Charge.E field and Faraday's cage

[fisico30]

Hello everyone.

If a static charge is placed inside a metallic box, there is going to be an E field inside the box, zero in the layer between the inner and outer walls of the box, and a field E outside the box, as if the box was not there.

If the same charge inside the metal box is accelerated instead, the E field that it produces will be zero outside the volume of the box (assume a perfectly conducting walls perfect reflection).

Why?
What about Gauss law? How does it apply in the second case for a Gaussian surface that surrounds the box?

thanks

 

[Ben Niehoff]

There are two mistakes. For simplicity, let's consider the box to be a conducting sphere.

1. For a static configuration, there will be some complicated field inside the sphere, zero field between the inner and outer layers of the sphere, and then outside the sphere, the electric field will be as though all the charge were concentrated at the center. This last bit is important.

2. For the accelerating charge, assuming the sphere's inner surface is perfectly reflective, you will still have a static field outside the sphere, as though all the charge were concentrated at the center. Note that the external field does not change, and so therefore does not radiate any energy; thus energy is conserved and Gauss's Law is also satisfied.

 

[granpa]

There are two mistakes. For simplicity, let's consider the box to be a conducting sphere.

1. For a static configuration, there will be some complicated field inside the sphere, zero field between the inner and outer layers of the sphere, and then outside the sphere, the electric field will be as though all the charge were concentrated at the center. This last bit is important.
.

WOW. never knew that. is this always true regardless of the shape of the box?

 

[granpa]

ok, I can see that for a sphere shaped box. the charges on the outer surface are free to go where ever they want so they spread out evenly. for a square box though the charges should want to go to the corners.

 

[atyy]

See Thorne's note at the bottom of the page:
http://www.feynmanlectures.info/flp_errata.html

 

[Meir Achuz]

The field outside depends on the shape of the conductor and does not depend on the position of the charge inside the cavity. The field outside is that of a point charge
ONLY if the conductor is a sphere.

 

[fisico30]

Great answer!

so, even before I read Thorne's notes, an accelerating charge creates inside the box a time-varying EM field, but outside the box a static field?

How is that possible?
thanks

 

[granpa]

because the field can't penetrate the conductor.

 

[fisico30]

thanks granpa,

I can see that, but which charge creates a "static" field outside the metal box?
There is only a surface current at the inner surface of the conductor.

Sure, perfect reflection at the inside wall. No field inside the conductor, but outside of it , I cannot get why there is a "static" field as Niehoff says.
A static field requires static charge.

 

[granpa]

if charge q is inside the sphere then charge -q will be on the inner surface of the conductor. the conductor is neutral so there must be a charge q floating around in the conductor somewhere. since its free to go wherever it wants it goes to the outside surface and spreads out evenly.

 

[fisico30]

Hi again granpa,

I continue to agree with you so far. On the outer surface there is that positive q. But it should be a time-changing charge and create a time-varying field outside, and not a static one...

 

[granpa]

what would cause it to vary? the field of the inner charge ends at the inner surface.

 

[fisico30]

true granpa,

but the inner charge "should" be related to the outer charge in the conductor, since they both derive from an induction process.
The inner surface has a time changing surface current, which implies a time changing surface charge density (continuity equation).
Therefore I would think that the outer charge density would continue to redistribute itself (therefore moving, accelerating) and causing a time chaging E field outside the box...

 

[granpa]

you're thinking to hard. its not that complicated. the field of the inner change and all its effects ends at the inner surface. end of story.

 

[fisico30]

got it! thanks for persevering.

 

[Ben Niehoff]

Note that if the conductor were NOT perfect (i.e., if it had a finite $\epsilon$), then you WOULD have a time-varying field outside the sphere. The trick is that you're using a perfect conductor.

 

[fisico30]

thanks Mr. Niehoff.

If an AC current source (instead of an accelerating charge) was inside the perfect metal box, then waves (E field ) would be generated inside the box, no field in the conductor, and no field outside the conductor or still a static field, right?
The net charge of the current source is zero.
If we used a Gaussian surface to surround the metal box, the E flux would be zero or not zero ?

 

[Ben Niehoff]

Gauss's Law is never violated.

 

[fisico30]

I believe so too.
If we surrounds a radiating antenna with a Gaussian surface, the flux has to be nonzero.
The net charge on the metal antenna is however zero...

 

[atyy]

If we surrounds a radiating antenna with a Gaussian surface, the flux has to be nonzero.

Calculate it. Or consider, for example, the direction of the E field, and the direction in which power is transported.

Gauss's law is really never violated.