Can an ungrounded conductive cavity provide electrostatic shielding?

[blackhead]

This is another question posted on Physics Stack Exchange querying the
claims in a physics book that a floating conductive cavity shields the
inside and outside regions from one another's electrostatic fields. I
would answer no, and the cavity must be held at a constant potential.


=====Can an ungrounded conductive cavity provide electrostatic
shielding?=====

This comes from Electromagnetic Fields and Waves by Lorrain et al,
page 77 on a Hollow, ungrounded conductor enclosing a charged body:
------
The surface charge density at a given point on the outside surface of
the conductor is independent of the distribution of Q in the cavity.
It is the same as if the conductor were solid and carried a net charge
Q. "
-----
1) I don't believe this. As I move +Q around the inside of the cavity,
this will affect the distribution of -Q on the inside surface, which
must also affect +Q on the outside surface to keep E=0 inside the
cavity conductor.

-----
Inversely, the field inside the cavity is independent of the field
outside the conductor. The conductor then acts as an *electrostatic
shield.
-----
2) I don't believe this either for similar reasons. The external
fields will affect the distribution of +Q on the outer surface, which
must affect the distribution of -Q on the inner surface and therefore
the electrostatic field inside the cavity, to maintain E=0 inside the
conductor.

So am I correct to disbelieve that an ungrounded conductive cavity can
provide electrostatic shielding for the reasons 1. and 2. above?

=====end of question=====

Regards, Larry.

[Jos Bergervoet]

On 2/1/2012 10:25 PM, blackhead wrote:
> This is another question posted on Physics Stack Exchange querying the
> claims in a physics book that a floating conductive cavity shields the
> inside and outside regions from one another's electrostatic fields. I
> would answer no,

But it makes only little difference whether the cage
is floating or not.. Outside a floating cage the field
you observe will only tell you the total amount of charge
that is present inside.

All other information of the inside field is unobservable.
Only one number comes through.. (the charge).

> and the cavity must be held at a constant potential.

Isn't that automatically the case since it is conducting?

> The surface charge density at a given point on the outside surface of
> the conductor is independent of the distribution of Q in the cavity.
> It is the same as if the conductor were solid and carried a net charge
> Q. "
> -----
> 1) I don't believe this. As I move +Q around the inside of the cavity,
> this will affect the distribution of -Q on the inside surface, which
> must also affect +Q on the outside surface

Why must it? The claim is that distribution on the
outside is unaltered. Can you give a counterexample?
Or disprove it on theoretical grounds?

...

> So am I correct to disbelieve that an ungrounded conductive cavity can
> provide electrostatic shielding for the reasons 1. and 2. above?

Hmm.. I would say that the shield always obtains a
constant potential (after some time) because if the
potential is not constant charge will continue to
redistribute itself. Doesn't that prove the shielding?

--
Jos

[blackhead]

On Feb 2, 9:28 pm, Jos Bergervoet <jos.bergerv...@xs4all.nl> wrote:
> On 2/1/2012 10:25 PM, blackhead wrote:
>
> > This is another question posted on Physics Stack Exchange querying the
> > claims in a physics book that a floating conductive cavity shields the
> > inside and outside regions from one another's electrostatic fields. I
> > would answer no,
>
> But it makes only little difference whether the cage
> is floating or not.. Outside a floating cage the field
> you observe will only tell you the total amount of charge
> that is present inside.

> All other information of the inside field is unobservable.
> Only one number comes through.. (the charge).

It will also tell you how the charge on the inner surface is
distributed. Change this distribution, and you change the distribution
of the field outside, even though the total flux remains equal to the
charge inside.

> > and the cavity must be held at a constant potential.
>
> Isn't that automatically the case since it is conducting?

In addition, a constant potential wrt say Earth.

> > The surface charge density at a given point on the outside surface of
> > the conductor is independent of the distribution of Q in the cavity.
> > It is the same as if the conductor were solid and carried a net charge
> > Q. "
> > -----
> > 1) I don't believe this. As I move +Q around the inside of the cavity,
> > this will affect the distribution of -Q  on the inside surface, which
> > must also affect +Q  on the outside surface

> Why must it? The claim is that distribution on the
> outside is unaltered. Can you give a counterexample?
> Or disprove it on theoretical grounds?

I claim that moving a charge Q inside the cavity will alter the
distribution of charge on the inner surface. This will change the
distribution of charge on the outer surface to maintain E=0 inside the
conductor.

> > So am I correct to disbelieve that an ungrounded conductive cavity can
> > provide electrostatic shielding for the reasons 1. and 2. above?
>
> Hmm.. I would say that the shield always obtains a
> constant potential (after some time) because if the
> potential is not constant charge will continue to
> redistribute itself. Doesn't that prove the shielding?

