Question

Two infinitely long thin straight parallel wires are kept a perpendicular distance $2R$ having uniform linear charge densities $+\lambda$ and $-\lambda$ respectively. The magnitude of electric field at a mid point between two wires will be ______.

• A. $\frac{\lambda}{\pi\varepsilon_0 R}$
• B. $\frac{\lambda}{2\pi\varepsilon_0 R}$
• C. $\frac{2\lambda}{\pi\varepsilon_0 R}$
• D. $\frac{\lambda}{4\pi\varepsilon_0 R}$

Hint:

If both wires had the same charge ($+\lambda$ and $+\lambda$), the electric field at the midpoint would be \textbf{zero} because the vectors would cancel out.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The electric field due to an infinitely long wire at a distance $r$ is $E = \frac{\lambda}{2\pi\varepsilon_0 r}$. We must use superposition to find the net field at the midpoint.
Step 2: Detailed Explanation:
The midpoint is at a distance $R$ from each wire. 1. Field due to $+\lambda$: $E_1 = \frac{\lambda}{2\pi\varepsilon_0 R}$ (directed away from the positive wire). 2. Field due to $-\lambda$: $E_2 = \frac{\lambda}{2\pi\varepsilon_0 R}$ (directed toward the negative wire). At the midpoint, both fields point in the same direction (from positive to negative). $$E_{net} = E_1 + E_2 = \frac{\lambda}{2\pi\varepsilon_0 R} + \frac{\lambda}{2\pi\varepsilon_0 R} = \frac{2\lambda}{2\pi\varepsilon_0 R} = \frac{\lambda}{\pi\varepsilon_0 R}$$
Step 3: Final Answer:
The correct option is (a).