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A charge experiencing simple harmonic motion

  1. Nov 25, 2015 #1
    1. The problem statement, all variables and given/known data
    Two positive charges +Q are held fixed a distance apart. A particle of negative charge -q and mass m is placed midway between them, then is given a small displacement x perpendicular to the line joining them and released. Show that the particle describes simple harmonic motion of period:

    ##\sqrt{\frac{\epsilon_0 m \pi^3 d^3}{Qq}}##

    2. Relevant equations
    ##T = 2\pi\sqrt{m/k}##
    ##F = \frac{1}{4\pi\epsilon_0} \frac{Qq}{r^2} = -kx##

    3. The attempt at a solution
    I wasn't getting it to work initially by working forwards so I worked backwards from the equation:
    ##2\pi\sqrt{m/k} = \sqrt{\frac{\epsilon_0 m \pi^3 d^3}{Qq}}##

    I knew that k was effectively F/x. so doing that I tried to get F by itself and got:

    ##F = \frac{2Q(-q)x}{4\pi\epsilon_0(\frac{d}{2})^3}##

    with some really creative algebra.

    I don't know how to get this from the original equation, but I know it's correct courtesy a hint from a friend. I know the significance of the 2 I got up top is because when you sum the forces in the x direction you get double one of the forces. I also know that at least 2 of the d/2 has to do with the distance between charges approaching that if the displacement it small. But I don't know where the x or the third d/2 come into the picture. I figure since the displacement is small we use some approximation method, but the only ones I know of are the small angle approximations for trig functions and Taylor series. Thoughts?
     
  2. jcsd
  3. Nov 25, 2015 #2

    Doc Al

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    Staff: Mentor

    Why don't you try writing an expression for the force as a function of displacement from the midpoint? The midpoint is a distance d/2 from each charge Q. So if you move a distance x towards one side, what's the new distance to each charge?

    To simplify the resulting expression, use the binomial theorem.
     
    Last edited: Nov 25, 2015
  4. Nov 25, 2015 #3
    I don't know if I'm confused or you are. The displacement is perpendicular to d. So if I build an equation for the force as a function of displacement I get:

    ##F(x) = \frac{-2Qq}{x^2 + \frac{d^2}{4}} \frac{1}{4\pi\epsilon_0}##

    What am I missing here? Am I supposed to treat x and d/2 as vector quantities?

    Here's a photograph of how I've interpreted the problem geometrically.:
     

    Attached Files:

  5. Nov 25, 2015 #4
    You have interpreted the problem correctly .

    No ,you are not supposed to treat x and d/2 as vector quantities . Instead , you need to find the net resultant force in the -x direction (i.e towards equilibrium position) . In other words find the restoring force .

    You can't add the two forces algebraically .
     
  6. Nov 25, 2015 #5
    Ok so if I define one force vector as:
    ##F(x) = \frac{-Qq}{x^2 + \frac{d^2}{4}} \frac{1}{4\pi\epsilon_0}##
    then break it into it's x component
    ##F_x = \frac{-Qq}{x^2 + \frac{d^2}{4}} \frac{1}{4\pi\epsilon_0}cos(tan^{-1}(\frac{d}{2x}))##
    which simplifies to:
    ##F(x) = \frac{-Qq}{x^2 + \frac{d^2}{4}} \frac{1}{4\pi\epsilon_0}\frac{1}{\sqrt{1+\frac{d^2}{4x^2}}}##
    and working some algebra in there:
    ##F(x) = \frac{-4Qq}{4x^2 + d^2} \frac{1}{4\pi\epsilon_0}\frac{2x}{\sqrt{4x^2 + d^2}}##
    more algebra:
    ##F(x) = \frac{-8Qqx}{(4x^2 + d^2)^{5/2}} \frac{1}{4\pi\epsilon_0}##
    Summing the x component doubles the force acting on it:
    ##F(x) = \frac{-16Qqx}{(4x^2 + d^2)^{5/2}} \frac{1}{4\pi\epsilon_0}##

    So that's as far as I can get with algebra on that simplification. Figure now would be a good time to check if I made a mistake.
     
    Last edited: Nov 25, 2015
  7. Nov 25, 2015 #6

    Doc Al

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    Staff: Mentor

    Well, I am often confused, but not here. (When I said " if you move a distance x towards one side" I mean a perpendicular distance.)

    Your equation for the force is almost right. Realize that only the component of the force from each charge in the direction of the displacement will contribute to the net force on the charge q. (This is the same point that Vibhor made.)
     
  8. Nov 25, 2015 #7

    Doc Al

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    Staff: Mentor

    Check that exponent (5/2) in the denominator.
     
  9. Nov 25, 2015 #8
    Awesome. Then for small x I get the equation I wanted at the start. Thanks guys.
     
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