37.553 pF

It's the self-capacitance of a charged conducting sphere in vacuum with a diameter of about 2/3rds of a meter...

Jon

It's 37.553 pF +/- 25%. Oh well.

Jeroen Belleman

Not really. It's only nominal resolution/range. Capacitance tolerance is

Klaus

There are a number of digital caps around. I found it interesting that three had the same max capacitance of 37.553 pF.

I was cleaning up my desk, a periodic necessity, and found an ad for a CTS digital cap, 37.553 pF. Then I found basically the same part from Ixys and Incide.

Looks like CTS tried to get into the IC business (including bunch of ECL parts) by reselling the Incide parts, but then Ixys acquired Incide and cut them off. Something like that.

The Pregrine and ST parts look nice. We used the Maxim "Flecap" part in a couple of products but they EOL'd it as usual. Never Buy Maxim.

Why does the sphere have to be charged?

Basically because in a closed system charge is conserved - you can't create charge out of nothing.

Imagine an empty universe that just contains a parallel plate capacitor and a voltage source. I can apply the voltage source to the parallel plates and increase the charge on one plate of the capacitor, while the charge on the other plate will decrease in kind to match it. The energy in a capacitor doesn't come from "excess electrons", but is stored in the electric field generated by the charge differential.

If I have an empty universe with just a conducting sphere, basically a single plate, where does the charge come from? The charge has to be there initially, or talking about the "capacitance" of this isolated system doesn't make any sense.

Where does the charge (electric or magnetic) come from in a propagating EM wave, then?

It's also like the thought experiment:

Suppose you have a (thermodynamically) closed cylinder of gas, one wall movable (a piston).

The gas consists of one molecule.

If the piston is moved out, how does the molecule cool adiabatically?

Suppose the piston is moved so quickly that the molecule doesn't touch it during that movement. Does it cool?

If the cylinder contains more than one molecule of gas, how does the gas cool?

I feel the answer depends upon continuity, because if the piston is moved very slowly, the molecule will have bounced off numerous times, and therefore been reflected with a small Doppler shift each time; the sum of those shifts gives the total energy change, and therefore the total cooling. This still works even if it only bounces off a few times during the motion.

It doesn't work if the piston moves supersonically (i.e., motion is completed without the molecule bouncing off as it moves), but perhaps there is the excuse that things behave differently in the supersonic regime; after all, one would expect this to produce a shockwave (in a normal body of gas) which would eventually dissipate and give thermal energy to the system.

Even though thermodynamics is only concerned with the initial and final states, it is fundamentally dependent upon the assumption that continuity is present between those states. It doesn't matter (at least within reason) how it was done, just that it was physically possible.

Anyway, E&M shares many similarities; you can't have [divergent] charge (E field) without charge (particles), and it will have taken some [displacement] current to create that charge [displacement]. It doesn't matter when, or how fast, or with what waveform, that current flowed; it's merely necessary that it happened continuously.

Tim

More curious is the hits you get for "241543903" (do image search).

Tim

What charge would that be? EM waves propagate just fine in a vacuum.

Same way as in gas collisions. Because of the net expansion the two colliding objects have lower radial velocity on average. Same idea as an orbital slingshot.

Then it isn't adiabatic.

By having the net radial velocity of colliding particles diminish, as above. It's slightly more complicated since you have to average over the impact parameter (offset from a perfectly central collision), but it's still a third-year undergraduate problem.

Right, IOW the motion is adiabatic.

Equilibrium thermodynamics depends on continuity and systems of large size. Nonequilibrium thermo doesn't--it can easily handle stuff like shock waves.

The E field in a propagating wave comes from dB/dt, not from rho. E and B continuously generate each other via the curl equations.

, and it will have taken some

I have no idea what you mean by that.

Cheers

Phil Hobbs

Right. Capacitance is twice the coefficient of V**2 in the expression for electrostatic energy, so you have to have some charge (at least notionally) for the concept to make any sense.

Cheers

Phil Hobbs

What charge are we talking about?

Classically, EM waves are propagating disturbances in an EM field. You need to accelerate a charge to create them, but once the disturbances are propagating they propagate out to infinity all by themselves.

In special relativity charge density is a relativistic quantity. Special relativity is not an area I have a ton of knowledge in outside the basics, but I guess you could imagine a situation where in an inertial frame you have a positron and electron sitting around generating electric fields. You then accelerate them and slam them into each other and they annihilate. But the acceleration generates a propagating EM wave, so if they're the only two particles in the universe then as far as the universe is concerned charge has ceased to exist.

But in some sense in a far away reference frame, the charges still "exist." The only way one has access to information about what's going on in the other frame is through the propagating disturbances, and the disturbances can't propagate instantaneously. The relative velocity of the two frames (in four-space) and the electric field at the source point, as a function of time, is all that's required to determine the electric field as a function of time at any other space.

In "reality", though, the energy and momentum of an accelerated charge isn't carried away by B and E fields, it's carried by photons; vector bosons, particles that are governed by the laws of quantum mechanics, i.e. have wavelike statistical properties. That "E and B field at perpendicular angles" thing is just a useful heuristic.

In this hypothetical single sphere universe, however, it's sort of an irrelevant exercise, as the "notional" charge and capacitance on the isolated sphere are simply a way of expressing Coloumb's law and the permittivity of free space of in a different form.

Heat capacity of the molecule is much less than the cylinder and walls.

(and I don't think you can apply thermo/ stat. mech. to one of something.)

George H.

Well, the classic British admonition "Oh, go boil your head, Bertie" was coined before refrigeration.

Oh, PID, as in STUPID.

Seems like the same as opening a valve to let the gas out. How does that cool?

Is the temperature only defined by the velocity of the molecules and not the density?

I'm still not sure the existence of capacitance depends on the presence of charge.

No. But do the experiment a million times, and the average will be the same as if you did it once with a million molecules (near enough).

Because molecules that collide with others which are moving towards the exit rebound with less energy than if there was not a net motion in that direction.

Neither. Temperature is a measure of the average kinetic energy of the molecules transmitted in collision with the walls. Note that different molecule geometries may have significant angular KE, which does not necessarily contribute to the temperature (hence different gases have different gammas).

If you drive into a parked car, do you do more damage than if you run up the back of someone driving away from you in a stream of traffic? If your car was an elastic rubber ball, would you rebound harder or less hard?

No-one said it did. Get a fluid mechanics text book and do some reading. Start with the explanation of why a bicycle pump gets hot.

The "wall" could be a notional one, but temperature is *sensed* by heat flow across a boundary; it could be the side of your temperature probe.

Clifford Heath.