Electrostatic shielding, grounded Faraday cage, Feynman

[J Jensen]

Hello everyone

I am trying to understand how an electrostatic field inside a cavity
in a conductor is shielded from a probe on the outside. For
simplicity, we could even work with a hollow metal sphere. Lets say we
had a bunch of positive charges inside, on a insulated stand, and the
conductor was not charged. A gaussian surface around the sphere would
show net flux, because of the net positive charge inside.

But if the sphere were grounded, then it is claimed that there is no
field outside. I can believe that if the conductor is grounded, there
would be an acumulation of negative charge on the inner surface, but I
can't see why that would prevent any E field lines from going out and
then looping back in to meet negative charge...maybe because all
points on the inner surface are negatively charged, so in any
direction the field lines die on contact? But that would also be true
in the non-grounded case, only then there would be a positive charge
on the outer surface. It seems hard to understand how the field could
be E=0 inside the metal of the conductor in both of those different
cases...!

Also, consider a charged parallel plate capacitor, insulated in the
cavity. Why would field lines from that not go outside the enclosing
conductor?

(I mentioned Feynman in the subject because this is famously discussed
in his Lectures on Physics, vol 2 ch 5 where he forgot to say that the
sphere had to be grounded).

--Jeff

[Joseph Warner]

The answer to your questions is in the Gaussian surface you draw
around.

If there is a charge inside of the a hollow metal sphere, for the
two cases you mention -- grounded and not grounded Gauss's Law
with E=0 inside the metal answers your questions.

Case grounded -- The E=0 outside the sphere then as you draw the
Gaussian surface around the sphere and let it shrink there is no
flux outside so you know that there has to be an equal number of
opposite charges inside the Gaussian surface as the charge inside
the hollow sphere to cancel. Now you need to determine where they
are. Shrink the Gaussian sphere to the diameter of the metal
sphere. You still have E=0. Shrink it just below the surface.
Knowing that E=0 inside metal then one can deduce there is no
surface charge on the outside of the metal. (The reason that E=0
inside of the metal is that if E <> 0 then there is a current and
a current implies loss of energy for a normal metal and since we
are in steady state and we have no power supply or battery that
cannot happen.) So now keep shrinking the Gaussian surface just
outside of the inner diameter of the metal sphere. Still E=0. Now
shrink it just inside the inner diameter. Now E<>0. One then can
deduce that there all the opposite charge must be surface charge
on the inner sphere.

One can use the same arguments for the non-grounded case to show
that there must be surface charge now on both the inner and outer
surface of the hollow metal sphere.

I think most people gets confused with trying to look at this
problem in the time domain. The solution is a steady state
solution after any currents have died out. For the first example,
if you have a hollow metal surface and then could instantaneously
place charge at the center of it there would be a current from
the ground up through the metal sphere to the inside surface
until the charge on the inner surface equals the charge at the
center.

"J Jensen" <jjensen14@hotmail.com> wrote in message
news:1191247215.550594.174250@19g2000hsx.googlegro ups.com...
> Hello everyone
>
> I am trying to understand how an electrostatic field inside a
> cavity
> in a conductor is shielded from a probe on the outside. For
> simplicity, we could even work with a hollow metal sphere. Lets
> say we
> had a bunch of positive charges inside, on a insulated stand,
> and the
> conductor was not charged. A gaussian surface around the
> sphere would
> show net flux, because of the net positive charge inside.

[J Jensen]

