Two identical particles, each having charge +q, are fixed in space and separated by a distance d. A third particle with charge -Q is free to move and lies initially at rest on the perpendicular bisector of the two fixed charges a distance x from the midpoint between the two fixed charges. See image.
a) Show that if x is small compared with d, the motion of -Q is simple harmonic along the perpendicular bisector. Determine the period of that motion.
b) determine the period of that motion
c) How fast will the charge -Q be moving when it is at the midpoint between the two fixed charges if initially it is released at a distance a<<d from the midpoint?
The Attempt at a Solution
After following another thread I resolved the Force on -Q in the y direction to be F = -2kqQ/(x2+d2/4) ⋅ x/√(x2+d2/4) I am unsure where to go from here