Step 1: Understanding the Concept:
This problem applies the principle of superposition to electric fields. The net electric field at any point in space due to multiple independent charges is the vector sum of the individual electric field intensities generated by each charge distribution on its own. We must track both the magnitudes and directions of these field vectors.
Step 2: Key Formula or Approach:
- The magnitude of the electric field ($E$) at a perpendicular distance $r$ from an infinite straight line charge with linear charge density $\lambda$ is derived via Gauss's Law as:
\[ E = \frac{\lambda}{2\pi \varepsilon_0 r} \]
- Direction Rule: Electric field lines point radially *away* from a positive charge distribution ($+\lambda$) and point radially *toward* a negative charge distribution ($-\lambda$).
Step 3: Detailed Explanation:
Let the two parallel line charges be situated along a coordinate axis separated by total distance $R$. We are evaluating the net field at the exact mid-way point.
The distance ($r$) from either line charge to this midpoint is:
\[ r = \frac{R}{2} \]
Let's find the individual vector fields at this central midpoint:
1. Field due to $+\lambda$ ($E_1$): The field points radially away from the positive line charge. If the positive charge is on the left, this field points to the right. Its magnitude is:
\[ E_1 = \frac{\lambda}{2\pi \varepsilon_0 \left(\frac{R}{2}\right)} = \frac{\lambda}{\pi \varepsilon_0 R} \]
2. Field due to $-\lambda$ ($E_2$): The field points radially toward the negative line charge. If the negative charge is located on the right, this field also points to the right. Its magnitude is:
\[ E_2 = \frac{\lambda}{2\pi \varepsilon_0 \left(\frac{R}{2}\right)} = \frac{\lambda}{\pi \varepsilon_0 R} \]
Since both electric field vectors $E_1$ and $E_2$ point in the exact same direction at the midpoint (away from positive, toward negative), we calculate the net electric field by adding their magnitudes directly together:
\[ E_{\text{net}} = E_1 + E_2 = \frac{\lambda}{\pi \varepsilon_0 R} + \frac{\lambda}{\pi \varepsilon_0 R} = \frac{2\lambda}{\pi \varepsilon_0 R} \]
Wait, let's re-verify the simplification:
\[ E_1 = \frac{\lambda}{2\pi\varepsilon_0 (R/2)} = \frac{2\lambda}{2\pi\varepsilon_0 R} = \frac{\lambda}{\pi\varepsilon_0 R} \]
\[ E_{\text{net}} = \frac{\lambda}{\pi\varepsilon_0 R} + \frac{\lambda}{\pi\varepsilon_0 R} = \frac{2\lambda}{\pi \varepsilon_0 R} \]
Therefore, option (B) matches our mathematical derivation perfectly. Let's correct the key indicator.
Step 4: Final Answer:
The net electric field exactly mid-way between the two charges is $\frac{2\lambda}{\pi \varepsilon_0 R}$, matching option (B).