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A few times I've been reminded that an object can have self capacitance and it's just not registering with me how this can be so. I'm sure there's a good explanation. Here's the explanation I am usually given but I'm still not grasping that capacitance can exist when there is only one "electrode": -

enter image description here

The implication of this is that if the outer sphere were massively distant from the inner sphere the formula reduces to: -

Capacitance = \$4\pi\epsilon_0\times a\$ because \$\dfrac{1}{b} = 0\$

This is used as the argument that the middle object has self-capacitance

See this as the follow on proof. Maybe I'm being stupid?


EDIT to ask a directly related question: -

I've correctly calculated that a 1mm radius ball will possess a so-called "self capacitance" of 0.111pF. If two such balls existed and were placed a million miles away from each other (in an empty universe) would the capacitance between them approximately equal 0.0555pF i.e. half of 0.111pF? This "guess" is based on the capacitance of a 1mm ball to an infinite sphere being 0.111pF and from the infinite sphere to the other ball is another 0.111pF.

Therefore, with 2 capacitors in series of the same value, the net capacitance is halved. I can't believe that this is true but I'm not really sure.

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  • \$\begingroup\$ Maybe I'm thinking wrong, but if you can put charge onto something, it must have a capacitance. But if there is no other electrode you wouldn't be able to put charge on something (how would you apply a voltage to make the charge go there). So maybe it's just the huge distance of the other electrode (earth?) to make the term 1/b negligible. \$\endgroup\$ – Arsenal Oct 29 '15 at 8:51
  • \$\begingroup\$ Are you looking for something that makes it more intuitive, or for something involving maths that prove it? For the earlier, imagine some cable, to make the voltage change on the other end, you have to push some current through it. \$\endgroup\$ – PlasmaHH Oct 29 '15 at 8:51
  • \$\begingroup\$ I'm guessing you're talking about the Wikipedia argument en.wikipedia.org/wiki/Capacitance#Self-capacitance: "Self-capacitance is the amount of electric charge that must be added to an isolated conductor to raise its electric potential by one unit (i.e. one volt, in most measurement systems). The reference point for this potential is a theoretical hollow conducting sphere, of infinite radius, centered on the conductor." \$\endgroup\$ – Fizz Oct 29 '15 at 9:09
  • \$\begingroup\$ @PlasmaHH I'm hoping for an intuitive answer - I have the maths in the question and I know I'm being stupid in not understanding this fundamental thing - I'm just not seeing something that I should see!! \$\endgroup\$ – Andy aka Oct 29 '15 at 9:10
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    \$\begingroup\$ I think the self capacitance is just useful to set a lower bound for the capacitance of a body. This body close to the earth, as in a real situation, will have then a bigger capacitance than the calculated "self" one. \$\endgroup\$ – Roger C. Oct 29 '15 at 9:48
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To expand on my comment, this notion of self-capacitance is a theoretical one, much like, say, an ideal operational amplifier is; the latter has infinite [open-loop] gain. Here you have a capacitor with infinite distance between plates and the 2nd, reference electrode/sphere also has infinite area. When one comes across such idealizations based on thought experiments, the question to ask is not whether it can exist, but ask why is it a useful notion.

Basically this notion is useful as a first-order approximation to problems where the distance is finite but large. For example, enter image description here

"Equation 5.4" there simply estimates the mutual capacitance of two [identical] conductors at large distance as one-half of the self-capacitance.

Similarly, a first-order approximation of the capacitance-to-ground (which is a mutual capacitance) of a conductor (placed at some significant distance from ground) is the self-capacitance of the conductor. Ground here is approximated to have infinite area.

Also worth noting here is that mutual capacitance of two finite-area electrodes differs from their self-capacitance, even with infinite distance between them.

Also, as you probably discovered already, "self-capacitance" is also used (e.g. in transformer contexts) to refer to the mutual, parasitic capacitance of the winding. These two are rather different notions. I don't think the former notion of self-capacitance helps much with estimating the latter in a transformer.


Actually, I can prove the last part even here. From Zangwill, Modern Electrodynamics (p. 140), the capacitance of a two-conductor capacitor is given by

$$C= \frac{C_{11}C_{22}-C^2_{12}}{C_{11}+C_{22}+2C_{12}}$$

Where \$C_{ij}\$ are the capacitance coefficients; in general, this is a symmetric matrix. So if \$C_{11} = C_{22}\$ and \$C_{12} = 0\$ then \$C=C_{11}/2\$, i.e. the capacitance of the two-spheres [at large distance] is half of the self-capacitance of a sphere.

