Suppose you have a metallic sphere

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Suppose you have a metallic sphere, that is not completly full from the middle, that is, the sphere is a shell.
Suppose we put another metallic sphere (but that is not a shell) in the middle of the metallic shell (and without touching the two spheres).
Then we charge the outer sphere with a certain charge, say +x (where x is positive).
Will the outer surface of the inner sphere get charged ?

The way i see it, the outer surface of the inner sphere will get charged by induction (and have a negative charge), and the center of the inner sphere will be charged by a positive charge, am i concluding right ?

Some people told me that this will not happen since the permitivity of the metal (epsilon) will be too high, what do you think ?

Thanks.
 
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drag

Science Advisor
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Re: Induction

You're right.
What does permitivity have to do with this ?
When you charge the shell positive you introduce
an electrostatic energy imbalance in the sphere. No
matter the metal's permitivity the free electrons in the
inner sphere will eventually gather on its outer
surface to remove this energy imbalance - nullify
the voltage inside the sphere.

Live long and prosper.
 

LURCH

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And if you charge the outer sphere enough, it will create so much imbalance that you can make lightning jump across the gap between the two! Not a terribly revealing insight, but it would just be really cool.
 
333
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Thanks.
What does permitivity have to do with this ?
Well, as we know, the charge that you put on the outer sphere will actually lie on the outer surface of the outer sphere.
We also know that electrical force is inversely proportional to the medium's electrical permitivity (epsilon).
Now, what lies between the charge on the outer sphere and the surface of the inner sphere is :
1-A layer of metal (Since we said the charge is on the outer surface of the outer sphere)
2-Another medium, or maybe space (between the two spheres).

So, if the permitivity of the metal is extremly high, the electrical force on the electrons making up the inner sphere will be extremly low (or maybe there will be no force if the epsilon can be considered +[oo]), so the inner sphere will no get charged.

I tried to explain that if the permitivity of this metal was so big then the charge will not spread over its surface from the first place, but with no success :smile:.

What do u think now ?
Thanks again.
 

vanesch

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The charge will only appear on the outside of the outer shell. The reason is simple: apply Gauss law on a spherical surface within the small sphere. Clearly, if you have a charge inside the massive sphere, you have an E-field within a conductor, so a current -> this is not possible here. So you cannot have a charge INSIDE de sphere, it has to be on the surface. By spherical symmetry, it has to have the same charge density everywhere. So the inner sphere has a NET charge. Where did it come from ? As the sphere was neutral, and not in contact with any other conductor before the shell got charged, it cannot have a net charge afterwards.
The reasoning is more complicated but has the same result when you consider other metal objects within the spherical shell. And that's the basic operational principle of a FARADAY CAGE: within a closed surface of conductors, NO E-FIELD from the outside can influence the E-FIELD on the inside (and hence the charge distribution).

cheers,
Patrick.
 

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