Can we arrange a finite number of point charges in a line so that they are all in equilibrium? Earnshaw's theorem proves that there cannot be a stable equilibrium point in an electric field, but what about unstable equilibrium? If not, how to prove it?
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$\begingroup$ What about the hydrogen atom? $\endgroup$– HolgerFiedlerCommented Jan 22, 2017 at 7:00
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If I put a charge of $+4q$ at $x=+x_0$ and $x=-x_0$ and have charge of $-q$ at the origin then the force on the charge at the origin is 0 by symmetry and the force on the outer charges is $$\frac{1}{4\pi\epsilon_0}\left(\frac{-4q^2}{x_0^2} + \frac{16q^2}{(2x_0)^2}\right) = 0$$ so the system is (unstable) equilibrium.
answered Jan 21, 2017 at 15:18
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$\begingroup$ Ahh yes, good spot. thank you, I've corrected then numbers $\endgroup$ Commented Jan 21, 2017 at 15:22
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$\begingroup$ You have not proven that the equilibrium is unstable. $\endgroup$ Commented Jan 21, 2017 at 16:58
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1$\begingroup$ @sammygerbil given that the OP mentioned Earnshaw's theorem and the fact that there are no stable equilibrium points in an EM field I felt that it was not necessary to prove it explicitly in this case. $\endgroup$ Commented Jan 21, 2017 at 20:25