Simple Harmonic Motion 3 Question 12

12. A thin fixed ring of radius 1m has a positive charge 1×105C uniformly distributed over it. A particle of mass 0.9g and having a negative charge of 1×106C is placed on the axis at a distance of 1cm from the centre of the ring. Show that the motion of the negatively charged particle is approximately simple harmonic. Calculate the time period of oscillations.

(1982, 5M)

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Answer:

Correct Answer: 12. 0.628s

Solution:

  1. Given, Q=105C

q=106C,R=1m and m=9×104kg

Electric field at a distance x from the centre on the axis of a ring is given by

E=14πε0Qx(R2+x2)3/2

If x«R,R2+x2R2 and EQx4πε0R3

Net force on negatively charged particle would be qE and towards the centre of ring. Hence, we can write

F=Qqx(4πε0)R3 or  acceleration a=Fm=Qqx(4πε0)mR3

as ax, motion of the particle is simple harmonic in nature. Time period of which will be given by

T=2π|xa| or T=2π(4πε0)mR3Qq

Substituting the values, we get

T=2π(9×104)(1)3(9×109)(105)(106)=0.628s



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