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Two charged beads, each with a charge of q and a mass of m, are positioned on a horizontal, frictionless, insulating circular ring of radius R. One bead is fixed to the ring, while the other oscillates slightly around its equilibrium position along the ring. What is the expression for the square of the angular frequency of these small oscillations? [ε₀ represents the permittivity of free space.]
- q² / (4πε₀R³m)
- q² / (32πε₀R³m)
- q² / (8πε₀R³m)
- q² / (16πε₀R³m)
Correct answer: q² / (32πε₀R³m)
Solution
The expression for the square of the angular frequency of these small oscillations is q² / (32πε₀R³m) because the force acting on the oscillating bead is the Coulomb force due to the fixed bead, which can be expressed as a function of the displacement from the equilibrium position, leading to a simple harmonic motion with the given angular frequency.
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