# Rolled paper capacitor is made from strips of metal foil

1. Apr 30, 2004

### denian

a rolled paper capacitor is made from strips of metal foil of dimensions 2 cm X 40 cm separated by paper of relative permittivity and thickness 0.002 cm. Estimate its capacitance.

the formula used is

2 X epsilon r X epsilon 0 X surface area
----------------------------------------
thickness

please explain why do we have to times 2 for rolled paper capasitor.

2. Apr 30, 2004

### gnome

First, thanks for your question. I never heard of &epsilon;r before. So now it seems that the "dielectric constant, &kappa;" which we learned about last year has had it's name changed to "relative permittivity &epsilon;r"? I wonder how universal that is.

In the process I found this site which has other useful information as well:
http://www.iupac.org/dhtml_home.html [Broken]

Back to your question:

Where did you find that formula? I think it is incorrect. For a capacitor with dielectric, I believe the correct formula is:
$$C = \frac{\epsilon_r\epsilon_0A}{d}$$
where A is the surface area and d is the distance separating them.

Last edited by a moderator: May 1, 2017
3. Apr 30, 2004

### turin

I learned it as &epsilon;r first.

4. Apr 30, 2004

### denian

$$C = \frac{\epsilon_r\epsilon_0A}{d}$$

that should be the formula. but my teacher say that for a rolled paper capacitor, we have to times two become

$$C = 2X\frac{\epsilon_r\epsilon_0A}{d}$$

i asked her, but she said, for rolled paper capacitor, we have to times 2 again...

5. May 1, 2004

### gnome

I finally found the answer. It seems your teacher is "almost" correct. The explanation (from David Swinscoe, "Roll Your Own Capacitors", IOP E-journals, Physics Education, vol 35 no 1, January, 2000) is that "both sides of each foil form capacitors and so effectively produce two capacitors in parallel."

I also found another article: S K Foong and C H Lim, "On the capacitance of a rolled capacitor", IOP E-journals, Physics Education, vol 37 no 5, September, 2002. They show that the capacitance of rolled capacitors approaches 2 x the capacitance of a flat capacitor with the same size (foil) plates as the number of turns increases. If you think about this you will see why: If you have 2 foil "plates" rolled just one complete turn, you just have a single "bent" capacitor, and there is none of this doubling-up effect at all. If you roll the same 2 plates (interleaved with 2 pieces of paper) two turns, there will now be 3 pairs of conductor surfaces separated by paper, but each of those pairs will be half the size of the original pair since the two-turn tube has half the diameter of a one-turn tube. Therefore, the capacitance of a two-turn rolled capacitor will be 3/2 the capacitance of the flat capacitor made with the same amount of foil. With three turns, the capacitance will be 5/3 times that of the flat capacitor.

The more turns, the closer it gets to double.

(However, I also looked it up in Serway, "Physics for Scientists and Engineers" which indicated that a rolled capacitor was effectively the same as a parallel plate capacitor and said nothing about this doubling effect. Oops.)

6. May 1, 2004

### denian

alrite. thank you.

7. May 3, 2004

### turin

Hmm. What edition do you have? I also have this book, but I couldn't find any detailed discussion of this type of capacitor. It is addressed briefly in a paragraph1 with no mention whatsoever of how to calculate the capacitance. I am still searching for further reference.

1 R. A. Serway. Physics: For Scientists and Engineers, 4th ed. (Saunders Coll. Publ., Philadelphia, 1996), pp. 753-4.

8. May 3, 2004

### gnome

I have the 5th edition.

You're right, there is no detailed discussion of the rolled capacitor. It's described briefly with an illustration of a capacitor with several (approx. 6) turns (Fig. 26.15a in this edition) and no mention there of how to calculate the capacitance. But at the end of the chapter, in Problem 43:
So, while they didn't exactly say that this was the correct way to calculate the capacitance, they certainly gave us that impression, and explicitly directed us to calculate it that way.

Last edited: May 3, 2004
9. May 3, 2004

### turin

gnome,
Our editions seem rather different. Yes, your edition certainly seems to give the impression, I agree. It is one of my pet peeves for a text book to use its excercises to introduce new material (i.e. Jackson's piece of crap E&M text or Arfken's convoluded math methods text).

denian,
If you're still hangin' around, my advice would be to not take the specific formula too seriously. In fact, if you decide to pursue physics, you will often run into order-of-magnitude formulas (I have found this to be particularly the case in QM) that are considered "good enough." What you're teacher has given you is one such example. Further support for your teacher's position is in the instruction to "estimate" the capacitance.

Well, this has been informative.

10. May 3, 2004

### gnome

Yes. I'm glad this question came up. But I don't want to leave the impression that I dislike the Serway book. In a work of this size there are bound to be some imperfections. In general, I think it's an excellent textbook, and I have no objection to the practice of introducing new material, or new ways of looking at the material, through problems. I think we get a more thorough understanding of the concepts if we have to figure out how to apply them in new ways rather than always mimicking the author's solutions. Anyway, it does include examples of how to solve a wide range of problems, and the solutions that are provided in the text are presented very clearly.

And if additional clues are needed, there's always Physics Forums.

11. May 4, 2004

### turin

I actually like the Serway book as well, my pet peeve notwithstanding.

Did you ever try to use the Jackson E&M book or the Arfken grad math methods book? If you are able to learn well from these texts, then you are a better man/woman than I.

12. May 5, 2004

### gnome

No, I've never used either of them. I'm just in an intro level modern physics course.

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