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A cavity inside a conductor is shielded from outside electric influences. But, if you put charges inside the cavity, the exterior of the conductor is not shielded from the fields by the inside charges. Why?
Also, a gravitational force on a particle inside a spherical shell is zero. However, the shell does not shield other bodies outside it from exerting gravitational forces on a particle inside? Why?
A cavity inside a conductor is shielded from outside electric influences.
Shielded from electrostatic effects, yes.
if you put charges inside the cavity, the exterior of the conductor is not shielded from the fields by the inside charges.
The exterior region know how much charge in is on and inside the conductor. But the exterior region does not know where the charge is.
Let's be clear. Imagine a world (world A) where everything was a conductor except that one cavity. Then a charge imbalance would develop (assuming an infinite universe) to form on that surface so that the electric field was zero everywhere in the conductor, i.e. everywhere except that cavity. The amount of charge on the surface will be equal and opposite the amount in the cavity.
Now imagine a world (world B) where the cavity is filled in, so it is a solid conductor. Now imagine that an amount of charge equal to the charge in the cavity is place on the conductor. It will naturally arrange itself so that the field of it and the field due to external fields is zero through the entire solid conductor.
Now if you take all the charges you get the outside charge plus outside surface charge produces zero field everywhere in the conductor and the cavity. And the field outside is due to both the charge on the outside surface and the external charge.
Now if you take all the charges you get the cavity charge plus inside surface charge produces zero field everywhere in the conductor and the outside. And the field in the cavity is due to both the charge on the inside surface and the cavity charge.
Each side (inside and outside) has its own surface and charge on its side and the field on that side is the field due solely to those things.
It is much less mysterious and much less asymmetrical than people make it seem. So why at all do people say the outside feels the inside? Remember when we put an charge on the outside equal to the charge on the cavity? That was merely so the total charge on both surfaces would be zero. Its from having a neutral conductor. Which I assumed merely to get the result you stated, which otherwise wouldn't hold.
So you didn't explicitly say the conductor had a net zero charge, but if it was allowed to have a charge that was equal and opposite to the cavity charge then the person outside would have no idea what either was. So if you can secretly place charge in the cavity and an equal and opposite amount on the surface no one on the outside would know. You could also move the cavity to any location inside or move the charge in the cavity in any location and no one outside knows.
The people outside only notice the charge on the outside. And that only is equal in magnitude to the cavity charge if the charge on the two surfaces came from rearranging charge from a conductor that was originally charge neutral. An assumption you didn't state.
So we talked a out how if you added more charge to he cavity and an equal and opposite to the inside surface then the people outside wouldn't know. Similarly you can ASD more charge to the outside surface and the people inside won't know.
And this turns out to be just like the gravity case. It is possible to make a shell of mass and the gravitational effect has no local effect on the things inside (it does make the things inside age slower than things on the outside but that is an effect from general relativity and the things inside won't notice unless they look outside since everything inside even their clocks will age slower).
Also, a gravitational force on a particle inside a spherical is zero. However, the shell does not shield other bodies outside it from exerting gravitational forces on a particle inside? Why?
Because now we are find exotic matter (with negative energy) to be so unusual that we refuse to just assume you intended the shell to be net energy neutral. With electric charge we were willing to assume you meant it to be net charge neutral and just either were being terse, lazy, or tricky. But with energy we culturally think you should say you want to have negative energy of you want it.
If you allowed negative energy you could do a similar thing, if you shell had parts with negative energy and parts with negative energy you could try to move them around so a negative energy shell countered the mass in the cavity and then put a positive energy shell on the outside surface. But no one is going to just assume that is what you meant.
So with the conductor we had to make assumptions that were reasonable to get your result. The corresponding assumptions for gravity are simply too unusually for us to be willing to just assume them.
If you stated the assumptions and made the same assumptions then you could get the same effects for electrostatics and gravity. Bit the assumptions will sound reasonable for electrostatics where we see neutral conductors all the time. And the assumptions will seem very strange for gravity because it is usual to even write equations that allow negative energy densities.
A conductor divides the whole space into two parts. (i) outside the conductor and (ii) inside the conductor.
We know that the field at a point in region due to charges outside the conductor is cancelled by the charges induced at the surface of the conductor. Suppose that we move the charges outside the conductor. The induced charge pattern will quickly change to nullify the field inside the conductor. So if a charge is kept inside the conductor, it will not feel any net force due to charges outside. This is called electrostatic shielding.
But as you can see the charge on the outside is acted upon by a force since there is no shielding effect on it.
A charged particle inside of a metal spherical shell is under a net force of zero if you are only considering the charge equally distributed on a shell. The same is true of gravity if the mass is equally distributed. This is a consequence of geometry.
The reason the shell doesn't shield from gravity is because gravity only attracts and there is no "freely moving mass" that is repelled by an outside mass which can move to the opposite side of the shell and counteract it. That is what happens with a charged conducting shell. It sort of shields from outside charge but really it just shifts its own charge distribution around so that it exactly cancels with outside influences. The electric fields of those outside influences still exist inside the shell, you cannot actually prevent an electric field from "penetrating" you can only interfere with it via the superposition of other electric fields.
