OK, so I know that charge flows from high potential to low potential, for eg. if there were $2$ spheres of $5V$ and $3V$, then charge would keep flowing until their potentials become equal, i.e. $4V$, so then even in the case of a sphere and wire, charge should flow from the sphere to wire until their potential becomes $V/2$ and then why does the wire develops a potential equal to the sphere i.e. $V$ ?
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3$\begingroup$ I don't understand your reasoning. Why would the two spheres end up with a potential of $4V$? Are you making additional assumptions? $\endgroup$– Alfred CentauriCommented Jun 8, 2020 at 2:04
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1 Answer
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This has to do with the capacitance of the wire and the sphere. As long as we are talking about wire with smaller dimensions than those of the sphere, when you connect an initially uncharged wire to the sphere, potential will be close to the initial potential of the sphere, but actually smaller in magnitude.
To understand why, if you neglect the effect of the nearby wire on the capacitance of the sphere and vice versa, the total charge must be conserved, so $$C_sV_0 = C_sV + C_wV$$ $$V = \frac{C_s}{C_s+C_w}V_0$$ where $V_0$ is the initial potential of the sphere, $V$ is the final potential, $C_s$ is the capacitance of the sphere and $C_w$ is the capacitance of the wire. So the wire does affect the final potential, the degree to which it does depends on its capacitance, i.e. its geometry and how it is positioned in relation to the sphere.
