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.
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:
(GR9677 #04)
Felectric = kqQ/r²
Fcentripetal = mv²/r = mω²r
Fe = Fc
kqQ/r² = mω²r
ω² = kqQ/mr³
with
k = 1/4πɛ0
r² = R² + x²
R ≫ x
r² ∼ R² → r³ ∼ R³
ω = √(qQ/4πɛ0mR3)
Answer: A
Notes: see problem GR9277 #65
No comments :
Post a Comment