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)?


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.

  • $\begingroup$ 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 '19 at 14:57
  • $\begingroup$ @SRS Yes, that's true. $\endgroup$ – knzhou Apr 4 '19 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 '19 at 15:37
  • $\begingroup$ @SRS The potential at a point charge is not defined (or you could say infinite) $\endgroup$ – Aaron Stevens Apr 4 '19 at 16:33
  • $\begingroup$ I have to think more about it and I'll get back. $\endgroup$ – SRS Apr 4 '19 at 16:37

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.