Two parallel infinite line charges with linear charge densities `+lambda C//m` and `-lambda` C/m are placed at a distance of 2R in free space. What is the electric field mid-way between the two line charges?

Options

A. `(lambda)/(2pi epsilon_(0) R)N//C`
B. zero
C. `(2lambda)/(pi epsilon_(0)R)N//C`
D. `(lambda)/(pi epsilon_(0)R)N//C`

Explanation

To find the electric field midway between two infinite parallel line charges with linear charge densities \( +\lambda \, \text{C/m} \) and \( -\lambda \, \text{C/m} \) separated by a distance of \( 2R \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Position**:
- The two line charges are placed at a distance of \( 2R \) apart. The point of interest is midway between them, which is at a distance of \( R \) from each line charge.

2. **Electric Field Due to a Line Charge**:
- The electric field \( E \) due to an infinite line charge with linear charge density \( \lambda \) at a distance \( r \) from the line charge is given by:
\[ E = \frac{2k\lambda}{r} \]
- Here, \( k = \frac{1}{4\pi \epsilon_0} \).

3. **Calculate the Electric Field from Each Line Charge**:
- For the positive line charge \( +\lambda \):
\[ E_1 = \frac{2k\lambda}{R} \]
This electric field points away from the line charge.
- For the negative line charge \( -\lambda \):
\[ E_2 = \frac{2k(-\lambda)}{R} = -\frac{2k\lambda}{R} \]
This electric field points towards the line charge.

4. **Determine the Direction of the Electric Fields**:
- The electric field \( E_1 \) from the positive line charge points to the right (away from the charge).
- The electric field \( E_2 \) from the negative line charge points to the right (towards the charge).

5. **Calculate the Net Electric Field**:
- Since both electric fields \( E_1 \) and \( E_2 \) are in the same direction (to the right), we can add them:
\[ E_{\text{net}} = E_1 + |E_2| = \frac{2k\lambda}{R} + \frac{2k\lambda}{R} = \frac{4k\lambda}{R} \]

6. **Substitute the Value of \( k \)**:
- Substitute \( k = \frac{1}{4\pi \epsilon_0} \):
\[ E_{\text{net}} = \frac{4 \cdot \frac{1}{4\pi \epsilon_0} \cdot \lambda}{R} = \frac{\lambda}{\pi \epsilon_0 R} \]

7. **Final Result**:
- The electric field midway between the two line charges is:
\[ E_{\text{net}} = \frac{\lambda}{\pi \epsilon_0 R} \, \text{N/C} \]