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Homework Help: Harmonic Motion with Electricity

  1. Sep 13, 2009 #1

    A uniform circular ring of charge Q= 4.50 microCoulombs and radius R= 1.30 cm is located in the x-y plane, centered on the origin. A point z is located along the Z axis. If z << R then E is proportional to z. (You should verify this by taking the limit of your expression for E for z << R.) If you place an electron on the z-axis near the origin it experiences a force Fz= -kz, where k is a constant. Obtain a numerical value for k. [I did this and obtained the value 2.946×10-9 N/m which I know is correct] What is the frequency of the small axial oscillations that the electron will undergo if it is released along the z-axis near the origin?

    2. Relevant equations

    So far I have looked at:

    \vec{F}_{net} = \Sigma \vec{F} = m \vec{a}

    KE= 1/2mv^2

    3. The attempt at a solution

    Ok so what haven't I thought about so far?
    My first approach was to use Fnet= ma and then substitute dz^2/d^2t for a and try separation of variables and integrate it, but I didn't really know how to make that work for a position dependent force as opposed to a velocity dependent force and I really couldn't get anything useful out of that. Then I thought about trying Potential energy and setting the potential energy at that point to the kinetic energy of the point in the middle of the ring of charge. Then I would have a velocity of that point I couldn't figure out if that would tell me anything, and I don't even think I'm supposed to use potential energy because we haven't even gotten close to learning that chapter yet. I figure I was closer with the Fnet=ma approach because the problem had us find the force and there should be a reason for that, right? So maybe I'm just messing up the calculus.
  2. jcsd
  3. Sep 20, 2009 #2


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    Welcome to PF!

    Hi nothilaryy! Welcome to PF! :smile:

    (I'm sorry nobody replied earlier: I hope you've done it by now, but if not …)

    Yes, it's z'' = -(k/m)z, which is a standard simple harmonic motion equation with general solution z = Acosωt + Bsinωt, where ω = … ? (work it out by differentiating twice! :wink:)

    (if you want to solve it properly, use the trick of putting v = dz/dt, so the chain rule gives you dv/dt = v dv/dz)