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Electron on ring axis performs SHM

Short answer

The axial field is E = λ R y / [2ε₀ (R² + y²)³ᐟ²]. For small y ≪ R, E ≈ λ y/(2ε₀ R²); the restoring force F = −eE ∝ −y gives SHM with period T = 2π √(2ε₀ m R²/(eλ)).

A thin ring made of plastic of radius R is uniformly charged with linear charge density λ. Calculate the electric field intensity at any point at an axial distance y from the center. If an electron is constrained to be in the axial line of the same ring, show that the motion of the electron is SHM.

Field on the axis of a charged ring

Ring of radius R, linear charge density λ (total Q = 2πRλ). Point P on the axis at distance y from the centre.

     dq (element on the ring)
      o
      | \
    R |   \  s = √(R² + y²)
      |     \
      O-------P-----> axis
          y
   (ring seen edge-on, centre O)
  1. Take element dq = λ dl. Its distance from P is s = √(R² + y²). dE = (1/4πε0) dq / s².
  2. Components perpendicular to the axis cancel by symmetry. The axial component is dE cosφ, with cosφ = y/s.
  3. E = ∫ (1/4πε0) (dq/s²)(y/s) = (1/4πε0) · y Q / (R² + y²)^(3/2)

E = λ R y / (2ε0 (R² + y²)^(3/2)), directed along the axis away from the ring (λ > 0).

SHM of an electron

  1. For λ > 0, the force on the electron (charge −e) is towards the centre: F = −eE.
  2. For small displacements y ≪ R, (R² + y²)^(3/2) ≈ R³, so

E ≈ λ y / (2ε0 R²), and F = −(eλ / 2ε0R²) y = −K y

  1. F is proportional to the displacement and opposite to it: this is the condition for simple harmonic motion.
  2. m d²y/dt² = −K y → ω = √(K/m) = √(eλ / (2ε0 m R²))

T = 2π √(2ε0 m R² / (eλ))

So the electron oscillates simple-harmonically about the centre of the ring, for small amplitudes.

✓ Answer checked by Claude Opus 5.5 (high effort)

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