I don't think it does. You have to show that the distribution of
charge on the inner surface isn't affected by charge on the outer
surface and I don't see how E=0 inside the conductor proves this.

If one has a solid conductor of any shape with an arbitary charge
distribution on the outer surface, then E=0 inside. Neither can there
be charge inside a conductor. We can then remove the conductor inside
to create a cavity which doesn't change a thing. Therefore, this
proves that the distribution of charge on the outside doesn't affect
the distribution of charge on the inside and so the cavity is shielded
from external electric fields, and I was wrong.

> Jos

[be...@iwaynet.net]

On 2/5/2012 5:29 PM, blackhead wrote:

> I don't think it does. You have to show that the distribution of
> charge on the inner surface isn't affected by charge on the outer
> surface and I don't see how E=0 inside the conductor proves this.
>
> If one has a solid conductor of any shape with an arbitary charge
> distribution on the outer surface, then E=0 inside. Neither can there
> be charge inside a conductor. We can then remove the conductor inside
> to create a cavity which doesn't change a thing. Therefore, this
> proves that the distribution of charge on the outside doesn't affect
> the distribution of charge on the inside and so the cavity is shielded
> from external electric fields, and I was wrong.

Yes, you were wrong but this doesn't quite explain it. And Jos sort of
had the idea too, but didn't quite get there either.

Consider a hollow sphere. What happens if you put a charge inside it? An
opposite charge appears on the inside of the sphere to cancel the fields
generated by the internal charge. Also charges appear on the OUTSIDE of
the sphere equal to those on the inside (conservation of charge)
However, the distribution of the charges on the INSIDE depend upon the
geometry of the charges you just inserted. If the charge distribution is
not uniform and symmetric the charge will adjust sideways to exactly
cancel the fields. It can do this because the sphere is conductive and
sideways fields move charges.

On the outside of the sphere is the distribution of the charge is again
determined by the fields and sideways forces. But OUTSIDE the sphere as
fields are headed to "infinity" and hence implies a symmetric terminal,
the sphere and space create a UNIFORM distribution of charge on the
outside of the sphere. Hence, inside and outside distributions are
independent of each other if internal and external charge distributions
are independent of each other.

Hence in one sense the internal cavity is shielded in that the internal
field does not apply to the external field. For fun just apply Gauss'
law. A surface around the internal charge gives +q as it should. But if
you enlarge the gaussian surface to BETWEEN the internal and external
shell of the sphere you get ZERO! Then outside the sphere you again get
+q. as you should.

And notice this odd thing: The metal conductive sphere actually forms a
CAPACITOR with + charge on one side and - charge on the other even
though it's metal!

[Jos Bergervoet]

On 2/5/2012 11:29 PM, blackhead wrote:
> On Feb 2, 9:28 pm, Jos Bergervoet<jos.bergerv...@xs4all.nl> wrote:

..

>> Hmm.. I would say that the shield always obtains a
>> constant potential (after some time) because if the
>> potential is not constant charge will continue to
>> redistribute itself. Doesn't that prove the shielding?
>
> I don't think it does. You have to show that the distribution of
> charge on the inner surface isn't affected by charge on the outer
> surface and I don't see how E=0 inside the conductor proves this.

So you don't believe shielding occurs?

> If one has a solid conductor of any shape with an arbitary charge
> distribution on the outer surface, then E=0 inside. Neither can there
> be charge inside a conductor. We can then remove the conductor inside
> to create a cavity which doesn't change a thing. Therefore, this
> proves that the distribution of charge on the outside doesn't affect
> the distribution of charge on the inside and so the cavity is shielded
> from external electric fields, and I was wrong.

So you _do_ believe shielding occurs? I'm afraid
I can't follow you.. Make up your mind!

--
Jos

[blackhead]

On Feb 7, 9:09 am, Jos Bergervoet <jos.bergerv...@xs4all.nl> wrote:
> On 2/5/2012 11:29 PM, blackhead wrote:
>
> > On Feb 2, 9:28 pm, Jos Bergervoet<jos.bergerv...@xs4all.nl>  wrote:
> ..
> >> Hmm.. I would say that the shield always obtains a
> >> constant potential (after some time) because if the
> >> potential is not constant charge will continue to
> >> redistribute itself. Doesn't that prove the shielding?
>
> > I don't think it does. You have to show that the distribution of
> > charge on the inner surface isn't affected by charge on the outer
> > surface and I don't see how E=0 inside the conductor proves this.
>
> So you don't believe shielding occurs?

