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Let's say I have two spherical conductors with different radii and different amount of positive charge on them. The spheres are far enough from each other. I connect them with a conducting wire. I'm told that the charge is going to flow until both of the conductors gain the same potential (on top/inside of them). Now, intuitively it seems fine but I do not quite understand a few things.

1) The intuition is that the electric field should be zero everywhere when they reach the same potential, because then the charge is not going to flow. But it is not always the case.

2) Shouldn't the wire also contain some charge after the rearrangement? Because if no, then its potential will be zero, and so there will be a difference in potentials between one of the conductors and the wire, so the charge should still flow into the wire.

I'll try to rephrase the question, because it seems people keep getting my question wrong - in general, I can't understand why charges should always flow until both of the conductors reach the same potential. I mean, the fact that the conductors have same potential doesn't really tell us that the field which drives charges is zero. What's the proof that connected conductors reach the same potential?

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    $\begingroup$ What is the orientation of spheres ?? Are they concentric ?? or just put apart by some distance ?? $\endgroup$ – ABC May 31 '13 at 12:28
  • $\begingroup$ They are far away from each other so we can neglect their effects on each other. $\endgroup$ – www May 31 '13 at 12:37
  • $\begingroup$ then the ideal wire seems to be your problem...... $\endgroup$ – ABC May 31 '13 at 12:41
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1) At the conductor surfaces, the tangential electric field is zero in equilibrium (otherwise charge would flow). The normal field need not be zero.

Zero tangential field is equivalent to the surface being an equipotential. (Potential is, after all, the line integral of the field (times -1).) The fact that the conductors have the same potential really does tell us that the field which drives charges is zero, and vice versa. If the tangential field is not zero, charges will flow until it is, or equivalently until the surface is an equipotential.

2) Note that a smaller sphere requires less charge than a larger one to achieve the same potential (you can see this by integrating the field from infinity for the two cases). The wire is really small (in radius), and needs correspondingly even less charge. In the limit of a 0 radius wire, the required charge goes to 0.

Differing charges do not necessarily imply different potentials. The capacitance, which depends on the geometry, gives the constant of proportionality between charge and potential. It's different for different shapes, so, at equal potentials, different shapes will contain different charges.

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  • $\begingroup$ 1) I do know that conductors have zero field inside them and have zero tangential field on the surfaces. This question is more about higher and lower potentials and interactions between two connected conductors, and how outer electric field is involved. 2) Then the wire has always a lower potential that a sphere, so charges can't be in equilibrium by this logic. If the spheres are far enough from each other, the only way charges will flow is because of the wire and the fact that it has different potential. Is it wrong? $\endgroup$ – www Jun 1 '13 at 11:51
  • $\begingroup$ I've edited my question, and rephrased a bit my question. Maybe now it is more clear. Thanks! $\endgroup$ – www Jun 1 '13 at 11:57
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When two charged spheres are joined by wire then charge flows from high potential to lower one until they come to same potential. The sphere larger in size acquires more charge as it has higher capacitance. After joining by wire, electric filed will not be zero but electric field will change according to its capacitance and charge shared. At one point electric field of one sphere may be canceled by electric field of another. The point is termed as null point. but other points electric field is not zero being same potential. Charge distributes on outermost surfaces of spheres so connecting wire only provides path for charge transfer. Finally no charge resides on it. charges always resides so that potential energy of system remains minimum. To maintain minimum energy the charges always reside on outer surfaces....

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    $\begingroup$ If these two charged spheres are far away from each other, the only reason charges pass from one sphere to another is because the wire has a zero potential. So as I see it, when we connect the wire, some charges are going through it until they reach another sphere and vice versa. But at the end, they must be in equilibrium, but if the wire still remains uncharged, meaning its potential is still zero, then how can such a situation be possible? $\endgroup$ – www May 31 '13 at 19:43
  • $\begingroup$ @WalterWhite If the wire has a negligible width, then it need not noticeably affect the potential around it. This answer makes sense in the context of that assumption. Yes, the wire is eqi-potential with the two spheres, but the potential falls as $-\ln{r}$ around it. $\endgroup$ – Alan Rominger Jul 3 '13 at 13:57
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At equilibrium the electric field inside the spheres and wire must be zero. The charges only on the spheres can not make the electric field zero everywhere in the wire so a charge distribution on the wire is also needed.

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It's actually pretty simple. The charges would repel each other and try to be as far as possible. The only way they can do this is by residing in the surface of the spheres

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