On Oct 2, 1:19 am, "Joseph Warner" <Joseph.D.War...@nasa.gov> wrote:
> The answer to your questions is in the Gaussian surface you draw
> around.
>
> If there is a charge inside of the a hollow metal sphere, for the
> two cases you mention -- grounded and not grounded Gauss's Law
> with E=0 inside the metal answers your questions.
>
> Case grounded -- The E=0 outside the sphere then as you draw the
> Gaussian surface around the sphere and let it shrink there is no
> flux outside so you know that there has to be an equal number of
> opposite charges inside the Gaussian surface as the charge inside
> the hollow sphere to cancel. Now you need to determine where they
> are. Shrink the Gaussian sphere to the diameter of the metal
> sphere. You still have E=0. Shrink it just below the surface.
> Knowing that E=0 inside metal then one can deduce there is no
> surface charge on the outside of the metal. (The reason that E=0
> inside of the metal is that if E <> 0 then there is a current and
> a current implies loss of energy for a normal metal and since we
> are in steady state and we have no power supply or battery that
> cannot happen.) So now keep shrinking the Gaussian surface just
> outside of the inner diameter of the metal sphere. Still E=0. Now
> shrink it just inside the inner diameter. Now E<>0. One then can
> deduce that there all the opposite charge must be surface charge
> on the inner sphere.
>
> One can use the same arguments for the non-grounded case to show
> that there must be surface charge now on both the inner and outer
> surface of the hollow metal sphere.
>
> I think most people gets confused with trying to look at this
> problem in the time domain. The solution is a steady state
> solution after any currents have died out. For the first example,
> if you have a hollow metal surface and then could instantaneously
> place charge at the center of it there would be a current from
> the ground up through the metal sphere to the inside surface
> until the charge on the inner surface equals the charge at the
> center.
>
> "J Jensen" <jjense...@hotmail.com> wrote in message
>
> news:1191247215.550594.174250@19g2000hsx.googlegro ups.com...
>
> > Hello everyone
>
> > I am trying to understand how an electrostatic field inside a
> > cavity
> > in a conductor is shielded from a probe on the outside. For
> > simplicity, we could even work with a hollow metal sphere. Lets
> > say we
> > had a bunch of positive charges inside, on a insulated stand,
> > and the
> > conductor was not charged. A gaussian surface around the
> > sphere would
> > show net flux, because of the net positive charge inside.

Joseph, thanks for your reply. You have indeed shown that all the
excess negative charge must lie on the inner surface. But the original
question is still open, which is: How can we show that there is no E
field external to the conductor?

Consider this. An amount Q of positive charge on one end of an
insulating rod and -Q at the other end. Its a dipole. Certainly there
are E field lines from the pos charge to the negative. If we enclose
this assembly in a Gaussian surface, there is certainly flux through
the surface, so the E field is not zero; only the net flux is zero.

Now if we enclose the dipole in a cavity in a conductor, like the
hollow metal sphere, is there any field outside? How do we see that?

--Jeff

[fleem]

Consider this. An amount Q of positive charge on one end of an
insulating rod and -Q at the other end. Its a dipole. Certainly there
are E field lines from the pos charge to the negative. If we enclose
this assembly in a Gaussian surface, there is certainly flux through
the surface, so the E field is not zero; only the net flux is zero.

Now if we enclose the dipole in a cavity in a conductor, like the
hollow metal sphere, is there any field outside? How do we see that?
--Jeff

Maybe this will help.

When field lines go through a conductor, they cause a force on the charge carriers in that conductor. Charge carriers are, by definition, free to move. So those field lines cause those charge carriers to move. Thus charge carriers will move until no field lines go through the conductor. The only field lines you'll ever see leaving or entering the surface of a conductor will, therefore, be caused by charges on that surface, and there's nothing that says charges can't bunch-up at various points on the surface in order to cancel field lines within the conductor.

Above is the classical explanation. Quantum mechanically speaking, "field lines" are seen as virtual photons from all charges, whether external or internal to the conductor, and they always freely go through the conductor. The charge carriers in the conductor move to cause the average field lines (virtual particles) to be zero within the conductor. However, it turns out even this may be wrong!

Here is a fascinating paper I wish I could understand better:

http://quantumfieldtheory.org/About.htm

It says a lot of things unrelated to this discussion which are very interesting. As far as this conversation goes, search it for the text "we find that the electromagnetic coupling grows with energy".

[Uncle Al]

J Jensen wrote:
>
> On Oct 2, 1:19 am, "Joseph Warner" <Joseph.D.War...@nasa.gov> wrote:
> > The answer to your questions is in the Gaussian surface you draw
> > around.
[snip]

> Joseph, thanks for your reply. You have indeed shown that all the
> excess negative charge must lie on the inner surface. But the original
> question is still open, which is: How can we show that there is no E
> field external to the conductor?
>
> Consider this. An amount Q of positive charge on one end of an
> insulating rod and -Q at the other end. Its a dipole. Certainly there
> are E field lines from the pos charge to the negative. If we enclose
> this assembly in a Gaussian surface, there is certainly flux through
> the surface, so the E field is not zero; only the net flux is zero.
>
> Now if we enclose the dipole in a cavity in a conductor, like the
> hollow metal sphere, is there any field outside? How do we see that?

