Two charges -q each are separated by dsitance 2d. A third charge +q is kept at mid-point O. find potential energy of +q as function of small distance x from 0 due to -q charges. Sketch PE Vs/x and convince yourself that the charge at 0 is in an unstable equilibrium.

Explanation

In fig, two charges `-q` each are shown at A and B, where `AB = 2d`. A charge `+q` is kept at mid point O of `AB, OP = x` is small displacement of `+q` charge. Therefore, potential energy of `+q` charge at P due to the two charges `-q` each is
`U = (1)/(4pi in_(0)) [-(q^(2))/(d + x) - (q_(2))/(d - x)] = (q^(2))/(4pi in_(0)) (2d)/(d^(2) - x^(2))` ...(i)
The potential energy (u) versus x graph is as shown in Fig.
Now `(dU)/(dx) = (-q^(2) (2d))/(4pi in_(0)) (2x)/(d^(2) - x^(2))` ..(ii)
At `x = 0`, From (i) `U_(0) = (-2q^(2))/(4pi in_(0) d)`, and from (ii), `(dU)/(dx) = 0`
`:. x = 0` is an equilibrium point
Now `(d^(2) U)/(dx^(2)) = ((-2dq^(2))/(4pi in_(0))) [(2)/((d^(2) - x^(2))^(2) - (8x^(2))/(d^(2) - x^(2))^(3))]`
`(d^(2) U)/(dx^(2)) = ((-2dq^(2))/(4pi in_(0))) (1)/((d^(2) - x^(2))^(3)) [2 (d^(2) - x^(2)) - 8x^(2)]`
At `x = 0, (d^(2) U)/(dx^(2)) = ((-2dq^(2))/(4pi in_(0))) (1)/(d^(6)) (2d^(2)) = - (q^(2))/(pi in_(0) d^(3))` which is less than zero.
Hence, the equilibrium of charge q at O is unstable.