table or unstable?
Let the four corners be located at the points (±a, ±a). Then the four distances from these points to the point (x,y) are (±a−x)2+(±a−y)2.
Ignoring the common factor of −Qq/4$\pi\epsilon$0, the code for the Mathematica Series expansion (to second order in x and y) for the energy U(x,y) of the −Q charge is as follows:
Series[
1/Sqrt[(a-x)^2+(a-y)^2]+
1/Sqrt[(a-x)^2+(-a-y)^2]+
1/Sqrt[(-a-x)^2+(a-y)^2]+
1/Sqrt[(-a-x)^2+(-a-y)^2],
{x,0,2},{y,0,2}]
This yields an energy of (to second order)
−Qq 2 $\sqrt$2 x2+y2
U(x,y)= + $\sqrt$ . (12.10)
4$\pi\epsilon$0 a 2 2a2
We see that the energy decreases with x and y, so the equilibrium is unstable for any direction of motion of the −Q charge in the xy plane. This instability is consistent with a theorem that we will prove in Section 2.12.
Note that U(x,y) reduces properly when x = y = 0, in which case the −Q charge is \sqrt2a away from the four q charges. Note also that the lack of terms linear in x or y is consistent with the fact that the force on the −Q charge is zero at the center of the square (the force involves the derivative of the energy). As far as motion in the xy plane near the origin is concerned, you can think of the charge roughly as a ball sitting on top of an inverted bowl.