A uniformly charged ring having total charge `Q` and of radius `R` is fixed in such a way so that plane of the ring is horizontal. A charged particle of charge `q` and mass `m` is in equilibrium at any point lying on `y`-axis under the combined forces of electric field and earth's gravitational field. Choose the correct statements

Explanation

To solve the problem of a charged particle in equilibrium under the influence of the electric field from a uniformly charged ring and the gravitational field, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have a uniformly charged ring with total charge \( Q \) and radius \( R \), fixed in a horizontal plane.
- A charged particle with charge \( q \) and mass \( m \) is placed along the y-axis (the axis perpendicular to the plane of the ring).

2. **Forces Acting on the Charged Particle**:
- The gravitational force acting on the particle is \( F_g = mg \) (downward).
- The electric force acting on the charged particle due to the ring can be expressed using the electric field \( E \) produced by the ring.

3. **Electric Field of the Ring**:
- The electric field \( E \) at a point on the y-axis at a distance \( y \) from the center of the ring is given by:
\[ E = \frac{1}{4\pi \epsilon_0} \cdot \frac{Qy}{(R^2 + y^2)^{3/2}} \]
- This electric field points away from the ring if \( Q \) is positive and towards the ring if \( Q \) is negative.

4. **Equilibrium Condition**:
- For the particle to be in equilibrium, the net force acting on it must be zero. Therefore, the upward electric force must balance the downward gravitational force:
\[ F_e = F_g \]
\[ qE = mg \]
- Substituting the expression for \( E \):
\[ q \cdot \frac{1}{4\pi \epsilon_0} \cdot \frac{Qy}{(R^2 + y^2)^{3/2}} = mg \]

5. **Analyzing the Charge Signs**:
- If both \( Q \) and \( q \) are positive, the electric force will act upward, which means the particle can be in equilibrium above the ring (on the positive y-axis).
- If both \( Q \) and \( q \) are negative, the electric force will also act upward, allowing equilibrium below the ring (on the negative y-axis).
- If \( Q \) and \( q \) have opposite signs, the electric force will act downward, which cannot balance the gravitational force acting downward.

6. **Specific Charge**:
- The specific charge \( \frac{q}{m} \) can be derived from the equilibrium condition:
\[ \frac{q}{m} = \frac{g(R^2 + y^2)^{3/2}}{4\pi \epsilon_0 Qy} \]
- This indicates how the charge-to-mass ratio influences the position of equilibrium.

### Conclusion:
- The charged particle can be in equilibrium at any point on the y-axis if the conditions regarding the signs of \( Q \) and \( q \) are satisfied. The specific charge \( \frac{q}{m} \) determines the nature of the equilibrium.