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A solid metal sphere of radius $R$ has charge $+2Q$. A hollow spherical shell of radius $3R$, concentric with the first sphere, has net charge $-Q$.

What would be the final distribution of the charge if the spheres were joined by a conducting wire?

Answer: $+Q$ on outer shell, $0$ on inner sphere.

Why is that? I would assume that the total charge should just evenly spread between the spheres.

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Using Gauss' law and the condition that there can be no electric field inside a conductor the initial charge distribution is as follows:

  • $+2Q$ charge on the outside of the inner sphere.

  • $-2Q$ charge on the inside of the outer sphere ($-Q$ original charge and $-Q$ induced charge)

  • $+Q$ induced charge on the outside of the outer sphere

There is an electric field between the two sphere and so there must be a potential difference between the two spheres.
Joining the two spheres with a conducting wire means that charges will flow until the potential difference between the two spheres becomes zero ie there is no electric field between the two conductors and inside the conducting wire.
So no charge can reside on the outside of the inner sphere and the inside of the outer sphere and this is achieved by the $-2Q$ on the inside of the outer sphere and the $+2Q$ on the outside of the inner sphere neutralising one another leaving charge $+Q$ on the outside of the outer sphere.

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We know that inside a conductor the electric field $E$ is always equal to $0$. Hence there cannot be any net charge inside the conductor. As like charges repel, the charges, being equal to their neighbors tend to go as far away from each other as possible. The only configuration where both the above conditions are met, is when the net charge resides on the outer surface of the arrangement. This means that $2Q+(-Q)=Q$ coulombs of charge resides at the surface of the arrangement.

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  • $\begingroup$ While the conclusion is right, the reasoning isn't complete. What if the inner sphere is never connected to the outer sphere? By your first sentence we must conclude that E between the spheres is 0, but it's not. $\endgroup$ – Bill N Aug 7 '15 at 14:54
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Something like that. In metal shel, charge distribute in the surface. So in this case, when you conect the center sphere with the wire, all the charges goes to the surface of the outer sphere. Then, the charge in the surface of the outer sphere is 2Q-Q=Q.

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