# How charge distribution takes place when a battery is connected to a conductor?

When one terminal of a battery say of 1.5 volt connected to a short length wire, few electrons get transferred from battery terminal to the wire raising the potential of the wire also to 1.5 volt. We know that raising the potential of a wire requires transferring certain numbers of electrons. Now consider the same battery is connected to very long wire of some thousand kilometer length. Will the battery be able to raise the potential of the wire?

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Please explain in detail what you think will happen. –  Alfred Centauri Jun 3 '13 at 12:12
not very sure! but i think the battery may not be able to supply the required number of electrons to raise the potential to 1.5 volt. –  user25355 Jun 3 '13 at 13:14

but i think the battery may not be able to supply the required number of electrons to raise the potential to 1.5 volt

One way to think about this problem is to consider that there is a capacitance between the unconnected terminal of the battery and the long wire connected to the other terminal.

This capacitance determines if the battery "has enough electrons" to charge the capacitor to 1.5V.

If the value of the capacitance is $C$, you know the amount of charge required:

$Q = 1.5V \cdot C$

So, to develop 1.5V between the long wire and the other battery terminal, the battery must have a charge capacity larger than Q.

Typically, a battery has an amp-hour rating which is a measure of electric charge. However, the unit of charge in the formula above is amp-seconds so you must convert amp-hours to amp-seconds.

I don't think that there will be much capacitance due to this wire. Let's see what kind of capacitance would be required to challenge a typical battery.

Consider the capacity of 1.5V AA battery for which a reasonable amp-hour rating is 1 Ah or 3600 Coulombs.

What is the value of the capacitance for which this amount of charge would be required to get to 1.5V?

$C = \dfrac{3600}{1.5} = 2400 F$

2400F is an enormous value of capacitance. I think it is unlikely that you could reasonably expect to configure the long wire in such a way as to achieve this amount of capacitance.

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+ Nice, although the long wire would also have inductance and would radiate. I'm not sure how you could take that effect out of the problem. Make it a twisted pair? –  Mike Dunlavey Dec 2 '13 at 16:24

Yes, although very less.

I=Q/t, =>I=ne/t, where q is charge, t is time, n is no. of electrons and e is charge of one electron. Transfer of electrons at a rate is current, which is bounded by V/R. If length of wire is high, that means R is also very high, which leaves a mere no. of electrons in the wire.

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when it is said wire, we assume negligible resistance! –  Saurabh Raje Jul 3 '13 at 13:56
@SaurabhRaje: Not necessarily. Nothing wrong with not assuming something that people are told to assume @ school. –  Dimensio1n0 Aug 2 '13 at 15:00