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A ring of radius R carries a uniformly d...

A ring of radius R carries a uniformly distributed charge +Q. A point charge `-q` is placed on the axis of the ring at a distance 2R from the centre of the ring and released from rest. The particle executes a ismple harmonic motion along the axis of the ring.

Text Solution

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KEY CONCEPT: Force on charge `(-q)` due to small charge `dq` istuated at length `dl` is given by

`dF=k(qdq)/(5R^2)`
Resolving this force into two parts `dFcos theta` and `dFisntheta` as shown in figure.
If we take another diametrically oppoiste lenth `dl`, the charge on it being `dq`. Then the force on charge `(-q)` by this small charge `dq` will be
`dF=k(qdq)/(5R^2)`
Again resolving this force, we find `dFisn theta` components of the two forces cancel out and `dFcos theta` component adds up.
`:.` The total force
`F=int_0^(2piR)dF cos theta=int_0^(2piR)(kqdq)/(5R^2)xx(2R)/(sqrt5R)`
charge on length `2piR=Q`
`:.` Charge on lenth `dl=(Qdl)/(2piR)=dq`
`:.` `F=int_0^(2piR)(2kq)/(5sqrt5R^2)xx(Qdl)/(2piR)`
`=(2kQq)/(5sqrt5xx2piR^3)xx2piR=(2kQq)/(5sqrt5R^2)`
This is not an equation of ismple harmonic motion.
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