# Electric charges and SHM

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 doesnt appear in the answer they want.. after all.

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

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

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.

ok lets 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??

Last edited:
stunner5000pt said:
ok lets 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.