Why Does Displacing a Charge Lead to Simple Harmonic Motion?

Click For Summary
SUMMARY

The discussion centers on the mechanics of a charge +q of mass m undergoing simple harmonic motion when displaced from its equilibrium position between two identical point charges +Q located at x = +b and x = -b. The correct angular frequency for this motion is derived as omega = (4kqQ/(mb^3))^(1/2). A participant mistakenly calculated the angular frequency as omega = (kqQ/(mb^3))^(1/2) by only considering the force from one point charge, which is incorrect because the net force acting on the charge at the origin must account for the contributions from both surrounding charges.

PREREQUISITES
  • Coulomb's Law for electric forces
  • Concept of equilibrium in electrostatics
  • Understanding of simple harmonic motion
  • Basic calculus for deriving angular frequency
NEXT STEPS
  • Study the derivation of angular frequency in simple harmonic motion
  • Review the principles of electrostatics, focusing on net forces
  • Explore the implications of equilibrium in multi-charge systems
  • Learn about the mathematical approximations used in physics, such as Taylor series expansions
USEFUL FOR

Students of physics, particularly those studying electromagnetism and mechanics, as well as educators seeking to clarify concepts of charge interactions and harmonic motion.

CrypticWeirdo
Messages
1
Reaction score
0

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?
 
Physics news on Phys.org
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.
 

Similar threads

  • · Replies 51 ·
2
Replies
51
Views
4K
Replies
3
Views
1K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 13 ·
Replies
13
Views
2K
Replies
1
Views
1K
Replies
4
Views
3K
Replies
23
Views
4K