Charges `5 muC and 10 muC` are placed 1 m apart. Work done to bring these charges at a distance 0.5 m from each other is `(k = 9 xx 10^(9) SI)`

Options

A. `9 xx 10^(4) J`
B. `18 xx 10^(4) J`
C. `45 xx 10^(-2)J`
D. `9 xx 10^(-1)J`

Answer

To solve the problem of calculating the work done to bring two charges from a distance of 1 meter to 0.5 meters apart, we can follow these steps:

### Step-by-Step Solution

1. **Identify the Charges and Constants**:
- Let \( Q_1 = 5 \, \mu C = 5 \times 10^{-6} \, C \)
- Let \( Q_2 = 10 \, \mu C = 10 \times 10^{-6} \, C \)
- The electrostatic constant \( k = 9 \times 10^9 \, N \cdot m^2/C^2 \)

2. **Calculate Initial Potential Energy**:
- The initial distance \( r_i = 1 \, m \)
- The formula for potential energy \( U \) between two point charges is given by:
\[ U = \frac{k \cdot Q_1 \cdot Q_2}{r} \]
- Substituting the values for initial potential energy:
\[ U_i = \frac{9 \times 10^9 \cdot (5 \times 10^{-6}) \cdot (10 \times 10^{-6})}{1} \]
- Calculate \( U_i \):
\[ U_i = \frac{9 \times 10^9 \cdot 50 \times 10^{-12}}{1} = 9 \times 50 \times 10^{-3} = 450 \times 10^{-3} \, J = 0.45 \, J \]

3. **Calculate Final Potential Energy**:
- The final distance \( r_f = 0.5 \, m \)
- Now calculate the final potential energy:
\[ U_f = \frac{9 \times 10^9 \cdot (5 \times 10^{-6}) \cdot (10 \times 10^{-6})}{0.5} \]
- Calculate \( U_f \):
\[ U_f = \frac{9 \times 10^9 \cdot 50 \times 10^{-12}}{0.5} = 9 \times 50 \times 10^{-3} \times 2 = 900 \times 10^{-3} \, J = 0.9 \, J \]

4. **Calculate Work Done**:
- The work done by the external agent is the change in potential energy:
\[ W = U_f - U_i \]
- Substitute the values:
\[ W = 0.9 \, J - 0.45 \, J = 0.45 \, J \]

5. **Convert to Standard Form**:
- The work done can also be expressed in microjoules:
\[ W = 0.45 \, J = 45 \times 10^{-2} \, J \]

### Final Answer
The work done to bring the charges from a distance of 1 meter to 0.5 meters apart is:
\[ W = 0.45 \, J \quad \text{or} \quad 45 \times 10^{-2} \, J \]