If you mean 'potential' by the term voltage, I believe tbe answer is already given by delta^{2} in #13. Wires are assumed to have zero resistance, so no electric field is required to maintain the current. No field, no potential 'difference', hence, 'constant' potential throughout the wire.
Edit: Sorry, the thread I linked was a 'featured thread' and not an insights article.

If you assume that the wire has a uniform voltage (which is the usual assumption) then you are automatically assured that the wire has a uniform voltage by that assumption!

What you may be asking is under what conditions that assumption is approximately valid when modeling a real circuit. That condition is when the resistance of the wire is very low. If the resistance is so low that the voltage drop is negligible then you are justified in making the assumption when modeling that circuit

Perhaps this will help. The picture below shows the equivalent circuit we use in circuit analysis to represent a real physical piece of wire. The OP is wondering about the resistance of a real wire R. If he continues his studies, he'll eventually discover that the L, C, and G terms are also nonzero in real life.

So, we can use this equivalent circuit to represent every segment of real life wire. But then we use Circuit Analysis (CA) to solve the whole thing. And the CA presumes that all the components are idealized and divorced from the physics. CA is a very powerful calculation technique. It is not physics.

Please read the article before posting more questions.

Say in a segment (current flowing from left to right) at a time to its left current is high and to its right its less..........
So to conserve charge, some charge stays at the junction.............
Now due to this piled up charge electric field will now push the charges on its left opposing their flow and push those on its right to support their flow........
The higher current eventually falls down and the lower current Eventually rises up....