A point charge `q_1 =1.0muC` is held fixed at origin. A second point charge `q_2 = -2.0 muC` and a mass `10^-4` kg is placed on the x-axis, 1.0 m from the origin. The second point charge is released from rest. What is its speed when it is 0.5 m from the origin?

Explanation

To solve the problem, we will use the principle of conservation of mechanical energy. The total mechanical energy (kinetic energy + potential energy) of the system remains constant if only conservative forces (like electric forces) are acting on it.

### Step 1: Write down the conservation of mechanical energy equation.
The conservation of mechanical energy states that the initial total energy equals the final total energy:

\[ KE_i + PE_i = KE_f + PE_f \]

Where:
- \(KE_i\) = Initial kinetic energy
- \(PE_i\) = Initial potential energy
- \(KE_f\) = Final kinetic energy
- \(PE_f\) = Final potential energy

### Step 2: Identify initial and final conditions.
- Initially, the second charge \(q_2\) is at a distance \(R_i = 1.0 \, m\) from the origin (where \(q_1\) is located).
- Finally, the second charge \(q_2\) is at a distance \(R_f = 0.5 \, m\) from the origin.
- The initial kinetic energy \(KE_i = 0\) (since it is released from rest).

### Step 3: Calculate the initial and final potential energies.
The electric potential energy \(PE\) between two point charges is given by:

\[ PE = k \frac{q_1 q_2}{r} \]

Where \(k\) is Coulomb's constant (\(k \approx 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2\)), \(q_1 = 1.0 \, \mu C = 1.0 \times 10^{-6} \, C\), and \(q_2 = -2.0 \, \mu C = -2.0 \times 10^{-6} \, C\).

- **Initial potential energy \(PE_i\)** at \(R_i = 1.0 \, m\):

\[ PE_i = k \frac{(1.0 \times 10^{-6})(-2.0 \times 10^{-6})}{1.0} = -2.0 \times 10^{-12} \, J \]

- **Final potential energy \(PE_f\)** at \(R_f = 0.5 \, m\):

\[ PE_f = k \frac{(1.0 \times 10^{-6})(-2.0 \times 10^{-6})}{0.5} = -4.0 \times 10^{-12} \, J \]

### Step 4: Substitute into the conservation of energy equation.
Substituting the values into the conservation of energy equation:

\[ 0 + (-2.0 \times 10^{-12}) = KE_f + (-4.0 \times 10^{-12}) \]

### Step 5: Solve for final kinetic energy \(KE_f\).
Rearranging gives:

\[ KE_f = -2.0 \times 10^{-12} + 4.0 \times 10^{-12} = 2.0 \times 10^{-12} \, J \]

### Step 6: Relate kinetic energy to speed.
The kinetic energy is also given by:

\[ KE_f = \frac{1}{2} mv^2 \]

Where \(m = 10^{-4} \, kg\). Setting the two expressions for kinetic energy equal gives:

\[ \frac{1}{2} mv^2 = 2.0 \times 10^{-12} \]

### Step 7: Solve for speed \(v\).
Substituting \(m\):

\[ \frac{1}{2} (10^{-4}) v^2 = 2.0 \times 10^{-12} \]

Multiplying both sides by 2:

\[ 10^{-4} v^2 = 4.0 \times 10^{-12} \]

Dividing by \(10^{-4}\):

\[ v^2 = \frac{4.0 \times 10^{-12}}{10^{-4}} = 4.0 \times 10^{-8} \]

Taking the square root:

\[ v = \sqrt{4.0 \times 10^{-8}} = 2.0 \times 10^{-4} \, m/s \]

### Step 8: Final answer.
The speed of the charge \(q_2\) when it is \(0.5 \, m\) from the origin is approximately:

\[ v \approx 0.0002 \, m/s \]