Electric potential of a spinning rod.

AI Thread Summary
The discussion revolves around calculating the angular velocity of an electric dipole consisting of charged spheres on a massless rod when aligned with a uniform electric field. The initial energy of the system is zero, and the potential energy is converted into kinetic energy as the dipole rotates. The correct approach involves using the potential energy of the dipole in the electric field and the moment of inertia to derive the angular velocity formula. After applying the relevant equations, the calculated angular velocity at the moment of alignment with the electric field is found to be 0.28 rad/sec. This solution highlights the importance of understanding energy conservation in the context of electric dipoles.
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Homework Statement


An electric dipole consists of 1.0 g spheres charged to +(-) 2.0 nC at the ends of a 10-cm-long massless rod. The dipole rotates on a frictionless pivot at it's center. The dipole is held perpendicular to a uniform electric field with field strength 1000 V/m, then released. What is the dipoles angular velocity at the instant it is aligned with the electric field?


Homework Equations


E= ΔVc/d
U = (Kq1q2)/r
qΔv = Kf + qΔV
ω = dθ/dt

The Attempt at a Solution


book answer = .28 rad/sec

I am at a loss as to where to start on this solution. The object shouldn't have any kinetic initial because it is motionless. I am supposed to calculate angular velocity after it passes 90°, yet do I use energy? I have tried using the following

U(int) = Kfinal + Ufinal

Kq1q2-r = kq1q2 + .5Iω^2

However, as you can see the Potentials cancel out to zero and give me a zero answer. There is an obvious flaw in my reasoning. Can someone please lead me in the right direction? Thanks.
 
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I think you're thinking of the potential of a single point charge - I think the formula for the potential of a dipole in an electric field given by

U = -\mathbf{p} \cdot \mathbf{E}
 
Last edited:
Alright guys, I went to the professor and got som help. For posterity in that class (I'm talking to you GC), I am posting the solution.

Using a bar chart, we find that we have zero initial energy. That is

0 = KE - Uq

.5Iω^2 - pE where p =qs

Given p = 2x10^-9 and "s" = .1m, then p = 2x10^-10

Rearrange your equation so that you solve for the ω.

[2(pE)/I]^.5 = ω

Where your moment of intertia is (5x10^-6)

(2(2x10^-10)(1000))/(5x10^-6) = .28 rad/sec
 
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