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The problem

There is an infinite chain (1d) of alternating charges that are in a distance a from each other. That is there is one charge +q and next to it -q and so on. Now calculate how much work it would take to insert one charge somewhere in the chain.

What I've tried

The work required has to be the potential after the insertion minus the potential energy before the insertion. The potential energy of the chain before the insertion is:

$E_{pot} = \frac{1}{4\pi\epsilon_0}\sum_{k=1}^{\infty}\left(\frac{q^2}{2ka}-\frac{q^2}{(2k-1)a}\right)$

This simplifies to:

$E_{pot} =\frac{q^2}{4\pi\epsilon_0 a}*(-log(2))$

Where I'm stuck

The problem is though that that is "from the view of one individual charge" in the sense that I took the potential energy between one specific charge and all other charges in the chain. The chain as a whole has of course a potential energy of infinity because it would take infinitely much work to seperate all those charges. Because there is no special point somewhere in the chain I felt that it was ok to just pick some arbitrary charge.

When you insert another charge though, there is one very special point in the chain: The point where those two same charges are next to each other. Therefore I feel like I cannot do the same with the chain after the insertion. I have no idea how to calculate the potential energy difference then.

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Alternate approach: since you feel that you understand the "one missing" case, compute the field for that case and find the work needed to bring the last charge in directly. (NB: I have not done this and can not comment on how much easier/harder the work might be; just suggesting other avenues that ought to work.) – dmckee May 21 '14 at 16:28

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