A stationary ring of radius 8.00 cm lies in the yz plane and has a uniform posit

Question

A stationary ring of radius 8.00 cm lies in the yz plane and has a uniform positive charge Q = 5.9 µC. A small particle that has mass m and a negative charge q = -5.9 µC is located at the center of the ring. When the particle is given a small displacement in the x direction, it oscillates about its equilibrium position at a frequency of 3.10 Hz.

(a) What is the particle's mass?
..... kg

(b) What is the frequency if the radius of the ring is doubled to 16.0 cm and all other parameters remain unchanged?
..... Hz

Explanation / Answer

the particle mass is

m = kqQ/ 4 pi^2 f^2 a^3

= (9 * 10 ^9 ) (5.9 * 10 ^-6 )^2/ 4 pi ^2 ( 3.10 Hz)^2 ( 0.08 m)^3

=1.61 kg

(b)

f = root 2kqQ/ ma^3

f' =  root 2kqQ/ m(2a)^3

f'/f = 1/ root 8 = f/ root 8 = 3.10/ root 8 = 1.096 Hz

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