capacitance and plates and frequency

[Gibson486]

So, I was thinking...

A higher dielectric material will increase capacitance....

However, what if there was a fixed air gap after the dielectric material? That air gap will get a lower electric field when a higher dielectric material is in series with it.  To me, that seems counter intuitive...because that would imply that the frequency is not that relevant, wouldn't it? Maybe i am  mixing up concepts, but doesn't lower frequencies penetrate low dielectric constants better? So, if I looked at this in terms of amplitude, to get the same amplitude on the e field, I could just lower it? or...am i just mixing two different concepts up? I would imagine that the gap would be a radiated wave, but maybe I am wrong?

[T3sl4co1l]

Frequency has nothing to do with it, or more particularly -- you don't want to think about frequencies where it does, because then you need to work with all the fields.

Confining the question to low frequencies, only electric field is needed, and the problem is electrostatic.  Literally, electric, not moving.

If you have four layers: metal, dielectric 1, dielectric 2, metal, then the total electric potential is the voltage between plates.  The average electric field is the potential over the distance between plates (assuming infinitely wide plates).  The electric field divides between the two dielectrics according to their thickness and constant.

A higher dielectric constant is more "conductive" (permittive) to electric field, thus drops less electric field than a higher K material, when those materials are connected in series.

You can draw an equivalent circuit as a resistor divider, where resistance is actually capacitance, and current is actually charge.  The usual capacitors-in-series formula then applies.

Which seems to be your intuition, and I'm not sure where you're coming from with the premise that it would be the other way around?

Tim

[David Hess]

The frequency is not relevant except that dielectric constant varies with frequency slightly, and sometimes alarmingly.  Where the electric field is intense, changes in dielectric constant can be a problem when they concentrate the electric field enough to cause breakdown, which is why air may be removed when potting high voltage assemblies.

[ejeffrey]

One way to think of it is to pretend there is a thin metal plate at the air-dielectric boundary.  Then you have two simple parallel plate capacitors in series and you can use the standard formulas to calculate everything you care about.  This works because the problem symmetry ensures there is no transverse electric field so the imaginary metal plate wouldn't have any currents flowing.

[Gibson486]

I was confused by this page....

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/dielec.html

The K is on the bottom. That means the higher your dielectric constant is, the lower the electric field....Sort of threw me off....but I guess it really means that the lower electric field means that more charge is actually moving around...

[TimFox]

One consequence of this is what happens in a bubble or other air gap in a dielectric between electrodes.  If you make a capacitor with two parallel plates and fill it almost all with dielectric (liquid or solid) except for a gap much thinner than the spacing, then the E field (voltage gradient) in the dielectric will be approximately equal to the voltage divided by the thickness of dielectric, but the E field in the air gap (K = 1) will be higher by the dielectric constant of the thick dielectric.  This is very important for high voltage insulation, since the breakdown field of the air is lower than the dielectric insulation.  Proper insulation design ensures that there are no such air-dielectric interfaces, for example by metallizing the ends of ceramic pillars or recessing the pillar into a counterbore in the electrode.  In electrostatics, this is due to continuity of the “D” field across a boundary.

[helius]

The K is on the bottom. That means the higher your dielectric constant is, the lower the electric field....Sort of threw me off....but I guess it really means that the lower electric field means that more charge is actually moving around...

Yes, that's to be expected. The law for capacitance
\$ C = {Q \over E} \$
means that with higher capacitance, more charge can be separated at the same electric potential. The capacitance is also defined by
\$ C = { \epsilon A \over d} \$
so by noting that the electric field gradient is not affected by epsilon, therefore we can store the same charge at lower E field density if we have a higher permittivity dielectric. This is equivalent to saying that higher permittivity stores more energy in a given field gradient. It permits or is susceptible to an electric field by entering a higher energy state. The lowest permittivity is vacuum, which has no polar molecules to react to the electric field by becoming polarized, so it stores the minimum energy.

[David Hess]

This is very important for high voltage insulation, since the breakdown field of the air is lower than the dielectric insulation.  Proper insulation design ensures that there are no such air-dielectric interfaces, for example by metallizing the ends of ceramic pillars or recessing the pillar into a counterbore in the electrode.

That is also why high voltage film capacitors are filled with oil and high voltage assemblies will be potted under a vacuum.  Breakdown in any bubbles, even tiny ones, will ionize the air inside producing oxidizers and high temperatures which gradually carve a channel through the dielectric or potting material and eventual failure.