A uniformly charged thin ring has radius `10.0 cm` and total charge `+12.0muC`. An electron is placed on the ring's axis a distance `25.0 cm` from the centre of the ring and is constrained to stay on the axis of the ring. The electron is then released from rest.
a. describe the subsequent motion of the electron
b. find the speed of the electron when it reaches the centre of the ring

Answer

A, D

Explanation

a. The eletron will be attracted towards the centre of C of the ring. At C net force is zero, but on reaching C electron has some kinetic enegy and due to inertia it crosses C, but on the other side it is further attracted towards C. Hence, motion of electron is oscillatory about point C.
b. As the electron approaches C,its speed (hence, kinetic energy) increases due to force of attraction towards the centre C. this increase in knetic energy is at the cost of electrostatic potential energy. Thus,
`1/2mv^2=U_i-U_f`
`=U_p-U_c=(-e)[V_P-V_C]` .........i
Here, V is the potential due to ring
`V_P=1/(4piepsilon_0).q/r` (q=charge on ring)
`=((9xx10^9)(12xx10^-9))/(sqrt((10)^2+(25)^2)xx10^-2)=401V`
`V_C=1/(4piepsilon_0).q/R`
`=((9xx10^9)(12xx10^-9))/(10xx10^-2)=1080V`
Substituting the proper values inn eqn i we have
`1/2xx9.1xx10^-31xxv^2=(-1.6xx10^-19)(401-1080)`
`=v=15.45xx10^6m//s`