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# Variable capacitor and discharge question?

I don't have a variable capacitor of my own to test this so I am relying on the instructable community.

What if you have a variable capacitor tuned so that it is at its maximum capacitance and then charge it in a circuit for 5 tau (essentially full charge). Then you completely removed it from any circuit so that you only had a variable capacitor with two (short) leads not connected to anything. Lastly, you turned the tuning knob so that the capacitor had less capacitance all the way to essentially 0 farads.

What happens to the stored energy?

## Discussions

I have to point out that the charge in a normal capacitor is NOT stored on the plates of a capacitor, but on the dielectric. Take a leyden jar for example. Charge it, separate the plates touch them measure them, no charge. Put the plates back on the dielectric and use tongs to discharge cap, POP you get a spark. You see there are a LOT of assumptions that are not exactly correct in this area of physics. Including the idea of doing work to separate plates. Its almost as though there are two types of capacitance. Air capacitors do not really have an effective dielectric, as you can blow away the air and still have an electrostatic charge on the plates, though this type of charge has VERY small capacitance's.

The defining formula for a capacitor is Q = C*V or V=(1/C)*Q. Just from this formula, if you assume Q is constant and C gets smaller, then V gets bigger.

Although I'd understand if you didn't quite trust the formula by itself. It'd be helpful to have a picture to go with it. So I decided to draw one.

Note that

for the kind of variable capacitor I've drawn here,That is to say the the plates move in a direction perpendicular to the electric forces pulling the charged plates together. Also note the spacing of the gap between the plates, d, remains constant. For this type of variable capacitor, where I don't do any work pulling the charged plates apart, the energy stored on the capacitor remains constant.I can change the capacitance without doing any work.The capacitance of this arrangement is assumed to be proportional to the overlapping area. As the plates are pulled apart, this area gets smaller, and so does the capacitance.

Now for a "hand waving" explanation of where that voltage increase is coming from:From the picture you can we can see that when the overlapping area gets smaller, the charges on the plates get crowded into a a smaller area. So we get an increase in charge density. From this increase in charge density we get an increase in electric field intensity. The voltage across the capacitor is essentially the electric fieldEbetween the plates times d, the constant width of the gap between the plates.Oops! Um... scratch that part about this not requiring any work. I think it actually does take work to crowd those like charges closer together. And the quantity of work required is (a1/a2) times the energy stored on the capacitor in its initial state with a1-overlapping. U1 = 0.5*C1*V1^2; U2 = 0.5*C2*V2^2; C1*V1 = C2*V2 = Q ; C1/C2 = a1/a2 Gives U2 = 0.5*C2*V2^2= 0.5*C2*(C1/C2)*V1*(a1/a2)*V1 = (a1/a2)*U1 U2=(a1/a2)*U1

No. No wait! The work required to change the capacitor from its old shape with a1 overlap to the new shape with a2 overlap is U2-U1:

U2 - U1 = ((a1/a2) -1)*U1 = ((c1/c2)-1)*U1

Example: If U1 = 10 J, then changing the capacitance to half of this would take (1/0.5 -1)=1times U1, or another 10 J of work. Changing the capacitance to a quarter of its initial value would take (1/0.25 -1)=3 times U1, or an additional 30 J of work.

If I just said that the

energy gainis U2/U1 = a1/a2, that probably would have made more sense.It's still stored. Energy is conserved. Since W = C V

^{2}/2, if you reduce C, V increases, just as Steve said. Since the relationship is quadratic, you can probably get above breakdown voltage fairly easily.At which point you get either a pretty spark across the outside leads, or a not-so-pretty spark between the plates, and your expensive variable cap has become an expensive Christmas-tree decoration.

Thanks for the explanation, how could a spark destroy a air variable capacitor

Vapourises parts of the plate.

It appears as an increased voltage across the plates, subject to leakage, as the air breaksdown.

It is an interesting question, net result being it arcs?

L

BIG time. Clearly one for the next meetup. That, and glowing Polo mints.

. Capacitors are (one of) my weak point(s), so I may be way off base, but wouldn't the potential stay the same? No electrons have been added?

A variable capacitor is usually an air variable and of not much capacity.

As such the stored charge is small and leaks off very quickly thru the air.

Take it as a thought experiment, the result is interesting. Its the basis of the "Electrophorous"....and can be simulated with two pieces of single sided PCB material (unetched).

Charge generously, by rubbing in bubble wrap.

Hand package to wife.

Ask her to separate boards, while accelerating in the opposite direction. Deny all knowledge of the effect after the screaming dies down....

Steve

OH that made me laugh.

Beeeg fat juicy (fed from a fair few pF) spark too, allegedly....

Ahem, "allegedly." I presume the consequences explain your current avatar (self-portrait?)...

It's still stored, assuming no losses. If you turn the dial back you still have all the excess electrons. I believe the answer is that there's a conversion between mechanical and electrical energy. Think about trying this with a simple flat-plate capacitor with variable spacing between its plates. If you charge them up when they're far away from each other, then bring them together, you'll have to do work against electrostatic repulsion; in separating them, that repulsion helps you and does some of that work. Other kinds of variable capacitor should behave much the same way. But static repulsion is a pretty weak force, and most tuning capacitors are not carrying a particularly high charge, so you're unlikely to notice that in most real-world situations.