mutual capacitance?

you'll see something like mutual capacitance in piezo-electric transformers.

There is. It's usually just called 'capacitance', unless you need to distinguish it from self-capacitance.

A 1-cm radius isolated sphere has a self-capacitance of 1 cm (Gaussian units), which is about 1.12 pF.

Cheers

Phil Hobbs

There are two commonly given values for the capacitance between the earth and the moon, 160uF and 3 uF. I think one is 2-wire capacitance and the smaller one is 3-wire.

Where is the universe's ground lug?

Well, there's some silly speculation about that we're living in a simulation, in which case it would be Node 0. ;)

Cheers

Phil Hobbs

We could connect to the celestial sphere, the one with all the track lights.

I picture the analog of two isolated coils, magnetically linked: two isolated capacitors, the flux of the 'primary' transmits through the 'secondary'. Why no such device?

Well, I have in mind your basic two plate capacitor. I don't recall self-capacitance -

What you're describing is precisely the mutual capacitance. It's the change of the charge on plate 1 due to the change in voltage on plate 2.

Cheers

Phil Hobbs

Self-capacitance is the capacitance of an object with respect to ground, when it is in a large grounded box. An infinite box, ideally.

Ground isn't necessary. An isolated conductor with a certain amount Q of free charge on it will have an E field. The voltage V is minus the line integral of E dot ds from the surface to infinity. The self-capacitance is Q/V.

Cheers

Phil Hobbs

Huh right, I had this picture of three hunks of stuff with various inter-capacitances. But that starts with one hunk!

George H.

The 3-wire measurement ignores Cy and Cz.

If the moon moved away from earth, Cem would approach zero, but Ceu and Cmu wouldn't change.

Most good c-meters will do 3-wire measurement, which allows a small cap to be measured at the ends of coaxial cables.

Just get a ball bearing floating in space and throw electrons at it.

OK we should be able to work out the earth-moon capacitance as a physics problem. Here it is using method of images.

(Hmm that is for sphere's of equal radius.)

As a first approximation we could guess that the Earth's C to the universe is decreased by the ratio of the field lines that hit the moon, to all of them... Which is pi*R_moon ^2/ (4*pi*Dist_E-M^2) R_earth ~6.4 x10^6 m C_earth ~ 640 uF R_moon ~1.7 x10^6 m and Dist_E-M ~3.8x10^8 m.

Putting that all in.. and hopefully making no mistakes I get a drcrease of 5x10^-6 or C_e-m ~3,200 pF .... 3.2 nF What was your number for C_e-m?

George H.

Of course this is going to only be true at low frequency... Speed of light and all.

Hmm there's this,

I'm going to say it's wrong. It does have your 160 uF and 3 uF numbers... but just engineers plugging in numbers. (first time I've called B. Pease, just an engineer :^)

GH

This is one of the solved problems in Maxwell's "Treatise on Electricity and Magnetism". Of course, this was before we had mks units, so the answer is given in meters instead of farads (or maybe he was jealous of Michael Faraday).

Well what's his answer? The cgs unit of capacitance is the cm. But we can convert it. (And if we make a mistake we'll beg Phil H. for help.)

George h.

I don't remember, and my copy of Maxwell's Treatise is packed away. It sho uld be easy to find a copy, so YCLIU (try Google). I vaguely recall that y ou needed a scaling factor of (1/4??0) to convert it to mks.

Cheers

Phil Hobbs