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 step by step, we will use the principle of conservation of energy.

### Step 1: Understand the initial conditions
We have two charged particles:
- Charge \( q_1 = 1 \, \mu C = 1 \times 10^{-6} \, C \)
- Charge \( q_2 = -1 \, \mu C = -1 \times 10^{-6} \, C \)
- Mass of each particle \( m = 50 \, g = 50 \times 10^{-3} \, kg \)
- Initial separation \( r_i = 2 \, m \)
- Final separation \( r_f = 1 \, m \)

### Step 2: Write the expression for potential energy
The electric potential energy \( U \) between two point charges is given by:
\[ U = k \frac{q_1 q_2}{r} \]
where \( k = 9 \times 10^9 \, N \cdot m^2/C^2 \).

### Step 3: Calculate the initial potential energy
At the initial separation of \( 2 \, m \):
\[ U_i = k \frac{(1 \times 10^{-6})(-1 \times 10^{-6})}{2} = 9 \times 10^9 \frac{-1 \times 10^{-12}}{2} = -4.5 \times 10^{-3} \, J \]

### Step 4: Calculate the final potential energy
At the final separation of \( 1 \, m \):
\[ U_f = k \frac{(1 \times 10^{-6})(-1 \times 10^{-6})}{1} = 9 \times 10^9 \frac{-1 \times 10^{-12}}{1} = -9 \times 10^{-3} \, J \]

### Step 5: Apply the conservation of energy
The total mechanical energy is conserved. Initially, the kinetic energy \( K_i = 0 \) because the particles are at rest. The total energy at the beginning is:
\[ E_i = U_i + K_i = -4.5 \times 10^{-3} + 0 = -4.5 \times 10^{-3} \, J \]
At the final state, the total energy is:
\[ E_f = U_f + K_f \]
where \( K_f = \frac{1}{2} m v^2 + \frac{1}{2} m v^2 = m v^2 \) (since both particles have the same speed \( v \)).

### Step 6: Set up the equation
Equating the total energy at the initial and final states:
\[ -4.5 \times 10^{-3} = -9 \times 10^{-3} + m v^2 \]
\[ m v^2 = -4.5 \times 10^{-3} + 9 \times 10^{-3} = 4.5 \times 10^{-3} \, J \]

### Step 7: Solve for speed \( v \)
Substituting \( m = 50 \times 10^{-3} \, kg \):
\[ 50 \times 10^{-3} v^2 = 4.5 \times 10^{-3} \]
\[ v^2 = \frac{4.5 \times 10^{-3}}{50 \times 10^{-3}} = \frac{4.5}{50} = 0.09 \]
\[ v = \sqrt{0.09} = 0.3 \, m/s \]

### Final Answer
The speed of each particle when their separation is \( 1 \, m \) is \( 0.3 \, m/s \) or \( \frac{3}{10} \, m/s \).

---