I don't believe shielding of the outside from the inside occurs

> > If one has a solid conductor of any shape with an arbitary charge
> > distribution on the outer surface, then E=0 inside. Neither can there
> > be charge inside a conductor. We can then remove the conductor inside
> > to create a cavity which doesn't change a thing. Therefore, this
> > proves that the distribution of charge on the outside doesn't affect
> > the distribution of charge on the inside and so the cavity is shielded
> > from external electric fields, and I was wrong.
>
> So you _do_ believe shielding occurs? I'm afraid
> I can't follow you.. Make up your mind!

I believe the inside is shielded from the outside.

> Jos

[blackhead]

If the charge inside is moved, that on the inner surface also moves,
so how can the field inside the conductor remain zero if that on the
outside doesn't change?

> Hence in one sense the internal cavity is shielded in that the internal
> field does not apply to the external field. For fun just apply Gauss'
> law. A surface around the internal charge gives +q as it should. But if
> you enlarge the gaussian surface to BETWEEN the internal and external
> shell of the sphere you get ZERO! Then outside the sphere you again get
> +q. as you should.

> And notice this odd thing: The metal conductive sphere actually forms a
> CAPACITOR with + charge on one side and - charge on the other even
> though it's metal!

There is a capacitance wrt infinity, but none between the inner and
outer surfaces since they're at the same potantial.

[Jos Bergervoet]

On 2/7/2012 3:41 PM, blackhead wrote:
> On Feb 7, 9:02 am, "BJAC...@teranews.com"<b...@iwaynet.net> wrote:

..

>> And notice this odd thing: The metal conductive sphere actually forms a
>> CAPACITOR with + charge on one side and - charge on the other even
>> though it's metal!
>
> There is a capacitance wrt infinity, but none between the inner and
> outer surfaces since they're at the same potantial.

At least it is a capacitor at zero voltage. But now
you believe they are at the same potential, so does
this mean you believe now that the whole metal
enclosure is at one single potential?

And in that case, don't you agree that from the inside
to the outside only one number gets through?! Whatever
is inside, you only see the potential of the (floating)
Faraday cage. That is equivalent to seeing just the
total charge that is present. (The two are related by
the capacitance to infinity)

So strictly speaking you are right: a floating cage
does not give complete shielding from inside to outside.
Just one real number leaks through.

But from outside to inside we have complete shielding.
So where do we disagree?

--
Jos

[blackhead]

On Feb 7, 3:20 pm, Jos Bergervoet <jos.bergerv...@xs4all.nl> wrote:
> On 2/7/2012 3:41 PM, blackhead wrote:
>
> > On Feb 7, 9:02 am, "BJAC...@teranews.com"<b...@iwaynet.net>  wrote:
>    ..
> >> And notice this odd thing: The metal conductive sphere actually forms a
> >> CAPACITOR with + charge on one side and - charge on the other even
> >> though it's metal!
>
> > There is a capacitance wrt infinity, but none between the inner and
> > outer surfaces since they're at the same potantial.
>
> At least it is a capacitor at zero voltage.

Yes, with it's termals shorted.

>But now
> you believe they are at the same potential, so does
> this mean you believe now that the whole metal
> enclosure is at one single potential?

I've always said E=0 inside the metal conductor which means it's at a
constant potential.

> And in that case, don't you agree that from the inside
> to the outside only one number gets through?!

Yes, but that number doesn't have associated with it from Gauss's law
one unique distribution of E on the outside.

>Whatever
> is inside, you only see the potential of the (floating)
> Faraday cage.

E is tangent to the outside surface which implies the potential will
vary along this normal.

>That is equivalent to seeing just the
> total charge that is present. (The two are related by
> the capacitance to infinity)

This case is more complicated in that there is an excess charge inside
a cavity, and a polarized charge on the inner/outer surfaces of the
conductor. The capacitance of a conductor is usually taken to be wrt
to a conductor at infinity where phi=0, and with the charge residing
on the outer surface of the conductor.

> So strictly speaking you are right: a floating cage
> does not give complete shielding from inside to outside.
> Just one real number leaks through.
>
> But from outside to inside we have complete shielding.
> So where do we disagree?

We agree here, but I would go further and emphasise that E at any
point outside the conductor will change as you move the charge around
in the cavity.

> --
> Jos

[be...@iwaynet.net]

Which is what I was saying too. I don't think we do disagree. We are
just clarifying the situation. I think the essence is the one number
"leaking through" thing while the field distributions don't.

[Jos Bergervoet]

On 2/7/2012 7:49 PM, blackhead wrote:
..

