A particle of mass m and positive charge q is attached to a light insulating thread of length L whose other end is fixed at O. A fixed positive charge Q is located a distance L vertically below O. The particle remains in equilibrium vertically above O with the thread taut, so the separation between q and Q is 2L. The minimum value of Q for which this inverted equilibrium is stable against a small horizontal displacement is:
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Question

A particle of mass m and positive charge q is attached to a light insulating thread of length L whose other end is fixed at O. A fixed positive charge Q is located a distance L vertically below O. The particle remains in equilibrium vertically above O with the thread taut, so the separation between q and Q is 2L. The minimum value of Q for which this inverted equilibrium is stable against a small horizontal displacement is:

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G1 Physics · Accepted Answer

For a small angular displacement φ from the upward vertical, U g U_g U g ​ = mgL cosφ and r = 2L cos(φ/2). Hence, U e U_e U e ​ = kqQ/[2L cos(φ/2)]. For small φ, cosφ ≈ 1 - φ²/2, sec(φ/2) ≈ 1 + φ²/8. Thus the coefficient of φ² in total potential energy is -mgL/2 + kqQ/(16L). For stable equilibrium this must be positive: kqQ/(16L) > mgL/2 ⇒ kqQ > 8mgL². Using k = 1/(4πε₀), Q m i n Q_min Q m ​ in = 32πε₀mgL²/q.