Faraday cage in a parallel plate capacitor

[AJ Bentley]

I would like to verify by myself if it's possible ? For me, the short distance of the Faraday surfaces compared to distance from capacitor would give energy, I would be happy to understand with calculations how this is false.

You are complicating the situation to the point of confusion. From the outside, a Faraday cage just looks electrically like a solid piece of metal.
If you place it between the plates of a capacitor, the exact position makes no difference so you may as well place it in contact with one of the plates. So all you have done is to make one plate thicker = closing the gap between the plates.

So you get exactly the same result as if you just moved one plate closer to the other.

 

[sophiecentaur]

A distance doesn't "give energy". Moving a force through a distance will involve work / energy transfer. The energy sums will give the right answer about work done and you can't necessarily trust an 'opinion' about the quantity of work. Not everything is intuitive.

If you think about the volts necessary to cause electrostatic forces, it is not surprising that the effects you are looking for will be small.

 

[Gh778]

With:

d=2mm thickness of Faraday cage
D=20 mm thickness of capacitor
S surface of capacitor and Faraday cage
U=voltage input for charged capacitor
Vacuum everywhere else

Could you at least give me:

Charge and field E of capacitor at start when Faraday cage are in it ?
Charge on the Faraday surfaces ?

 

[AJ Bentley]

I'm sure there are plenty of people who can work this for you. But you'll learn more if you work it out yourself.

Here's a calculator to help you.
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/pplate.html

 

[sophiecentaur]

With:

d=2mm thickness of Faraday cage
D=20 mm thickness of capacitor
S surface of capacitor and Faraday cage
U=voltage input for charged capacitor
Vacuum everywhere else

Could you at least give me:

Charge and field E of capacitor at start when Faraday cage are in it ?
Charge on the Faraday surfaces ?

Why should the charges "on the Faraday surfaces" be any different from the charges on the capacitor plates? Kirchoff 2 tells you that charge is conserved.

 

[Gh778]

Why should the charges "on the Faraday surfaces" be any different from the charges on the capacitor plates? Kirchoff 2 tells you that charge is conserved.

So, the charge on each surface (capacitor or Faraday cage) is the same ? Even thickness of Faraday cage is 2 mm compared to thickness of capacitor 20 mm ? It's not 1/10 of the charge because Faraday cage cancel only field which seeing by itself ?

 

[sophiecentaur]

I think it would be better to think in terms of polarisation than "charge on each surface" because there will always be a finite depth to the region of high charge at the edges of a conductor or dielectric.

A metal block, which is all your Faraday cage is behaving like (did you not accept that earlier post?) will have zero field across it. It will polarise with the same amount of displaced charge as the unbalanced charge on the capacitor plates. That is the equilibrium situation, surely. What would happen if the charges were different? Some more charges would move to make them equal.
Have you stopped looking elsewhere for enlightenment about this? Look at some other sources now we have discussed it at length. They may make more sense to you.

 

[Gh778]

Ok, you're right the charge on surface of the Faraday cage is the same than capacitor. In this case if I collapse cage to itself it's easy to see this give energy: Ff > Fl+Fr

 

[willem2]

Ok, you're right the charge on surface of the Faraday cage is the same than capacitor. In this case if I collapse cage to itself it's easy to see this give energy: Ff > Fl+Fr

I didn't catch until now, that the cage was the whole width of the capacitor. If that is the case the charges on the surface of the cage are indeed the same for the capacitor (for an ideal capacitor where we can ignore the edges). But in this case it is easy to compute the electric fields, and you can't use coulombs law like you did. For a charge attracted by an infinite plate the force is independent of the distance of the plate, which you can see by using gauss' law.

http://farside.ph.utexas.edu/teaching/316/lectures/node27.html

or just doing the integral.

http://www.physlink.com/education/askexperts/ae544.cfm.

There is in fact a force on the surface charges of the cage that pulls them away from each other towards the plates, and only the fact that they can't leave the metal keeps them where they are.