From Banerjee you can see that the following happens as the distance gets large:

enter image description here

Those dimensionless capacitance coefficients plotted there are simply:

enter image description here

which means exactly what I wrote above in terms of the non-dimensionless ones.

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  • \$\begingroup\$ I've added an edit section to my question that asks if it is true that two 1mm balls placed a million miles away from each other (in a part of the universe that is physically empty) will have a mutual capacitance that is approximately half their self capacitance. \$\endgroup\$ – Andy aka Oct 29 '15 at 10:47
  • \$\begingroup\$ @Andy aka: If you want to see precise calculations, not engineering ones, (and for spheres rather than circles), you'll have to look at physics paper, for example fmf.uni-lj.si/~podgornik/download/Lekner-attraction.pdf See section 2 for large/infinite distance. \$\endgroup\$ – Fizz Oct 29 '15 at 11:01
  • \$\begingroup\$ No I don't think so. I think I'll ask if this can be migrated to physics. \$\endgroup\$ – Andy aka Oct 29 '15 at 11:42
  • \$\begingroup\$ I think you can just ask the last part (your edited-in question) there, since it won't be really a duplicate. I don't know that anyone would want to get into those calculations here. Here's a simpler calculation for equal radii spheres: electrostatics.org/images/ESA_2014_G_Banerjee_et_al.pdf \$\endgroup\$ – Fizz Oct 29 '15 at 11:48
  • \$\begingroup\$ I think that's a good document but it produces an infinite series and doesn't confidently proclaim much that I could take away in the 2 minutes I was reading it for. \$\endgroup\$ – Andy aka Oct 29 '15 at 11:53
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You have to keep in mind that the whole field of electronics is an abstraction of the underlying physics of charged particles attracting and repelling each other.

Lets look at an isolated ball from this perspective.

Start with neutrally charged ball. There are equal numbers of positively and negatively charged particles on the sphere. They balance each other out.

If we want to make current flow into the ball, we are going to need to push charged particles onto it. This will take work because these charged like-charged particles repel each other. The more charged particles we push onto the sphere, the harder it gets to push more because there is more force repelling the new ones. The more we add to the ball, the more force it takes to add even more charge.

If we make the ball bigger, there is more room for the charged particles to spread out. Therefore, for a given number of charged particles stored on the ball, the amount of force it take to add a given additional amount of charge will be less for a large ball than for a small ball.

The force we are talking about here is voltage, and the size of the ball is its capacitance.

Seen this way, hopefully it makes sense why an isolated object can have capacitance.

A standard electronics capacitor is really just two balls (or more likely plates) placed a certain distance apart from each other. IN the limit where the two balls are infinitely far from each other, the above view still works perfectly- keeping in mind that it will take double the force to move a charged particle from one plate to the other because you not only have to over come the repulsion of the ball you are adding to, you also have to overcome the attraction from the ball you are taking from. As you move the balls closer together, they start to interact and make it easier to move the charges because the imbalances start to cancel out. The closer the balls get, the less force it take to move a charge from one to the other (or the more charge you can move with a given force). So, as the plates of a capacitor get closer, the capacitance (amount of charge to push between the two plates for a given voltage) goes up!

All the math works out perfectly and flows directly from Coulomb's law with the help of some geometry and calculus.

I'd highly recommend this book...

enter image description here

Matter and Interactions, Volume II: Electric- and Magnetic Interactions

(good paper giving a taste of this approach here)

It completely changed the way I see wire, batteries, capacitors, & resistors and made so many otherwise confusing things finally make sense. It is all just charged particles attracting and repelling each other.

When you are ready to peek under the covers of inductors and antennas so to see that they are also just charged particles attracting and repelling each other (albeit in relativistic frames), then I'd highly recommend this book...

enter image description here

Berkeley Physics Course: Electricity and Magnetism v. 2

(older edition is in public domain and free here)

...which again profoundly changed the way I see the world.

