# 'Rolled' homemade capacitors

mathsciguy
By rolled, I mean this kind of basic homemade capacitors:
http://www.ehow.com/how-does_4852927_building-a-capacitor.html

Kind sirs, I have few questions regarding this kind of capacitor:
1.) Would I still be able to get the 'theoretical free-space' capacitor of the same dimensions using $C_o = \frac{A \epsilon _o}{d}$? I'm guessing A should be the area of the paper dielectric.
2.) For those who didn't bother looking up the link, the parallel plate capacitor is made by sandwiching aluminum foils and papers; in this case it's: foil, paper, foil, paper, in that order. My question is, this is just one capacitor right? The last paper layer is there just so the two conductors doesn't touch and cause a short circuit, which also doubles as a solid dielectric.
3.) What would be a good substitute (and cheap) dielectric if we are still to use the same set-up besides paper?

Thanks in advance.

Last edited:

## Answers and Replies

Staff Emeritus
Science Advisor
Gold Member
It is a single cap.

Try using saran (plastic) wrap.

mathsciguy
Is that plastic polyethylene? Sorry, I'm not familiar with materials. I'm asking because I'd also like to look up the dielectric constant of the materials I'm using.

mathsciguy
Oops, it's polyvinylidene chloride I could've looked that up earlier. Anyways, anyone willing to answer 1?

Staff Emeritus
Science Advisor
Gold Member
A is the area of your plates.

mathsciguy
I tried doing this, and I tried to read the potential difference across the plates after I charged it using a voltage source. I did get some reading, but it diminishes steadily over time, am I doing it right?

Last edited:
Science Advisor
Gold Member
If the volts are dropping slowly then it must be working (well done) and it is discharging slowly through the meter. You can find an approximate value for the Capacitor you have made as follows.
Charge it and find how long (seconds) it takes to discharge to 1/e of the start value (1/2.7); this is called the time constant. Assume the input resistance of the meter is about 10MΩ.
The Capacity will be approximately 1/(10e7 X time constant) in Farads.

Try applying the meter after a short while. If the volts have dropped without the meter there, there is another 'leakage path'. Not unlikely but make sure you are not causing a path through your fingers - they can easily cause some leakage. Also, some damp can be to blame.

mathsciguy
By discharge, do you mean the actual charge of the capacitor or its voltage? Also, by start value, is it the applied voltage or the initial charge?

Also, you are talking about $\tau = RC$ where tau is the time constant, R is the resistance, and C is the capacitance. Right?

Last edited:
Science Advisor
Gold Member
The voltage is proportional to the charge in it. (Q = CV)
The discharge follows an exponential law. The volts drop by the same proportion in the same interval, whatever voltage you start with. You could choose any time and read the meter, then wait till the volts drop to the 1/2.7 value. But it would probably be easier to measure the voltage with the battery (?) connected and calculate the 1/2.7 value before you start - then disconnect the battery and watch the volts fall. How long, approximately are we talking about, for it to go to half its initial value? A stopwatch (an app on your smartphone, perhaps?) would be handy.

mathsciguy
Ok I get it. I have a digital multitester here do you think it's safe to assume 10MΩ as the deflecting resistance of the voltmeter?

Science Advisor
Gold Member
Mega Ohms!!!!

mathsciguy
Sorry, I edited it just now lol. I'm quite groggy at this time already. Anyway, your posts were really helpful, thank you.

Edit: I really just want to make things clear, I can use the resistance of the voltmeter which is 10 Mega Ohms as the resistance for the time constant equation right? (My knowledge of how these meters works is still shaky.)

Last edited:
Science Advisor
Homework Helper
Gold Member
Deleted

Science Advisor
Gold Member
Sorry, I edited it just now lol. I'm quite groggy at this time already. Anyway, your posts were really helpful, thank you.

Edit: I really just want to make things clear, I can use the resistance of the voltmeter which is 10 Mega Ohms as the resistance for the time constant equation right? (My knowledge of how these meters works is still shaky.)

Yep.
But check that there is no decay when the meter is disconnected. Dab the meter on after a while to see. Then you can be sure that the dominant (lowest) resistance in the RC time constant is the R of the meter.

Be careful with your decimal points and expect a 'low value' then let us know how well you have done.

mathsciguy
The voltage drops really quick the last time I tried it, I don't think I can measure the time with a stopwatch. Would having a larger capacitor help?

Science Advisor
Gold Member
The voltage drops really quick the last time I tried it, I don't think I can measure the time with a stopwatch. Would having a larger capacitor help?

Of course, double the Capacitance, double the time.
To get a noticeable difference, though, you would probably need ten times the value of C.
But if you can see any delay at all, compared with the meter without a capacitor, your time constant will be several tens of ms. You could do your calculation on that assumption.

Digital displays are not too good for this sort of exercise. Ideally, you'd use an oscilloscope which could give you a nice trace or an electronic meter with a needle display (an old valve voltmeter would do the job fine!)

