What are the consequences of connecting a non-ideal conductor to a battery in open loop?

Question

Suppose I have a battery with $\Delta V=5[\mathrm{V}]$. Now I connect a piece of a metal wire to the "+" side of the battery only.

Let's assume that the ambient air is not conductive at all:

1. Are any charges going to reposition themselves between the battery and the attached conductor in this open-loop configuration?
2. If no charges are repositioning themselves between the battery and the conductor, why the potential of the conductor is immediately equal to that of the "+" of the battery upon attachment ($5[\mathrm{V}]$, minus possible voltage drop due to non-ideality of the conductor)?
3. Since this conductor is not ideal, can the whole open-loop system of battery+conductor be treated as an effective plate capacitor in closed-loop with the battery?

Answer 1

In answer

1) Yes, charge will flow to make the potential of the wire the same as the + terminal of the battery given that the wire is neutral or at earth prior to making contact. The ammount of charge that would flow would depends on the difference in potential between the wire before you attach it to the battery and the battery - we could calculate that with $Q=CV$ where $Q$ is the charge, $V$ is the potential difference and $C$ is the capacitance of the wire (no charge flows if $V=0$, which would mean the wire is at the exact potential of the +terminal before it is attached).

2) The answer is there will probably be a bit of shifting of charges... but note that there will be no voltage drop - because we have open circuit no current is flowing. Voltage drop would be calculated from $V = IR$, but $I$ is zero and so the voltage drop will be zero.

3) Yes the wire is like a capacitor, but the capacitance is very low - less than $10^{-12}~F$ typically.

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