A positively charged thin metal ring of radius R is fixed in the x-y plane with its centre at the or

Q: Why doesn't the particle go to $-\infty$?

Because the force is always attractive towards the ring. Once it crosses the origin and moves to the negative $z$-axis, the electric field points downwards, so the force on the negative charge points upwards, pulling it back to the origin.

Q: Why is the motion not SHM for all values of $z_0$?

For Simple Harmonic Motion, the restoring force must be strictly proportional to the displacement ($F \propto -x$). Here, the force has a complex dependence on $z_0$ due to the $(R^2 + z_0^2)^{3/2}$ term in the denominator. It only becomes linear when $z_0$ is very small.

Q: What happens at the exact center of the ring?

At the exact center ($z_0 = 0$), the electric fields from all parts of the ring cancel out perfectly, making the net electric field and the net force zero. This is the equilibrium position.

Q: How do we know the force is restoring?

A restoring force always points towards the equilibrium position. Since the ring is positively charged and the particle is negatively charged, they attract each other, constantly pulling the particle towards the center of the ring.

Q: What is the time period of the SHM for small $z_0$?

By comparing the restoring force $F = -k z_0$ with the standard SHM equation $F = -m \omega^2 z_0$, we can find the angular frequency $\omega = \sqrt{\frac{Qq}{4\pi\epsilon_0 m R^3}}$. The time period is then $T = \frac{2\pi}{\omega}$.

Question

A positively charged thin metal ring of radius $R$ is fixed in the $x$-$y$ plane with its centre at the origin $O$. 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

Options:

(A) periodic for all values of $z_0$ satisfying $0 < z_0 < \infty$

(B) simple harmonic for all values of $z_0$ satisfying $0 < z_0 \le R$

(C) approximately simple harmonic provided $z_0 \ll R$

(D) such that $P$ crosses $O$ and continues to move along the negative $z$-axis towards $z = -\infty$

Sigma Percentile · Accepted Answer

### Visualizing the Setup
Imagine a positively charged metal ring lying perfectly flat in the $x$-$y$ plane. Right above its center, along the $z$-axis, we place a negatively charged particle at a distance $z_0$. This setup is a classic example of electrostatic interaction where symmetry plays a beautiful role.

### The Electric Field on the Axis
Because the ring is uniformly charged, the horizontal components of the electric field produced by opposite segments of the ring cancel each other out perfectly. The net electric field $\vec{E}$ points straight along the $z$-axis, away from the center of the ring.

Mathematically, the magnitude of this electric field at a distance $z_0$ is given by the standard formula:

$$E = \frac{1}{4\pi\epsilon_0} \frac{Q z_0}{(R^2 + z_0^2)^{3/2}}$$

### The Restoring Force
Since our particle has a negative charge ($-q$), the electrostatic force acting on it will be opposite to the direction of the electric field. The force pulls the particle downwards, directly towards the origin $O$. The magnitude of this force is:

$$F_e = \frac{1}{4\pi\epsilon_0} \frac{Qq z_0}{(R^2 + z_0^2)^{3/2}}$$

Because this force is always directed towards the center, the particle will accelerate downwards, cross the origin, slow down on the negative $z$-axis, and then return. This means the motion is strictly **periodic** for any starting position $z_0 > 0$. It will never escape to $-\infty$.

### The Small Displacement Approximation
But what if the particle is released from very close to the center? If $z_0 \ll R$, the value of $z_0^2$ becomes negligibly small compared to $R^2$. We can approximate the denominator:

$$(R^2 + z_0^2)^{3/2} \approx (R^2)^{3/2} = R^3$$

Substituting this back into our force equation, we get a beautifully simplified expression:

$$F_e \approx \frac{1}{4\pi\epsilon_0} \frac{Qq}{R^3} z_0$$

### Simple Harmonic Motion
Look closely at the simplified force equation. The restoring force is directly proportional to the displacement $z_0$. In physics, whenever a restoring force is proportional to the negative of the displacement ($F \propto -x$), the resulting motion is **Simple Harmonic Motion (SHM)**.

Therefore, for small distances ($z_0 \ll R$), the particle executes SHM. For larger distances, the motion remains periodic but is no longer simple harmonic due to the complex $(R^2 + z_0^2)^{3/2}$ term in the denominator.

**Conclusion:** The motion is periodic for all $z_0 > 0$, and approximately simple harmonic for $z_0 \ll R$.

Animated Solution for Physics - Electrostatics: A positively charged thin metal ring of radius $R$ is fixed in the $x$ - $y$ plane with its centre at the origin $O$ . 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

Select Answer:

Visualized Solution

$Visualizing the Setup$

$Electric Field on the Axis$

$Expression for Electric Field$

$Force on the Particle$

$Nature of Motion$

$Approximation for Small z_{0}$

$Simple Harmonic Motion$

$Conclusion$

The Sigma Insight: Electric Field

Visualizing the Setup Imagine a positively charged metal ring lying perfectly flat in the $x$ - $y$ plane

The Electric Field on the Axis Because the ring is uniformly charged, the horizontal components of the electric field produced by opposite segments of the ring cancel each other out perfectly

The Restoring Force Since our particle has a negative charge ( $- q$ ), the electrostatic force acting on it will be opposite to the direction of the electric field

The Small Displacement Approximation But what if the particle is released from very close to the center? If $z_{0} ≪ R$ , the value of $z_{0}^{2}$ becomes negligibly small compared to $R^{2}$

Simple Harmonic Motion Look closely at the simplified force equation

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Electric Field on the Axis

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Expression for Electric Field

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Force on the Particle

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Nature of Motion

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Approximation for Small z0​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Simple Harmonic Motion

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conclusion