Electric dipole's maximum angular velocity

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SUMMARY

The discussion focuses on calculating the maximum angular velocity (\(\omega_{max}\)) of an electric dipole in an electric field. The dipole consists of charges +q and -q separated by a distance D, with a moment of inertia I about its center of mass. The relevant equations include the dipole moment \(p = qD\), torque \(\vec{\tau} = \vec{p} \times \vec{E}\), and potential energy \(U = -\vec{p} \cdot \vec{E}\). The solution involves equating potential energy to kinetic energy to derive the maximum angular velocity when the dipole aligns with the electric field.

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  • Understanding of electric dipole moment and its calculation
  • Familiarity with torque and its relation to angular motion
  • Knowledge of potential energy in electric fields
  • Basic principles of rotational dynamics, including moment of inertia
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Homework Statement


Consider an electric dipole located in a region with an electric field of magnitude [tex]\vec{E}[/tex] pointing in the positive y direction. The positive and negative ends of the dipole have charges +q and -q, respectively, and the two charges are a distance D apart. The dipole has moment of inertia I about its center of mass. The dipole is released from angle [tex]\theta[/tex], and it is allowed to rotate freely.

What is [tex]\omega_{max}[/tex], the magnitude of the dipole's angular velocity when it is pointing along the y axis?

Homework Equations



dipole moment p= qd

[tex]\vec{\tau}[/tex]=[tex]\vec{p}[/tex]X[tex]\vec{E}[/tex]

U= -[tex]\vec{p}[/tex] [tex]\cdot[/tex][tex]\vec{E}[/tex]

The Attempt at a Solution



I attempted to use energy, but I am not sure how to do it correctly - does potential energy equal kinetic? is the potential energy the one described in the above equation?
 
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dipole in a E field gives torque, torque is a force and can then be related to moment of inertia and angular acceleration
 
You can use your expression for U and (1/2)I\omega^2 for the KE.
 

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