Understanding capacitance formula

[Femme_physics]

Understanding "capacitance" formula

In this formula

http://img810.imageshack.us/img810/9127/capacitance.jpg 1) What does it mean the area of overlap of the two plates? Does it mean the "area" of the plates? And, is the area of one plate always the same size of the other plate?

And also,

2) The distance between the plates is measured is what, normally? mm?

And they also said that "εr is the relative static permittivity (sometimes called the dielectric constant) of the material between the plates (for a vacuum, εr = 1);"

So my last question:

3) In a vacuum there is no static permittivity or that there is maximum static permittivity? If what I understand static permittivity is pretty much like the ability to store charge...? So, vacuum has no ability to store charge...shouldn't it be zero?

PS and while I'm at it, is the time constant measured only in seconds?

 

[Dmytry]

1: the area where you have both plates separated by this distance. Think of a variable capacitor, where this overlap area varies.
http://en.wikipedia.org/wiki/Variable_capacitor
or imagine you have a piece of foil, covered by piece of plastic wrapper, covered by another piece of foil. The area where foil overlaps.
The formula is approximate and ignores capacitance between parts that do not overlap.

2: same unit in which you measure the electric constant. That is, meters.
3: relative static permittivity of vacuum is 1 . 'Relative static permittivity' of anything else is relative to the vacuum.

edit: time constant, same units that you use for other quantities (or else you have to convert). Farad is s⁴A²kg^-1m^-2 so that'd be seconds. In general, SI uses seconds, kilograms, meters, etc.

 

[sophiecentaur]

If you state everything in SI units then you should have no trouble with getting the answer right. It's just a matter of remembering to!

The formula you are quoting only works for an arrangement where the dimensions of the plates is a lot more than the separation. 'Edge effects' come into it when you have small plates and large separation.
For a small plate over a large plate, the area would be that of the small plate; i.e. the bit of the small plate that overlaps the large plate.
Re "vacuum". Why should the fact that there is no charge 'in' a vacuum stop extra electrons appearing on one plate and a scarcity of electrons on the other plate? That's what happens when a capacitor is charged. For a capacitor with a dielectric, there is polarisation of the molecules (but no flow) within the intervening insulator which allows a greater charge displacement for the same PD.

 

[Femme_physics]

For a small plate over a large plate, the area would be that of the small plate; i.e. the bit of the small plate that overlaps the large plate.

Ahh...there's the punchline :)

Re "vacuum". Why should the fact that there is no charge 'in' a vacuum stop extra electrons appearing on one plate and a scarcity of electrons on the other plate? That's what happens when a capacitor is charged. For a capacitor with a dielectric, there is polarisation of the molecules (but no flow) within the intervening insulator which allows a greater charge displacement for the same PD.

Hmm. I'll indulge in this fact later and repost here.

edit: time constant, same units that you use for other quantities (or else you have to convert). Farad is s4A2kg-1m-2 so that'd be seconds. In general, SI uses seconds, kilograms, meters, etc.

Thanks :)

3: relative static permittivity of vacuum is 1 . 'Relative static permittivity' of anything else is relative to the vacuum.

So a vacuum is not a maximum or a miminum, right? It's just a relative point?

 

[MisterX]

So a vacuum is not a maximum or a miminum, right? It's just a relative point?

The static permittivity is at minimum for a vacuum, at least for typical situations.