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Question

Two charges each of magnitude Q are fixed apart at a distance of '2a'. A third charge (−q of mass ′m′) is placed at the midpoint of the two charges. Now if −q charge is slightly displaced perpendicular to the line joining the charges then, find its time period.

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Solution

For a small displacement x perpendicular to the line joining the chargesthe potential energy , U ( r ) = Qq4πϵ0r , of the displaced charge is now :U(x)=2×−Qq4πϵ0√a2+x2The force vector acting on the particle is in the negative derivative w.r.t x of this conservative potential function. Form symmetry we know that it will act "perpendicular to the line joining the charges".→F = - dudx = Qq2πϵ0x(a2+x2)12By taylor expansion, for small x we have: x(a2+x2)32 = xa3(1+x2a2)−32In other words, →F linearises as:→F = −Qq2a3πϵ0x , and it is this linearise restorative force that characterises the harmonic oscillaator.i.e. x+Qq2πma3ϵox=0 can be written in the standrad SHM wave form x+ω2x=0Therefore; T=2πw = 2π√2ma3πϵ0Qq

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