The electrostatic potential energy of two point charges, `1muC` each, placed 1 metre apart in air is :

Options

A. `9 xx10^(3)J`
B. `9 xx10^9J`
C. `9xx10^(-3)J`
D. `9 xx10^(-3)J`

Explanation

To find the electrostatic potential energy (U) of two point charges, we can use the formula:

\[ U = \frac{k \cdot q_1 \cdot q_2}{r} \]

Where:
- \( U \) is the electrostatic potential energy,
- \( k \) is Coulomb's constant (\( 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \)),
- \( q_1 \) and \( q_2 \) are the magnitudes of the charges,
- \( r \) is the distance between the charges.

### Step-by-step Solution:

1. **Identify the values**:
- Both charges \( q_1 \) and \( q_2 \) are \( 1 \, \mu C = 1 \times 10^{-6} \, C \).
- The distance \( r \) between the charges is \( 1 \, m \).
- The value of \( k \) is \( 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \).

2. **Substitute the values into the formula**:
\[ U = \frac{9 \times 10^9 \cdot (1 \times 10^{-6}) \cdot (1 \times 10^{-6})}{1} \]

3. **Calculate the product of the charges**:
\[ (1 \times 10^{-6}) \cdot (1 \times 10^{-6}) = 1 \times 10^{-12} \]

4. **Substitute this back into the equation**:
\[ U = \frac{9 \times 10^9 \cdot 1 \times 10^{-12}}{1} \]

5. **Perform the multiplication**:
\[ U = 9 \times 10^{-3} \, \text{J} \]

6. **Final result**:
The electrostatic potential energy of the two point charges is:
\[ U = 0.009 \, \text{J} \quad \text{or} \quad 9 \, \text{mJ} \]