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Electric Potential with oscillation

  • Thread starter lCaelestis
  • Start date
  • #1

Homework Statement


A proton moves along the x-axis, where an arrangement of source charges has created the electric potential V=6000x^2, where V is in volts and x is in meters.

(a) Graph the potential between x = -5.0 and x = +5.0.

(c) By exploiting the analogy with the potential energy of a mass on a spring, determine the "effective spring constant" of the electric potential.

(d) What is the proton's oscillation frequency (in Hz)?

Homework Equations



U=qV
[tex]U=\frac{1}{2}kx^2[/tex]


The Attempt at a Solution



(a) The potential is just a parabola

(c) [tex]
U=\frac{1}{2}kx^2=qV=6000qx^2[/tex]
[tex]k=2 \times 6000q[/tex]

(d) I can't even grasp what this means. Does the proton oscillate? The voltage only goes to zero once.
 

Answers and Replies

  • #2
kuruman
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The proton oscillates just as if it were a spring-mass system with a "spring constant" k that you found in (c). Yes, the voltage goes to zero once. What does that mean about the potential and kinetic energies of the proton?

What expression relates the frequency of a spring-mass system to the spring constant k?
 
  • #3
Thanks, now I understand.

I got the equation
[tex]f=\frac{\sqrt{\frac{k}{m}}}{2 \pi}[/tex]
 

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