When -q is displaced by y perpendicular to the line joining the charges, what happens to the components of the two attractive forces?

Components along the joining line cancel; components along y add toward O

Components along y cancel; those along the line add

Why must y be small to get SHM?

So that (d^2/4 + y^2)^(3/2) ≈ (d/2)^3, making F proportional to y

So that the force becomes constant

So that the charge -q is not attracted

Why does the motion qualify as SHM?

The force is proportional to displacement and directed toward O

The force is constant and directed toward O

The force is proportional to y^2 and directed away from O

The force is proportional to displacement and directed away from O

With ω^2 = 4 q q0/(π ε0 m d^3), what is T = 2π/ω?

T = sqrt(ε0 m π^3 d^3/(q q0))

T = sqrt(ε0 m π^2 d^3/(q q0))

T = sqrt(ε0 m π d^3/(4 q q0))

T = 2π sqrt(q q0/(π ε0 m d^3))

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SHM of charge between two positive charges

Short answer

For small perpendicular displacement y, the net restoring force is F = −(4 q q₀/(π ε₀ d³)) y, giving ω² = 4 q q₀/(π ε₀ m d³). The period is T = 2π/ω = √(ε₀ m π³ d³/(q q₀)).

A particle of charge −q and mass m is placed midway between two equal positive charges q0 of separation d. If the negative charge −q is displaced in a perpendicular direction to the line joining them and released, show that the particle describes a SHM with a period T = √(ε0 m π³ d³ / (q q0)).

Setup: two fixed positive charges q₀ are a distance d apart. A particle of charge −q and mass m starts at the midpoint O. It is displaced by a small distance y perpendicular to the line joining the charges.

          -q  (y)
           o
          /|\
         / | \ F (towards each q0)
  q0 o---/--O--\---o q0
        <---- d ---->

Derivation

The distance of −q from each fixed charge is s = √(d²/4 + y²).
Each positive charge attracts −q with force F = q q₀ / (4πε₀ s²), along the line to that charge.
The components of the two forces along the line joining the charges cancel by symmetry. The components along y (towards O) add: F_y = 2 F (y/s) = 2 q q₀ y / [4πε₀ (d²/4 + y²)^(3/2)]
For small displacement, y ≪ d, neglect y² compared to d²/4: F_y ≈ 2 q q₀ y / [4πε₀ (d/2)³] = 16 q q₀ y / (4πε₀ d³) = 4 q q₀ y / (π ε₀ d³)
This force is directed towards O, opposite to the displacement, and proportional to y. So the equation of motion is m d²y/dt² = − (4 q q₀ / π ε₀ d³) y, which is the equation of simple harmonic motion, d²y/dt² = −ω² y.
Angular frequency: ω² = 4 q q₀ / (π ε₀ m d³).
Period: T = 2π/ω = 2π √[π ε₀ m d³ / (4 q q₀)] = √[4π² · π ε₀ m d³ / (4 q q₀)]

T = √[ ε₀ m π³ d³ / (q q₀) ]

This is the required result.

✓ Answer checked by Claude Opus 5.5 (high effort)

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SHM of charge between two positive charges

Derivation

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