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I have to prove that a negative charge will execute an harmonic motion when placed in the line that separates two positive charges $Q$. (The negative charge $-q$ will not be placed on the center).
I clearly understand that it will be an harmonic motion, but i can't show it mathematically: i've tried the traditional way:
Let the two positive charges be $Q_1$ and $Q_2$, and the position of the $-q$ charge is at an $X$ distance from the center:
$$F_1 - F_2 = ma$$ Where $$F_1 = \frac{kQ_1q}{(d/2 -x)^2}$$ And $$F_2 = \frac{kQ_2q}{(d/2 + x)^2}$$ The problem is that im not being able to change this equation in order to get the traditional simple harmonic motion differential equation.
Any help would be great.
Thanks!

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  • $\begingroup$ This is a really good question. No idea why this was closed... $\endgroup$
    – Soham
    Commented Apr 16, 2018 at 11:00

1 Answer 1

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The motion cannot be along the line joining the +ve charges because such motion is unstable : the -ve charge would accelerate away from the centre point and towards the nearest +ve charge.

I think it must be assumed that the -ve charge moves along the perpendicular bisector of the two +ve charges, and is constrained to remain on this line.It must be constrained because this motion is again unstable : the smallest deviation either side will result in acceleration towards one of the charges.

Suppose the two +ve charges are $2d$ apart and the -ve charge is distance $y$ from their midpoint along the perpendicular bisector. I am using $y$ to distinguish transverse displacement from longitudinal displacement $x$ along the line joining the +ve charges.

Then the distance $r$ between the -ve and +ve charges is given by $r^2=d^2+y^2$ and the force of attraction to either charge is $kQq/r^2$. The total of the components of these forces along the bisector is the restoring force
$F=2(kQq/r^2)(y/r)=2kQq(y/r^3)$.

For small values of $y<<d$ we have $r^3 \approx d^3$ so the equation of motion is :
$m\ddot y = -F = -2kQq(y/d^3)$
$\ddot y + \omega^2 y = 0$
where $\omega^2=2kQq/md^3$.

The oscillation is only harmonic for small displacements $y<<d$.

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  • $\begingroup$ Stability is a mathematical concept - can you demonstrate that it is unstable mathematically? "Because I say so" works sometimes, but math always has the final word. (I'm trying to but I'm rusty). Need to integrate over $(\frac{d}{2} \pm x) \ / \ \lvert \frac{d}{2} \pm x \rvert^3 $. $\endgroup$
    – uhoh
    Commented Aug 11, 2016 at 5:59
  • $\begingroup$ @uhoh : I don't think mathematics is necessary to recognise instability here. In this scenario if the -ve charge is closer to one of the +ve charges than the other, the net force on it will be away from the central position, and will increase as that asymmetry increases. $\endgroup$ Commented Aug 11, 2016 at 10:50
  • $\begingroup$ If they are supposed to be co-linear in the problem, could the negative charge just pass through the other charge, then turn around and pass through again? The problem doesn't say "proton" or "ion", just charges Q and -q, which are purely mathematical constructs. Anyway I have a hunch your (and @knzhou 's) interpretation might be right - the problem might assume the initial position is displaced perpendicularly to the line connecting the other two. $\endgroup$
    – uhoh
    Commented Aug 11, 2016 at 11:00
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    $\begingroup$ @uhoh : That's an interesting suggestion. I don't know how that would work mathematically, passing through a point of infinite potential energy $-kQq/r$ as $r \to 0$. If it does work I think it would be difficult to calculate the period. If all 3 charges were +ve then co-linear oscillation is possible, but again unstable against lateral perturbations. $\endgroup$ Commented Aug 11, 2016 at 11:10
  • $\begingroup$ The question is good, this is what a solution looks like from Scholarpedia. $\endgroup$
    – uhoh
    Commented Aug 24, 2016 at 10:11

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