Electric potential of a spinning rod.

[Lusos]

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.

 

[clementc]

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

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

 

[Lusos]

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