> I've always said E=0 inside the metal conductor which means it's at a
> constant potential.
>
>> And in that case, don't you agree that from the inside
>> to the outside only one number gets through?!
>
> Yes, but that number doesn't have associated with it from Gauss's law
> one unique distribution of E on the outside.

There we disagree! The Laplace equation for the potential
outside a conductor with one constant value at the surface
and zero at infinity has only one unique solution. This
also fixes the solution for E.

You can multiply this solution with a real number, to match
any value we want for the potential at the surface.

>> Whatever
>> is inside, you only see the potential of the (floating)
>> Faraday cage.
>
> E is tangent to the outside surface which implies the potential will
> vary along this normal.

But in exactly the same way for any arbitrary content
inside the cage (up to one real constant determined
by the total charge inside).
...

>> That is equivalent to seeing just the
>> total charge that is present. (The two are related by
>> the capacitance to infinity)
>
> This case is more complicated in that there is an excess charge inside
> a cavity, and a polarized charge on the inner/outer surfaces of the
> conductor.

But the result (outside) is the same for all those
complicated situations (up to a constant).

> The capacitance of a conductor is usually taken to be wrt
> to a conductor at infinity where phi=0, and with the charge residing
> on the outer surface of the conductor.

Capacitance = (total charge present) / V_surface

..

>> So strictly speaking you are right: a floating cage
>> does not give complete shielding from inside to outside.
>> Just one real number leaks through.
>>
>> But from outside to inside we have complete shielding.
>> So where do we disagree?
>
> We agree here, but I would go further and emphasise that E at any
> point outside the conductor will change as you move the charge around
> in the cavity.

It cannot change. Laplace!

For a non-perfect conductor you have transient
effects, but the steady state still will become
the same. For an incomplete shield (with holes
in it) you would indeed have change.

--
Jos

[blackhead]

On Feb 8, 8:56 am, Jos Bergervoet <jos.bergerv...@xs4all.nl> wrote:
> On 2/7/2012 7:49 PM, blackhead wrote:
>   ..
>
> > I've always said E=0 inside the metal conductor which means it's at a
> > constant potential.
>
> >> And in that case, don't you agree that from the inside
> >> to the outside only one number gets through?!
>
> > Yes, but that number doesn't have associated with it from Gauss's law
> > one unique distribution of E on the outside.
>
> There we disagree! The Laplace equation for the potential
> outside a conductor with one constant value at the surface
> and zero at infinity has only one unique solution. This
> also fixes the solution for E.
>
> You can multiply this solution with a real number, to match
> any value we want for the potential at the surface.

What about the charge inside?
Remember I'm talking about a charge inside affecting the field outside
so we need to be using Poissons's equation which has a unique solution
for a *fixed* charge distribution and boundary condition.

No, Poisson! and the charge distribution isn't fixed ;)

[Jos Bergervoet]

On 2/8/2012 11:54 PM, blackhead wrote:
> On Feb 8, 8:56 am, Jos Bergervoet<jos.bergerv...@xs4all.nl> wrote:
>> On 2/7/2012 7:49 PM, blackhead wrote:
>> ..
>>
>>> I've always said E=0 inside the metal conductor which means it's at a
>>> constant potential.
>>
>>>> And in that case, don't you agree that from the inside
>>>> to the outside only one number gets through?!
>>
>>> Yes, but that number doesn't have associated with it from Gauss's law
>>> one unique distribution of E on the outside.
>>
>> There we disagree! The Laplace equation for the potential
>> outside a conductor with one constant value at the surface
>> and zero at infinity has only one unique solution. This
>> also fixes the solution for E.
>>
>> You can multiply this solution with a real number, to match
>> any value we want for the potential at the surface.
>
> What about the charge inside?

I only solve the Laplace equation outside the shield so
that is in a charge-free region. I have enough information
since the shield is constant V. I do not have to look
inside to do the mathematics. That is precisely the
essential reason why the shielding occurs (up to one
single real number).

> Remember I'm talking about a charge inside affecting the field outside

I do not need your talking to solve the Laplace equation!
The solution is unique (up to a constant) and the shape
of the field does therefore not depend on the inside.

> so we need to be using Poissons's equation

No we need not. We finished the job without that. If you
make a hole in the shield you have to do the complete
calculation but a closed shield allows this short-cut.

--
Jos

[blackhead]

OK, won't the distribution of the charge in the cavity affect V of the
shield?

If yes, then won't this affect the scaling of the shape of E, and
therefore affect the value of E at some point outside?

If no, then the shield does indeed shield the outside from the inside.

