The figure shows a dumbbell consisting of two identical small particles, each of

Question

The figure shows a dumbbell consisting of two identical small particles, each of mass m, attached to the ends of a thin, massless rod of length a that is pivoted at its center. The particles have charges of +q and -q, and the dumbbell is located in a uniform electric field E. Show that for small values of the angle theta between the direction of the dipole and the direction of the electric field, the system displays a rotational form of simple harmonic motion, and obtain an expression for the period of that motion.

Explanation / Answer

We can apply Newton’s second law in rotational form to obtain the differential equation of motion of the dipole and then use the small angle approximation sin to show that the dipole experiences a linear restoring torque and, hence, will experience simple harmonic motion.

Apply = I to the dipole:

-pE sin = I(d2/dt2)

where is negative because acts in such a direction as to decrease .

For small values of , sin =======>    -pE = I(d2/dt2)

Express the moment of inertia of the dipole: I = 0.5ma2

Relate the dipole moment of the dipole to its charge and the charge separation: p = qa

Substitute for p and I to obtain   0.5ma2(d2/dt2) = -qaE

or   (d2/dt2) = -2qE/ma

the differential equation for a simple harmonic oscillator with angular frequency = sqrt(2qE/ma)

Express the period of a simple harmonic oscillator: T = 2/

Substitute for and simplify to obtain: T = 2*sqrt(ma/2qE)

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