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 (in N / m ) acting on a unit length of each linear cherge will be

Answer

To find the force acting on a unit length of each infinite linear charge, we can follow these steps:

### 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. **Electric Field Due to a Single Infinite Line Charge:**
The electric field \( E \) created by an infinite line charge at a distance \( r \) from it can be calculated using the formula:
\[ E = \frac{2k\lambda}{r} \]
where \( k \) is Coulomb's constant, \( k \approx 9 \times 10^9 \, N \cdot m^2/C^2 \).

3. **Calculate the Electric Field:**
Substituting the known values into the formula:
\[ E = \frac{2 \times (9 \times 10^9) \times (5 \times 10^{-6})}{0.1} \]
\[ E = \frac{90 \times 10^3}{0.1} = 900000 \, N/C \]

4. **Force per Unit Length on Each Charge:**
The force \( F \) acting on a unit length of the linear charge can be calculated using:
\[ F = \lambda \cdot E \]
Since we are considering the force per unit length, we can directly use the linear charge density \( \lambda \) and the electric field \( E \):
\[ F = \lambda \cdot E = (5 \times 10^{-6}) \cdot (900000) \]
\[ F = 4.5 \, N/m \]

### Final Answer:
The force acting on a unit length of each linear charge is **4.5 N/m**.

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