Simple harmonic motion of charged particles

  • #1

Homework Statement


Two identical particles, each having charge +q, are fixed in space and separated by a distance d. A third particle with charge -Q is free to move and lies initially at rest on the perpendicular bisector of the two fixed charges a distance x from the midpoint between the two fixed charges. See image.

a) Show that if x is small compared with d, the motion of -Q is simple harmonic along the perpendicular bisector. Determine the period of that motion.

b) determine the period of that motion

c) How fast will the charge -Q be moving when it is at the midpoint between the two fixed charges if initially it is released at a distance a<<d from the midpoint?

v_max=?

Homework Equations




The Attempt at a Solution



After following another thread I resolved the Force on -Q in the y direction to be F = -2kqQ/(x2+d2/4) ⋅ x/√(x2+d2/4) I am unsure where to go from here
 

Answers and Replies

  • #2
Ray Vickson
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Homework Statement


Two identical particles, each having charge +q, are fixed in space and separated by a distance d. A third particle with charge -Q is free to move and lies initially at rest on the perpendicular bisector of the two fixed charges a distance x from the midpoint between the two fixed charges. See image.

a) Show that if x is small compared with d, the motion of -Q is simple harmonic along the perpendicular bisector. Determine the period of that motion.

b) determine the period of that motion

c) How fast will the charge -Q be moving when it is at the midpoint between the two fixed charges if initially it is released at a distance a<<d from the midpoint?

v_max=?

Homework Equations




The Attempt at a Solution



After following another thread I resolved the Force on -Q in the y direction to be F = -2kqQ/(x2+d2/4) ⋅ x/√(x2+d2/4) I am unsure where to go from here
The question tells you exactly what to do: look at the case of small ##|x|## (that is, ##|x| \ll d##). Do you know HOW to do that?
 
  • #3
Wouldn't the fraction on the right go to zero?
 
  • #4
Ray Vickson
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Wouldn't the fraction on the right go to zero?
Small ##|x|##, not zero!

Think of it this way: what sort of force equation would you need in order to have Newton's laws give you simple harmonic motion? That is, if ##\text{Force} = F(x),## for some function ##F(x)##, what type of function ##F## do you need? Can you get such a function of ##x## in your "electrical" case, at least if ##|x|## is small enough?
 

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