# Electrical Field Inside a Charged Metallic Conductor

STAii
I have recently seen a proof (using Gauss' law) showing that there is no electrical field inside a charged metallic conductor (due to the carge on its surface). (i can provide this proof if needed)
If that proof was right, then this must mean that the inner sphere in the original question (see the link at the beginning of the post) will not be charged.
So, will or won't be the inner sphere get charged ?

STAii
So ... no takers ?
*Bump*

Greetings STAii !
Originally posted by STAii
If that proof was right, then this must mean that the inner sphere in the original question (see the link at the beginning of the post) will not be charged.
So, will or won't be the inner sphere get charged ?
Hmm... I'm not quite certain what you're talking
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/gaulaw.html#c1

The POTENTIAL INSIDE (the inner sphere in this case) should
be zero - which is going to be the case in the inner (conducting)sphere. Also, because it's conducting (in your previous example)
the permativity is not going to be high. Further more, practicly
there is no such thing as an infinite permativity since the
atoms or molecules of every material have some level of
potential electrical energy which they can not surpass and
once it is reached the electrons will be separated from
the ions as a discharge current.

Live long and prosper.

STAii
Here is a the proof (i hope my translation is ok).
Take a spherical conductor, charge it.
The charge is now on the outer surface of the sphere.
Now, we will use Gauss' law to calculate the electric field inside the conductor.
Take Gauss' surface to be another (imaginary) sphere that has a radius smaller than the original sphere's radius (and they both have the same center).
According to Gauss' law, the electrix flux on this surface will equal Zero (since there is no charge inside the conductor, remember that the charge goes on the outer surface of the sphere).
But flux=EAcos[the], so :
EAcos[the] = 0
But we know that A [x=] 0 (A = 4[pi]r, and r is not zero).
So : Ecos[the] = 0
Since we find symetry in the point at which we are calculating the electric field, than [the] will not equal [pi]/2 (cause, how will you determine its direction then ?), therefore we conclude that E = 0

Now back to the original question.
If this proof is right, then the electrons in the inner metallic sphere (in the original question) will not be affected by an electric force from the outer sphere, so the inner sphere will not have a negatively charged surface. Right ? (the question now is no more about the permitivity of the metal).

Thanks a lot.
(edited for formatting mistake)

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But, when the outer shell is separated by a gap from the
inner sphere you have a non-unform permitivity within.
Further more, in a conductor the potential is 0 throughout
the volume of the sphere when the charge is ditributed on
the surface, but here, due to non-uniform permitivity
and as long as the charge on the shell is not high
enough to cause a discharge from the sphere, the potential
inside is not zero.

Live long and prosper.

Whoa. I looked at the original thread you referred to and it's WRONG. I don't know who the mentor was when the original thread appeared, but he must have been asleep at the switch. A spherical metallic shell completely shields its contents from electrical fields. You can charge it as much as you like; there will be no effect on the contents. In fact, the shell needs not be spherical. This is called a Faraday cage. Look at the entry say in wikipedia. Faraday cages are in common use at accelerator labs.

Originally posted by krab
Whoa. I looked at the original thread you referred to and it's WRONG. I don't know who the mentor was when the original thread appeared, but he must have been asleep at the switch. A spherical metallic shell completely shields its contents from electrical fields. You can charge it as much as you like; there will be no effect on the contents. In fact, the shell needs not be spherical. This is called a Faraday cage. Look at the entry say in wikipedia. Faraday cages are in common use at accelerator labs.
Hmm... So, if I charge a shell with a huge electrical
charge the sphere, separated by the gap inside, will not
discharge ?

Live long and prosper.

Originally posted by drag
Hmm... So, if I charge a shell with a huge electrical
charge the sphere, separated by the gap inside, will not
discharge ?

We have a Faraday cage at our lab, which we charge to 300,000 volts. Inside this cage live many power supplies, vacuum pumps, gauges, control system, etc. They operate as normal; there are no sparks jumping from the cage (the shell) to the components inside at any time.

Rather than provide you with a proof, I urge you to go to

STAii
So (i read the link, just want to make sure if i understood well).
If we had a faraday cage, and we did not charge it, but we put it under the effect of an electric field (generated by another object), will the faraday cage still shield its contents from the field generated by the other object ?
Thanks (this seems interesting).

Originally posted by STAii

If we had a faraday cage, and we did not charge it, but we put it under the effect of an electric field (generated by another object), will the faraday cage still shield its contents from the field generated by the other object ?

Yes. The electrons in the metal surface of the cage, since they are mobile, will move until there is no more electric field inside. In this way, they redistribute to exactly cancel the external electric field, for points inside the cage. So even though you did not charge the cage, there is a charge on it, which varies from + at one end, through 0 to - at the other end, and if you add it up over the whole surface you get zero.

STAii
Got it, thanks.

Greetings !
Originally posted by krab
We have a Faraday cage at our lab, which we charge to 300,000 volts. Inside this cage live many power supplies, vacuum pumps, gauges, control system, etc. They operate as normal; there are no sparks jumping from the cage (the shell) to the components inside at any time.
I'm sorry, but I'm somewhat confused here.
What exactly do you charge at 300,000 Volts ?

More to the point, in the case mentioned here you
are not dealing with an external electric field.
The charge is on the outer shell itself and the inside
is not hollow but contains a sphere. I do not understand
why you say that the sphere will no be affected ?
After all, inside the sphere there must be no electric
fields, so it must cancel the external electric field
which will exist at its surface due to its presense
inside the shell, without which it would of course be 0.
Am I seriously missing something here ?

Peace and long life.

drag:
I'm sorry, but I'm somewhat confused here. What exactly do you charge at 300,000 Volts ? Also, isn't your Faraday cage earthed ?

The Faraday cage is charged to 300 kV. There is a DC power supply between the FC and ground. Other labs have similar setups with 750 kV. These cages are large. One could stay in the cage while it is at voltage, but I don't know of anyone who has done it. A tandem vandegraaff typically runs at 25MV, and the stripping terminal contains many sensitive components that are in no danger from this incredibly high voltage.

Recall the question from STAii:
I have recently seen a proof (using Gauss' law) showing that there is no electrical field inside a charged metallic conductor

I'm saying that not only can you prove this statement using simple electrostatic laws, but also that it has practical applications which are in daily use.

Greetings !
Originally posted by krab
The Faraday cage is charged to 300 kV. There is a DC power supply between the FC and ground. Other labs have similar setups with 750 kV. These cages are large. One could stay in the cage while it is at voltage, but I don't know of anyone who has done it. A tandem vandegraaff typically runs at 25MV, and the stripping terminal contains many sensitive components that are in no danger from this incredibly high voltage.
What's the voltage for ?
Originally posted by krab
I'm saying that not only can you prove this statement using simple electrostatic laws, but also that it has practical applications which are in daily use.
But this just applies to hollow conductors, doesn't it ?

Live long and prosper.