# Stability and Laplace's equation

Consider four positive charges of magnitude $q$ at four corners of a square and another charge $Q$ placed at the origin. What can we say about the stability at this point?

My attempt goes like this. I considered 4 charged placed at $(1,0)$, $(0,1)$, $(-1,0)$, $(0,-1)$ and computed the potential and and it's derivatives. When using the partial derivative test, the result was that origin is a stable equilibrium position. $$V(x,y)=k[\frac{1}{\sqrt{(x-1)^2 + y^2}}+\frac{1}{\sqrt{(x+1)^2 + y^2}}+\frac{1}{\sqrt{x^2 + (y-1)^2}}+\frac{1}{\sqrt{x^2 + (y+1)^2}}]$$ $$\partial_x V= -k[\frac{x-1}{((x-1)^2 + y^2)^\frac{3}{2}}+ \frac{x+1}{((x+1)^2 + y^2)^\frac{3}{2}} + \frac{x}{(x^2 + (y-1)^2)^\frac{3}{2}} + \frac{x}{(x^2 + (y+1^2)^\frac{3}{2}}]$$

$$\partial_{xx} V= k[\frac{2(x-1)^2 -y^2}{((x-1)^2 + y^2)^\frac{5}{2}} + \frac{2(x+1)^2 -y^2}{((x+1)^2 + y^2)^\frac{5}{2}} + \frac{2x^2 -(y-1)^2}{(x^2 + (y-1)^2)^\frac{5}{2}} +\frac{2x^2 -(y+1)^2}{(x^2 + (y+1)^2)^\frac{5}{2}}]$$

$$\partial_{yx} V= 3k[\frac{(x-1)y}{((x-1)^2 + y^2)^\frac{5}{2}} + \frac{(x+1)y}{((x+1)^2 + y^2)^\frac{5}{2}} + \frac{x(y-1)}{(x^2 + (y-1)^2)^\frac{5}{2}} +\frac{x(y+1)}{(x^2 + (y+1)^2)^\frac{5}{2}}]$$

$\partial_{yy}$ is same as $\partial_{xx}$ except for x and y exchange by symmetry. At the origin(equilibrium point), $\partial_{xx}$ and $\partial_{yy}$ are positive while $\partial_{yx}=\partial_{xy}=0$ Hence by the partial derivative test for stability I have a stable equilibrium. A local minimum of potential.

Now starts my confusion. According to what I learnt in Laplace's equation $\Delta V=0$ for potential, in a charge free region(I take the region without charges with origin in it) there can never have a local minima or maxima within the boundary. This contradicts with the above conclusion that we have a minimum of potential.

• alternative way to see it: substituting $y = m x$ leads to a Taylor expansion $V(x) = k(4 + 2 (m^2 + 1) x^2 + \dots)$ Commented Aug 6, 2015 at 20:20
The Coulomb Potential is a solution to Laplace's equation in 3 dimensions. In 2 dimensions the equivalent solution is a logarithmic potential. You have written down the Coulomb potential for 4 charges but then treat the problem as 2 dimensional, which is causing your problems. To resolve this you need to add a load of $z^2$s to your potential.