Consider the analogous magnetic dipole. Start with a seamless
thick-walled hollow sphere fabricated of persistent ferromagnetic
material such as ferrite, Sm-Co, or Fe-Nd-B. Magnetize it across its
radius (its thickness not its diameter). The south (north) pole is
then wholly the inside surface and the north (south) pole is wholly
the outside surface.

Is there any detectable magnetic field external to the outside
surface? If you can diddle the construct so there is you will be
awarded a Prize for presenting a magnetic monopole and fully
symmetrizing Maxwell's equations. You need only demonstrate the
existence of one magnetic monopole in the entire universe.

High sensitivity testing for external field is facile. Drop it
through a superconducting loop (solenoid) with a SQUID detector. If
there is reproducibly even a single quantum of net induced current you
get the Prize. All even-pole contributions entering the loop will
exactly cancel exiting the loop. Mount the magnetic monpole in a
hopper car of a model train set and continuously circulate the
horizontal train through the vertical loop to continuously build flux.

http://arxiv.org/pdf/physics/0512220
<http://en.wikipedia.org/wiki/Magnetic_monopole#Attempts_to_find_monopoles>
http://prola.aps.org/abstract/PRL/v53/i22/p2067_1
http://adsabs.harvard.edu/abs/1989ITM....25.1208H

--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
http://www.mazepath.com/uncleal/lajos.htm#a2

[Joseph Warner]

Jensen,

I missed your previous post. Uncle Al's analog to a magnetic
system is correct.

To show you the explanation for the electric case requires more
time than I would like to spend on the subject. This especially
true with Uncle Al's explanation. But if you want to try it
yourself, the argument will be that E=0 inside a conductor and
that for a normal metal in equilibrium there cannot be any
current. With these conductions one should be able to convince
oneself that there cannot be an external field associated with
the charge at the center of the sphere.

Sometime it is too bad that we can't have a chalk board online.
:-)

Good luck

[Vladimir Dergachev]

J Jensen wrote:

> Hello everyone
>
> I am trying to understand how an electrostatic field inside a cavity
> in a conductor is shielded from a probe on the outside. For
> simplicity, we could even work with a hollow metal sphere. Lets say we
> had a bunch of positive charges inside, on a insulated stand, and the
> conductor was not charged. A gaussian surface around the sphere would
> show net flux, because of the net positive charge inside.
>

A simple way to see that is to consider not a sphere, but a spherical
shell - i.e. an enclosure with some thickness to it.

If there was an electric field in the material of the shell than the
electrons in that material would move along it - and the stuff would heat
up.

So in a static case, when everything has stabilized long ago, we cannot have
an electric field inside metal - or anything conducting.

What we can have though is charge on the surface of the conductor - but it
all must be at the same potential regardless of what else goes in the
system as otherwise we would have electric field parallel to the surface
(which would get electrons moving).

Thus our system now looks like this:

* Outside
* surface charge determined from requirement of constant potential = U
* metal - zero E
* surface charge determined from requirement of constant potential = U
(same one !)
* inside

Thus our entire system has split up in two, connected with a single common
variable U which is a number (not a field !). Thus isolation: the outside
does not know the shape of the charge inside.

Note, however, that when field change and conductors have finite
conductivity than the picture is not perfect. You still get isolation, but
due to electrons "sloshing about" some field leaks through. What's more,
changing field means changing U - and thus RF emition.

In practice, it is relatively easy to construct a shielding enclosure that
attenuates RF by a factor of 10x, but doing better is not easy.

best

Vladimir Dergachev

[J Jensen]

