A positively charged thin metal ring of radiu | vedclass

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Question

(d) Here $E = \frac{1}{{4\pi {\varepsilon _0}}}.\frac{{Q{z_0}}}{{{{({R^2} + z_0^2)}^{3/2}}}}$
where $Q$ is the charge on ring and ${z_0}$is the dista

Vedclass · Accepted Answer

Both $(a)$ and $(c)$

Vedclass · Answer

(d) Here $E = \frac{1}{{4\pi {\varepsilon _0}}}.\frac{{Q{z_0}}}{{{{({R^2} + z_0^2)}^{3/2}}}}$
where $Q$ is the charge on ring and ${z_0}$is the distance of the point from origin.
Then $F = qE = \frac{{ - \,Qq{z_0}}}{{4\pi {\varepsilon _0}{{({R^2} + z_0^2)}^{3/2}}}}$
When charge $-q$ crosses origin, force is again towards centre i.e., motion is periodic.
Now if ${z_0} < < R$
 $F = - \frac{1}{{4\pi {\varepsilon _0}}}.\frac{{Qq{z_0}}}{{{R^2}}}$ $&rArr;$$F \propto - {z_0}$ i.e., motion is $S.H.M.$

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

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