Two charged particles having charges `1 and -1 mu C` and of mass `50 g` each are held at rest while their separation is `2 m`. Now the charges are released. Find the speed of the particles when their separation is `1 m`.

Options

A. `(1)/(5) ms^-1`
B. `(3)/(5) ms^-1`
C. `(3)/(10) ms^-1`
D. `(2)/(7) ms^-1`

Explanation

To solve the problem, we will use the principle of conservation of energy. Initially, the two charged particles have potential energy due to their separation, and when they are released, this potential energy is converted into kinetic energy as they move apart.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Charge of particle 1, \( Q_1 = 1 \, \mu C = 1 \times 10^{-6} \, C \)
- Charge of particle 2, \( Q_2 = -1 \, \mu C = -1 \times 10^{-6} \, C \)
- Mass of each particle, \( m = 50 \, g = 0.05 \, kg \)
- Initial separation, \( r_i = 2 \, m \)
- Final separation, \( r_f = 1 \, m \)

2. **Calculate Initial Potential Energy (U_i):**
The potential energy between two point charges is given by the formula:
\[ U = \frac{k \cdot Q_1 \cdot Q_2}{r} \]
where \( k = \frac{1}{4 \pi \epsilon_0} \approx 9 \times 10^9 \, N \cdot m^2/C^2 \).

For the initial separation:
\[ U_i = \frac{(9 \times 10^9) \cdot (1 \times 10^{-6}) \cdot (-1 \times 10^{-6})}{2} = \frac{-9 \times 10^{-3}}{2} = -4.5 \times 10^{-3} \, J \]

3. **Calculate Final Potential Energy (U_f):**
For the final separation:
\[ U_f = \frac{(9 \times 10^9) \cdot (1 \times 10^{-6}) \cdot (-1 \times 10^{-6})}{1} = -9 \times 10^{-3} \, J \]

4. **Apply Conservation of Energy:**
The total mechanical energy is conserved. The initial potential energy will equal the sum of the final potential energy and the total kinetic energy (K) of the two particles.
\[ U_i + K = U_f \]
Rearranging gives:
\[ K = U_f - U_i \]
Substituting the values:
\[ K = (-9 \times 10^{-3}) - (-4.5 \times 10^{-3}) = -9 \times 10^{-3} + 4.5 \times 10^{-3} = -4.5 \times 10^{-3} \, J \]

5. **Calculate Kinetic Energy:**
Since both particles have the same mass and they move apart with the same speed \( v \), the total kinetic energy is:
\[ K = 2 \cdot \frac{1}{2} m v^2 = m v^2 \]
Therefore:
\[ m v^2 = 4.5 \times 10^{-3} \, J \]
Solving for \( v^2 \):
\[ v^2 = \frac{4.5 \times 10^{-3}}{0.05} = 0.09 \]
Taking the square root gives:
\[ v = \sqrt{0.09} = 0.3 \, m/s \]

### Final Answer:
The speed of each particle when their separation is 1 m is \( 0.3 \, m/s \).