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Question

(A) Let the negatively charged particle be at $P$ with $OP = Z_0$.The electric field at $P$ due to the ring is $E = \frac{1}{4 \pi \epsilon_0} \frac{Q

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It is periodic for all values of $Z_0$ satisfying $0 < Z_0 < \infty$.

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(A) Let the negatively charged particle be at $P$ with $OP = Z_0$.The electric field at $P$ due to the ring is $E = \frac{1}{4 \pi \epsilon_0} \frac{Q Z_0}{(R^2 + Z_0^2)^{3/2}}$,where $Q$ is the charge on the ring.Since the ring is positively charged,$E$ is always directed away from $O$.Consequently,a negatively charged particle experiences an attractive force towards $O$ and undergoes periodic motion.The equation of motion is $ma = -\frac{q Q Z_0}{4 \pi \epsilon_0 (R^2 + Z_0^2)^{3/2}}$.For $Z_0 \ll R$,the equation becomes $ma \approx -\frac{Q q Z_0}{4 \pi \epsilon_0 R^3}$. Since the acceleration $a$ is proportional to the displacement $Z_0$ and directed towards $O$,the particle undergoes approximate simple harmonic motion $(SHM)$.As the electric field is always directed away from $O$,when the particle $P$ crosses $O$,an attractive force acts on it in the opposite direction,causing it to oscillate rather than move to $Z = -\infty$.

$A$ thin,positively charged metal ring of radius $R$ is fixed in the $X-Y$ plane with its center at the origin $O$. $A$ negatively charged particle $P$ is released from rest at the point $(0, 0, Z_0)$,where $Z_0 > 0$. Describe the motion of the particle.

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