ElectrostaticsHard
Question
A positively charged thin metal ring of radius R is fixed in the x-y plane with its centre at the origin. A negatively charged particle P is released from the rest at the point (0, 0 z0) where z0 > 0. Then the motion of P is
Options
A.periodic for all values of z0 satisfying 0 < z0 < ∞
B.simple harmonic for all values of z0 satisfying 0 < z0 ≤ R
C.approximately simple harmonic provoded z0 << R
D.such that P crosses O and continues to move along the negative z-axis towards z = - ∞
Solution
Let Q be the charge on the ring, the negative charge - q is released from point P(0,0,z0).
The electric field at P due to the charged ring will be along positive z-axis and its magnitude will be
E = 0 at centre of the ring because z0 = 0
Forece on charge at P will be towards centre as shown, and its magnitude is
......(i)Similarly, when it crosses the origin, the force ia again towards centre O.
Thus, the motion of the particle is periodic for all values of z0 lying between 0 and ∞
Secondly, if z0 << R, (R2 + z02)3/2
R3 
[From Eq. (i)] i.e., the restoring force Fe ∝ - z0. Hence, the motion of the particle will be simple harmonic. (Here negative sign implies that the force is towards its mean position.)
Create a free account to view solution
View Solution FreeMore Electrostatics Questions
The figure shows some of the electric field lines corresponding to an electric field. The figure suggests :-...In x - y coordinate system a particle P of charge -q and mass m enters at x = 0 as shown in figure. In this region poten...The electric field due to a uniformly charged non-conducting sphere of radius R as a function of the distance from its c...The electric field in a certain region is acting radially outward and is given by E = Ar. A charge contained in a sheper...Two infinite plane, parallel sheets, separated by a distance d have equal and opposite uniform charge densities s. Eletr...