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A particle of mass m and charge -Q can move only along the axis of a ring of radius a. The ring carries a uniform positive linear charge density +lambda. The particle is released slightly displaced from the centre of the ring (its equilibrium point on the axis). Find the period of its small oscillations.
- T = 4*pi*sqrt(epsilon0*m*a² / (lambda*Q))
- T = 2*pi*sqrt(2*epsilon0*m*a² / (lambda*Q))
- T = 2*pi*sqrt(4*epsilon0*m*a² / (lambda*Q))
- T = 2*pi*sqrt(epsilon0*m*a² / (2*lambda*Q))
Correct answer: T = 2*pi*sqrt(2*epsilon0*m*a² / (lambda*Q))
Solution
The ring's axial field at distance x from the centre is E = q*x/(4*pi*epsilon0*(a²+x²)^(3/2)), which for x << a reduces to E = q*x/(4*pi*epsilon0*a³) with total charge q = lambda*2*pi*a. The force on the charge -Q is restoring: F = -[Q*q/(4*pi*epsilon0*a³)]*x, giving omega² = Q*q/(4*pi*epsilon0*m*a³). Substituting q = 2*pi*a*lambda yields omega² = Q*lambda/(2*epsilon0*m*a²), so T = 2*pi/omega = 2*pi*sqrt(2*epsilon0*m*a²/(lambda*Q)).
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