Review / Chapter 1 — Electrostatics

Electric Flux

Chapter 1 — Electrostatics

Flux is a way to quantify "how much field pokes through a surface." Picturing the field as a bundle of field lines, flux literally counts the number of lines crossing a surface. It's the bridge between Coulomb's law and Gauss's law.

Definition

On a surface \(S\), at each point choose an outward (or otherwise consistent) normal vector and form a differential area vector \(d\vec A = \hat n\, dA\). The flux of \(\vec E\) through \(S\) is

\[\Phi_E = \int_S \vec E\cdot d\vec A.\]

If the surface is closed, a circle on the integral reminds you: \(\oint_S \vec E\cdot d\vec A\). Only closed surfaces give an unambiguous outward normal.

Sign and meaning

\(\vec E\cdot d\vec A = E\cos\theta\, dA\) where \(\theta\) is the angle between the field and the normal. So:

For a closed surface, positive net flux means more field "coming out" than going in.

Simple cases

Uniform field through a flat surface: \(\Phi = \vec E\cdot \vec A\) where \(\vec A\) is the total area vector.

Radial field from a point charge through a concentric sphere: \(\vec E\) is everywhere parallel to \(d\vec A\), so

\[\Phi = E\cdot 4\pi r^2 = \frac{kq}{r^2}\cdot 4\pi r^2 = 4\pi k q = \frac{q}{\epsilon_0}.\]

The radius canceled — an early hint at Gauss's law.

Flux through deformed surfaces

A nice qualitative picture: if the field is produced by a point charge inside a closed surface, then no matter how you deform the surface (bumps, dents, stretch it into a potato), every field line leaving the charge crosses the surface exactly once. Closed surfaces with no enclosed charge have every inbound line matched by an outbound line, giving net flux zero.

Practice Problems

Problem 1easy
A uniform field \(\vec E = 200\,\hat z\) N/C passes through a flat rectangle of area \(0.5\) m\(^2\) tilted so that its normal makes \(60°\) with \(\hat z\). What is the flux?
Hint
\(\Phi = EA\cos\theta\).
Solution

\(\Phi = (200)(0.5)\cos 60° = 50\) V·m (equivalently N·m\(^2\)/C).

Answer: 50 N·m\(^2\)/C.

Problem 2easy
A uniform field points horizontally. A cubical box has its faces parallel/perpendicular to the field. What is the net flux through the (closed) box?
Hint
Flux in one face = flux out the opposite face; side faces have no flux.
Solution

The field enters one face with \(\vec E\cdot d\vec A = -EA\) (inward normal is outward there flipped) and exits the opposite face with \(+EA\); they cancel. Top/bottom/side faces contribute zero because \(\vec E\perp d\vec A\).

Answer: \(\Phi = 0\).

Problem 3medium
A point charge \(+q\) sits at the center of a cube. What is the flux through any one of its six faces?
Hint
By symmetry the total flux \(q/\epsilon_0\) is shared equally across the six faces.
Solution

Total flux through the cube is \(q/\epsilon_0\) (Gauss). By symmetry each face gets \(q/(6\epsilon_0)\).

Answer: \(\Phi_{\text{face}} = q/(6\epsilon_0)\).

Problem 4medium
A hemispherical bowl of radius \(R\) sits with its opening facing the \(+\hat z\) direction in a uniform field \(\vec E = E_0\hat z\). Compute the flux through the curved part of the bowl (not the flat opening).
Hint
Close the surface with the flat disk and use the fact that net flux through a closed surface in a uniform field is zero.
Solution

Close with the flat disk (area \(\pi R^2\), normal in \(-\hat z\)). Net closed flux in a uniform field = 0. Flux through the flat disk (outward from the closed surface, so pointing \(-\hat z\)) is \(-E_0 \pi R^2\). So flux through the curved part is \(+E_0\pi R^2\).

Answer: \(\Phi = E_0\pi R^2\).