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## Homework Statement

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?

v_max=?

## Homework Equations

## The Attempt at a Solution

After following another thread I resolved the Force on -Q in the y direction to be F = -2kqQ/(x

^{2}+d

^{2}/4) ⋅ x/√(x

^{2}+d

^{2}/4) I am unsure where to go from here