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  • \$\begingroup\$ +1 for Purcell's E&M book. (Berkeley) I heard they have a new edition out that uses MKS rather than cgs units. \$\endgroup\$ – George Herold Oct 29 '15 at 15:35
  • \$\begingroup\$ Agreed! Even though I already have a copy of the 1984 edition and the original is free, I happily played $60 for the new edition to get the MKS units and the additional problem sets. \$\endgroup\$ – bigjosh Oct 29 '15 at 15:54
  • \$\begingroup\$ I've actually looked in Purcell's first, but alas I could not find the basic equation of the two-conductor capacitor that I had to use. Ironically, the [self-advertised as] graduate-level textbook by Zangwill was a bit more "for dummies" in this respect. Purcell does cover coefficients of capacitance (p. 109), but he just never seems to get to that obvious particularization. \$\endgroup\$ – Fizz Oct 29 '15 at 17:51
  • \$\begingroup\$ For capacitance & resistance stuff you want to start with M&E II. Written for high school students so super simple without fancy math or vocabulary, but the ideas and explanations are beautiful. Will change your life. I'll try to post a link to a lecture covering this, but the book is the best. I have extras to lend out if you are anywhere near NYC! \$\endgroup\$ – bigjosh Oct 29 '15 at 18:31
  • \$\begingroup\$ M&E Lecture on Capacitors is Chapter 20 here, but not nearly as good as reading the book. \$\endgroup\$ – bigjosh Oct 29 '15 at 19:28
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Self capacitance is very real. My favorite example of this is the capacitance of your body. So you should think of the other electrode as ground. Which here on Earth is never very far away. But if it's further away than the size of the object, the capacitance formula you posted is a decent approximation. (Well I'm a physicist and have no problem treating the body as a sphere :^) So try this as a way to measure your self capacitance.

Get out your x10 (10 meg) scope probe, charge up your body by rubbing on some fabric, my chair works fine. Now with the 'scope set for single shot touch the end of of the probe and look at the decay. From the time constant determine your capacitance. (I'm about 200 pF, I am growing a bit of a beer belly)

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  • \$\begingroup\$ @Andyaka, You just need to see ground as the other terminal for the capacitance... even if it's very far away. \$\endgroup\$ – George Herold Oct 30 '15 at 14:08
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I'll answer this part of the question:

I've correctly calculated that a 1mm radius ball will possess a so-called "self capacitance" of 0.111pF. If two such balls existed and were placed a million miles away from each other (in an empty universe) would the capacitance between them approximately equal 0.0555pF i.e. half of 0.111pF?

As you know C=Q/V, i.e. C is the number of Coulombs that will accept the capacitor per each Volt (or a given "effort").

Imagine first that you have a power supply that has the small ball in one of its terminals and the huge ball (infinite sphere) in the other terminal. The two balls are separated by a huge distance and therefore the electric field influence between them can be neglected. While this capacitor charges, the necessary effort for putting charges into the small ball will increase (because the already put charges in the small ball are repelling the new ones). In contrast getting charges from the huge ball is almost free (you might think in terms of charge density). Therefore the capacitance of this capacitor only depends on the small ball. In contrast if the small ball approaches the huge ball, the electric influence between them cannot be neglected anymore. And the capacity of this new capacitor will be bigger than the Cself, which is therefore a lower bound.

For the two small balls the same argument applies. When they are infinitely separated the power supply has to provide an effort to put charges to one of the balls, but also has to provide an effort to get charges from the other small ball. At the end, for a given effort or voltage, the number of coulombs will be the half, therefore C=Cself/2. In the other hand, if the two small balls gets closer the electric influence between them cannot be neglected anymore, and the capacitance will be bigger than Cself/2, which is the lower bound.

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  • \$\begingroup\$ Thanks Roger, so is it a yes, no or a nearly? \$\endgroup\$ – Andy aka Oct 30 '15 at 8:05
  • \$\begingroup\$ Yes, for a separation of zillion miles the capacitance would be Cself/2. Theoretically it would be very very slightly above this value, but no one could measure the difference I guess (you would need too many digits). \$\endgroup\$ – Roger C. Oct 30 '15 at 8:47
  • \$\begingroup\$ Thanks - that deserves an upvote because it's starting to gel and my mind is starting to accept the stupid sphere at infinity!! \$\endgroup\$ – Andy aka Oct 30 '15 at 8:51
  • \$\begingroup\$ Great!! When spheres are separated by infinite they don't interact, electrically but you still you need to provide effort to get charges from one of them and more effort to put it into the other one. \$\endgroup\$ – Roger C. Oct 30 '15 at 11:39

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