A negative charge q on the left side of the cage will feel an attractive force from the positive plate of the capacitor to the left, an attractive force from the positive side of the cage and a repulsive force from the negative plate of the capacitor which points to the left.
These forces are all equal and a force of

\frac {q \sigma} {\epsilon_0}

pointing to the left results, where \sigma is the charge density on the plates of the capacitor and the cage.

 

[Gh778]

For a charge attracted by an infinite plate the force is independent of the distance of the plate, which you can see by using gauss' law.

I don't understand, could you say each surface of Faraday cage is not attracted from nothing ? Note the capacitor has not an infinite surface, considered it like 40mm*40mm for example, this is considered like infinite ?

Even there is no force, the capacitor has gain energy when the Faraday has collapse. And we can recover energy from "dielectric" (iron) when discharge it.

 

[willem2]

I don't understand, could you say each surface of Faraday cage is not attracted from nothing ?

I don't understand what this means.

Note the capacitor has not an infinite surface, considered it like 40mm*40mm for example, this is considered like infinite ?

If the distance between the plates is small compared to the size of the plates the capacitor is of considered as an ideal capacitor where the field is the same everywhere between the plates. In a non-ideal capacitor, the induced charges would be smaller. In any case "You can't use coulombs law like you did" remains valid. Coulombs law only works for point charges or spherically symmetric charges.

Even there is no force, the capacitor has gain energy when the Faraday has collapse. And we can recover energy from "dielectric" (iron) when discharge it.

I don't understand what this means either.

 

[Gh778]

If I use your links, I considered circular surfaces for capacitor and Faraday cage. I integrate from 0 to 40 mm I find: 950-173-133 for the sum of force (at a factor near).
From 0 to 40 mm for 2 mm distance = 950
From 0 to 40 mm for 9 mm distance = 173
From 0 to 40 mm for 11 mm distance = 133

Like the force is attracted, how can I integrate with the real field ?

Even there is no force, the capacitor has gain energy when the Faraday has collapse. And we can recover energy from "dielectric" (iron) when discharge it.

I don't understand what this means either.

When the cage has disappear the capacity of capacitor go down so the energy go up. It's what we say before.

 

[sophiecentaur]

This thread seems to be going in circles now and the Faraday cage is just getting in the way of understanding the basics. Altering the spacing of two plates has exactly the same effect as changing the size of the Faraday cage so why not just consider that simpler model? That simple model gives a very reasonable result for Capacity and, hence, Energy stored that can't be a surprise to anyone.
Physically removing the Faraday cage (or shrinking it) will involve putting energy into the system. What more is there to say?

 

[Gh778]

This thread seems to be going in circles now

I'm trying to understand why my intuition is false and in the same time I'm learning physics because I like it:)

For now, like I don't know how to calculate I do some simulations with FEMM 2D software. You're right with my dimensions the force need energy. If I modify dimensions the force decrease and if I take dimensions like :

Height of capacitor = 2 mm (y axis)
Thickness of capacitor (where air is) = 215 mm (x axis)
Height of Faraday cage = 190 mm (y axis)
Thickness of Faraday cage = 3 mm (x axis)

Proof is 1 mm for all

With these values I have the charge of Faraday cage at 50 % of capacitor charge.

The force give energy for collapse Faraday cage

I give extra large surface for the simulation and the mesh is very small and I don't forget boundaries conditions.

After, the force give energy if I collapse Faraday cage but maybe with a cage like that the theory say it's like that*?

NB: I have read on internet a lot of examples with the energy for remove dielectric and some say the energy for remove equal to the energy the capacitor win. With a Faraday cage (use like a dielectric with air) if we consider the energy for remove the Faraday cage is the energy the capacitor win, after remove the Faraday cage we can recover energy from collapse cage and the energy of the dielectric too. This last energy is lost by source but the energy from collapsing cage come from where ?

 

[Gh778]

With a circular capacitor and a semi circular Faraday cage, the force to move up (not collapse) the cage is up (like drawing) and this go down the capacity. Sure we can't move a high distance. I give 2 femm files for test. The "az4" file is the capacitor at start when I'm charging it. We can see the force at start positive at up. The "az5" file is the capacitor when the cage is moved, like that it's possible to see the charge and like that the capacity.