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 Ehrenfest theorem (or it doesn't)?

Edit: I meant Earnshaw's theorem.

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)?

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 Ehrenfest theorem (or it doesn't)?

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 Ehrenfest theorem (or it doesn't)?

Edit: I meant Earnshaw's theorem.