## period of an electric dipole rotating in an external electric field

A simple electric dipole in an external magnetic field performs harmonic motion (for small angles between E and L) due to the moments by the forces acting on the respective charges. Assuming there is no system to dissipate energy that is. The mechanic moment is given by p x E and the energy of the system is given by -p.E
We were asked to calculate the period of this harmonic motion. I tried using conservation of energy to find it and I found something but I'm suspecting it's not correct.
What I've found:

At theta Pi/2, both the kinetic energy and the energy of the dipole rotation are zero, which means that their changes are equal.
When looking at theta = 0 this gives:
p*E = I * omega^2 / 2 which gives us omega after one fourth of the period.
Alpha = d omega / dt and I * Alpha = p x E .
So Alpha * (t2-t1) = omega2-omega1 (approximately) and Alpha = p x E /I
Putting these together and using omega1 = 0:
omega2/(T/4) = sqrt(p*E*2/I)/(T/4) = p x E / I which gives:
T= sqrt(p*E*2/I)*4*I)/(p x E)

It's too easy to be correct.. Where did I make an error (if indeed I did :)) and how could I find the real solution?

Thanks !
 It's looks very close to being right. For small angle theta, (alpha, or angular acceleration) = p*E/I which is what you have. Not sure if I understand the question, but I think you will find this is true: T(period in seconds) = 2*Pi*SQRT(I(dipole moment of inertia)/p(dipole charge moment) X E(electric field strength) if Theta is << 5-10 degrees (this is a small angle approximation, assuming that the dipole may rotate between 10 degrees and 350 degrees along the designated axis). That dipole will oscillate with a predictable period according to the properties found similar to other systems that involve symple harmonic motion of oscillations...ovbiously if your angle theta is higher than 10 degrees, as you may have mentioned, the dipole will oscillate with a different period. Clean: T=2*Pi*[(I/p*E)]^1/2 Check your units, they do check out: ***C=J/V, thus J = N*m or kg * m^2/s^2 and E can be V/m or N/C T(s)=2*Pi*[I(kg*m^2)/p(C*m OR J*m/V) * E(V/m)]^1/2 s = 2*pi* kg m^2 _______________ which leaves you sec^2 in the top.... Jm/V * V/m taking sq rt gives you seconds. Hope this helps you a little bit....it might not, just trying to help....

Recognitions:
Homework Help
Yes, I would say don't use an argument like this on an exam or in homework. It will work out dimensionally, but it actually can't fail to do so...

 Quote by wimvd At theta Pi/2, both the kinetic energy and the energy of the dipole rotation are zero, which means that their changes are equal.
If by "energy of dipole rotation", you mean $$\frac{1}{2} I \omega^{2}$$, then this is tautologically true, because the only kinetic energy involved here is the rotational kinetic energy of the dipole. So the changes are equal because these are two ways of talking about the same energy...

 When looking at theta = 0 this gives: p*E = I * omega^2 / 2 which gives us omega after one fourth of the period. Alpha = d omega / dt and I * Alpha = p x E .
You can get away with this dimensionally, but not quantitatively: since |p x E| = $$pE sin\theta$$, with theta being the angle between the dipole vector and the magnetic field, this plainly can't be a constant. So it isn't strictly legal to take this next step, except in a very rough way:
 So Alpha * (t2-t1) = omega2-omega1 (approximately)
So the rest will follow, since you are ignoring whatever dimensionless constant would go in there, said constant depending on the initial angle you chose.

Here's where you're on the right track:

 Alpha = d omega / dt and I * Alpha = p x E .
When you put these together, you get

$$I \cdot \frac{d \omega}{dt} = -pE sin\theta$$

taking care to use the correct measure for angular displacement, which will get you the minus sign on the right-hand side.

As seekingaeolus says, you want to use the "small angle approximation" (which works pretty well out to 30º and very well for angles smaller than 10º) to write $$sin \theta \approx \theta$$ (in radians). If you also use the definition of angular speed, $$\omega = \frac{d \theta}{dt}$$ , you end up with

$$I \cdot \frac{d^2 \omega}{dt^2} \approx -pE \cdot \theta$$

or $$\frac{d^2 \omega}{dt^2} + \frac{pE}{I} \cdot \theta = 0$$

which, at this stage in your physics courses, you are asked to recognize as a "simple harmonic motion differential equation", from which you can read off what $$\omega^2$$ is for this dipole oscillator, and thus find its period.