Permittivity of Pyrex/Water

twinsen · Mar 23, 2007

Hey
As some of you may have seen i have beeen doing an experiment measuring the permittivity of water in an attempt to calculate the refractive index.
This has produced some odd readings as my experiments concentrate in the static range of permittivity i should be seeing a permittivity of water of about 80.
Reference - http://www.lsbu.ac.uk/water/explan3.html#diel [Broken]
An interesting thing happened the other day though when i was redoing the experiment with the capacitance plates outside a pyrex container filled with water my permittivity came out as about 4-5.
My question is do the plates carry the charge through the pyrex around the container or through the water.
It seems permittivty of pyrex is in this range and as far as i know does not vary as much with frequency unlike water.

It is quite hard to describe without a diagram but id like you guys oppinions on this.

Alex

lpfr · Mar 24, 2007

If I understand your description, when you measured with the plates outside the pyrex, between a plate and the other you found pyrex-water-pyrex (neglecting the two thin air gaps between the plates and the pyrex).
You can see this measure as three capacitors in series: one pyrex, one water and a second pyrex. I suspect that you calculated the permittivity ignoring the permittivity and the thickness of the two pyrex capacitors. Am I wrong?

twinsen · Mar 24, 2007

that sounds right but how do capacitors behave when in series?
I used a cube like pyrex container with water inside. I suppose the easiest way t work this out is to see what results i get with just the container with air inside.
Its weird that the relative permittivity of the whole container+water is so close to that of pyrex.
This is why i thought that the capacitor might be transfering charge so to say through the container and not the water.

Thanks
Alex

lpfr · Mar 25, 2007

Hi,
The inverse of the capacitance of several capacitors in series is the sum of the inverses of capacitances of all capacitors:
(1/Ct) = (1/C1)+ (1/C1)+ ....+ (1/Cn)
Here the three capacitors:
(1/Ct) = (1/Cp) + (1/Cw) + (1/Cp)
For simplicity, let's assume that the capacity of each capacitor can be approximated with the plate capacitor formula. To be valid, dielectric thickness must be much less than dimensions of plates. If the object you have is a cube the approximation is rather coarse, but it helps for the explanation.
Capacity of a plane capacitor is:
C= epsr * epsz * S / th
Here:
epsz = permittivity of vacuum (8.85 10^-12).
epsr = relative permittivity of dielectric (also called dielectric constant).
S = surface of plates.
th = thickness of dielectric.
Now if you write the capacity of all three capacitors and work a little the formulas you will obtain:
Ct= epsz * S / (2 thp / epsrp + thw / epsrw)
Where thp and thw are the thickness of the pyrex and the water.
And epsrp and epsrw are de dielectrics constants of pyrex and water.
The "weird" result is due to the fact that the term thw/epsrw is far smaller than the other: thp/epsrp. Then the total capacity is mainly determined by the pyrex capacitors. From an electrical point of view, the impedance of the water capacitor is almost a short-circuit compared to the impedance of the pyrex capacitors. From an electrostatic point of view, the potential difference is much smaller in the water than in the pyrex.
If instead water you put just air, the situation somewhat inverses and you will mostly see the air. The capacity of the air condenser is less than that of pyrex. But beware, all this depends on the thickness of each dielectric.
Hope that helps.

twinsen · Mar 25, 2007

That sounds good but i still get odd values for the permittivity of water maybe my data is rubbish as when i measure the permittivity of air which should be close to permittivity of vaccuum (relative permittivity of 1) i get a relative permittivity of 35.6 whats more is tat this varies with frequency from 0.66 @ 100khz to 267 @ 1kHz and 22.4 @ 10kHz when i am using a variable AC supply this is when using eps0 = C*d/A where
d = distance between plates
A = area of plate

for the reading of 35.6 i used a multi meter which had a capacitance setting. Do you think this would use an AC voltage to check capacitance and it would seem to be 10kHz<>1kHz

Here is a worksheet detailing the experiment for the other readings
http://www.fyslab.hut.fi/kurssit/Tfy-3.15xx/Tp_ohjeet/30en.pdf
I have also tried this experiment as it says on the sheet by varying the distances and then finding the gradient of the line. This gives me a permittivity for air of 1.3E-13
When i try putting the plates in the container instead of outside i get U1>U2 giving a negative number in the square root so i cant calculate the capacitance.
It is only with a larger gap that U1<U2 and the capacitance can be found. I will try with a bigger container next week to see if i can get a gradient that i can measure for water.

I think i am starting to understand it though.

Thanks
Alex

lpfr · Mar 25, 2007

Hi,
First thing, verify your measuring procedure.
Do an approximate estimation of the value of capacitance you are going to measure.
Take a real capacitor with a similar value and measure it with your installation. repeat at other frequencies.
If this works, it will exonerate your multimeter. Then you must examine the geometry of your pyrex capacitor, and also other parasitic capacitances due to the wiring.
Remember that for the formulas to work, "d" must be much smaller than the diameter or any dimension of the plates.
Another possibility is that your multimeter has an internal impedance to low. It must be at least 100 times the value of R. This could cause your negative number under the square root.

I do not understand how you can get U1 > U2. To measure U2 just short-circuit your capacitor with a bit of wire without touching any other part.

Good look.

twinsen · Mar 25, 2007

I would have though that the water would have short circuited the setup anyway. The value for U2 was just a tad bigger than U1 i thought maybe due to the way the multimeter measures AC current.

From an electrical point of view, the impedance of the water capacitor is almost a short-circuit compared to the impedance of the pyrex capacitors. From an electrostatic point of view, the potential difference is much smaller in the water than in the pyrex

How much smaller should the potential be this may be fluctuating reading on the multimeter causing U2>U1

Anyway i have until next friday to write up my report and finish off any experiments so i havnt got an endless amount of time.
Not even sure if im meant to be asking on forums and things but if im learning something its worth it.

Thanks
Alex

lpfr · Mar 25, 2007

I will not give you any indication that your own teachers wont have given to you.

Tell me: output resistance of your signal generator.
Value of R.
Dimension of your capacitor (approximate).

lpfr · Mar 25, 2007

Did you used tap water or distilled o deminaralized water?

twinsen · Mar 25, 2007

I used just tap water the plates are 0.0159 square meters each and the container is 0.186m each side.
For the resistance i attatched a variable resistor and set it at 100kohms and then measured the resistance across it for U1.

Thanks
Alex

lpfr · Mar 26, 2007

If you compute de capacitance of your pyrex capacitor, you must find something near 1 nanofarad. the resistance of your capacitor in tap water can be as low as 100 ohms (conductivity of tap water varies from town to town and even from tap to tap). Compute the modulus of the impedance (1/(2 pi F C)) and you will conclude that you must use non conducting water. That is either distilled or demineralized water. You can buy the last in food and hardware stores and even in gas stations (for lead-acid batteries).

Measure U2 short-circuiting the capacitor with a short conductor. Keep yours hands off before reading any measure: your body is resistive and capacitive.

Keep all mains power supply cables as far as possible from the capacitor and multimeter.

Use a high input impedance multimeter. All recent digital multimeters are high impedance.

Test your measuring device with a commercial capacitor, but do not expect to find the nominal value. Real values of most electronic capacitors are within 20% or the nominal value.

Hope that helps.