# Capacitor with conductive metal inbetween plates

#### seza

Nov 6, 2015
40
Hi,

As always this is great forum, I am learning great deal here

1. Conductive parallel plates - lets say they are made of Aluminium
2. An insulator made up of two layers, can be anything - lets say paper
3. A very thin metal sheet such as Aluminium foil sandwiched inbetween the insulating layers

Lets assume my manufacturing skills allow me to make the cap with these dimensions:
:: the surface area of the cap will be 100 x 100mm
:: the thickness of each insulating layer will be 50microns
:: the thickness of the sandwiched Al sheet will be 5microns

If we only apply DC current - what effect would that setup have on: Capacitance? Voltage? etc ...

thanks

#### Harald Kapp

##### Moderator
Moderator
Nov 17, 2011
13,179
Only the insulating layers will contribute to the effective distance between the two outer plates. The inner plate will effectively short-circuit (if one can use the term in this context) the electrical field between the outer plates. Thus capacitance can be calculated from the area of the outer plates, the thickness of the insulation and the permittivity of the insulator.
This is assuming the inner plate has the same size as the outer plates and is placed centered.

#### seza

Nov 6, 2015
40
So are you saying the metal sheet will have no impact?
As the metal sheet is highly conductive, in the presence of the electric field does that not cause the Alum atoms to polarise and therefore contribute to increasing the dielectric constant? Or the presence of the al sheet in the middle of the field cause the collapse of the field totaly?

#### Alec_t

Jul 7, 2015
3,354
So are you saying the metal sheet will have no impact?
If it is unconnected it will reduce the capacitance slightly. Without the sheet the two outer plates are 2 x 50 = 100u apart. With the sheet the plates are 105u apart. Capacitance is inversely proportional to plate separation.

#### Harald Kapp

##### Moderator
Moderator
Nov 17, 2011
13,179
As the metal sheet is highly conductive, in the presence of the electric field does that not cause the Alum atoms to polarise and therefore contribute to increasing the dielectric constant? Or the presence of the al sheet in the middle of the field cause the collapse of the field totaly?
No it doesn't. It will act as a short circuit for the electric field. Compare this to two plate capacitors in series where each capacitor has only one dielectric separator. This is essentially the same as having the conductive plate in the center of a single capacitor with twice the dieletric thickness:

#### hevans1944

##### Hop - AC8NS
Jun 21, 2012
4,826
You might want to investigate how supercapacitors are made. They are essentially two capacitors in series, but with the common electrode inaccessible because of the design. Therefore the effective capacitance is half that of either of the two capacitors in series. If these two capacitors are 100 farads each, then the effective capacitance of the supercapacitor is 50 farads. Goggle "supercapacitor construction" for the initial entrance to this fascinating rabbit hole of technology.

#### seza

Nov 6, 2015
40
Thank you all - guys for your input on this,
wherever possible, I do like to experiment,
in this case I made a rudimentary cap in two configuration:
:: Two Al foil sheets measuring 5 x 5cm, for separator/insulator
I used a plastic sleeve (the shiny birthday card cover)
:: the same as above with the addition of an al foil inside the sleeve

Just simple electrostatic cap no electrolyte, measured the capacitance
with multimeter, found a marked increase after inserting the al inbetween,
a good 25% increase what's more is by adding another sheet inbetween the
capacitance increased further,
Just to clarify - the Al sheet inbetween is separated/insulated from both plates,

How is that possible?

#### Nanren888

Nov 8, 2015
622
C = k A/D - first capacitance, where "k" relates to the dielectric. Here air.
Later two in series as Harald Kapp explained above.

C = C1 in series with C2, but each now has a much smaller "D" as you are filling the space up with conductor.
.
So let's say I half fill the space, D, by choosing an Al plate that is D/2 thick, so the total air space is now D/2, and let's say I put the Al in the middle, so I make two capacitors that are the same, by symmetry. So each new capacitor gets one half of the remaining space, one quarter of the space I had before.
Each new capacitor has capacitance, Cnew = k*A/(D/4) = 4*k*A/D, so each new capacitor is 4 times the capacitance of the starting one. Makes sense as the distance between plates is one quarter what it was before.
Now they are in series as explained by Harald Kapp, above.
So Ctot = (Cnew * Cnew)/(Cnew + Cnew) = (4*C*4*C)/(4C+4C) = (16*C^2)/(8*C) = 16C/8 = 2C
.
We have made a capacitor twice the value by effectively making the plates closer together.
Maybe we could have just moved the original plates closer?

#### Harald Kapp

##### Moderator
Moderator
Nov 17, 2011
13,179
Maybe we could have just moved the original plates closer?
The 'correct' method would have the capacitor with 2 plates and a single sheet of foil in between compared to a capacitor with 3 plates and a single piece of foil in between the plates.
As described by @seza the 2 plate capacitor had 2 foils in between which reduces the capacity as elaborated by @Nanren888 .

#### seza

Nov 6, 2015
40
Thank you Nanren888,

Also thank you to Mr Kapp for the diagram above,
helped me realised the obvious fact that I have made
two caps in series! Might be a bit early for me here have not had a coffee yet!

I must admit I did not measure the voltage,
so what you are saying is that we have an increase in capacitance at the expense
of voltage! so new voltage will be 1/2 original V?

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