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The formula of electric potential is $$V=\frac{kQ}{r}$$ and the voltage across two points is \$a\$ and \$b\$ that determines the work of the electric field on a charge \$q_1\$ from \$a\$ to \$b\$.

If, say, we have short wire and make a circuit with a power source, then there would be some voltage. Now if we use a long wire instead of short wire, isn't now the distance increased between \$a\$ and \$b\$? So why does the voltage remain same if we consider the wire conductivity 100%? If we use the above formula then it means that the voltage must decrease. So how does all this happen?

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If we use the above formula then it means that the voltage must decrease.

You're attempting to apply a result outside of the context in which it was derived.

Outside of a static spherically symmetric charge distribution with total charge \$Q\$, the electric potential is indeed given by the formula you quote.

But, in the circuit context you give, we're not outside of a static spherically symmetric charge distribution so this formula, and any conclusion you draw from it, isn't valid in this context.

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If you assume the wire conductivity is 100% (no resistance), then there cannot be any voltage across the wire (E=IR but R=0). Therefore, it doesn't make any difference how long the wire is as the voltage will still be 0.

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Your formulae $$V=\frac{kQ}{r}$$ is for the potential of a point charge in an infinite non-conductive medium. Simply put,it doesn't apply to the case within an electric wire which in the ideal case is what is known as a equipotential i.e. "at the same voltage". Of course in the real world you can have a voltage drop along the length of the wire when current flows.

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