mathsciguy
I want to make a parallel plate capacitor without dielectric between it, that is, a capacitor with free space.

Can I do a working capacitor with just a two parallel cardboard coated in aluminum? I figured it should look like this:

Also, if there's anyone who's got a better idea on how I could make this, I'd really appreciate it if you could share that.

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Gold Member
To get yourself a measurable amount of capacity, the plates need to be close together and have a large area (also the cardboard may not be a good insulator).
It may be best to use your two bent sheets of aluminium (nice and rigid and flat) and hold them with four nylon screws and nuts, with one or two nuts as spacers. You can then adjust the capacity by altering the number of nuts/spacers. If you can't get access to a drill, some thin blocks of plastic could be stuck to the plates but make sure you have the plates well insulated so the glue must not bridge over the insulating plastic spacers.

mathsciguy
You mean I'd connect two aluminum sheets together directly by using nylon screws? Sorry, I can't quite visualize that, I'd be very thankful I you could provide an image with that. I'm sorry if this sounds so imposing.

Gold Member
The sheets need a small gap between them. I was suggesting a nylon nut (or two) in between them, to keep them apart. Thus:
Nylon screw head - plate - nylon spacing nut - plate - nylon nut (at each corner)
If the plates are reasonably thick, the single nut spacing should keep them far enough apart to allow for any small amount of bending of the plates
If you use four screws (near the corners) the arrangement would be nice and strong. Of course, you need to avoid the thing flexing and causing a short between the plates so thick(ish) metal would be appropriate. Solder lugs could be attached to the plates using the nylon screws too, to make a good connection. (Don't solder them whilst in place or the screw will melt!). Is that description better?
What was the capacitor to be used for? Are you hoping to charge it up with a high voltage source?

harrylin
I want to make a parallel plate capacitor without dielectric between it, that is, a capacitor with free space.

Can I do a working capacitor with just a two parallel cardboard coated in aluminum? I figured it should look like this: [..]

Also, if there's anyone who's got a better idea on how I could make this, I'd really appreciate it if you could share that.
Old radios were tuned with these:

Here the capacity was changed by changing the surface that the plates are facing each other.

A much simpler example is here:

(both from Wikipedia)

So yes, I agree with sophiecentaur. Basically you can make two plates like that, which you can make to face each other at short distance by means of nylon screws with spacers (nuts or rings). You can change the capacity simply by changing the spacers (or turning the bolts).

And yes you could make the plates by sticking aluminium foil on carton: that makes them more rigid. If they don't bend too much under gravity, you can just have a bottom plate on the table with a top plate above it, held up with the four screws. If necessary you can add a few insulating spacers in-between them.

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mathsciguy
Is that description better?

Yes, it is, thank you. I basically need nylon screws (which I think are really good insulators) to put the two plates together.

Another question, are aluminum foils (this is what I have at the moment) the optimal, cheapest, and most convenient conductor for this kind of set up?

As embarrassing as it is, I'm not sure what voltage reading is high or not but I'm planning to use something that generates not more than 12V.

Also, will I be able to slip a dielectric inside? What I'm actually planning is to read the voltage (potential difference) of a capacitor in free space and one with a dielectric.

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... a capacitor in free space and one with a dielectric.

I do understand what you mean, but just FYI, your "free space" is air and air IS a dielectric. Look in any table of dielectric constants and you will find air listed with a value of 1 (it's the standard/baseline). Perhaps what you mean is a solid dielectric.

Note that the radio cap in Harylin's post uses air as the dielectric.

mathsciguy
I do understand what you mean, but just FYI, your "free space" is air and air IS a dielectric. Look in any table of dielectric constants and you will find air listed with a value of 1 (it's the standard/baseline). Perhaps what you mean is a solid dielectric.

Note that the radio cap in Harylin's post uses air as the dielectric.

Oops yeah, sorry, I know that. Though that was rather, pedantic of you haha. Thanks anyway.

Gold Member
Yes, it is, thank you. I basically need nylon screws (which I think are really good insulators) to put the two plates together.

Another question, are aluminum foils (this is what I have at the moment) the optimal, cheapest, and most convenient conductor for this kind of set up?

