A circular ring of radius R with uniform positive charge density lambda per unit length is located i

Q: Why does the particle never return once it crosses the center?

Once the particle crosses the center of the ring, the repulsive electrostatic force changes direction. Instead of opposing the motion, it now pushes the particle further away towards negative infinity.

Q: Why is the potential maximum at the center?

For a positively charged ring, the potential on its axis is given by $V = \frac{kQ}{\sqrt{R^2+x^2}}$. The denominator is smallest when $x=0$, making the potential maximum at the center.

Q: Can we solve this using force and integration?

Yes, you can integrate the variable force $F = qE$ from $x = R\sqrt{3}$ to $x = 0$ to find the work done, which equals the change in kinetic energy. However, using the potential difference is much faster and less prone to calculation errors.

Q: What if the particle had a negative charge?

If the particle were negatively charged, it would be attracted to the ring. It would accelerate towards the center, cross it, slow down, stop, and then return, executing oscillatory motion.

Q: Does the mass of the ring matter here?

No, the mass of the ring is irrelevant because we assume the ring is fixed in place. Only the mass of the projected particle matters for its kinetic energy.

Question

A circular ring of radius $R$ with uniform positive charge density $\lambda$ per unit length is located in the $y$-$z$ plane with its centre at the origin $O$. A particle of mass $m$ and positive charge $q$ is projected from the point $P (R \sqrt{3}, 0, 0)$ on the positive $x$-axis directly towards $O$, with an initial speed $v$. Find the smallest (non-zero) value of the speed $v$ such that the particle does not return to $P$.

Sigma Percentile · Accepted Answer

### The Setup: A Dance of Charges

Imagine a beautifully symmetric scenario: a circular ring of radius $R$ sits perfectly in the $y$-$z$ plane, centered right at the origin $O$. This isn't just any ring; it carries a uniform positive charge density $\lambda$ per unit length.

Now, picture a tiny particle of mass $m$ and positive charge $q$ positioned on the $x$-axis at a specific point $P$, where the distance from the origin is $x = R\sqrt{3}$. We give this particle a sharp push directly towards the center of the ring with an initial speed $v$.

Because both the ring and the particle carry positive charges, they despise each other. As the particle moves closer to the ring, it faces an ever-increasing repulsive electrostatic force. It's like trying to push two identical magnetic poles together. The question is: what is the absolute minimum speed $v$ we need to give the particle so that it never returns to point $P$?

### The Potential Barrier: The Mountain to Climb

To understand the condition for the particle to never return, we must look at the electric potential landscape. The electric potential $V(x)$ on the axis of a uniformly charged ring is given by the elegant formula:

$$V(x) = \frac{1}{4\pi\varepsilon_0} \frac{Q}{\sqrt{R^2 + x^2}}$$

Here, $Q$ is the total charge on the ring. Notice the denominator. As $x$ approaches zero (the center of the ring), the denominator becomes as small as possible, which means the potential $V(x)$ reaches its absolute maximum at the center $O$.

Think of this potential as a physical hill. The center of the ring is the peak of the mountain. If the particle has just enough kinetic energy to reach the peak, it will cross over to the other side ($x < 0$). Once it crosses the center, the repulsive force from the ring will push it *away* towards negative infinity. It will never return to $P$!

So, our goal is simple: give the particle enough kinetic energy to overcome the potential difference between its starting point $P$ and the peak at $O$.

### Calculating the Potential Difference

Let's calculate the exact height of this "potential mountain" relative to our starting point.

First, the potential at the starting point $P$ (where $x = R\sqrt{3}$):

$$V_P = \frac{1}{4\pi\varepsilon_0} \frac{Q}{\sqrt{R^2 + (R\sqrt{3})^2}} = \frac{1}{4\pi\varepsilon_0} \frac{Q}{\sqrt{R^2 + 3R^2}} = \frac{1}{4\pi\varepsilon_0} \frac{Q}{2R}$$

Next, the potential at the center $O$ (where $x = 0$):

$$V_O = \frac{1}{4\pi\varepsilon_0} \frac{Q}{R}$$

The potential barrier the particle must overcome is the difference between these two:

$$\Delta V = V_O - V_P = \frac{1}{4\pi\varepsilon_0} \left( \frac{Q}{R} - \frac{Q}{2R} ight) = \frac{1}{4\pi\varepsilon_0} \frac{Q}{2R}$$

### Conservation of Energy: The Master Key

In the absence of any non-conservative forces like friction, the total mechanical energy of the particle is conserved. The loss in kinetic energy as it slows down must exactly equal the gain in electrostatic potential energy.

To find the minimum initial speed $v_{	ext{min}}$, we assume the particle just barely makes it to the center, meaning its final kinetic energy at $O$ is zero.

$$\frac{1}{2}mv_{	ext{min}}^2 = q \Delta V$$

Substitute our expression for $\Delta V$:

$$\frac{1}{2}mv_{	ext{min}}^2 = q \left( \frac{1}{4\pi\varepsilon_0} \frac{Q}{2R} ight)$$

### The Final Elegance

We are almost there! The problem gives us the linear charge density $\lambda$, not the total charge $Q$. But we know that the total charge is simply the charge density multiplied by the circumference of the ring:

$$Q = \lambda (2\pi R)$$

Let's plug this into our energy equation:

$$\frac{1}{2}mv_{	ext{min}}^2 = q \left( \frac{1}{4\pi\varepsilon_0} \frac{\lambda (2\pi R)}{2R} ight)$$

Watch how beautifully the terms cancel out. The $2\pi R$ in the numerator cancels with parts of the denominator:

$$\frac{1}{2}mv_{	ext{min}}^2 = \frac{q\lambda}{4\varepsilon_0}$$

Multiply both sides by 2 and divide by $m$:

$$v_{	ext{min}}^2 = \frac{q\lambda}{2\varepsilon_0 m}$$

Taking the square root gives us our final, elegant answer:

$$v_{	ext{min}} = \sqrt{\frac{q\lambda}{2\varepsilon_0 m}}$$

This is the exact minimum speed required to conquer the electrostatic mountain and ensure the particle never looks back!

Visualized Solution

\text{Visualizing the Setup}

\text{Condition to Not Return}

\text{Potential at } P \text{ and } O

\text{Potential Difference}

\text{Conservation of Energy}

\text{Final Calculation}

\text{The Way Forward}

The Sigma Insight: Electric Potential and Potential Difference

The Setup

The Potential Barrier

Calculating the Potential Difference

Conservation of Energy

The Final Elegance

Similar Questions

Two thin wire rings each having a radius $R$ are placed at a distance $d$ apart with their axes coinciding. The charges on the two rings are $+ q$ and $- q$ . The potential difference between the centres of the two rings is

Two identical thin rings, each of radius $R$ , are coaxially placed a distance $R$ apart. If $Q_{1}$ and $Q_{2}$ are respectively the charges uniformly spread on the two rings, the work done in moving a charge $q$ from the centre of one ring to that of the other is

A charge $Q$ is distributed over two concentric hollow spheres of radii $r$ and $R$ ( $> r$ ) such that the surface densities are equal. Find the potential at the common centre.

A thin spherical conducting shell of radius $R$ has a charge $q$ . Another charge $Q$ is placed at the centre of the shell. The electrostatic potential at a point $P$ at a distance $R /2$ from the centre of the shell is

A charge $Q$ is distributed over three concentric spherical shells of radii $a, b, c$ ( $a < b < c$ ) such that their surface charge densities are equal to one another.\nThe total potential at a point at distance $r$ from their common centre, where $r < a$ would be