# Faraday cage in a parallel plate capacitor

1. Nov 1, 2012

### Gh778

Hi,

A parallel plate capacitor is charged from fixed voltage, after I cut voltage source.
Now, what's happened if I place a Faraday cage inside capacitor (without touch capacitor) ? Capacity changed ? So energy of capacitor changed ?

2. Nov 1, 2012

### Gordianus

In that case the capacitance would go up and the stored energy would go down.

3. Nov 1, 2012

### Gh778

ok, like the energy go down I think we win energy from move Faraday cage, but there is no charge in the cage, how energy is win ?

Edit: I just understand that there is charges in cage, so we win energy from attraction.

In this case, I can place Faraday cage, cut it (in the length) inside plane capacitor and move out the cage, the capacitor recover all its energy, and I have another capacitor charged (maybe with another greater dielectric inside), so in this case where the energy is lost ?

Last edited: Nov 1, 2012
4. Nov 1, 2012

### sophiecentaur

Any apparent energy imbalance is reconciled by mechanical work done - even if you think you 'snuck in from the side'.

5. Nov 1, 2012

### Gh778

A plate capacitor has a dielectric of 60 inside with 1 mm of thickness. We charge it from external voltage source. When charged cut the source. Move in a Faraday cage which envelop the dielectric (without touch conductor), this decrease the thickness to 0.1 mm for example but the capacity go down because dielectric go down at 1. A final, the capacity go down and the energy go up ?

6. Nov 1, 2012

### Gordianus

To be honest I couldn't understand the last post

7. Nov 1, 2012

### sophiecentaur

Are you sure the capacity goes down? The polarisation around the Faraday cage will be greater than the polarisation of the dielectric it is surrounding so will you not be left with two high value capacitors in series (the gaps either side of the cage are the same as {or less than} the gaps on either side of the piece of dielectric. )? That implies that the Capacity will increase. The dielectric constant will only go to 1 if you remove the dielectric.

8. Nov 1, 2012

### Gh778

Sorry, I'm not fluently in english
Charge a plate capacitor which has 1mm of thickness and a dielectric like teflon (relative permittivity of 60). Cut source voltage. Envelop all teflon of thin iron. The iron reduce thickness to 0.1 mm but now the relative permittivity is 1 not 60. At final the capacity is down to 1/6 and energy go up ?

No, I'm not. I'm trying to understand what's happened exactly. With a closed cage, the cage will create a reverse field, this compensate all external fields. Field in dielectric will see less field and cage will go up its field for compensate this. At final, all the field will be in the cage not in the dielectric I think. If I discharge capacitor, why electric field will not pass through iron, it's easier ?

I think about another case: Take a plate capacitor thickness of 10 mm with air inside. We have a Faraday cage in it of 2 mm at the center of the capacitor. We charge the capacitor and cut voltage source after. It's possible to move closed surface of the Faraday cage because the force is 1/2²-1/4²-1/6² > 0 like that the charge in the Faraday cage could be cancel. After discharge capacitor will recover more energy because voltage is more important. The Faraday cage could be 2 surfaces and links between. So I think I don't understand something important. Edit: like Faraday cage charged at % of capacitor use 20 mm for capacitor and 2 mm for Faraday cage.

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Last edited: Nov 2, 2012
9. Nov 2, 2012

### willem2

A faraday cage will act as a dielectric with infinite permittivity. A dielectric with a permittivity of 60 in an elictric field will polarize and produce an elictric field of its own that will cancel 59/60 of the outside field.
A faraday cage will cancel ALL of the field, so there is no field inside it.
It seems obvious, that if you replace some of the dielectric in a capacitor and replace it with a dielectric with higher permittivity that the capacity will go up.

The field will go up in the space between the cage and the plates of the capacitor. The charges on the surface of the cage will attract more charge from the plate, so there will be more charges on the plate close to the cage, and less elsewhere.
(if the capacitor isn't connected, so the total charge is fixed)

The

10. Nov 2, 2012

### Gh778

Ok for the infinite permittivity of Faraday cage but where energy is lost when I move close surfaces of Faraday cage ? If the cage has a thin thickness, it's possible to recover energy from move close its surfaces and in this cage capacity of capacitor decrease and so energy go up ?

11. Nov 2, 2012

### sophiecentaur

You have lost NO energy. As I said earlier, Work is done by or on (depending) the system by the action of moving bits around. The faraday cage will be attracted into place and if you let it move freely, it would 'pendulum' from side to side for ever if it didn't collide with something. If you use your hands to move the cage then you would not be aware of the small amount of work done by the pulling of the cage into place.

12. Nov 2, 2012

### Gh778

I don't understand where I'm wrong, I details:

1/ A plate capacitor has like dielectric a Faraday cage with thin wall, all is fixed. The capacitor is 20 mm and the Faraday cage is 2 mm in the center of the capacitor, elsewhere there is air
2/ Charge with U voltage, all energy lost from source is on capacitor
3/ Cut voltage source, no need energy
4/ Cut Faraday cage for have only 2 surfaces, no need energy
5/ Move close 2 surfaces of the Faraday cage, this no need energy, in fact with datas this give little energy: 1/2²-10/9²-10/11² >0
6/ Touch 2 surfaces of Faraday cage, no need energy
7/ The capacity go down of the capacitor, this increase energy of the capacitor

If you see where is my error ?