I have a DMM which happens to have a switch position with capacitance measurements - that would be good, of course. I used to use it when I had students doing the same exercise as yours. Could you borrow one, perhaps? C measurement is quite common on the more expensive meters you can get.

mathsciguy
Of course, double the Capacitance, double the time.
To get a noticeable difference, though, you would probably need ten times the value of C.
But if you can see any delay at all, compared with the meter without a capacitor, your time constant will be several tens of ms. You could do your calculation on that assumption.

Digital displays are not too good for this sort of exercise. Ideally, you'd use an oscilloscope which could give you a nice trace or an electronic meter with a needle display (an old valve voltmeter would do the job fine!)

I have a DMM which happens to have a switch position with capacitance measurements - that would be good, of course. I used to use it when I had students doing the same exercise as yours. Could you borrow one, perhaps? C measurement is quite common on the more expensive meters you can get.

I also have a valve voltmeter though, but I don't think I could barrow that kind of DMM you're talking about. I'm quite interested how could we do this: "But if you can see any delay at all, compared with the meter without a capacitor, your time constant will be several tens of ms." assumption? I'd be really really thankful if you could expound on that.

harrylin
The voltage drops really quick the last time I tried it, I don't think I can measure the time with a stopwatch. Would having a larger capacitor help?
If you have video camera or photo camera with video mode, then you can record at least 15 frames/second. However, not sure if the volt meter is fast enough...

Science Advisor
Gold Member
I also have a valve voltmeter though, but I don't think I could barrow that kind of DMM you're talking about. I'm quite interested how could we do this: "But if you can see any delay at all, compared with the meter without a capacitor, your time constant will be several tens of ms." assumption? I'd be really really thankful if you could expound on that.

I just meant that, if it was too fast for you to think of using a stopwatch yet it was a 'visible' delay, then that would fit in with that sort of time constant. Any quicker and you'd not spot it at all. 10ms is the period of flashing of a mains filament bulb (UK @ 50Hz) and you can only just see flicker. I would have thought that the eye could pick up on several times that time period. Perhaps 100ms would be a more likely value for the threshold of perceiving the decay. But it's only a ball park figure and could give you an idea of just how good your capacitor is. You could also work out the probable capacitance if you could measure / estimate the thickness of the film (measure many thickenesses and divide by n, rather than trying just one thickness) Look up the formula for a parallel plate capacitor and can calculate the area of foil (double it because it's a rolled sandwich) you can also find the relative permittivity of the PVC. See what agreement you get. It won't be very close because the spacing will be far from uniform inside the roll - but don't worry. Even a vague agreement will be satisfying (as in calculating the Mass of the Universe lol).

mathsciguy
Well, the problem is, what I'd like to do is compute for the dielectric constant experimentally so It'd be good if I could get an accurate result for calculating the resulting capacitance with a dielectric. Also, the last time I tried it with a 3V potential difference, the voltage drops straight to about 0.90~80V which is quite lesser than the 1/e of the initial value.

I guess I'd try doing it with the valve meter now and look at what figures I can get.

Science Advisor
Gold Member
Well, the problem is, what I'd like to do is compute for the dielectric constant experimentally so It'd be good if I could get an accurate result for calculating the resulting capacitance with a dielectric. Also, the last time I tried it with a 3V potential difference, the voltage drops straight to about 0.90~80V which is quite lesser than the 1/e of the initial value.

I guess I'd try doing it with the valve meter now and look at what figures I can get.

You have chosen a very difficult experiment if you don't have some fairly specialist electronics equipment, I'm afraid. If you want to 'measure' the dielectric constant, you will need to have a capacitor with a removable dielectric (to compare C with air and with dielectric). That means it needs to have an air gap so it will need to be self supporting. The mechanics dictate that it must have a pretty low capacity (small area and relatively large spacing) so you would need to measure RF characteristics and not just 'discharge'. That's because the time constant in any circuit will be in the order of microseconds rather than nearly a second.
It's definitely not a kitchen table top job.

That means your time constant must be much less than 100ms.

mathsciguy
You have chosen a very difficult experiment

Yes, I have realized this the hard way, one of the reasons I'm playing with the idea of doing theoretical work (I don't know on which field yet). Doing experiments just isn't my thing, at least for now that's how I see it.

Thank you Sophiecentaur, you have given me so much insight on this, I'd still probably continue the experiment some other time, though, with proper equipments and supervision.

Last edited:
Science Advisor
Gold Member
Actually. I only said it was hard if you don't have access to the equipment. All you need is an RF bridge - or equivalent - which you would expect to find in any RF electronics lab. Then you could easily measure the capacity of a suitable air-spaced capacitor, prior to inserting your dielectrics. It's just not a DIY exercise. But that applies to many practical experiments and it shouldn't put you off the idea of experimenting. Grab any opportunity for practical work. I think you are far too young to be saying that any aspect of learning "just isn't my thing". If you want an insight into any part of Science, it will involve the analysis of practical results and, if you have even an inkling about how they were produced, you have a great advantage. It's much easier to smell a rat and question data if you know how it was obtained - even if someone else can get it for you in less time or cheaper.