Two equal positive point charges 'Q' are fixed at points B (a, 0) and A(-a, 0). Another test charge `q_(0)` is also placed at O (0, 0). Show that the equilibrium at 'O' is
(i) Stable for displacement along X-axis
(ii) Unstable for displacement along Y-axis.

Explanation

(i) When charge is shifted along y-axis :
Let x-y direction as :-

After resolving into components, net force will be along negative y-axis so the particle will return to its original position. So, it is stable equilibrium
(ii) When negative charge `q_(0)` is shifted along x-axis.

Initially, `vec(F)_(AO)+vec(F)_(BO)=vec(0)" "implies" "|vec(F)_(AO)|=|vec(F)_(BO)|=(KQq_(0))/d^(2)`
When charge `q_(0)` is slightly shifted towards + x axis by small distance `Deltax` then `|vec(F)_(BO)| gt |vec(F)_(AO)|`
also, these forces are attractive forces (due to negative charge)
Therefore, the particle wil move towards positive x-axis and will not return to its original position So, it is unstable equilibrium for negative charge.