Behavior of a capacitor with varying capacitance

[FireStorm000]

There was a thread a while ago regarding a capacitor where the capacitance was changed while it was part of a circuit. I won't link the thread here as it was kind of a train wreck...

Now the explanation given was that in order to change the capacitance work would have to be done to cause that, which would allow the voltage to remain fixed. I initially questioned that explanation and was left wondering what forces the voltage to remain fixed. Let's say we charge a capacitor, disconnect it, and hook up a voltage probe. We then double the capacitance(say by moving the plates closer). There isn't a path from one terminal to the next, so no charge can flow between the plates, and as such we must hold charge fixed. So that leaves energy and voltage that can vary.

Now, I think I have an answer, but I'm a little iffy on it. Say we move the two plates closer; you then have a stronger E field and a smaller distance. IIRC, Voltage between two points involves integrating E field over the distance, so 2X stronger E field times 1/2 the distance still gives the same voltage, but work was done by moving the plates with the force from the charge separation? Is that close, or way off?

 

[Bobbywhy]

Don't be limited by a limited viewpoint. Variable capacitors can be both mechanically controlled (as you describe) and electronically controlled. What work is done during this variance? In the case of a mechanical capacitor, it seems obvious: overcoming the friction of the mechanism. Can you imagine if "work is being done" by adjusting the varactor diode? This Wiki entry gives a good overview of various types of variable capacitors and answers your questions fairly well. If you study this, and some of the references given at the bottom, and still have some questions or doubts, then do post them here. Members of Physics Forums are ready and willing to help any true searcher.
http://en.wikipedia.org/wiki/Variable_capacitor

 

[FireStorm000]

The wiki article does have some useful information, yes, but doesn't really answer the question unless I missed a section. Put more clearly, what happens to the energy already in the capacitor when you change the capacitance *after* charging it. There's obviously quite a few ways to create such a device, but I'm more asking about the function of an "ideal", simplified test case.

 

[AlephZero]

The strength of the electric field between the plates does not depend on the distance between the plates. It only depends charge per unit area on each plate.
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elesht.html#c2

The energy stored in the capacitor depends on the electric field strength and the volume enclosed between the plates.
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capeng.html#c2

So, if you move the plates closer together, the energy stored in the capacitor decreases. Where does the energy go as it decreases? You have to opposite charges on the plates that attract each other, and they move closer together. So the internal energy in the capacitor is converted into mechanical work (force x distance) that moves the plates closer together.

You should be able to work out the change in the capacitance and the voltage for yourself, from the information on the hyperphysics web pages.

"Friction in the mechanism" has nothing to do with the question.

For a variable capacitor where you change the area of the plates, not their separation, the math is different. If you reduce the area, you increase the charge per unit area, so you increase the electric field. You also reduce the volume between the plates, of course, Again, you can work through the details using the information on the hyperphysics pages.

 

[FireStorm000]

Ah, that makes more sense. I was trying figure it out in my head but it's apparent I don't remember some of these equations well enough. It makes sense thinking about it though. Work is done when the plates move in the plate separation changing version; I chose that as my example because of how easy it is to see where the energy goes if it enters or leaves the capacitor mechanically - the first definition of work one usually sees, Force dot product with displacement. The answer to my question then would be that both voltage and energy stored vary with capacitance. I'll break out the scratch paper after this and try to do the rotary/variable plate area capacitor math.

 

[nasu]

There are two different situations possible:

1. The capacitor is charged and then disconnected from the battery.
In this case the charge on the plates does not change but the voltage does, when you change the capacitance (by reducing the distance between plates).
Using
U=(1/2)Q^2/C
you can see that the energy decreases when the capacitance increases.

2. The capacitor remains connected to the battery.
In this case the voltage indeed remains the same when you change the capacitance but the charge on the plates changes.
Using
U=(1/2)CV^2
can be seen that the energy increases when the capacitance is increased.

Which one are you describing in the OP?

 

[FireStorm000]

There are two different situations possible...
Which one are you describing in the OP?

First one:

Let's say we charge a capacitor, disconnect it {from the power source}, and hook up a voltage probe.

Am I correct in saying that the answer to my question depends on *how* I change the capacitance? IE: Plate distance/plate area/dielectric constant? That energy and voltage are determined by the specifics of the parameters determining capacitance?

I hadn't initially realized the implications of this:

What work is done during this variance?

But i realized after some sleep that, while mechanical work must be done to change the plate distance, changing the effective plate size by shifting the plates sideways wouldn't require mechanical work because the Columb force is acting orthogonal to the motion. So already there's a critical difference based on how I change the capacitance.

 

[nasu]

And what is your question?
I cannot see a clearly formulated question in the OP.

Regarding the change in capacitance due to lateral motion, the field is perpendicular to the plates only for ideal, infinite plates.
In this case moving the plates laterally will make no difference.
For real plates, the field near the edges is not perpendicular to the plates. Some work will be done.

 

[FireStorm000]

And what is your question?
I cannot see a clearly formulated question in the OP.

Regarding the change in capacitance due to lateral motion, the field is perpendicular to the plates only for ideal, infinite plates.
In this case moving the plates laterally will make no difference.
For real plates, the field near the edges is not perpendicular to the plates. Some work will be done.

The question was if my understanding of how the plates would behave was correct; it was not, and as such we have moved on to me trying to figure out what *IS* correct. I'm currently thinking about if the final values of voltage and stored energy depend on how I change the capacitance rather than just on the final and initial capacitance. Would help if I could focus enough to do the math...

EDIT: Stuff with end effects and the details of what happens on the edges of the plates wasn't covered in my physics classes. I assume the E field looks something like a cross between the normal parallel plate capacitor and that between two opposite point charges, but not clear what that would do as far as mechanical work done.