FAD1022 L4 — Electric Flux and Gauss Law

Course: FAD 1022 – Basic Physics 2
Semester: 2 2025/2026
Lecturer: Dr. Hashlina Rusdi
Location: No 29, Aras 3 Kompleks PASUM
Topic: Electric Flux and Gauss's Law

Summary

This lecture introduces the concepts of electric flux and Gauss's Law — powerful tools for calculating electric fields in situations with high symmetry. The lecture covers electric flux through surfaces, Gauss's Law formulation, and applications to line charges, plane sheets, and parallel conducting plates.

Prerequisites

Students should understand:

  • Charge in a Uniform Electric Field
  • Basic electric field concepts from Lectures 1-3

01 — Electric Flux

Definition

Electric flux ($\Phi_E$) measures the number of electric field lines passing through a surface.

Formula

$$\Phi_E = EA\cos\theta$$

Where:

  • $E$ = electric field magnitude
  • $A$ = area of the surface
  • $\theta$ = angle between $\vec{E}$ and the normal to the surface

Flux Through a Closed Surface

For a closed surface, we consider the net flux entering or leaving.

02 — Gauss's Law

Statement

The net number of field lines through a closed surface (Gaussian surface) is proportional to the charge enclosed divided by $\varepsilon_0$.

Mathematical Form

$$\Phi_E = \frac{Q_{enclosed}}{\varepsilon_0}$$

Or equivalently: $$\oint \vec{E} \cdot d\vec{A} = \frac{Q_{enclosed}}{\varepsilon_0}$$

Where:

  • $\varepsilon_0 = 8.85 \times 10^{-12}$ C² N⁻¹ m⁻² (permittivity of free space)

Key Insight

Gauss's Law relates the electric flux through a closed surface to the charge enclosed within that surface.

03 — Applications of Gauss's Law

1. Field of a Line Charge

Setup: Infinite line of charge with linear charge density $\lambda$

Gaussian Surface: Cylindrical surface coaxial with the line

Result: $$E = \frac{\lambda}{2\pi r \varepsilon_0}$$

Where:

  • $\lambda$ = charge per unit length (C/m)
  • $r$ = distance from the line

Key point: $\vec{E}$ is perpendicular to the curved surface and parallel to the end caps (no flux through end caps).

2. Field of an Infinite Plane Sheet of Charge

Setup: Infinite plane with surface charge density $\sigma$

Result: $$E = \frac{\sigma}{2\varepsilon_0}$$

The field is:

  • Uniform (independent of distance from the plane)
  • Perpendicular to the plane
  • Same magnitude on both sides

3. Field Between Oppositely Charged Parallel Conducting Plates

Setup: Two parallel plates with equal and opposite charge densities

Result between plates: $$E = \frac{\sigma}{\varepsilon_0}$$

Outside the plates: $E = 0$

Explanation:

  • Outside (points a and c): Fields from both plates cancel ($E_1$ and $E_2$ in opposite directions)
  • Between plates (point b): Fields add ($E_1$ and $E_2$ in same direction)
  • Since each plate contributes $\frac{\sigma}{2\varepsilon_0}$, the total is $\frac{\sigma}{\varepsilon_0}$

Key Formulas Summary

Configuration Electric Field
Point charge (at distance r) $E = \frac{kQ}{r^2}$
Line charge (at distance r) $E = \frac{\lambda}{2\pi r \varepsilon_0}$
Infinite plane sheet $E = \frac{\sigma}{2\varepsilon_0}$
Parallel plates (between) $E = \frac{\sigma}{\varepsilon_0}$
Parallel plates (outside) $E = 0$

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