A charged particle having some mass is resting in equilibrium at a height `H` above the center of a uniformly charged non-conducting horizontal ring of radius ` R`. The equilibrium of the particle will be stable .

Options

A. (a)For all values of H
B. (b)Only if `H gt (R)/(sqrt(2))`
C. (c)Only if `H lt (R)/(sqrt(2))`
D. (d)Only if `H = (R)/(sqrt(2))`

Explanation

To solve the problem of a charged particle resting in equilibrium above the center of a uniformly charged non-conducting horizontal ring, we will analyze the forces acting on the particle and determine the conditions for stable equilibrium.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have a charged particle (let's denote its charge as \( Q \)) positioned at a height \( H \) above the center of a uniformly charged ring of radius \( R \).
- The charged ring creates an electric field \( E \) at the height \( H \).

2. **Electric Field Due to the Ring**:
- The electric field \( E \) at a distance \( H \) from the center of the ring (along the axis) is given by the formula:
\[ E = \frac{kQ}{R^2 + H^2} \]
where \( k \) is Coulomb's constant and \( Q \) is the total charge on the ring.

3. **Forces Acting on the Particle**:
- The gravitational force acting on the particle is \( mg \) (downward).
- The electric force acting on the particle due to the electric field is \( QE \) (upward).

4. **Equilibrium Condition**:
- For the particle to be in equilibrium, the electric force must balance the gravitational force:
\[ QE = mg \]

5. **Stability of Equilibrium**:
- To determine whether the equilibrium is stable or unstable, we need to analyze the behavior of the forces when the particle is slightly displaced from its equilibrium position.
- We consider two cases based on the height \( H \) relative to \( \frac{R}{\sqrt{2}} \).

6. **Case 1: \( H < \frac{R}{\sqrt{2}} \)**:
- If the particle is displaced slightly downward, the electric field \( E \) will decrease, leading to a decrease in the electric force \( QE \).
- The gravitational force \( mg \) remains constant. Thus, the net force will act downward, causing the particle to move further away from equilibrium. This indicates **unstable equilibrium**.

7. **Case 2: \( H > \frac{R}{\sqrt{2}} \)**:
- If the particle is displaced slightly upward, the electric field \( E \) will also decrease, leading to a decrease in the electric force \( QE \).
- However, since the particle is now above the equilibrium position, the gravitational force \( mg \) will pull it back down. Thus, the net force will act downward towards the equilibrium position, indicating **stable equilibrium**.

8. **Conclusion**:
- The equilibrium of the charged particle will be stable if the height \( H \) is greater than \( \frac{R}{\sqrt{2}} \).

### Final Answer:
The equilibrium of the particle will be stable if \( H > \frac{R}{\sqrt{2}} \).

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