A positively charged thin metal ring of radius R is fixed in the xy plane with its centre at the origin O. A negatively charged particle P is released from rest at the point `(0, 0, z_0)` where `z_0gt0`. Then the motion of P is

Options

A. pariodic for all values of `z_(0)` satisfying `0 lt z_(0) lt oo`
B. simple harmonic for all values of `z_(0)` satisfying `0 lt z_(0) le R`
C. approximately simple harmonic provided `z_(0) lt lt R`
D. such that P crosses O and continues to move along negative z-axis towards `z=-oo`

Answer

A, C

Explanation

Let Q be the charge on the ring, the negative charge -q is released from point `P (0, 0, z_(0))`. The electric field at P due to the charged ring will be along positive z-axis and its magnitude will be
`E=1/(4piepsilon_(0))(Qz_(0))/((R^(2)+z_(0)^(2))^(3//2))`
`E=0` at centre of the ring because `z_(0)=0`
Therefore, force on charge P will be towards centre as shown, and its magnitude is
`F_(e)=qE=1/(4piepsilon_(0))(Qq)/((R^(2)+z_(0)^(2))^(3//2)).z_(0)` ...(i)

Similarly, when it courses the origin, the force is again towards centre O.
Thus, the motion of the particle is periodic for all values of `z_(0)` lying between 0 and `oo`.
Secondly, if `z_(0) lt lt R, (R^(2)+z_(0)^(2))^(3//2)~~R^(3)`
`z_(0) lt lt R, (R^(2)+z_(0)^(2))^(3//2)~~R^(3)`
`Fe~~-1/(4piepsilon_(0)).(Qq)/R^(3).z_(0)` (From Eq. 1)
i.e., the restoring force `F_(e) prop -z_(0)`. Hence, the motion of the particle will be simple harmonic. (Here negative sign implies that the force is towards its mean position).