Electric Potential of Two Spheres

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When two insulated conducting spheres are connected by a wire, they reach the same electric potential due to their conductive nature. The potential for each sphere can be expressed as V = kQ/r, where Q is the charge on each sphere and r is their respective radius. The total charge is conserved, meaning Q = Q1 + Q2, where Q1 is the charge on the first sphere and Q2 is the charge on the second sphere. After connection, the relationship kQ1/r1 = kQ2/r2 holds true, indicating that the charges redistribute until the potentials equalize. Understanding this balance is crucial for solving problems involving electric potential in conductive systems.
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Problem:

An insulated spherical conductor of radius r1 carries a charge Q. A second conducting sphere of radius r2 and initially uncharged is then connected to the first by a long conducting wire. After the connection, what can you say about the electric potential of each sphere? How much charge is transferred to the second sphere? Assume the connected spheres are far apart compared to their radii. (Why make this assumption?)


I think I am confused about potential in general, so bare with me here. I know that for a point charge Q the potential at a point P is kQ/r. In class my teacher derived the potential for a sphere, in the case of the first sphere it would be kQ/r. The charge then distributes to the other sphere, but I am not sure how to derive the balance. I also know that the potential will be the same on both spheres since they are conducting, the electric field is 0. Let the first sphere have charge Q1 and the second Q2, then Q = Q1 + Q2. V = kQ1/r = kQ2/r since the potentials are equal...I don't think is right though. Someone want to clear this up for me? Thanks. A lot.
 
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anyone know how to do this?
 
Yeah that's right, the potential of each sphere is equal after they're connected, so kQ1/r1=kQ2/r2.
 

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