Question

The value of electric field at a distance of 1 m from an infinite line charge density 1 C/m is ________.

• A. \( 2\pi\varepsilon_0 \)
• B. \( \frac{1}{2\pi\varepsilon_0} \)
• C. \( \frac{\varepsilon_0}{2\pi} \)
• D. \( \frac{2\pi}{\varepsilon_0} \)

Hint:

For an infinite line of charge, use Gauss's law with a cylindrical surface: \( E = \frac{\lambda}{2\pi \varepsilon_0 r} \). The field is radial and only depends on distance \( r \), not length.

Answer 1

The Correct Option is B

We are given:
- A line charge with linear charge density \( \lambda = 1\,\text{C/m} \)
- Distance from the line: \( r = 1\,\text{m} \)
We are to calculate the electric field \( E \) at this distance from the line using electrostatics.
Step 1: Understand the system
The charge is distributed uniformly along an infinite straight line. Due to the symmetry:
- The electric field lines will be radial (perpendicular to the line).
- The magnitude of the field depends only on radial distance \( r \) from the wire.
Step 2: Apply Gauss’s Law
Gauss’s Law states: \[ \oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enclosed}}}{\varepsilon_0} \] Step 3: Use cylindrical symmetry
Choose a cylindrical Gaussian surface of:
- Radius \( r \)
- Length \( L \)
The surface area of the curved side (only this contributes to flux):
\[ A = 2\pi r L \] Step 4: Total enclosed charge
The line has charge density \( \lambda \), so for length \( L \), total charge enclosed is: \[ Q = \lambda \cdot L \] Step 5: Apply Gauss’s Law
From Gauss’s law: \[ E \cdot 2\pi r L = \frac{\lambda L}{\varepsilon_0} \] Simplifying: \[ E = \frac{\lambda}{2\pi \varepsilon_0 r} \] Step 6: Substitute known values
Given: \[ \lambda = 1\,\text{C/m}, r = 1\,\text{m} \] So, \[ E = \frac{1}{2\pi \varepsilon_0} \] \[ \boxed{E = \frac{1}{2\pi \varepsilon_0}} \]