Harmonic Motion with Electricity

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SUMMARY

The discussion focuses on the analysis of harmonic motion of an electron influenced by a uniform circular ring of charge Q = 4.50 microCoulombs and radius R = 1.30 cm. The force experienced by the electron along the z-axis is given by Fz = -kz, where the constant k is calculated to be 2.946×10-9 N/m. The participants discuss the application of Newton's second law (Fnet = ma) and the formulation of the motion equation z'' = -(k/m)z, which leads to the identification of the system as undergoing simple harmonic motion.

PREREQUISITES
  • Understanding of Newton's second law (F = ma)
  • Familiarity with simple harmonic motion equations
  • Basic calculus for differentiation and integration
  • Knowledge of electric fields and forces from charge distributions
NEXT STEPS
  • Learn how to derive the frequency of oscillation in simple harmonic motion
  • Study the concept of potential energy in electric fields
  • Explore the application of the chain rule in calculus for motion equations
  • Investigate the behavior of forces in position-dependent systems
USEFUL FOR

Students in physics, particularly those studying electromagnetism and mechanics, as well as educators seeking to understand the dynamics of charged particles in electric fields.

nothilaryy
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Homework Statement



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?

Homework Equations



So far I have looked at:

[tex] \vec{F}_{net} = \Sigma \vec{F} = m \vec{a}[/tex]

KE= 1/2mv^2


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.
 
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Welcome to PF!

Hi nothilaryy! Welcome to PF! :smile:

(have a mu: µ and try using the X2 tag just above the Reply box :wink:)
nothilaryy said:
… 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?

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 …

(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)
 

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