SHM in electric fields

So you must use both of them, not just one.In summary, the conversation discusses the motion of a charge in equilibrium at the origin between two identical point charges located on the x-axis. The charge is displaced a small distance and released, resulting in simple harmonic motion with an angular frequency of (4kqQ/(mb^3))^(1/2). One person asks for an explanation of their method for solving the problem, and another person explains that the force on the charge must be compared with the forces of both point charges, not just one, in order to find the correct answer.
  • #1

Homework Statement


A charge +q of mass m is free to move along the x axis. It is in equilibrium at the origin, midway between a pair of identical point charges, +Q, located on the x axis at x = +b and x = -b. The charge at the origin is displaced a small distance x << a and released. Show that it can undergo simple harmonic motion with an angular frequency

omega=(4kqQ/(mb^3))^(1/2)

Homework Equations


E=ke(q/r2)
(1+c)n is approximately equal to 1+nc

a=x(omega)^2


The Attempt at a Solution



Well, I'm not really asking for a solution per se. I get the question, got the correct answer, how it was done; what I want to know is why my method is wrong.

I got it by first using Coulomb's law to set up a force comparison, between the point-charge in the origin, and one of the point charges next to it. So...

F=kqQ/b2=ma

Where I substituted a for x(omega)^2.

Solving for omega got me close to the correct answer, but my TA could not explain why my method was wrong...so I'm curious why.

My answer was omega=(kqQ/(mb^3))^(1/2)

Any takers?
 
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  • #2
CrypticWeirdo said:
I got it by first using Coulomb's law to set up a force comparison, between the point-charge in the origin, and one of the point charges next to it. So...

The charge is in equilibrium at the origin, the forces of repulsion due to two other charges cancel out. You cannot compare with only one other charge. The force on it is due to the other 2 charges.
 

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