Two equal positive charges Q are fixed at x = ±a. A particle of mass m and charge q is constrained to move on the perpendicular bisector y-axis. Under gravity it has two equilibrium positions A and B with OA : OB
= 1 : 3√3. Which equilibrium position is stable?

Question

Two equal positive charges Q are fixed at x = ±a. A particle of mass m and charge q is constrained to move on the perpendicular bisector y-axis. Under gravity it has two equilibrium positions A and B with OA : OB

= 1 : 3√3. Which equilibrium position is stable?

G1 Physics · Accepted Answer

On the y-axis, E(y) = 2kQy/(a²+y²)³ᐟ². This field first increases with y, reaches a maximum at y=a/√2, and then decreases. The two equilibrium points have the same electric force qE=mg. Solving with OB/OA=3√3 gives OA = a/(2√3), OB = 3a/2. Thus A lies on the increasing branch and B on the decreasing branch. At A, an upward displacement increases electric force, so the particle moves farther away: unstable. At B, an upward displacement decreases electric force below mg, producing a restoring dow…