bridge.
At some point you will need to make a capacitance measurement and this part of the problem may defeat you.
It is for all these reasons that I asked for a bit more info.
As far as I see it, there are only two bare, solid, parallel wires immersed in a dielectric substance, all around with no gaps for a considerable distance. The equation should be a pretty good approximation in that case for low frequency excitation. Of course, the reality is that there are numerous considerations in practice that impact the capacitance, unaccounted for in an ideal only equation.
Good on you Jim for the reference.
I understood the problem to be about capacitance of pair type cable. Your observation indicates it may be different.
From the perspective of wire drawing, I can say that the 'characteristics' of wire drawing solutions are related to the 'soap' content and that an optical refractive index type of measurement is typically used.
We do grape growing and wine making, and we use an optical refractive index measuring gauge which is calibrated in 'degrees Brix'. This is for measuring sugar concentration in grape juice. I am aware that similar gauges are used for the measurement of the properties of wire drawing solutions.
I note from your reference, the use of a pair of wires to measure the lubricant characteristics.
It will be interesting to see what comes back.
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