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Question

Let the negative charged particle is at $P$. Thus $O P=z_0$

The electric field is at $P$ due to ring is $E=\frac{1}{4 \pi \epsilon_0} \frac{Q z_0}{

Vedclass · Accepted Answer

$0 < Z_0 < $$\infty$ શરતને સંતોષે તેવી $Z_0$ ની બધી જ કિમતો માટે આવર્તીંય છે.

Vedclass · Answer

Let the negative charged particle is at $P$. Thus $O P=z_0$

The electric field is at $P$ due to ring is $E=\frac{1}{4 \pi \epsilon_0} \frac{Q z_0}{\left(R^2+z_0^2\right)^{3 / 2}}$ where $Q$ is the charge on the ring.

As the ring is positively charged so $E$ is always directed away from $O$.

Hence a negatively charged particle is accelerated towards $O$ and undergoes periodic motion.

i.e $m a=-\frac{ q }{4 \pi \epsilon_0} \frac{ Qz _0}{\left( R ^2+ z _0^2\right)^{3 / 2}}$

For $z_0 \ll R, m a=-\frac{Q q z_0}{4 \pi \epsilon_0 R^3}$. Thus the acceleration $(a)$ is proportional to the $z-$ coordinate, and the particle undergoes approximate SHM. As $E$ is directed always away from $O$ so when particle $P$ will cross $O$, again some attractive force will act on it and hence it will not continue to move along negative $z$-axis towards $z=-\infty$.

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