> Jos- Hide quoted text -
>
> - Show quoted text -

[Jos Bergervoet]

If yes, then the whole solution becomes just multiplied by
a constant, to fit the new value of V on the shield. If
you apply Gauss law on a sphere outside the shield then
it would tell you the total charge inside has changed.
But your redistribution did not change the charge, so
the answer is no. Redistribution is invisible outside.

--
Jos

[Timo Nieminen]

On Feb 2, 7:25 am, blackhead <larryhar...@softhome.net> wrote:
> This is another question posted on Physics Stack Exchange querying the
> claims in a physics book that a floating conductive cavity shields the
> inside and outside regions from one another's electrostatic fields. I
> would answer no, and the cavity must be held at a constant potential.
>
> =====Can an ungrounded conductive cavity provide electrostatic
> shielding?=====
>
> This comes from Electromagnetic Fields and Waves by Lorrain et al,
> page 77 on a Hollow, ungrounded conductor enclosing a charged body:
> ------
> The surface charge density at a given point on the outside surface of
> the conductor is independent of the distribution of Q in the cavity.
> It is the same as if the conductor were solid and carried a net charge
> Q. "
> -----
> 1) I don't believe this. As I move +Q around the inside of the cavity,
> this will affect the distribution of -Q  on the inside surface, which
> must also affect +Q  on the outside surface to keep E=0  inside the
> cavity conductor.

The external charge distribution won't depend on where the charge is
internally. The charge -Q on the interior surface will be such that
the field in the conductor is zero. So, beyond conservation of charge
giving +Q as the total exterior charge, the interior surface has no
effect: E=0 is the conductor means that the inside charges exert no
force on the exterior charges, no matter where on the exterior surface
the charge might be.

> -----
> Inversely, the field inside the cavity is independent of the field
> outside the conductor. The conductor then acts as an *electrostatic
> shield.
> -----
> 2) I don't believe this either for similar reasons. The external
> fields will affect the distribution of +Q  on the outer surface, which
> must affect the distribution of -Q  on the inner surface and therefore
> the electrostatic field inside the cavity, to maintain E=0  inside the
> conductor.

This one goes back to Franklin for the experimental observation, and
Priestly for the conclusion that it implies an inverse square force
law (equivalently, Gauss's law). It's the same thing as the first
case: the charge on the exterior surface re-arranges itself to shield
the conductor from the external field. Again, in the conductor E=0.
So, no effect on the interior surface, so no re-arrangement of charge
on the inside.

> So am I correct to disbelieve that an ungrounded conductive cavity can
> provide electrostatic shielding for the reasons 1. and 2. above?

No, but it is a surprising result. Certainly surprised Ben Franklin
(and it's worth reading his letter to Priestly).

[Szczepan Bialek]

Uzytkownik "Timo Nieminen" <ti...@physics.uq.edu.au> napisal w wiadomosci
news:497fb6f0-9270-42eb...@rk3g2000pbb.googlegroups.com...

On Feb 2, 7:25 am, blackhead <larryhar...@softhome.net> wrote:
>

> E=0 in the conductor means that the inside charges exert no

force on the exterior charges, no matter where on the exterior surface
the charge might be.

It is the version for kids.

The Maxwell's definition is different (Part III)
http://en.wikisource.org/wiki/On_Physical_Lines_of_Force
"When a difference of tension exists in different parts of any body, the
electricity passes, or tends to pass, from places of greater to places of
smaller tension. If the body is a conductor, an actual passage of
electricity takes place; and if the difference of tensions is kept up, the
current continues to flow with a velocity proportional inversely to the
resistance, or directly to the conductivity of the body."

If the charge is than it "tends to pass". So E is not zero.
S*

[Jos Bergervoet]

On 2/9/2012 9:32 AM, Timo Nieminen wrote:
> On Feb 2, 7:25 am, blackhead<larryhar...@softhome.net> wrote:

..

>> So am I correct to disbelieve that an ungrounded conductive cavity can
>> provide electrostatic shielding for the reasons 1. and 2. above?
>
> No, but it is a surprising result. Certainly surprised Ben Franklin
> (and it's worth reading his letter to Priestly).

In that sense it's a lot like the "no hair theorem".

--
Jos

[Timo Nieminen]

On Feb 9, 6:55 pm, "Szczepan Bialek" <sz.bia...@wp.pl> wrote:
>
> The Maxwell's definition is different (Part III)http://en.wikisource.org/wiki/On_Physical_Lines_of_Force

> "When a difference of tension exists in different parts of any body, the
> electricity passes, or tends to pass, from places of greater to places of
> smaller tension. If the body is a conductor, an actual passage of
> electricity takes place; and if the difference of tensions is kept up, the
> current continues to flow with a velocity proportional inversely to the
> resistance, or directly to the conductivity of the body."
>
> If the charge is than it "tends to pass". So E is not zero.
> S*

Yes, we know that if you connect a potential difference across a
conductor, we get a current. That's irrelevant to electrosatic
shielding.

