Potential due to a system of charges

All four corners are at the same distance x from the centre.
$V = \frac{k(-Q)}{x} + \frac{k(-q)}{x} + \frac{k(2q)}{x} + \frac{k(2Q)}{x} = 0$
$-Q - q + 2q + 2Q = 0 \Rightarrow Q + q = 0$
$Q = -q$

Work done in moving a charge Q from D to E: $W = Q(V_E - V_D)$
BC = AC = 2a. D is the midpoint of BC, so DB = DC = a; E is the midpoint of CA, so EC = EA = a.
By the symmetry of the isosceles triangle about the bisector of angle C, the distance of D from A equals the distance of E from B: AD = BE.
$V_D = \frac{q}{4\pi\varepsilon_0}\left(\frac{1}{a} + \frac{1}{a} + \frac{1}{AD}\right)$
$V_E = \frac{q}{4\pi\varepsilon_0}\left(\frac{1}{a} + \frac{1}{a} + \frac{1}{BE}\right)$
Since AD = BE, $V_D = V_E$.
$W = Q(V_E - V_D) = 0$

A is the midpoint of the side joining the two +q charges, so each +q is at a distance L from A.
Each −q charge is at the far corners: its distance from A is $\sqrt{(2L)^2 + L^2} = \sqrt{5}L$
Potential at A due to the two +q charges: $V_+ = 2\times\frac{1}{4\pi\varepsilon_0}\frac{q}{L} = \frac{1}{4\pi\varepsilon_0}\frac{2q}{L}$
Potential at A due to the two −q charges: $V_- = -2\times\frac{1}{4\pi\varepsilon_0}\frac{q}{\sqrt{5}L} = -\frac{1}{4\pi\varepsilon_0}\frac{2q}{\sqrt{5}L}$
$V_A = V_+ + V_- = \frac{1}{4\pi\varepsilon_0}\frac{2q}{L}\left(1 - \frac{1}{\sqrt{5}}\right)$