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Consider a one-dimensional situation on a straight line (say, $x$-axis). Let a charge of magnitude $q$ be located at $x=x_0$, the potential satisfies the Poisson's equation $$\frac{d^2V}{dx^2}=-\frac{\rho(x)}{\epsilon_0}=-\frac{q\delta(x-x_0)}{\epsilon_0}.$$ If $q>0$, $V^{\prime\prime}(x_0)<0$, and if $q<0$, $V^{\prime\prime}(x_0)>0$. Therefore, it appears that the potential $V$ does have a minimum at $x=x_0$, for $q<0$. Does this imply that $x=x_0$ is a point of stable equilibrium? I must be missing something because this appears to violate Earnshaw's theorem (or it doesn't)?

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Your example does not contradict Earnshaw's theorem for electrostatics, because it rules out stable equilibrium in a region without charge, possibly containing fields made by charges outside that region. Here you're doing the exact opposite, looking at the only point in your situation with charge.

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  • $\begingroup$ Yes. I meant Earnshaw's theorem. Thanks. Does it mean that there must be an Earnshaw's theorem for Newtonian gravitation? Because in a massless region, again one has $V^{\prime\prime}(x)=0$? $\endgroup$ – SRS Apr 4 at 14:57
  • $\begingroup$ @SRS Yes, that's true. $\endgroup$ – knzhou Apr 4 at 14:58
  • $\begingroup$ I am not yet totally comfortable with this. If you have a charge at some point $x=x_0$, is it not correct to look at the behaviour of the potential at that point? @knzhou $\endgroup$ – SRS Apr 4 at 15:37
  • $\begingroup$ @SRS The potential at a point charge is not defined (or you could say infinite) $\endgroup$ – Aaron Stevens Apr 4 at 16:33
  • $\begingroup$ I have to think more about it and I'll get back. $\endgroup$ – SRS Apr 4 at 16:37
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So technically $V''(x_0)$ doesn't have an actual value, since $\delta(x-x_0)\to\infty$ as $x\to x_0$. However, if you understand the Dirac delta distribution to be a limit of a function whose peak "gets narrower" with its integral remaining constant, then this is fine and you could say there is a minimum at $x_0$ for $q<0$

This can be more easily understood by just thinking about the motion of a positive charge in this potential. It will move towards the negative charge, i.e. towards the minimum of the potential.

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