In the electrostatic limit, a conductor is at a uniform potential:
"when a difference of tension exists in different parts of any body"
does not apply. So, E=0.

For external sources not connected to the conductor, the steady-state
case is electrostatic.

[Jos Bergervoet]

On 2/9/2012 10:37 AM, Timo Nieminen wrote:
> On Feb 9, 6:55 pm, "Szczepan Bialek"<sz.bia...@wp.pl> wrote:
>>
>> The Maxwell's definition is different (Part III)http://en.wikisource.org/wiki/On_Physical_Lines_of_Force
>> "When a difference of tension exists in different parts of any body, the
>> electricity passes, or tends to pass, from places of greater to places of
>> smaller tension. If the body is a conductor, an actual passage of
>> electricity takes place; and if the difference of tensions is kept up, the
>> current continues to flow with a velocity proportional inversely to the
>> resistance, or directly to the conductivity of the body."
>>
>> If the charge is than it "tends to pass". So E is not zero.
>> S*
>
> Yes, we know that if you connect a potential difference across a
> conductor, we get a current. That's irrelevant to electrosatic
> shielding.
>
> In the electrostatic limit, a conductor is at a uniform potential:

But it is always interesting to see where the approximation
breaks down! So what can go wrong here?

1) In a real material the charge is *not* exactly on the
surface. There is an "accumulation layer" or a "depletion
layer" with finite thickness. In a metal it is not very
thick but in a semiconductor it might be a few micrometers.
If the accumulation and depletion layer on the inside and
outside do not overlap (if the shield is thick enough) this
will not be a problem. But a very thin shield may fail to
give complete shielding (even if it *isn't* an insulator!
Just because the necessary electrons don't fit in).

2) The potential on the outside may be not constant if the
shield is made of different materials welded together or
if different crystal-lattice orientations are seen from
different sides. The work-function may then be position
dependent. This will not change the fact that the shielding
occurs, but the constant potential at the outside boundary
is an oversimplification.

3) Even without the complications of point 2, the potential
exactly at the boundary has strong microscopic fluctuations
and you have to move away slightly from the atomic structure
to let it average out to one value (which then still depends
on the work function, but let's assume that to be constant!)

I think only point 1) really affects the shielding: for weak
fields it will still work but strong fields will "saturate"
the shielding capability of such a very thin shield..

--
Jos

[blackhead]

I don't understand this. The sum of all the electric forces inside
might be zero, but those individual forces from the interior charges
still exist.

Maybe one can show that the energy stored in a polarized electric
field within a closed conductor enclosing a constant charge in a
cavity, is independent of where that charge is; giving a constant
potential V inside the conductor and therefore constant E at any point
outside.

> > -----
> > Inversely, the field inside the cavity is independent of the field
> > outside the conductor. The conductor then acts as an *electrostatic
> > shield.
> > -----
> > 2) I don't believe this either for similar reasons. The external
> > fields will affect the distribution of +Q  on the outer surface, which
> > must affect the distribution of -Q  on the inner surface and therefore
> > the electrostatic field inside the cavity, to maintain E=0  inside the
> > conductor.

> This one goes back to Franklin for the experimental observation, and
> Priestly for the conclusion that it implies an inverse square force
> law (equivalently, Gauss's law). It's the same thing as the first
> case: the charge on the exterior surface re-arranges itself to shield
> the conductor from the external field. Again, in the conductor E=0.
> So, no effect on the interior surface, so no re-arrangement of charge
> on the inside.

> > So am I correct to disbelieve that an ungrounded conductive cavity can
> > provide electrostatic shielding for the reasons 1. and 2. above?
>
> No, but it is a surprising result. Certainly surprised Ben Franklin

> (and it's worth reading his letter to Priestly).- Hide quoted text -

[Szczepan Bialek]

Uzytkownik "Jos Bergervoet" <jos.ber...@xs4all.nl> napisal w wiadomosci
news:4f33a8cc$0$6894$e4fe...@news2.news.xs4all.nl...

