Electromagnetism – Oscillation

Question 3-4: refer to a thin, nonconducting ring of radius R, as shown below, which has a charge Q uniformly spread out on it.

 photo Screenshots_2014-11-06-17-08-30_zpsa12ff792.png

A small particle of mass m and charge –q is placed at point P and released. If R ≫ x, the particle will undergo oscillations along the axis of symmetry with an angular frequency that is equal to:

A. (qQ/4πɛ0mR³)½
B. (qQx/4πɛ0mR⁴)½
C. qQ/4πɛ0mR³
D. qQx/4πɛ0mR
E. (qQx/4πɛ0m)½ [1/(R² + x²)]

(GR9677 #04)

Solution:

Felectric = kqQ/r²
Fcentripetal = mv²/r = mω²r

FFc
kqQ/r² = mω²r
ω² = kqQ/mr³

with
= 1/4πɛ0
r²  = R² + x²
R ≫ x
r² ∼ R²  → r³ ∼ R³

ω = (qQ/4πɛ0mR³)½

Answer: A

Notes: see problem GR9277 #65

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