A particle of mass M charged minus Q is constrained to move along the axis of a of a ring of radius a the ring carries a uniform charge density + Lambda along its circumference initially the particle lies in the Plane of the ring at a point where no net force acts on it the period of oxidation of the particle when it is displaced slightly from its equilibrium position is

eA particle of mass m and charge -Q is constrained to move along the axis of a ring of radius R .The ring carries a uniform charge density +λ+λ along its length . Initially the particle is in the centre of the ring . Now it is displaced slightly along the axis of ring it`s time period of oscillation is(a)2π∈0mR2λQ−−−−−−−√(b)π2∈0mR2λQ−−−−−−−−√(c)2π2∈0mR2λQ−−−−−−−−√(d)4π2∈0mR2λQ−−−−−−−−√(a)2π∈0mR2λQ(b)π2∈0mR2λQ(c)2π2∈0mR2λQ(d)4π2∈0mR2λQjeemain physics class12 ch2 electrostatics electrostatic-potential-and-capacitance electrostatics-of-conductors difficultflag Answer comment Clay6 Question Share Shareasked Feb 18, 2014 by yamini.v recategorized Aug 22, 20141 AnswerAnswer : (c) 2π2∈0mR2λQ−−−−−−−−√2π2∈0mR2λQExplanation :Electric potential on the axis of ring at a distance x is =kq(x2+R2)12kq(x2+R2)12and Electric field on the axis of ring is Ex=−∂V∂xi^Ex=−∂V∂xi^Ex=−(−12kq2x(x2R2)32)Ex=−(−12kq2x(x2R2)32)Ex=kqx(x2+R2)32=λRx2∈0(x2+R2)32Ex=kqx(x2+R2)32=λRx2∈0(x2+R2)32Restoring force Fres=−QExFres=−QEx=−QλRx2∈0(x2+R2)32=−QλRx2∈0(x2+R2)32Since x