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A positively charged thin metal ring of radius R is fixed in the xy- plane with its centre at the o. A negatively charged particle P is released from rest at the point  (0,0,z_{0}) , where. z_{0}>0 Then the motion of P is 

  • Option 1)

    Periodic for all values of  z_{0} satisfying  0<z_{0}<\infty

     

     

     

  • Option 2)

    Simple harmonic for all values of satisfying  0<z_{0}<R

  • Option 3)

    Approximately simple harmonic provided  z_{0} \ll R

  • Option 4)

    Both a and c

 

Answers (1)

best_answer

As we learned

 

If x>>R -

E=\frac{kQ}{x^{2}}  ,  V=\frac{kQ}{x}

-

 

 Here  E=\frac{1}{4\pi \varepsilon _{0}}\cdot \frac{Qz_{0}}{(R^{2}+z_{0}^{2})^{3/2}}

where Q is the charge on ring and z_{0} is the distance of the point from origin.

Then F=qE=\frac{-Qqz_{0}}{4\pi \varepsilon _{0}(R^{2}+z_{0}^{2})^{3/2}}

When charge – q crosses origin, force is again towards centre i.e., motion is periodic.

Now if z_{0}\ll R

\therefore F=-\frac{1}{4\pi \varepsilon _{0}}\cdot \frac{Qqz_{0}}{R^{2}}\Rightarrow F\alpha -z  i.e., motion is S.H.M.


Option 1)

Periodic for all values of  z_{0} satisfying  0<z_{0}<\infty

 

 

 

Option 2)

Simple harmonic for all values of satisfying  0<z_{0}<R

Option 3)

Approximately simple harmonic provided  z_{0} \ll R

Option 4)

Both a and c

Posted by

Avinash

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