Applications of Gauss' Theorem > Electric Field due to an Infinite Line of Charge

• The wire is infinitely long, thin, and has uniform linear charge density λ.
• The charge distribution has cylindrical symmetry — the field magnitude depends only on the perpendicular distance r from the wire.
• Field direction is radially outward (for positive λ) or radially inward (for negative λ).

Step 1: Choose the Gaussian Surface

Select a coaxial cylinder of radius r and length l, with the wire along its axis.

Step 2: Analyse Flux Contributions

• Curved surface: field \[\vec E\] is radial and perpendicular to the surface at every point → flux = E × Area.
• Flat end caps: field is parallel to the caps (no perpendicular component) → flux = 0.

Why end caps contribute zero flux: Since \[\vec E\] is perpendicular to the wire's axis and the end caps' area vectors are along the axis, \[\vec E\] . \[d\vec A\] = 0 at every point on the caps.

Step 3: Apply Gauss's Law

Φ_E​ = E(2πrl)

Enclosed charge: q = λl

Step 4: Solve for E

By Gauss's Law, Φ_E = q/ε₀​:

E(2πrl) = \[\frac {λl}{ε_0}\]

Final Result: E = \[\frac {λ}{2πε_0r}\]

Vector form: \[\vec{E}=\frac{\lambda}{2\pi\varepsilon_0r}\hat{r}\]

Q. The electric field at a perpendicular distance of 0.5 m from an infinite line charge is 3.6 × 10³ V/m. Find the linear charge density.

Solution:
Using E = \[\frac {λ}{2πε_0r}\]: