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Three equal charges Q are fixed at the vertices of an equilateral triangle of side a, and a fourth charge q (of mass m) sits at the centroid in equilibrium. For small displacements of the fourth particle along the plane, what is the period of its oscillation?

1. 2*pi*sqrt((4*pi*e0*m*a³)/(9*sqrt(3)*Q*q))
2. pi*sqrt((4*pi*e0*m*a³)/(9*sqrt(3)*Q*q))
3. 2*pi*sqrt((2*pi*e0*m*a³)/(27*sqrt(3)*Q*q))
4. 2*pi*sqrt((pi*e0*m*a³)/(27*sqrt(3)*Q*q))

Correct answer: 2*pi*sqrt((4*pi*e0*m*a³)/(9*sqrt(3)*Q*q))

Solution

The centroid is at distance r = a/sqrt(3) from each vertex. For a small in-plane displacement the three Coulomb forces give a net linear restoring force F = -k*x with k = 9*sqrt(3)*Q*q/(4*pi*e0*a³) (after summing the radial-component derivatives of the three interactions and using r³ = a³/(3*sqrt(3))). The SHM period is T = 2*pi*sqrt(m/k) = 2*pi*sqrt((4*pi*e0*m*a³)/(9*sqrt(3)*Q*q)).

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