Question

Two parallel infinite line charges \(+\lambda\) and \(-\lambda\) are placed with a separation distance \(R\) in free space. The net electric field exactly mid-way between the two charges is:

• A. zero
• B. \(\dfrac{2\lambda}{\pi\varepsilon_0 R}\)
• C. \(\dfrac{\lambda}{\pi\varepsilon_0 R}\)
• D. \(\dfrac{\lambda}{2\pi\varepsilon_0 R}\)

Hint:

Remember: \[ E=\frac{\lambda}{2\pi\varepsilon_0 r} \]

• Electric field due to \(+\lambda\) points away from the charge
• Electric field due to \(-\lambda\) points towards the charge
• At midpoint between opposite line charges, fields add up

Answer 1

The Correct Option is B

Concept: Electric field due to an infinite line charge is given by: \[ E=\frac{\lambda}{2\pi\varepsilon_0 r} \] where: \[ \lambda = \text{Linear charge density} \] \[ r = \text{Perpendicular distance from the line charge} \]

Step 1: Find the distance of midpoint from each line charge. The separation between the line charges is: \[ R \] Hence midpoint is at: \[ r=\frac{R}{2} \] from each line charge.

Step 2: Calculate electric field due to each line charge. Field due to one line charge: \[ E_1=\frac{\lambda}{2\pi\varepsilon_0(R/2)} \] \[ E_1=\frac{\lambda}{\pi\varepsilon_0 R} \] Similarly: \[ E_2=\frac{\lambda}{\pi\varepsilon_0 R} \]

Step 3: Determine direction of fields. At the midpoint:

• Field due to \(+\lambda\) is away from positive charge
• Field due to \(-\lambda\) is towards negative charge

Both fields act in the same direction. Therefore: \[ E_{\text{net}}=E_1+E_2 \] \[ E_{\text{net}} = \frac{\lambda}{\pi\varepsilon_0 R} + \frac{\lambda}{\pi\varepsilon_0 R} \] \[ E_{\text{net}} = \frac{2\lambda}{\pi\varepsilon_0 R} \] Therefore, the correct answer is: \[ \boxed{\frac{2\lambda}{\pi\varepsilon_0 R}} \]