Review / Chapter 1 — Electrostatics

Field of an Infinite Line

Chapter 1 — Electrostatics

A long, straight, uniformly charged wire is the canonical cylindrically symmetric source. With Gauss's law the field drops out in two lines: it points radially outward from the wire and falls off as \(1/r\) — slower than a point charge, faster than an infinite plane.

Setup and symmetry

Take an infinite straight line of charge along the \(z\)-axis with uniform linear density \(\lambda\) (C/m). Symmetries:

Left with \(\vec E = E(r)\hat r\) in cylindrical coordinates.

Gauss's law cylinder

Coaxial cylinder of radius \(r\), length \(L\). Flux contributions:

Enclosed charge: \(\lambda L\). Gauss:

\[E\cdot 2\pi r L = \frac{\lambda L}{\epsilon_0} \implies \boxed{E = \frac{\lambda}{2\pi\epsilon_0 r}}.\]

Direction: outward for \(\lambda > 0\), inward for \(\lambda < 0\).

Why \(1/r\) and not \(1/r^2\)?

Field lines from a line of charge spread into a cylinder, not a sphere. Total flux through any cylinder is constant (\(\lambda L/\epsilon_0\)), while the cylinder's curved area grows as \(r\), so \(E\propto 1/r\). You can check this by brute-force integrating Coulomb's law over the wire too — it's a calculus exercise, but Gauss is vastly easier.

Variants

Infinite cylindrical shell (radius \(R\), uniform \(\sigma\)): zero field inside; \(E = \lambda_{\text{eff}}/(2\pi\epsilon_0 r)\) outside, where \(\lambda_{\text{eff}} = 2\pi R\sigma\).

Solid infinite cylinder (radius \(R\), uniform volume density \(\rho\)): for \(r < R\), \(Q_{\text{enc}} = \rho \pi r^2 L\) gives \(E = \rho r/(2\epsilon_0)\). For \(r > R\), point-line behavior \(E = \rho R^2/(2\epsilon_0 r) = \lambda/(2\pi\epsilon_0 r)\).

Practice Problems

Problem 1easy
A long wire carries \(\lambda = 2\,\mu\)C/m. Find \(E\) 10 cm from the wire.
Hint
\(E = \lambda/(2\pi\epsilon_0 r)\).
Solution

\(E = (2\times 10^{-6})/(2\pi\cdot 8.854\times 10^{-12}\cdot 0.1) \approx 3.6\times 10^5\) N/C.

Answer: \(\approx 3.6\times 10^5\) N/C, radially outward.

Problem 2medium
Two parallel infinite wires, each with linear density \(\lambda\), separated by distance \(d\). Find the field at a point equidistant between them.
Hint
Superpose; by symmetry the fields from the two wires are antiparallel at the midpoint.
Solution

Each wire gives \(\lambda/(2\pi\epsilon_0 (d/2)) = \lambda/(\pi\epsilon_0 d)\). At the midpoint, the two contributions point in opposite directions (each away from its own wire) and cancel.

Answer: \(\vec E = 0\).

Problem 3medium
Coaxial cable: inner wire has linear density \(+\lambda\), outer cylindrical shell (radius \(b\)) has \(-\lambda\) total linear density. Find \(E\) for \(r < b\) and \(r > b\).
Hint
Gauss at radius \(r\).
Solution

\(r < b\) (between inner wire and outer shell): \(Q_{\text{enc}} = \lambda L\), \(E = \lambda/(2\pi\epsilon_0 r)\). \(r > b\): \(Q_{\text{enc}} = 0\), \(E = 0\).

Answer: \(E = \lambda/(2\pi\epsilon_0 r)\) inside; \(E = 0\) outside.

Problem 4hard
A solid infinite cylinder (radius \(R\)) has uniform volume density \(\rho\). Find \(E(r)\) for \(r < R\) and \(r > R\), and verify the two results match at \(r = R\).
Hint
Charge per length: \(\lambda = \rho\pi R^2\).
Solution

Inside (\(r < R\)): \(E\cdot 2\pi r L = \rho\pi r^2 L/\epsilon_0\), so \(E = \rho r/(2\epsilon_0)\). Outside: \(E = \rho R^2/(2\epsilon_0 r)\). At \(r = R\), both give \(\rho R/(2\epsilon_0)\). Check.

Answer: \(E = \rho r/(2\epsilon_0)\) inside, \(E = \rho R^2/(2\epsilon_0 r)\) outside.