Multiple parallel plate capacitor question.

[mrspeedybob]

Consider a capacitor with 4 parallel plates and 3 dielectric insulators arranged like this...+I-I+I-

The insulator in the center has an equal amount of + and - charge on each side of it so shouldn't the electric field across this insulator be almost zero? If I constructed this capacitor with the first and last insulators capable of holding 100 volts and this center insulator only capable of holding 10 volts could I actually charge the capacitor to 100 volts without penetrating the center insulator?

 

[Baluncore]

I think you should rephrase your question in terms of three discrete capacitors in series.

 

[mrspeedybob]

I think you should rephrase your question in terms of three discrete capacitors in series.

Sorry, I guess my question left room for interpretation other then what I intended. My +I-I+I- diagram was intended to represent only the geometric layout of the capacitor. Each + represents a metal plate, each - represents a metal plate, and each I represents a dielectric insulator. Electrically, both + plates are connected to one side of a source or load and both - plates are connected to the other side so it's just a single multi-plate capacitor. Nothing is in series. If anything it may be modeled as 2 capacitors in parallel but that modeling ignores the particular effect of the geometry of the construction that I am asking about.

 

[davenn]

OK so your ascii diag suggests this layout... yes ?

I still see that as the series arrangement that baluncore suggested

As for how it would charge with an applied voltage ... I'll leave that to others to comment on

further pondering, I also tend to agree with baluncore that you can view it as 3 separate caps in series
like this ...

if my first diagram is a correct representation of your idea
then the lower part of the second pic showing 3 caps in series is just the same thing
all that is being done is sub-dividing the 2 inner plates and putting a bit of wire between them
( they are each effectively still a single plate)

Dave

 

[Baluncore]

+I-I+I-.
If the two (+)s are connected and the two (–)s are connected we have three (possibly identical) capacitors in parallel.
The breakdown voltage of the combination will be the breakdown voltage of the capacitor with the lowest breakdown voltage.

 

[mikeph]

You're talking about a single capacitor with interdigitated electrodes. I don't understand your reasoning to say the electric field would be zero in the dielectric, and I don't know how you mean an insulator can "hold" volts. Nothing holds volts. You charge the capacitor which generates an electric field between the terminals. This is mathematically equivalent to setting two different voltages on the terminal surfaces, and Maxwell's equations reduce to Poisson's law for the voltage, so you get a voltage gradient (==electric field) within the dielectrics.

It's no different to a parallel plate capacitor, just a different geometry that increases the capacitance per unit length.

 

[mrspeedybob]

You're talking about a single capacitor with interdigitated electrodes. I don't understand your reasoning to say the electric field would be zero in the dielectric,

If each plate is charged with a certain number of charges (electrons or holes) and we call that number Q then the charge on the first plate would +Q and the charge on the second plate would be -Q. It seems to me like +Q-Q=0. The same reasoning would apply to the 3'rd and 4'th plates. This would make the net charge on either side of the center insulator 0.

I don't know how you mean an insulator can "hold" volts. Nothing holds volts.

What I mean is the maximum voltage that can be applied across the insulator before it breaks down and allows significant current flow.

 

[berkeman]

If each plate is charged with a certain number of charges (electrons or holes) and we call that number Q then the charge on the first plate would +Q and the charge on the second plate would be -Q. It seems to me like +Q-Q=0. The same reasoning would apply to the 3'rd and 4'th plates. This would make the net charge on either side of the center insulator 0.

But your figure shows +/- across each insulator: "+I-I+I-"

Where are you getting that there is no separation of charge across the center insulator?

 

[mrspeedybob]

But your figure shows +/- across each insulator: "+I-I+I-"

Where are you getting that there is no separation of charge across the center insulator?

I can think of 2 different ways to further explain the question.

Plates 2 and 3 create a potential across the center insulator. Plates 1 and 4 also create a potential across the center insulator except in the opposite direction. Shouldn't they cancel each other out?

or

The 3 elements to the left of the center insulator are +I-. Since there are just as many + charges as - charges these 3 elements combine to form a neutral object. The 3 elements to the right of the center insulator are identical and so also form a neutral object. Since the center insulator has neutral objects on either side of it why should there be any net electric field across it?

 

[berkeman]

I can think of 2 different ways to further explain the question.

Plates 2 and 3 create a potential across the center insulator. Plates 1 and 4 also create a potential across the center insulator except in the opposite direction. Shouldn't they cancel each other out?

or

The 3 elements to the left of the center insulator are +I-. Since there are just as many + charges as - charges these 3 elements combine to form a neutral object. The 3 elements to the right of the center insulator are identical and so also form a neutral object. Since the center insulator has neutral objects on either side of it why should there be any net electric field across it?

Ah! I finally understand what you are asking. No, the metal plates 2 & 3 shield any field from 1 & 4 from showing up at the center insulator in your diagram.

As mentioned previously in this thread, this is how real-world capacitors are constructed. You either roll up an inter-digitated foil+insulator stack, or you use a number of inter-digitated foil plates separated by insulators. Each pair of foil plates is independent of the others in the stack, by virtue of the shielding effect of the other plates. Hope that makes sense now.

 

[mikeph]

If each plate is charged with a certain number of charges (electrons or holes) and we call that number Q then the charge on the first plate would +Q and the charge on the second plate would be -Q. It seems to me like +Q-Q=0. The same reasoning would apply to the 3'rd and 4'th plates. This would make the net charge on either side of the center insulator 0.