Last edited: Nov 2, 2012
13. Nov 2, 2012

### willem2

Point 5. moving the 2 surfaces closer will cost energy when the cages is still in the capacitor.

14. Nov 2, 2012

### Gh778

Ok, if the Faraday cage is large but with 2mm of thickness for the cage and 20 mm for the capacitor. One surface of the cage is at 2 mm to another but at 9 mm to capacitor surface (other at 11 mm). Like force is in 1/d² we don't have something like that for one surface:

1/2²-10/9²-10/11² ? this is an attraction force not a repulsive

15. Nov 2, 2012

### willem2

If you really want to do that, you'll need to integrate over the surfaces of the plate and the cage, wich both won't have an uniform charge distribution. Your simple approximation doesn't take into account that there will be a lot more charge on the plates of the capacitor than induced charge on the cage.

The induced charges only form, because there is a force on them. If you want to eliminate them by collapsing the cage, you will have to to work against this force.

16. Nov 2, 2012

### Gh778

It's for that I wrote 10 times charges for capacitor in the simple formula. If cage is 2 mm and capacitor 20 mm, the charge on cage is not 1/10 of charge on capacitor ?

yes, I want to see energy when collapsing the cage. I don't understand your sentence, there are another forces than cage or capacitor ?

17. Nov 2, 2012

### sophiecentaur

It's all to do with 6/ and 7/
7/: The capacity goes UP not down. Think of it like this: Initially,you have three capacitors in series, comprising an air gap between one plate and half of the Faraday cage, the dielectric, the other half of the Faraday cage, and another air gap. The final arrangement is just two narrow air gaps in series, between plates and the sides of the Faraday cage, corresponding to a much higher capacity altogether.

6/:
When you touch the two halves of the Faraday cage together, current will flow. Neglecting resistance and EM radiation, for a start. The current will induce a magnetic field, which will store energy. The Faraday cage is a conductor of finite dimensions so it has Inductance. So you will have an LC tuned circuit which will resonate (ring) for ever if there is no loss. However, there will be finite resistance in the metal and also there will be EM radiation so the ringing will die down eventually and 'your' lost energy will have been accounted for in the form of losses.
What you are describing is a form of a famous old 'paradox' problem in which a charged capacitor is connected to an identical uncharged capacitor and there is a difference in stored energy before and after.

Last edited: Nov 2, 2012
18. Nov 2, 2012

### Gh778

Imagine the capacitor without Faraday cage is has 20 mm of air. Now, the same capacitor with Faraday cage don't see 18 mm of air because relative permittivity of iron is high ? At first, we have 18 mm of air at final we have 20 mm of air, this don't go down capacity ?

You're right there is another energy here ;)

19. Nov 2, 2012

### willem2

Since the field of the capacitor is able to form the induced charges, it has to be stronger than the field that from the induced charges themselves. If it weren't so, there wouldn't be any induced charges, and you would not have a faraday cage.

20. Nov 2, 2012

### sophiecentaur

I don't know what picture you have in your mind about this experiment but how can you think that introducing a conducting box in between the parallel (fixed) plates of a capacitor can produce a greater gap between them? The Faraday cage is just the equivalent to putting a plate close to the top plate and connecting it, with a wire, to another plate, close to the bottom capacitor plate. This constitutes one high value capacitor in series with another high value capacitor. This will have a much higher capacity than a compound capacitor made of these two with another low value capacitor connected between them.
Just to check - do you remember the correct formula for connecting capacitors in series? Could this be the problem?

21. Nov 2, 2012

### sophiecentaur

Owch! You may have got that right but the double / triple negative in the middle has thrown me completely. Talking in terms of fields, the Faraday cage has zero field across it (no PD) and has just increased the fields across the remaining gaps and the total PD appears just across them. This means more charges can be displaced for a given PD at the terminals - i.e. a higher capacitance

22. Nov 2, 2012

### willem2

If you eliminate the faraday cage, the capacity goes down. You replace an area with essentially infinite permittivity with an area with a smaller permittivity, so the capacity goes up.

His point was that it doesn't cost energy to connect the two halves of the faraday cage, wich is true. The problem is with step 5. You can get energy out of the system by inserting the faraday cage into the capacitor, so you have to do work, either to get the induced charges to cancel while the cage is inside the capacitor or to pull out the cage out of the capacitor first.

You don't get any resonation, because the separated faraday cage won't discharge while it's still in the field of the capacitor

23. Nov 2, 2012

### Gh778

When the Faraday cage is in the capacitor the gap is lower for me because it's like have 18 mm. But after collapse Faraday cage we have 20 mm, like d increase the capacity is lower so the energy greater ? I'm trying to understand.

24. Nov 2, 2012

### willem2

The capacity is indeed lower and this means that (with an unconnected capacitor) the energy will be greater. This should tell you that you have to do work to do step 5.
The same happens if you pull the plates of a capacitor apart.

25. Nov 2, 2012

### Gh778

Ok, how I could calculate the work need to move close surfaces of Faraday cage ? Could you help me for that ?