Dipoles angular velocity when aligned with electric field?

AI Thread Summary
In the discussion about the angular velocity of an electric dipole aligned with an electric field, the dipole consists of charged spheres on a massless rod, initially held perpendicular to the field. The force acting on the dipole is calculated using F=qE, resulting in a force of 7 µN. The initial calculation for angular velocity yields approximately 0.121 rev/s, but concerns arise regarding the torque's dependency on the dipole's angle relative to the electric field. It is suggested that an energy conservation approach could provide a more accurate assessment by comparing potential energy differences when the dipole is horizontal versus vertical. The discussion emphasizes the need to account for varying torque based on the dipole's orientation in the electric field.
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Homework Statement



An electric dipole consists of 2.0 g spheres charged to 5.0 nC (positive and negative) at the ends of a 12 cm long massless rod. The dipole rotates on a frictionless pivot at its center. The dipole is held perpendicular to a uniform electric field with field strength 1400V, then released. What is the dipole’s angular velocity at the instant it is aligned with the electric field?

Homework Equations



F=mω^2r
F=qE

The Attempt at a Solution



F=qE
F=(5*10^-9)(1400)
F=7*10^-6N

7*10^-6=mω^2r
ω=√(7*10^-6/(.004*.12))
ω=.121rev/s?

I'm just unsure if I did this correctly, I thought it seemed to easy that way.
 
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The force that you calculated is only perpendicular to the rod when the rod is horizontal. Only a portion of that force will result in torque on the rod when it is at some other angle. That is to say, the torque is going to vary with the rod's angle with respect to the field. You could integrate the torque*dθ to find the work done, or,...

You might consider an energy conservation approach. What's the difference in potential energy between the rod horizontal and the rod vertical, given that the field strength is 1400 V/m?
 
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