A thin fixed of radius 1 metre has a positive charge `1xx10^-5` coulomb uniformly distributed over it. A particle of mass 0.9 gm and having a negative charge of `1xx10^-6` coulomb 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 approaximately simple harmonic. Calculate the time period of oscillations.

Answer

B, C

Explanation

KEY CONCEPT: The electric field due a uniformly charged ring of radius r at a point distant x from its center on its axis is given by
`E=k(Qx)/((r^2+x^2)^(3//2))`

`r=1m`
`Q=10^-5C`
mass of particle `m=0.9xx10^-3kg`
charge on particle `q=-10^-6C`
`:.` Force on the negative charge q will be `F=qE`
`:.` `F=(-kQq)/((r^2+x^2)^(3//2))xx x` or, `mA=(-kQq)/((r^2+x^2)^(3//2))xx x`
or, `A=-k(Qq)/(m(r^2+x^2)^(3//2))xxx`
For `x lt lt r` `A=-(kQq)/(r^3)xxx`
`implies` The motion is simple harmonic in nature.
Comparing the above equation with `A=-omega^2x` we get
`:.` `omega^2=(kQq)/(mr^3)` or `omega=sqrt((kQq)/(mr^3))`
`:.` `(2pi)/(T)=sqrt((kQq)/(mr^3))impliesT=2pisqrt((mr^3)/(kQq))`
`T=2xx3.14[(0.19xx10^-3xx1^3)/(9xx10^9xx10^-5xx10^-6)]^(1//2)`
`=6.28[0.01]^(1//2)=6.28[0.1]`
`T=0.628sec`