# Is an infinite set of positive and negative charges next to each other $… - + - + …$ a position of equilibrium?

Suppose we have an infinite set of positive and negative charges next to each other: $$... + - + - + - ...$$

I am wondering if this a position of equilibrium.

Intuitively I would say that it is a position of equilibrium since the negative charge on the left of $$+$$ repeal $$+$$ in the right direction as much as the $$-$$ on the right which repeal $$+$$ on the left. Moreover if I take one $$+$$ and get it off this infinite set then we will have the following position of equilibrium :

$$... + - + - - + ...$$ Thus the two $$-$$ are going to move onto each other.

I don't really know if what I am saying is false, if it's a good justification...

So is this infinite set a position of equilibrium and how to "prove" it ?

• Thank you. But as @R.W.Bird said applying point 2) shows that the initial configuration is an unstable equilibrium since if I move just by a tiny amount the position of a proton it will go away from the chain. We will then have $... + - + - - + ...$. Moreover the total force on each charge is $0$ just by symmetry right ? – ZingZong Jun 16 '20 at 16:54
• @ZingZong I agree that it is unstable. What symmetry are you invoking in order to argue that the $...+ - + - - + ...$ configuration has a zero force on each charge? If you zoom in on the part that's $...+ - - ...$, you see that the middle charge is tugged to the left by the positive charge and also repelled to the left by the negative. Therefore $F\neq=0$. – Stratiev Jun 16 '20 at 17:17
• Sorry I wasn't clear. I was talking about the initial configuration : $... + - + - + ...$ – ZingZong Jun 16 '20 at 17:21