Two identical coaxial rings each of radius R are separated by a distance of `sqrt3R`. They are uniformly charged with charges `+Q` and `-Q` respectively. The minimum kinetic energy with which a charged particle (charge `+q`) should be projected from the centre of the negatively charged ring along the axis of the rings such that it reaches the centre of the positively charged ring is

Options

A. `(Qq)/(4piepsilon_0R)`
B. `(Qq)/2piepsilon_0R`
C. `(Qq)/(8piepsilon_0R)`
D. `(3Qq)/(4piepsilon_0R)`

Explanation

To solve the problem, we need to determine the minimum kinetic energy required for a charged particle \( +q \) to be projected from the center of the negatively charged ring to reach the center of the positively charged ring.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have two identical coaxial rings, each with radius \( R \).
- The distance between the two rings is \( \sqrt{3}R \).
- The first ring (let's call it Ring A) has a charge of \( -Q \) and the second ring (Ring B) has a charge of \( +Q \).

2. **Identifying the Forces**:
- When the charged particle \( +q \) is projected from the center of Ring A, it will experience an attractive force due to the negatively charged ring and a repulsive force from the positively charged ring as it approaches.

3. **Using Conservation of Energy**:
- The total mechanical energy at the starting point (center of Ring A) must equal the total mechanical energy at the final point (center of Ring B).
- The equation can be expressed as:
\[ K_A + U_A = K_B + U_B \]
- Here, \( K_A \) is the initial kinetic energy, \( U_A \) is the potential energy at the center of Ring A, \( K_B \) is the kinetic energy at the center of Ring B (which we want to be zero for minimum energy), and \( U_B \) is the potential energy at the center of Ring B.

4. **Calculating Potential Energies**:
- The potential energy \( U_A \) at the center of Ring A (due to Ring B) is given by:
\[ U_A = -\frac{kQq}{\sqrt{R^2 + (\sqrt{3}R)^2}} = -\frac{kQq}{2R} \]
- The potential energy \( U_B \) at the center of Ring B (due to Ring A) is given by:
\[ U_B = \frac{kQq}{\sqrt{R^2 + (\sqrt{3}R)^2}} = \frac{kQq}{2R} \]

5. **Setting Up the Energy Equation**:
- Since we want \( K_B = 0 \) for minimum kinetic energy, we can substitute:
\[ K_A - \frac{kQq}{2R} = 0 + \frac{kQq}{2R} \]
- Rearranging gives:
\[ K_A = \frac{kQq}{R} \]

6. **Final Result**:
- Thus, the minimum kinetic energy with which the charged particle \( +q \) should be projected from the center of the negatively charged ring is:
\[ K = \frac{kQq}{R} \]