As embarrassing as it is, I'm not sure what voltage reading is high or not but I'm planning to use something that generates not more than 12V.

Also, will I be able to slip a dielectric inside? What I'm actually planning is to read the voltage (potential difference) of a capacitor in free space and one with a dielectric.

What experiment do you actually plan to do? If you connect 12V across your C and look with a multimeter, then the voltage will drop pretty quickly when you disconnect your 12V. (I doubt that you could tell the difference with an without the C, aamof because the time for the volts to drop to 1/e of start volts is RC. C, in this case will be no more than, perhaps 1000 pF (10e-9F) and the resistance of your meter will be, say 10e6Ω. That is a time constant of 10e-3s.
This design of Capacitor is more suited to gathering charges which have been built up electrostatically and then producing small sparks with high voltages.
Adding a (commonly available) solid dielectric won't increase C by more than a factor of 5 so that won't help you.
This is one of those times when the actual values count if you want a result. For a parallel plate capacitor, the formula for C is
c= ε0εrA/d
which means it's proportional to area and inversely proportional to separation.
The thing about Capacitors is that they have to try quite hard to make them with values that are 'useful' for this sort of (low frequency) application. If you want a high enough value of capacitance for your purpose then the way to do it is using a sandwich of aluminium foil and 'cling film', rolled into a 'swiss roll'. Several metres of the sandwich, combined with the small gap which the cling film provides perhaps 1000times the value of your design. (You get a doubling of the area too! For free! This should give a time constant of about 1s, which will allow you to watch the volts drop at a sensible rate.
Just imagine the trouble they have to go to to produce a 10,000μF capacitor!

mathsciguy
^
Shoot, I didn't take the discharging into account, this is getting more and more complicated. What I actually want to do is to try to measure some dielectric constants from:
K=C/C_o
I got: K=V_o/V

Gold Member
^
Shoot, I didn't take the discharging into account, this is getting more and more complicated. What I actually want to do is to try to measure some dielectric constants from:
K=C/C_o
I got: K=V_o/V

I think you have this a bit muddled. To measure C from measuring V, you would need to put a given charge onto your Capacitor. How would you do that? - OR, if you knew the V, how would you measure the charge?
You have reached the Land of Measurement and it has some very difficult places, one of which you have found yourself in! You would expect to use special equipment for this job. Have you looked up methods of measuring Capacitance or Dielectric Constant?

The easiest ways to measure a C are to use the time constant with a known value of R or to use a source of AC and measure the Reactance, in one of various ways.
If all you need is εr then you can compare time constants for a Capacitor with and without a dielectric (you can treat Air as a vacuum because there's no way you can measure its εr). But you will still need a decent size of capacitor and that would be difficult unless you have a way to measure short times or make a vast air-gap capacitor (difficult). Alternatively, you can use an RF source and then a relatively small Capacitor can make its presence felt.

There is a really cool capacitor experiment I saw once as a kid. If you take a large tetra pak and a second one that you have cut open and turned inside out and stick it into the first. They are electronically separated by the thin layers of plastics and form a capacitor (you have to contact the aluminium layer with crocodile clamps or something similar). If you move the inner tetra pak in and out you can adjust the capacitance.

Gold Member
There is a really cool capacitor experiment I saw once as a kid. If you take a large tetra pak and a second one that you have cut open and turned inside out and stick it into the first. They are electronically separated by the thin layers of plastics and form a capacitor (you have to contact the aluminium layer with crocodile clamps or something similar). If you move the inner tetra pak in and out you can adjust the capacitance.
How was this demonstrated? I mean how was the value of this small C established?

It was a project where they build a radio from trash. So they said they used it as a trimmer. I don't know if they actually used it alone of if they had help from a real trimmer harvested from some old circuit board. Maybe it was enough to detune the radio. They might have just put it in for the optical effect though.

Gold Member
Maybe it was enough to detune the radio. They might have just put it in for the optical effect though.

That's probably what it was. A trimmer of a few tens if pF and something to make the set up nicely Heath Robinson.
Which is good fun but would be hard to use in a comparative study of dielectrics. Lol.