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CR Drost
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If the wires are perfect conductors and you connect the terminals sufficiently in the past from when you touch the edges, then yes: there is a slightly higher density of electrons in the one wire, leading to a higher voltage, and the electrons will travel from the higher density to the lower density without realizing that they have to "come from" this battery that's far away.

Of course your conductor might not propagate electric signals quite at the speed of light and it may in theory be feasible to get on a spaceship and outrun the signal that is raising the voltage, touching the opposite side of the wires without getting hurt. Two parallel lines like that form a sort of big RC-circuit with capacitors and resistors in series and parallel, so it does take some time to "fill up" the extra electron density on the one side. Conversely it takes also some time to "drain off" the extra electrons through you, and therefore this assumption I'm making above that $R \to 0$ is crucial for getting the maximum "oomph" on you from the "capacitor" that's in front of you.

If the wires are perfect conductors and you connect the terminals sufficiently in the past from when you touch the edges, then yes: there is a slightly higher density of electrons in the one wire, leading to a higher voltage, and the electrons will travel from the higher density to the lower density without realizing that they have to "come from" this battery that's far away.

Of course your conductor might not propagate electric signals quite at the speed of light and it may in theory be feasible to get on a spaceship and outrun the signal that is raising the voltage, touching the opposite side of the wires without getting hurt. Two parallel lines like that form a sort of big RC-circuit with capacitors and resistors in series and parallel, so it does take some time to "fill up" the extra electron density on the one side.

If the wires are perfect conductors and you connect the terminals sufficiently in the past from when you touch the edges, then yes: there is a slightly higher density of electrons in the one wire, leading to a higher voltage, and the electrons will travel from the higher density to the lower density without realizing that they have to "come from" this battery that's far away.

Of course your conductor might not propagate electric signals quite at the speed of light and it may in theory be feasible to get on a spaceship and outrun the signal that is raising the voltage, touching the opposite side of the wires without getting hurt. Two parallel lines like that form a sort of big RC-circuit with capacitors and resistors in series and parallel, so it does take some time to "fill up" the extra electron density on the one side. Conversely it takes also some time to "drain off" the extra electrons through you, and therefore this assumption I'm making above that $R \to 0$ is crucial for getting the maximum "oomph" on you from the "capacitor" that's in front of you.

Source Link
CR Drost
  • 39k
  • 3
  • 43
  • 116

If the wires are perfect conductors and you connect the terminals sufficiently in the past from when you touch the edges, then yes: there is a slightly higher density of electrons in the one wire, leading to a higher voltage, and the electrons will travel from the higher density to the lower density without realizing that they have to "come from" this battery that's far away.

Of course your conductor might not propagate electric signals quite at the speed of light and it may in theory be feasible to get on a spaceship and outrun the signal that is raising the voltage, touching the opposite side of the wires without getting hurt. Two parallel lines like that form a sort of big RC-circuit with capacitors and resistors in series and parallel, so it does take some time to "fill up" the extra electron density on the one side.