Final answer:
For two spherical conductors connected by a wire, their surface potentials equalize due to charges redistributing until they reach an equipotential state, calculated by kq1/r1 = kq2/r2. The other sphere's charge is not included in the potential calculation for each sphere when they are connected because the system's combined potential is what equalizes.
Step-by-step explanation:
When two conductors with charges q1 and q2, and radii r1,r2 are connected by a metal rod, the charges redistribute until the electric potentials of the two conductors are equal. This occurs because the system reaches equilibrium only when the potential is the same throughout the conductor, turning the entire system into an equipotential. The formula to calculate the potential of a spherical conductor is based on the assumption that the conductor's charge distribution is equivalent to a point charge at its center; therefore, outside the conductor, the potential at its surface is kq/r. When two spherical conductors are connected, the potentials on their surfaces must equalize, which can be expressed as kq1/r1 = kq2/r2.
The concern you raised regarding whether the potential of one sphere should also include the influence of the other sphere's charge is valid. However, initially when the two spheres are sufficiently separated (before they are connected with a wire), each sphere can be treated as if it were isolated, because the influence of the other sphere's charge is not significant compared to the sphere's own charge. Furthermore, after being connected by a wire, the spheres can still be treated as isolated in terms of calculating each one’s potential due to their charges equalizing and reaching the same potential throughout the equipotential system.