Ok, but why is "net charge on either side of the insulator" a relevant quantity here?

What physical significance does the net charge have?

You're essentially elaborating on the law of current conservation.

 

[Baluncore]

If each plate is charged with a certain number of charges

Capacitance is defined as the ratio of charge to voltage. C = Q / V.

The cross connection of (+) with (+) and (–) with (–), that is not shown in the OP ASCII diagram, "+I–I+I–", makes for a “three capacitors in parallel” situation. Parallel capacitors all share the same voltage, but allocate charge in proportion to their capacitance, q = c * V.

Without cross connection, the three insulators shown can be modeled as “three independent capacitors in series”. In series, each capacitor is subjected to the same current and so has the same accumulated charge, but the voltage across each is inversely proportional to their capacitance, v = Q / c.

It is a mistake to model all the capacitors as being in parallel and subjected to the same voltage, while at the same time expecting to treat the current and charge as if they are in series.

In order to understand the situation it is necessary to clearly specify the interconnection of the conductors shown by “+” or “–” on either side of the insulators symbolised in the original "+I–I+I–".

 

[mrspeedybob]

I did an experiment recently related to this question.

I wanted to compare a 2 plate capacitor to a 4 plate capacitor. If the energy in the capacitor is stored on the plates then the 4 plate capacitor should have twice the capacity of the 2 plate capacitor. If the energy is stored in the dielectric as deformed electron orbitals then the 4 plate capacitor should store 3 times the energy of the 2 plate capacitor because it uses 3 insulators instead of 1.

I made a simple capacitor using 2 sheets of aluminum foil and a sheet of paper. I measured its capacitance using a Fluke multimeter and got a value of 3.7 nF. Adding 2 more sheets of paper and 2 more sheets of foil to create a 4 plate interdidgitated capacitor increased the capacitance to 9.0 nF.

So I got an increase of about 2.4 times. This doesn't support either of my 2 hypothesis.

What is the correct explanation of my experimental results?

 

[Baluncore]

To test the theory you should have an odd number of plates and an even number of insulators. That makes it possible to ground the two outer plates as a screen or shield. That will eliminate stray capacitance with the environment that you may be including in your measurement.

 

[sophiecentaur]

I did an experiment recently related to this question.

I wanted to compare a 2 plate capacitor to a 4 plate capacitor. If the energy in the capacitor is stored on the plates then the 4 plate capacitor should have twice the capacity of the 2 plate capacitor. If the energy is stored in the dielectric as deformed electron orbitals then the 4 plate capacitor should store 3 times the energy of the 2 plate capacitor because it uses 3 insulators instead of 1.

I made a simple capacitor using 2 sheets of aluminum foil and a sheet of paper. I measured its capacitance using a Fluke multimeter and got a value of 3.7 nF. Adding 2 more sheets of paper and 2 more sheets of foil to create a 4 plate interdidgitated capacitor increased the capacitance to 9.0 nF.

So I got an increase of about 2.4 times. This doesn't support either of my 2 hypothesis.

What is the correct explanation of my experimental results?

I suspect that your 'manufacturing process' may be introducing inaccuracies into your experiment. You can assume that the formula for the Capacitance of an ideal plate capacitor so, if your results don't agree, it's not due to the theory being wrong. The spacing is critical and you can't be sure that this is under control inside the device. I guess it's just possible that you are not counting all the spaces in your calculation for the interleaved construction(?).

 

[mrspeedybob]

I suspect that your 'manufacturing process' may be introducing inaccuracies into your experiment. You can assume that the formula for the Capacitance of an ideal plate capacitor so, if your results don't agree, it's not due to the theory being wrong. The spacing is critical and you can't be sure that this is under control inside the device. I guess it's just possible that you are not counting all the spaces in your calculation for the interleaved construction(?).

I know that formula would apply to the capacitor with 2 plates and one insulator. I wasn't sure if it applied to the capacitor with 4 plates and 3 insulators since that geometry would have more dielectric material in relation to the conductor's surface area. That's kind of what I was trying to determine with my experiment.

I'm still not sure if it applies.

 

[Baluncore]

If the energy in the capacitor is stored on the plates then the 4 plate capacitor should have twice the capacity of the 2 plate capacitor. If the energy is stored in the dielectric as deformed electron orbitals then the 4 plate capacitor should store 3 times the energy of the 2 plate capacitor

Since capacitance is dependent on the dielectric constant of the insulation and not on the metal or conduction of the plates the energy must be stored between the plates rather than in them.

The external fringing fields through the air add slightly to the capacitance. You are not accounting for that component of the field.

 

[sophiecentaur]

I know that formula would apply to the capacitor with 2 plates and one insulator. I wasn't sure if it applied to the capacitor with 4 plates and 3 insulators since that geometry would have more dielectric material in relation to the conductor's surface area. That's kind of what I was trying to determine with my experiment.

I'm still not sure if it applies.

Each 'gap' will contribute more or less equally to the capacitance. Three plates should give you twice the capacitance, four plates should give you three times etc.. When you have two foils, wrapped / rolled around each other (the normal method of construction) there are two effective gaps involved and this could be affecting your measured ratios.

But this only applies when you have the 'plates' wired to give parallel capacitance. Merely splitting the dielectric and putting a plate between two existing plates (as in some of the earlier pictures on this thread) will make no difference because you are just connecting two, double value Cs in series.