A particle of mass 2 g and charge `1 muC` is held at rest on a frictionless surface at a distance of 1m from a fixed charge of 1 mC. If the particle is released it will be repelled. The speed of the particle when it is at distance of 10 m from fixed charge is :

Options

A. `10 "ms"^(-1)`
B. `20 "ms"^(-1)`
C. `60 "ms"^(-1)`
D. `90 "ms"^(-1)`

Answer

To solve the problem, we will use the principle of conservation of energy. The initial potential energy of the charged particle will be converted into kinetic energy as it moves away from the fixed charge.

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Mass of the particle, \( m = 2 \, \text{g} = 2 \times 10^{-3} \, \text{kg} \)
- Charge of the particle, \( q_1 = 1 \, \mu\text{C} = 1 \times 10^{-6} \, \text{C} \)
- Fixed charge, \( q_2 = 1 \, \text{mC} = 1 \times 10^{-3} \, \text{C} \)
- Initial distance from the fixed charge, \( r = 1 \, \text{m} \)
- Final distance from the fixed charge, \( d = 10 \, \text{m} \)

2. **Write the Conservation of Energy Equation**:
The total mechanical energy is conserved, so we have:
\[ \text{Initial Potential Energy} + \text{Initial Kinetic Energy} = \text{Final Potential Energy} + \text{Final Kinetic Energy} \]
Since the particle is initially at rest, the initial kinetic energy is zero:
\[ U_i + 0 = U_f + K_f \]
Where:
- \( U_i = \frac{k \cdot q_1 \cdot q_2}{r} \)
- \( U_f = \frac{k \cdot q_1 \cdot q_2}{d} \)
- \( K_f = \frac{1}{2} m v^2 \)

3. **Substituting the Values**:
Using \( k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \):
\[ \frac{k \cdot q_1 \cdot q_2}{r} = \frac{k \cdot q_1 \cdot q_2}{d} + \frac{1}{2} m v^2 \]
Substituting the values:
\[ \frac{9 \times 10^9 \cdot (1 \times 10^{-6}) \cdot (1 \times 10^{-3})}{1} = \frac{9 \times 10^9 \cdot (1 \times 10^{-6}) \cdot (1 \times 10^{-3})}{10} + \frac{1}{2} (2 \times 10^{-3}) v^2 \]

4. **Calculating Potential Energies**:
\[ U_i = 9 \times 10^9 \cdot 1 \times 10^{-6} \cdot 1 \times 10^{-3} = 9 \times 10^{-9} \, \text{J} \]
\[ U_f = \frac{9 \times 10^{-9}}{10} = 9 \times 10^{-10} \, \text{J} \]

5. **Substituting into the Energy Equation**:
\[ 9 \times 10^{-9} = 9 \times 10^{-10} + \frac{1}{2} (2 \times 10^{-3}) v^2 \]
Simplifying:
\[ 9 \times 10^{-9} - 9 \times 10^{-10} = 10^{-3} v^2 \]
\[ 8.1 \times 10^{-9} = 10^{-3} v^2 \]

6. **Solving for \( v^2 \)**:
\[ v^2 = \frac{8.1 \times 10^{-9}}{10^{-3}} = 8.1 \times 10^{-6} \]

7. **Finding \( v \)**:
\[ v = \sqrt{8.1 \times 10^{-6}} = 0.00284 \, \text{m/s} \approx 2.84 \, \text{m/s} \]

### Final Answer:
The speed of the particle when it is at a distance of 10 m from the fixed charge is approximately **2.84 m/s**.