# 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?

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?

active| newest | oldestAlthough 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.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.

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

L

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

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