A charge experiencing simple harmonic motion

[PhotonSSBM]

Homework Statement

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:

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

Homework Equations

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

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?

 

[Doc Al]

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.

 

[PhotonSSBM]

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.

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.:

 

[Vibhor]

I don't know if I'm confused or you are. The displacement is perpendicular to d.

You have interpreted the problem correctly .

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?

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 .

 

[PhotonSSBM]

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 .

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.

 

[Doc Al]

I don't know if I'm confused or you are. The displacement is perpendicular to d.

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.)

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?

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.)

 

[Doc Al]

more algebra:
$F(x) = \frac{-8Qqx}{(4x^2 + d^2)^{5/2}} \frac{1}{4\pi\epsilon_0}$

Check that exponent (5/2) in the denominator.

 

[PhotonSSBM]

Awesome. Then for small x I get the equation I wanted at the start. Thanks guys.