Question

A negatively charged particle -q is placed at the center of a uniformly charged ring, where the ring has a total positive charge Q as shown in Example 19.5. The particle, confined to move along the x axis, is displaced a small distance x along the axis (x is much smaller than a) and released. Show that the particle oscillates in simple harmonic and calculate its frequency.

Answer 1

The solution would be like this for this specific problem:

F=−k∗x∗q∗Q/(+)F−≈k∗x∗q∗Q/R3[(1−3/2*x2/R3]
F=−k∗x∗q∗/QR3
F=ma
−k∗q∗Q/R3*x=ma
−k∗x=m∗a
a==ω2x
ω=(k/m)1/2

ω=(kqQ/R3)1/2

I am hoping that this answer has satisfied your query and it will be able to help you in your endeavor, and if you would like, feel free to ask another question.