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Here, the blue sphere (positively charged with +Q charge) is at a lower potential and the pink sphere (positively charged with +Q charge) is at a higher potential, even though they are equally positively charged, as the radius of the pink sphere is smaller.

TLDR: the blue sphere is at a lower potential and the pink sphere is at a higher potential

Scenario 1:

After we connect the two spheres by an uncharged conducting wire, current will flow from the pink sphere to the blue sphere until the electric potentials of the blue and pink spheres become equal. After equilibrium is reached, no charge will be found on the conducting wire; the uncharged conducting wire will remain uncharged after equilibrium. The uncharged conducting wire is then easily removed.

Scenario 2:

After we connect the two spheres by an uncharged conducting wire, current flows from higher potential to lower potential. The charges redistribute themselves to achieve an equipotential surface: after equilibrium is reached, the electric potentials of the pink ball, blue ball, and wire are equal. It's more appropriate to view the three objects, the blue ball, the pink ball, and the wire as a singular object after the balls have been connected by the wire. In this scenario, however, the conducting wire also becomes charged after equilibrium, even though it was uncharged initially.

My question:

  1. Which scenario is correct?
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    $\begingroup$ If the spheres are conducting material, then all the charge will distribute until all electrons are "equally spaced" according to the overall shape of the electric field gradient. I'd expect that the wire, being part of a single conducting object, will contain some of the electrons. I don't want to try to calculate the shape of the E-field :-) $\endgroup$ May 5 at 13:47

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There will be charges on the wire, because if there were no charges on the wire and positive charges on the balls, then some charges would still want to go in the wire, so it wouldn't be an equilibrium.

But the idea behind this experiment is to consider a wire with infinitly small diameter, so the charge inside the wire can be neglected, because charges confined in a small volume would create a high electrical potential and we wouldn't be in equilibrium.

So it is a good approximation to say that the total charge on both spheres remains $2 \cdot Q$ and the charge on the wire is $0$. Scenario $1$ would hence be a good approximation of this system although it is never really possible to achieve it since wires have a finite diameter.

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  • $\begingroup$ Sorry I meant with an infinitely small diameter. So the radius of the cable should be negligible in comparison with the radius of the spheres $\endgroup$ May 5 at 15:09
  • $\begingroup$ "But the idea behind this experiment is to consider a wire with infinity small diameter, so the charge inside the wire can be neglected, because charges confined in a small volume would create a high electrical potential and we wouldn't be in equilibrium." - I'm having some trouble understanding this paragraph, unfortunately. Do you mean that as the wire has an infinitely small diameter, its volume will also be infinitely small. So, the amount of charge present on the infinitely small surface area of the wire will be very less. So, we can neglect it. $\endgroup$ May 5 at 15:29
  • $\begingroup$ Needless to say, charges don't accumulate inside a wire ever; they stay on the surface of the wire. "because charges confined in a small volume would create a high electrical potential and we wouldn't be in equilibrium."- are you trying to explain why charges don't stay inside a wire by this sentence? $\endgroup$ May 5 at 15:32
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    $\begingroup$ Yes you got it right. Also if you have a very small volume, even a few charges would already be pretty near from another, so we would have a high electric field at the surface of this volume, hence a high potential in the wire. But since everything must have the same potential this cannot be, so the charge inside a cable we infinit small volume must be infinitely small. $\endgroup$ May 5 at 15:36
  • $\begingroup$ For your second question, of course there will only be charge on the surfaces. I was trying to say that since we are in equilibrium, all the surfaces (from cable and spheres) must have exactly the same potential. This can only be achieved if the charges inside the wire are infinitely small (because of the infinitely small diameter of the wire) $\endgroup$ May 5 at 15:42

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