A charge ` +Q` is uniformly distributed over a thin ring with radius ` R`. A negative point charge `-Q` and mass `m` starts from rest at a point far away from the centre of the ring and moves towards the centre. Find the velocity of this particle at the moment it passes through the centre of the ring .

Explanation

To find the velocity of the negative point charge `-Q` as it passes through the center of the ring, we can use the principle of conservation of energy. Here’s the step-by-step solution:

### Step 1: Understand the Initial and Final States
Initially, the charge `-Q` is at rest at a point far away from the ring, where the potential energy is considered to be zero. As it moves towards the center of the ring, it will gain kinetic energy.

### Step 2: Calculate the Potential Energy at Infinity
At a point far away from the ring, the potential energy (PE) of the charge `-Q` is zero:
\[ PE_{\text{initial}} = 0 \]

### Step 3: Calculate the Potential Energy at the Center of the Ring
When the charge `-Q` is at the center of the ring, it experiences a potential due to the uniformly distributed charge `+Q`. The potential \( V \) at the center of the ring due to a ring of charge is given by:
\[ V = \frac{kQ}{R} \]
where \( k \) is Coulomb's constant.

The potential energy (PE) of the charge `-Q` at the center of the ring is:
\[ PE_{\text{final}} = -Q \cdot V = -Q \cdot \frac{kQ}{R} = -\frac{kQ^2}{R} \]

### Step 4: Apply Conservation of Energy
According to the conservation of energy, the total mechanical energy remains constant. Thus, the initial potential energy plus the initial kinetic energy equals the final potential energy plus the final kinetic energy:
\[ PE_{\text{initial}} + KE_{\text{initial}} = PE_{\text{final}} + KE_{\text{final}} \]

Since the charge starts from rest:
\[ 0 + 0 = -\frac{kQ^2}{R} + \frac{1}{2} mv^2 \]

### Step 5: Rearranging the Equation
Rearranging the equation gives:
\[ \frac{1}{2} mv^2 = \frac{kQ^2}{R} \]

### Step 6: Solve for Velocity \( v \)
Multiplying both sides by 2 and solving for \( v \):
\[ mv^2 = \frac{2kQ^2}{R} \]
\[ v^2 = \frac{2kQ^2}{mR} \]
\[ v = \sqrt{\frac{2kQ^2}{mR}} \]

### Final Answer
The velocity of the particle at the moment it passes through the center of the ring is:
\[ v = \sqrt{\frac{2kQ^2}{mR}} \]

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