Maximum Excursion

The particle will initially move downward under the influence of the other three charges, but will come to a halt and move upward again some time after the net force changes sign. When the particle begins to move upward again, it has zero speed, which is to say that the other charges have done zero net work on it since it started moving. This is equivalent to saying that it has gone through an electric potential difference of zero. Therefore, we can do the following:

1) Write an expression for the electric potential energy of the free charge at the origin
2) Write a general expression for the electric potential energy of the free charge when it has moved downward by a distance $d$
3) Equate the two expressions and solve for $d$

Since we are equating two potential energy expressions, the constants typically associated with such an expression cancel out and are irrelevant. I will therefore not include them.

At the origin, the normalized potential energy is $(\frac{1}{1} - \frac{1}{1} - \frac{1}{1}) = -1$

At a distance $d$ downward, the expression is $\frac{1}{1+d} - \frac{1}{\sqrt{1+d^{2}}} - \frac{1}{\sqrt{1+d^{2}}} = \frac{1}{1+d} - \frac{2}{\sqrt{1+d^{2}}}$

The equation we solve for $d$ is therefore $\frac{1}{1+d} - \frac{2}{\sqrt{1+d^{2}}} + 1 = 0$

Solving this equation for $d$ yields $d \approx 0.81$