Time period of a oscillatory dipole
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There is a specific formula for the dipole moment and torque but we can sorta derive that as we go along:

We start with Newton's 2nd Law for circular motion: τ=Iα ⋆
We place the origin at the mid point and exploit the symmetry and for minimum time we arrange the positive and negative charges as shown. The torque is:
τ=2×(r×F)
r=(L2cosθL2sinθ) and F=(Eq0)
⇒τ=−LEqsinθ ez
The inertia for 2 point masses is 2×M(L2)2=ML22
And so from ⋆:
−LEqsinθ ez=ML22..θ ez
We make the usual linearisation assumption, θ≈sinθ for small θ and we get this DE:
..θ+2EqMLθ=0 which is in usual form: ..θ+ω2θ=0
So period T=2πω=2π√ML2Eq
This has been linearised to simple harmonic motion (we have a linear restorative torque), and the minimum time τ for the arrangement to first go through the equilibrium point (ie where θ=0) is τ=T4=π2√ML2Eq

We start with Newton's 2nd Law for circular motion: τ=Iα ⋆
We place the origin at the mid point and exploit the symmetry and for minimum time we arrange the positive and negative charges as shown. The torque is:
τ=2×(r×F)
r=(L2cosθL2sinθ) and F=(Eq0)
⇒τ=−LEqsinθ ez
The inertia for 2 point masses is 2×M(L2)2=ML22
And so from ⋆:
−LEqsinθ ez=ML22..θ ez
We make the usual linearisation assumption, θ≈sinθ for small θ and we get this DE:
..θ+2EqMLθ=0 which is in usual form: ..θ+ω2θ=0
So period T=2πω=2π√ML2Eq
This has been linearised to simple harmonic motion (we have a linear restorative torque), and the minimum time τ for the arrangement to first go through the equilibrium point (ie where θ=0) is τ=T4=π2√ML2Eq
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