Oscillating Charge (No Springs Attached)

Electricity and Magnetism Level 5

There are three identical collinear particles of charge ( q = 1 0 − 6 C ) (q = 10^{-6}\text{ C}) and mass ( m = 1 kg ) (m = 1\text{ kg}) . Two of the particles are held in place at ( x = 0 m ) (x = 0\text { m}) and ( x = 1 m ) (x = 1\text{ m}) respectively. The third particle is free to move, and is initially positioned at ( x = 1 4 m ) (x = \dfrac{1}{4}\text{ m}) .

If the Coulomb constant is ( k = 9 × 1 0 9 Nm 2 C − 2 ) (k = 9\times10^{9}\text{ Nm}^{2}\text{ C}^{-2}) , determine the oscillation period of the moving charge (rounded to the nearest half second).

While solving, neglect all forces except for the basic Coulomb force.


The answer is 9.5.

Steven Chase by Steven Chase , 37, USA

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2 solutions

Steven Chase
Aug 1, 2016

In order to ensure a good result, I solved in two very different ways:

Method 1) Potential Energy Method / Hand Analysis / Numerical Integration of Hand-Derived Integral

Method 2) Force Method / Small-time-step computer simulation / No advanced math

Method 1:

Method 2:

Simulation Strategy for each loop iteration:

1) Update position based on velocity from the last loop iteration

2) Update velocity based on acceleration from the last loop iteration

3) Calculate force based on position

4) Calculate acceleration based on force and mass

5) Go back to step 1

Code:

Result Plot:

Note that the charge oscillates between x = 0.25 x = 0.25 and x = 0.75 x = 0.75 , as expected. And the oscillation period is again roughly 9.5 seconds.

3 Helpful 1 Interesting 1 Brilliant 0 Confused
K T
Nov 20, 2020

I got a period of 9.4724278 s (also using a sim code, will share below, although I think nobody is interested)

Two remarks:

  • The problem would be clearer if you add that the particle starts without velocity.
  • In a real experiment the particle would not oscillate much, it would very soon escape away from the axis. 
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    import math;

    c=0.009 # = kq^2 = 9×10^9 × (1×10^6)^2
    t=0; x=0.25; Ek=0.0; dt=0.0001
    while x<0.5:
        a=c*(1/x**2 - 1/(1-x)**2)
        v = math.sqrt(2*Ek) + 0.5*a*dt  # estimate of mean velocity
        x+=v*dt
        t+=dt
        Ek=c*(16/3 -1/x - 1/(1-x))   # kinetic energy at x
    print ("t={} x={} Ek={} v={} a={}".format(t,x,Ek,v,a))

    print ("period = {}".format(4*(t-(x-0.5)/v)))

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