Natural Science Forum / Physics / Research / October 2007

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.

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