Proving SHM in a System of Electric Charges

[stunner5000pt]

Two postitive charges +Q are held at fixed distance d from each other. A particle of charge -q is placed at the mid point between them and given a small displacement y, on the line perpendicular to the line joining them and released. Show that hte particle described SHM of period $$(\frac{\epsilon_{0} m \pi^3 d^3}{qQ})^\frac{1}{2}$$

i drew a figure of what this might look like.

I was wondering if i had to prove that this particle is going to undergo SHM. I can easily say that at points +y and -y the force on the particle is the maximum and is $$\pm \frac{Qqd}{4 \pi \epsilon_{0} tan^2 \theta}$$

would i then simply plug the acceleration of this force into th4e SHM equation??

But how would i get rid of the tan theta?? I doesn't appear in the answer they want.. after all.

 

[JoAuSc]

Can't you just replace tan(theta) with (d/2)/y?

 

[stunner5000pt]

that would produce a y in the final answer. Look at hteh answer i want. I doesn't have any Ys involved in it. Thats why tan theta cannot be replaced like that

 

[JoAuSc]

The force from each charge has two components. Using symmetry arguments you can see the vertical components cancel out, and that the horizontal components (the y-components) of the two forces from the two charges add up. Let psi be the angle between the diagonal and the vertical axis (it seems to be a more convenient angle to work from than theta).

To find the y-components of the forces, we need to make two approximations: 1.) because the distance y is small, the change in the distance r between q and each Q will not change much from d/2; 2.) the force component in the y-direction is F*sin(psi), and for small angles sin(psi) = psi. (Zoom in on the portion of the sin(x) vs. x curve around the origin and you'll see it resembles a line with slope 1 crossing the origin). So F*sin(psi) ~ F*psi. Don't forget to change psi to the distance y that it corresponds to.

The equation for simple harmonic motion is

m*a - (restoring force) = 0

where the restoring force is proportional to x (else it wouldn't be "simple"). If you do the algebra with F*psi you'll find you get some expression that's proportional to y. Let that expression be k. According to formulas about SHM, the angular velocity omega is equal to sqrt(m/k), and the period is equal to 2pi/omega. Do the algebra, and you should end up with the right answer.

 

[stunner5000pt]

ok let's see

force on each end of the Y axis (horizontal as in the diagram) would be due to both charges

$$F = \frac{2}{4 \pi \epsilon_{0}} \frac{Qq}{y^2}$$

$$F = \frac{1}{2 \pi \epsilon_{0}} \frac{Qq}{y^2}$$

the acceleration due to this forcei s
$$a = \frac{Qq}{2 \pi \epsilon_{0} y^2 m}$$

plugging into $$ma - ky = 0$$ where k is restoring force

$$k = \frac{Qq}{2 \pi \epsilon_{0} y^3}$$

$$T = 2 \pi \sqrt{\frac{m}{k}}$$

$$T = 2 \pi \sqrt{\frac{2 \pi \epsilon_{0} y^3 m}{Qq}}$$

$$T = \sqrt{\frac{8 \epsilon_{0} m \pi^3 y^3}{Qq}}$$
now $$y = \frac{d}{2} \tan{\theta}$$

$$T = \sqrt{\frac{\epsilon_{0} m \pi^3 d^3 \tan^3{\psi}}{Qq}}}$$

how do i eliminate the tan psi?? do i use the assumption that tan psi = 1 for small angles??

 

[JoAuSc]

ok let's see

force on each end of the Y axis (horizontal as in the diagram) would be due to both charges

$$F = \frac{2}{4 \pi \epsilon_{0}} \frac{Qq}{y^2}$$

Actually, it should be divided by r^2 = y^2 + (d/2)^2 ~ (d/2)^2.