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A question specifies two identical conducting spheres, both of negligible radius (I take this to mean they're to be treated as point charges), and with charges of opposite sign. They're separated by some distance, and the force with which they attract each other is also provided.

They are now connected by a conducting wire, and the force with which they repel each other at equilibrium is also provided. We're required to find the initial charges on the spheres.

The solution treats the final charges on both spheres to be the average of the initial charges on them.

How do you get there? The usual calculations of equating potentials to find the final situation can't be done here, since they're point charges. I don't really see how you can calculate the charges flowing from and to point objects, since charge, as I understand it, is defined to be a point entity itself.

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  • $\begingroup$ Apparently the solution assumes the two spheres to have the same radius. $\endgroup$
    – SoterX
    May 3 at 13:36
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When connected by a wire the two spheres will have the same potential, at equilibrium. As they have the same radius, they have the same capacitance so they must have the same charge. This condition, plus the condition that the total charge must be conserved, provides the formula for the final charges in terms of original charges. In this case is the arithmetic mean but this is not a general result. It works only for identical spheres. Otherwise you need to take into account the different capacitances if the spheres. The small radius condition is given so that you can use Coulomb's law for the force. Otherwise the charge distributions are not spherically symmetric as they are conductors and the charge distributions are influenced by the forces between them.

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  • $\begingroup$ So we're supposed to both use and ignore the small-radius condition howsoever we like? $\endgroup$
    – harry
    May 3 at 14:11
  • $\begingroup$ No, you don't ignore it. You assume that the spheres are identical. This gives you the final charge distribution no matter what is the shape or size of the objects. But to use Coulomb' law for the forces you have to assume that the spheres are point-like. This is not a real-life situation, I guess it is a textbook problem. They have to make solvable by elementary means. $\endgroup$
    – nasu
    May 3 at 16:21
  • $\begingroup$ I understand that. But isn't talking about a charge distribution on a point object... meaningless? $\endgroup$
    – harry
    May 3 at 23:16

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