Two point charges of `+ 1.0 mu C ` are kept stationary 2m apart. How much work is needed to be done to bring them 1m apart ?

Options

A. 4.5 mJ
B. 9 mJ
C. 45 mJ
D. 90 mJ

Answer

To solve the problem of how much work is needed to bring two point charges of +1.0 µC from a distance of 2 meters apart to 1 meter apart, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the concept of potential energy between point charges**:
The potential energy (U) between two point charges is given by the formula:
\[ U = k \frac{q_1 q_2}{r} \]
where \( k \) is Coulomb's constant (\( k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \)), \( q_1 \) and \( q_2 \) are the magnitudes of the charges, and \( r \) is the distance between them.

2. **Calculate the initial potential energy (U1)**:
For the initial distance of 2 meters:
\[ U_1 = k \frac{(1.0 \times 10^{-6} \, \text{C})^2}{2 \, \text{m}} = 9 \times 10^9 \frac{(1.0 \times 10^{-6})^2}{2} \]
\[ U_1 = 9 \times 10^9 \frac{1.0 \times 10^{-12}}{2} = 4.5 \times 10^{-3} \, \text{J} \]

3. **Calculate the final potential energy (U2)**:
For the final distance of 1 meter:
\[ U_2 = k \frac{(1.0 \times 10^{-6} \, \text{C})^2}{1 \, \text{m}} = 9 \times 10^9 \frac{(1.0 \times 10^{-6})^2}{1} \]
\[ U_2 = 9 \times 10^9 \times 1.0 \times 10^{-12} = 9 \times 10^{-3} \, \text{J} \]

4. **Calculate the work done (W)**:
The work done to bring the charges from 2 meters to 1 meter apart is equal to the change in potential energy:
\[ W = U_2 - U_1 \]
\[ W = 9 \times 10^{-3} \, \text{J} - 4.5 \times 10^{-3} \, \text{J} = 4.5 \times 10^{-3} \, \text{J} \]

5. **Convert to millijoules**:
Since \( 1 \, \text{J} = 1000 \, \text{mJ} \):
\[ W = 4.5 \times 10^{-3} \, \text{J} = 4.5 \, \text{mJ} \]

### Final Answer:
The work needed to be done to bring the two charges 1 meter apart is **4.5 mJ**.