Two infinite linear charges are placed parallel to each other at a distance `0.1 m` from each other. If the linear charge density on each is `5 mu C//m`, then the force acting on a unit length of each linear charge will be:

Options

A. `2.5 N//m`
B. `3.25 N//m`
C. `4.5 N//m`
D. `7.5 N//m`

Explanation

To find the force acting on a unit length of each infinite linear charge, we can use the formula for the force per unit length between two parallel line charges. The formula is given by:

\[ F = \frac{2k \lambda^2}{r} \]

where:
- \( F \) is the force per unit length,
- \( k \) is Coulomb's constant (\( k \approx 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \)),
- \( \lambda \) is the linear charge density,
- \( r \) is the distance between the two charges.

### Step-by-step Solution:

1. **Identify the given values**:
- Linear charge density, \( \lambda = 5 \, \mu C/m = 5 \times 10^{-6} \, C/m \)
- Distance between the charges, \( r = 0.1 \, m \)

2. **Substitute the values into the formula**:
\[ F = \frac{2 \cdot (9 \times 10^9) \cdot (5 \times 10^{-6})^2}{0.1} \]

3. **Calculate \( (5 \times 10^{-6})^2 \)**:
\[ (5 \times 10^{-6})^2 = 25 \times 10^{-12} \, C^2/m^2 \]

4. **Substitute this back into the force equation**:
\[ F = \frac{2 \cdot (9 \times 10^9) \cdot (25 \times 10^{-12})}{0.1} \]

5. **Calculate the numerator**:
\[ 2 \cdot 9 \cdot 25 = 450 \]
Thus, the numerator becomes:
\[ 450 \times 10^{-3} = 0.45 \times 10^3 = 450 \, N \]

6. **Now divide by \( 0.1 \)**:
\[ F = \frac{450}{0.1} = 4500 \, N/m \]

7. **Final result**:
\[ F = 4.5 \times 10^3 \, N/m = 4.5 \, N/m \]

### Conclusion:
The force acting on a unit length of each linear charge is \( 4.5 \, N/m \).