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I am doing this homework question on Mastery Physics and I do not understand why this field would make a particle have Simple Harmonic Motion behaviors. Here is the question:

Imagine a small metal ball of mass m and negative charge $−q_0$. The ball is released from rest at the point $(0,0,d)$ and constrained to move along the $z$ axis, with no damping. If $0 < d ≪a$, what will be the ball's subsequent trajectory?

This is the Picture of the Diagram

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  • $\begingroup$ Can you clarify why you don't understand? Do you know how to apply Coulomb's law? Are you having trouble with the vector math? Are you confused about the equation of motion? How to recognize whether or not it supports SHO? $\endgroup$
    – Brionius
    Commented Jan 9, 2016 at 21:32

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If the ring is negatively charged, then it will repel the ball and you won't have SHM - the ball will just fly off.

If on the other hand, the ring is positively charged then the force is always attractive and towards the center of the ring (horizontal components cancel). If the ball is above, then the ring pulls it down and if the ball is below then the ring pulls it up. This is acting exactly like a restoring force, so you will end up with oscillatory behavior.

To definitely be SHM you need to show that $F = -kz$ is true, so I'd start by trying to find the electric force $F$ as a function of $z$. Hyperphysics can help you out, although you may want to apply the approximation that $d << a$ to get SHM.

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    $\begingroup$ Without a detailed calculation it is obvious from symmetry that the force is zero in the plane of the ring, and non-zero (and towards the center) when it's not. For small enough displacements this must mean a linear term and voila! SHM appears. $\endgroup$
    – Floris
    Commented Jan 9, 2016 at 22:29
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Electric field at a height $z$ on the axis of a ring carrying charge $Q$ uniformly distributed on it can be shown to be $$ {E(0,0,z)}=\frac{1}{4\pi\epsilon_0}\frac{Qz}{(a^2+z^2)^{3/2}} \hat{z}$$ if $d<<a$ the in the denominator $z$ can be neglected and force on the charge at a height $z$ will be $$F(0,0,z)=-\frac{1}{4\pi\epsilon_0}\frac{q_0Qz}{a^3}\hat{z}=-kz\hat{z}$$ Or $$m\ddot{z}+kz=0$$ Where $k=\frac{1}{4\pi\epsilon_0}\frac{q_0Q}{a^3}$ and frequency of oscillation will be $\omega=\sqrt{\frac{k}{m}}$.

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  • $\begingroup$ To the extent that the question is about how to crunch through the math it would be subject to our policy about homework-like questions, where as the conceptual aspect would be a good question for the site from the beginning. None of which is your problem except that the answer you've written here answers the former (off-topic) interpretation rather than the latter (on-topic) interpretation. $\endgroup$ Commented Jan 9, 2016 at 23:37

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