# Facing a paradox: Earnshaw's theorem in one dimension

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)?

• 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 $V^{\prime\prime}(x)=0$? – SRS Apr 4 '19 at 14:57
• 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 – SRS Apr 4 '19 at 15:37
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 $$q<0$$