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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 ?

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In order to show that a configuration is an equilibrium configuration of charges, you need to show two things.

  1. You need to compute the forces at the configuration you have drawn and show that the total force on each charge is 0.
  2. You need to also show that small perturbations around it don't grow. This means that you need to show that placing one of the charges slightly off, would result in a net force on that charge in the direction of the initial position.

Try solving the problem in one dimension first.

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  • $\begingroup$ In this case, you may also need to show that a global compression of the charge distribution wouldn't change the potential energy density of the configuration. $\endgroup$ – S. McGrew Jun 16 at 13:17
  • $\begingroup$ Applying your conditions makes the initial configuration an unstable equilibrium. $\endgroup$ – R.W. Bird Jun 16 at 16:06
  • $\begingroup$ 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 ? $\endgroup$ – ZingZong Jun 16 at 16:54
  • $\begingroup$ @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 $F\neq=0$. $\endgroup$ – Stratiev Jun 16 at 17:17
  • $\begingroup$ Sorry I wasn't clear. I was talking about the initial configuration : $... + - + - + ...$ $\endgroup$ – ZingZong Jun 16 at 17:21

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