> On 2/9/2012 10:37 AM, Timo Nieminen wrote:
>> On Feb 9, 6:55 pm, "Szczepan Bialek"<sz.bia...@wp.pl> wrote:
>>>
>>> The Maxwell's definition is different (Part
>>> III)http://en.wikisource.org/wiki/On_Physical_Lines_of_Force
>>> "When a difference of tension exists in different parts of any body, the
>>> electricity passes, or tends to pass, from places of greater to places
>>> of
>>> smaller tension. If the body is a conductor, an actual passage of
>>> electricity takes place; and if the difference of tensions is kept up,
>>> the
>>> current continues to flow with a velocity proportional inversely to the
>>> resistance, or directly to the conductivity of the body."
>>>
>>> If the charge is than it "tends to pass". So E is not zero.
>>> S*
>>
>> Yes, we know that if you connect a potential difference across a
>> conductor, we get a current. That's irrelevant to electrosatic
>> shielding.
>>
>> In the electrostatic limit, a conductor is at a uniform potential:
>
> But it is always interesting to see where the approximation
> breaks down! So what can go wrong here?
>
> 1) In a real material the charge is *not* exactly on the
> surface. There is an "accumulation layer" or a "depletion
> layer" with finite thickness. In a metal it is not very
> thick

The charges for the kids are like a fluid. Such are only on the surface of
metal.
For you the charges are electrons. They always ocupy the whole volume with
the proper density gradient.

>but in a semiconductor it might be a few micrometers.
> If the accumulation and depletion layer on the inside and
> outside do not overlap (if the shield is thick enough) this
> will not be a problem. But a very thin shield may fail to
> give complete shielding (even if it *isn't* an insulator!
> Just because the necessary electrons don't fit in).
>
> 2) The potential on the outside may be not constant if the
> shield is made of different materials welded together or
> if different crystal-lattice orientations are seen from
> different sides. The work-function may then be position
> dependent. This will not change the fact that the shielding
> occurs, but the constant potential at the outside boundary
> is an oversimplification.
>
> 3) Even without the complications of point 2, the potential
> exactly at the boundary has strong microscopic fluctuations
> and you have to move away slightly from the atomic structure
> to let it average out to one value (which then still depends
> on the work function, but let's assume that to be constant!)
>
> I think only point 1) really affects the shielding: for weak
> fields it will still work but strong fields will "saturate"
> the shielding capability of such a very thin shield..

It is imposible to explain the electrostatic phenomena with the charge in
the form of fluid.
S*

[Jos Bergervoet]

But if they sum up to zero isn't by definition their force
on the exterior charges zero? I think you understand the
claim perfectrly well. You just don't want to believe it.
Test1:
"Consider all interior charges together with the
charge layer on the inner surface of the shield"
You understand that? Test2:
"the E-field of those charges, evaluated at a point
on the outer surface is 0. Evaluated on a point
inside the metal of the shield it is also zero.
And evaluated anywhere outside it is zero"
Do you understand that claim?

There was no claim that in the interior the E-field
is zero, so why are you complaining that the forces
there still exist? No-one said they didn't exist!

You have to be more clear, blackhead!

--
Jos

[blackhead]

ok

>If
> you apply Gauss law on a sphere outside the shield then
> it would tell you the total charge inside has changed.
> But your redistribution did not change the charge, so
> the answer is no. Redistribution is invisible outside.

This tells me that the total electric flux on a surface outside won't
change with constant charge and is independent of distribution of the
charge, which I'm fine with; but will E at a fixed point outside the
sphere change as the charge is moved within the cavity? Laplace's
equation will say yes, if V is constant; but how do we show V is
constant?

> Jos- Hide quoted text
>
> - Show quoted text -

[Timo Nieminen]

1) is a nice student exercise to include as a final part of an
exercise in electrostatics. "Find the capacitance of a cube. How much
does it depend on point 1)?". Nice, because it can be adequately
answered qualitatively, on the basis of some numbers from the
literature.

The extended version of 1) is "how real are surfaces".

[Timo Nieminen]

A Maxwellian would say that there are no individual forces from the
interior charges; only the E field exerts a force on a static charge.
(Priestly perhaps would have disagreed.)

How about a mathematical approach? Inside the conductor - between the
interior and exterior surfaces - we have E=0. We know this, because we
have static sources, and therefore must have, after sufficient time to
reach steady-state, J=0. (Else we would have Ohmic losses, but no
source of energy.)

So, take a mathematical surface within the conductor. (We don't need
to do this next step, but let me inlcude it: From E=0, the flux
through the surface is zero, and the charge in the interior surface
must be -Q in response to the internal charge +Q.) Since the have a
zero-E, and thus constant potential, closed surface, the only solution
satisfying the Laplace equation outside (until we run into other
sources) is constant. One can, e.g., prove this by finding the
multipole expansion of the external field by integration of the
potential on the surface. More simply, one can note that the potential
on this boundary is a constant + the potential due to no interior
sources at all (i.e., constant + zero). So, we can't see any interior
sources.

