Field due to Infinite Line Charge

Field at perpendicular distance r from an infinite line charge with linear charge density λ. Direction: radially outward from the line.

Choosing the Gaussian surface

An infinite line charge with linear charge density $\lambda$ has cylindrical symmetry. The field must be directed radially outward (for $\lambda > 0$) and have the same magnitude at all points equidistant from the line.

Choose a coaxial cylinder of radius $r$ and length $L$ as the Gaussian surface.

Applying Gauss's law

The closed surface has three parts:

• Curved surface: $\vec{E} \perp d\vec{A}$ is false here — actually $\vec{E} \parallel \hat{n}$ on the curved surface, so contribution = $E \cdot 2\pi r L$
• Two flat end-caps: $\vec{E} \perp \hat{n}$ (field is radial, caps face axially), so contribution = $0$

Total flux:

$$\Phi = E \cdot 2\pi r L$$

Enclosed charge:

$$q_{enc} = \lambda L$$

By Gauss's law:

$$E \cdot 2\pi r L = \frac{\lambda L}{\varepsilon_0}$$ $$\boxed{E = \frac{\lambda}{2\pi\varepsilon_0 r}}$$

Direction: radially outward from the line (for $\lambda > 0$).

Key feature

The field falls as $1/r$ — slower than a point charge ($1/r^2$). This is a consequence of the one-dimensional extension of the source.