On Oct 4, 10:49 am, Uncle Al <Uncle...@hate.spam.net> wrote:
> J Jensen wrote:
>
> > On Oct 2, 1:19 am, "Joseph Warner" <Joseph.D.War...@nasa.gov> wrote:
> > > The answer to your questions is in the Gaussian surface you draw
> > > around.
>
> [snip]
>
> > Joseph, thanks for your reply. You have indeed shown that all the
> > excess negative charge must lie on the inner surface. But the original
> > question is still open, which is: How can we show that there is no E
> > field external to the conductor?
>
> > Consider this. An amount Q of positive charge on one end of an
> > insulating rod and -Q at the other end. Its a dipole. Certainly there
> > are E field lines from the pos charge to the negative. If we enclose
> > this assembly in a Gaussian surface, there is certainly flux through
> > the surface, so the E field is not zero; only the net flux is zero.
>
> > Now if we enclose the dipole in a cavity in a conductor, like the
> > hollow metal sphere, is there any field outside? How do we see that?
>
> Consider the analogous magnetic dipole. Start with a seamless
> thick-walled hollow sphere fabricated of persistent ferromagnetic
> material such as ferrite, Sm-Co, or Fe-Nd-B. Magnetize it across its
> radius (its thickness not its diameter). The south (north) pole is
> then wholly the inside surface and the north (south) pole is wholly
> the outside surface.
>
> Is there any detectable magnetic field external to the outside
> surface? If you can diddle the construct so there is you will be
> awarded a Prize for presenting a magnetic monopole and fully
> symmetrizing Maxwell's equations. You need only demonstrate the
> existence of one magnetic monopole in the entire universe.
>
> High sensitivity testing for external field is facile. Drop it
> through a superconducting loop (solenoid) with a SQUID detector. If
> there is reproducibly even a single quantum of net induced current you
> get the Prize. All even-pole contributions entering the loop will
> exactly cancel exiting the loop. Mount the magnetic monpole in a
> hopper car of a model train set and continuously circulate the
> horizontal train through the vertical loop to continuously build flux.
>
> http://arxiv.org/pdf/physics/0512220
> <http://en.wikipedia.org/wiki/Magnetic_monopole#Attempts_to_find_monop...>http://prola.aps.org/abstract/PRL/v53/i22/p2067_1http://adsabs.harvard.edu/abs/1989ITM....25.1208H
>
> --
> Uncle Alhttp://www.mazepath.com/uncleal/
> (Toxic URL! Unsafe for children and most mammals)http://www.mazepath.com/uncleal/lajos.htm#a2

Thanks for your reply, Uncle Al. I'm not quite sure the magnetic
monople analogy has any bearing though. Too bad you don't have any
links on this electrostatic shielding problem (I can't find any
either).

--Jeff

[Grouchy]

On Oct 8, 6:47 am, J Jensen <jjense...@hotmail.com> wrote:
> On Oct 4, 10:49 am, Uncle Al <Uncle...@hate.spam.net> wrote:
>
>
>
> > J Jensen wrote:
>
> > > On Oct 2, 1:19 am, "Joseph Warner" <Joseph.D.War...@nasa.gov> wrote:
> > > > The answer to your questions is in the Gaussian surface you draw
> > > > around.
>
> > [snip]
>
> > > Joseph, thanks for your reply. You have indeed shown that all the
> > > excess negative charge must lie on the inner surface. But the original
> > > question is still open, which is: How can we show that there is no E
> > > field external to the conductor?
>
> > > Consider this. An amount Q of positive charge on one end of an
> > > insulating rod and -Q at the other end. Its a dipole. Certainly there
> > > are E field lines from the pos charge to the negative. If we enclose
> > > this assembly in a Gaussian surface, there is certainly flux through
> > > the surface, so the E field is not zero; only the net flux is zero.
>
> > > Now if we enclose the dipole in a cavity in a conductor, like the
> > > hollow metal sphere, is there any field outside? How do we see that?
>
> > Consider the analogous magnetic dipole. Start with a seamless
> > thick-walled hollow sphere fabricated of persistent ferromagnetic
> > material such as ferrite, Sm-Co, or Fe-Nd-B. Magnetize it across its
> > radius (its thickness not its diameter). The south (north) pole is
> > then wholly the inside surface and the north (south) pole is wholly
> > the outside surface.
>
> > Is there any detectable magnetic field external to the outside
> > surface? If you can diddle the construct so there is you will be
> > awarded a Prize for presenting a magnetic monopole and fully
> > symmetrizing Maxwell's equations. You need only demonstrate the
> > existence of one magnetic monopole in the entire universe.
>
> > High sensitivity testing for external field is facile. Drop it
> > through a superconducting loop (solenoid) with a SQUID detector. If
> > there is reproducibly even a single quantum of net induced current you
> > get the Prize. All even-pole contributions entering the loop will
> > exactly cancel exiting the loop. Mount the magnetic monpole in a
> > hopper car of a model train set and continuously circulate the
> > horizontal train through the vertical loop to continuously build flux.
>
> >http://arxiv.org/pdf/physics/0512220
> > <http://en.wikipedia.org/wiki/Magnetic_monopole#Attempts_to_find_monop...>http://prola.aps.org/abstract/PRL/v53/i22/p2067_1http://adsabs.harvar...
>
> > --
> > Uncle Alhttp://www.mazepath.com/uncleal/
> > (Toxic URL! Unsafe for children and most mammals)http://www.mazepath.com/uncleal/lajos.htm#a2
>
> Thanks for your reply, Uncle Al. I'm not quite sure the magnetic
> monople analogy has any bearing though. Too bad you don't have any
> links on this electrostatic shielding problem (I can't find any
> either).
>
> --Jeff

Jeff,

See if you can dig up a copy of "Noise Reduction Techniques in
Electrical Systems" by Henry W. Ott (ISBN 0-471-85068-3). The last
chapter (12) in the edition I have, along with the chapter
bibliography covers this subject (as an overview). I don't know if
prof Ott is still alive (I hope so), or if his work has been
digitized. He taught an off hours course at bell labs back in the
day, and the above text came from those lecture notes. As a result,
it's very 'power point friendly' and just defers directly to technical
papers on the particulars, most of which should be available...

Here's an additional link to the IBM journal of Research and
Development back issues dating all the way to 1957:

http://www.math.utah.edu/pub/tex/bib/toc/ibmjrd.html

Grouchy

[kmpatel]

I know this is 2 years too late, but the other replies didn't really answer the question the confusion Jeff had about E-field inside the metal.

The E-field in the perfectly conducting metal will always be ZERO. The fundamental misunderstanding, in all the response, is assumption that the sphere is always CHARGELESS and that the charge on the outside surface must be the same as that of the inside surface for the E-field is zero. Neither of these assumptions are correct. This misconception is derived from the intuitive example of a parallel plate, but it should be realized the sphere is a closed geometry, completely surrounding the inner space where the other charge is located. It should be noted that shape of the hollow metal does NOT matter. It can be a square box, and the results will still be the same. The answer has absolutely nothing to do with heat generation and current flow.

Certainly, for the UNGROUNDED metal case, the charge of the conductor will remain the same at all time (presumably 0, but not necessarily). Let's assume the sphere has +Q1 charge and there is +Q2 charge in the space within the sphere, the inner surface of the conductor will naturally accumulate -Q2 charge. As a result on the inner surface charge, the E-field perfectly terminates at the inner surface. The E-field is zero beyond the inner surface, UNTIL the outer surface is encountered. In order to maintain the same charge as before, the outer surface of the sphere will have accumulated +Q1+Q2 charge (the total charge on the sphere is +Q1+Q2-Q2=+Q1). A "new" E-field emits from the total charge on the outer surface, according to Q1+Q2. Regardless of the value of Q1 or Q2 and any other external fields, in the metal E=0 and the charge in the sphere remains constant.

On the GROUNDED sphere, the charge on the sphere is neither constant nor always zero. Indeed it can gather and dissipate charge from/to the GROUND at will. Without any charge in inner space, the charge on the metal will be ZERO by virtue of the GROUND, which has infinite capacity to store charge. When a charge of +Q2, is inserted in the inner space, the inner surface will collect -Q2 from the GROUND and distributed around. This is fundamentally different than the UNGROUND case, where there was no supply of charge and charge-pairs had to be split apart to generate necessary charge on the inside surface. In either case, the E-field perfectly terminate at the charge on the inside surface; however, since the GROUND provided the charge, there are no charge of opposite polarity to collect on the outer surface, and thus there is no field outside the sphere (at least generated by the Q2 inside the sphere). Again, the charge on the outer surface is different than on the inner surface, but that is okay. The charge on the sphere is NON-ZERO. The field in the metal is still ZERO.

The irony is that, in the case where the metal sphere starts uncharged, the addition of charge to the inner space will cause the GROUNDED sphere actually accumulate charge, where as the UNGROUNDED sphere will always maintain a net ZERO charge. And that is why the field outside can be different between the grounded and ungrounded case, while in the metal itself, E is always ZERO.

This is why, for a Faraday cage to be effective on STATIC fields, the cage must be grounded.