> Maybe one can show that the energy stored in a polarized electric
> field within a closed conductor enclosing a constant charge in a
> cavity, is independent of where that charge is; giving a constant
> potential V inside the conductor and therefore constant E at any point
> outside.

Maybe. Only if there is no force on the enclosed charge (and therefore
no work required to move it). From the above extra point, the charge
on the interior surface must be -Q, regardless of how the enclosed +Q
is distributed. Is it sufficient to then say that the capacitance C of
the interior surface then only depends on its geometry, and thus with
charge -Q, it must have potential V=-Q/C?

[blackhead]

Yes. But Timo suggested that because the sum of the interior-cavity,
interior-surface and exterior-surface electric forces are zero inside
the conductor, then this means the sum of the interior electric forces
must be zero on the exterior surface. I don't get this.

>I think you understand the
> claim perfectrly well. You just don't want to believe it.

I want to understand how it's possible.

> Test1:
>    "Consider all interior charges together with the
>     charge layer on the inner surface of the shield"
> You understand that? Test2:

So far so good.

>    "the E-field of those charges, evaluated at a point
>     on the outer surface is 0. Evaluated on a point
>     inside the metal of the shield it is also zero.
>     And evaluated anywhere outside it is zero"
> Do you understand that claim?

Yes, I understand the claim, but don't understand why it's true.

> There was no claim that in the interior the E-field
> is zero, so why are you complaining that the forces
> there still exist? No-one said they didn't exist!
>
> You have to be more clear, blackhead!

Yes, I do need to be clear about the three charge distributions:
inside the cavity, on the interior surface, and that on the exterior
surface. And the three regions: cavity, conductor, outside the
conductor.

I will admit one silly mistake I have made in this discussion:
Forgetting that the change in the cavity charge distribution can
compensate for the change in the interior-surface charge distribution.
So it's possible to keep the exterior-surface charge distribution
constant while still maintaining E=0 inside the conductor even though
the interior-surface charge distribution has changed.

> Jos- Hide quoted text -

[Szczepan Bialek]

"Timo Nieminen" <ti...@physics.uq.edu.au> napisal w wiadomosci

news:e0bd587f-a2cb-4ddc...@qt7g2000pbc.googlegroups.com...

On Feb 10, 2:46 am, blackhead <larryhar...@softhome.net> wrote:
>
>
>> I don't understand this. The sum of all the electric forces inside
> might be zero, but those individual forces from the interior charges
> still exist.

>A Maxwellian would say that there are no individual forces from the
interior charges; only the E field exerts a force on a static charge.
(Priestly perhaps would have disagreed.)

>How about a mathematical approach? Inside the conductor - between the
interior and exterior surfaces - we have E=0. We know this, because we
have static sources, and therefore must have, after sufficient time to
reach steady-state, J=0. (Else we would have Ohmic losses, but no
source of energy.)

After sufficient time the "tends to flow" still exist. So E is not zero.
S*

[Jos Bergervoet]

On 2/10/2012 3:27 AM, blackhead wrote:
...

>>> I don't understand this. The sum of all the electric forces inside
>>> might be zero, but those individual forces from the interior charges
>>> still exist.
>>
>> But if they sum up to zero isn't by definition their force
>> on the exterior charges zero?
>
> Yes. But Timo suggested that because the sum of the interior-cavity,
> interior-surface and exterior-surface electric forces are zero inside
> the conductor, then this means the sum of the interior electric forces
> must be zero on the exterior surface. I don't get this.

Well, I also would say that I'm not sure I understand what
Timo meant.. Two separate things can be stated:
1) The sum of the interior forces is zero inside the conductor.
2) The sum of the exterior forces is zero inside the conductor.
What you wrote above is the sum of all of them, which is of
course also zero, but that fact would not (by logic alone)
imply that both are zero individually..

...

> I will admit one silly mistake I have made in this discussion:
> Forgetting that the change in the cavity charge distribution can
> compensate for the change in the interior-surface charge distribution.

That is of course in a way remarkable.. (A two-dimensional
distribution would seem to have insufficient degrees of freedom
to always be able to compensate a three-dimensional distribution!)
But as modern physicists we are familiar with to the holographic
principle (Susskind-'t Hooft), so yes, it was silly of you to
forget that! :-)

> So it's possible to keep the exterior-surface charge distribution
> constant while still maintaining E=0 inside the conductor even though
> the interior-surface charge distribution has changed.

OK, so you believe it is possible. But you don't see the proof
that it actually happens?!

--
Jos

[blackhead]

I think it can be proven from the conservation of energy. I know that
charge within the cavity is not affected by external electric fields.
So charge inside the cavity cannot do work on external charges and
therefore cannot affect the electric field outside.




> Jos