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A positively charged thin metal ring of ...

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

A

periodic 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 contiude to move along the negative z-axis towards `z = - oo`

Text Solution

Verified by Experts

The correct Answer is:
A, C
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)/(4piepsi_(0)) (qz_(0))/((R^(2)+z_(0)^(2))^(3//2))`
E=0 at centre of the ring because `z_(0)=0`. Force on charger at P will be towards centre as shown, & its magnitude is
`F_(e)=qE= (1)/(4piepsi_(0). (Qq)/((R^(2)+z_(0)^(2))^(3//2)) z_(0) ........(i)`
Similarly, when it crosses 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)`
`F_(e)=1/(4piepsi_(0)) . (Qq)/(R^(3)).z_(0)` [From Eq. (i)]
i.e., the restoring force `F_(e) alpha -z_(0)`. Hence, the motion of the particle will be simple harmonic. (Here negative sign implies that the force is towards its mean position.)
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