An electron inside a Faraday cage

  • B
  • Thread starter etotheipi
  • Start date
  • #1
etotheipi
Gold Member
2019 Award
2,744
1,675
I came across a problem that seemed fairly interesting; it asks what would be the trajectory of an electron released from rest inside a Faraday cage (which is itself within a uniform gravitational field ##\vec{g} = -g\hat{y}##). I didn't quite understand their explanation.

They say that electrons inside the walls of the Faraday cage move freely until they reach equilibrium. We end up with a small net negative charge on the lower side of the Faraday cage and a net positive charge on the upper side, which sets up an electric field inside the walls, pointing downward, of magnitude ##\frac{m_eg}{e}##. They then say that this same homogenous electric field must also exist inside the Faraday cage (otherwise we would have a non-zero line integral of ##\vec{E}## along a path through the wall and the interior), and as a result the electric force on the internal electron exactly balances the gravitational force... so the electron floats (i.e. the Faraday cage effectively "shields the gravitational field").

My problem with that explanation is that if the walls of the Faraday cage are conducting, there should be no electric field inside them (see red part). I wondered if someone could clarify what's going on? Thanks :smile:
 

Answers and Replies

  • #2
vanhees71
Science Advisor
Insights Author
Gold Member
2019 Award
16,017
7,321
Do you have the reference?
 
  • Like
Likes etotheipi
  • #4
etotheipi
Gold Member
2019 Award
2,744
1,675
Here's the given solution in full:

new doc 2020-07-13 16.25.37_1.jpg

new doc 2020-07-13 16.25.37_2.jpg
 
  • Like
Likes vanhees71
  • #5
A.T.
Science Advisor
10,650
2,241
My problem with that explanation is that if the walls of the Faraday cage are conducting, there should be no electric field inside them (see red part).
I think this assumption ignores gravity and the mass of the electrons.
 
  • Like
Likes etotheipi
  • #6
vanhees71
Science Advisor
Insights Author
Gold Member
2019 Award
16,017
7,321
I think they simply mean within the cavity not within its walls. At least this would make sense and corresponds to what's calculated.

What I didn't understand is the remark about the positron. Here they seem to rather consider a single positron close to a practically infinite conducting half-space. If there'd be a positron within a gas of electrons it would rather quickly get annihilated with an electron leaving two photons ;-).
 
  • Like
Likes etotheipi
  • #7
etotheipi
Gold Member
2019 Award
2,744
1,675
That would indeed make more sense, thank you! I also didn't understand what they were trying to get at with the positrons, but the setup is fairly unrealistic anyway, so I thought I'd just roll with it :wink:
 
  • #8
A.T.
Science Advisor
10,650
2,241
I think they simply mean within the cavity not within its walls.
They explicitly say "in the wall" several times. The electrons in a conductor need to be supported against gravity too, so they sink down until their repulsion balances the gravity on them.
 
  • Like
Likes davenn
  • #9
Paul Colby
Gold Member
1,216
298
They explicitly say "in the wall" several times.
That's the way I read it as well. Personally, I have a hard time with a conductor developing a charged distribution because electrons sink to the bottom a bit in a gravitational field. Having expressed my somewhat baseless opinion, I look forward to the discussion.
 
  • #10
A.T.
Science Advisor
10,650
2,241
Personally, I have a hard time with a conductor developing a charged distribution because electrons sink to the bottom a bit in a gravitational field.
Seems like the puzzle is based on this paper:
https://www.sciencedirect.com/science/article/abs/pii/0003491668903011

But as the abstract states, it is very idealized and theoretical:
Whether the necessary assumptions apply to real conductors is not certain.
 
  • Informative
  • Like
Likes Paul Colby and etotheipi
  • #11
Paul Colby
Gold Member
1,216
298
I hate paywalls.
 
  • Like
Likes davenn and etotheipi
  • #12
Nugatory
Mentor
12,988
5,699
there should be no electric field inside them
The electric field within the walls is zero. You may choose, as the author of this problem has, to write this zero as the sum of two contributions with opposite sign: the contribution from the non-uniform distribution of charged particles within the walls, and the contribution from the charged particle inside the cage.
 
  • Like
Likes etotheipi
  • #13
etotheipi
Gold Member
2019 Award
2,744
1,675
The electric field within the walls is zero. You may choose, as the author of this problem has, to write this zero as the sum of two contributions with opposite sign: the contribution from the non-uniform distribution of charged particles within the walls, and the contribution from the charged particle inside the cage.
This sort of makes sense, but the equilibrium condition inside the walls is derived by balancing the weight with the electric field produced by the separation of charges. If the electric field inside is indeed zero (as we would expect), then we would need to assume the charges in the walls have zero mass (as @A.T. alluded to earlier)
 
  • #14
vanhees71
Science Advisor
Insights Author
Gold Member
2019 Award
16,017
7,321
They explicitly say "in the wall" several times. The electrons in a conductor need to be supported against gravity too, so they sink down until their repulsion balances the gravity on them.
Ok, then I misunderstood the idea.

So you just look at a piece of a conductor in the gravitational field of the Earth. For a static situation then of course there must be a little negative charge density at the bottom of the conductor, because of gravity (the positive lattice is solidly bound and thus is (almost) unaffected by the gravitational field).

At each point in the material the electric force ##\vec{F}_{\text{em}}=\mathrm{d}^3 x E (-e) n \vec{e}_z## due to the built-up charge density must cancel the gravitational force ##\vec{F}_{\text{grav}}=-\mathrm{d}^3 x m g n \vec{e}_z##, i.e.,
$$E=-m/e g.$$
It's now an interesting task to calculate the actual ##n(z)## and ##E(z)## (which I've not done yet).

[EDIT:] I just got the idea that the idea of the book is wrong, because everywhere in the interior of course you must have ##\vec{\nabla} \cdot \vec{E}=\rho##, but ##\vec{E}=\text{const}## and thus ##\rho=-e n + Z e n_{lattice}=0##. So there's only a negative net-surface charge density on the bottom and a positive net-surface charge density on the top, with ##\sigma_{\text{top}}=-\sigma_{\text{bottom}}## as in a capacitor. So (neglecting fringe fields) there's no electric field outside and thus also none in the Faraday cage.
 
Last edited:
  • Informative
Likes etotheipi

Related Threads on An electron inside a Faraday cage

Replies
1
Views
1K
  • Last Post
Replies
8
Views
3K
  • Last Post
Replies
14
Views
6K
  • Last Post
Replies
19
Views
3K
  • Last Post
Replies
12
Views
4K
Replies
0
Views
3K
Replies
2
Views
2K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
17
Views
5K
  • Last Post
Replies
4
Views
2K
Top