🎯⭐ INTERACTIVE LESSON

Gauss's Law

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Gauss's Law - Complete Interactive Lesson

Part 1: Electric Flux

⚡ Electric Flux

Part 1 of 7 — Electric Flux

$Phi_E = \oint \vec{E} \cdot d\vec{A} = \int E\cos\theta,dA$

For a uniform field through a flat surface: $Phi_E = EA\cos\theta$

$\theta$ is the angle between $\vec{E}$ and the outward normal $\hat{n}$

Worked Example

$E = 500$ N/C passes through a $0.2$ m² surface perpendicular to it. Find $\Phi$ .

$\Phi = EA\cos 0° = 500(0.2)(1) = 100$ N·m²/C ✅

Concept Check 🎯

Electric Flux 🧮

$E = 500$ N/C, $A = 0.2$ m², $\theta = 0°$ . Flux (N·m²/C)?
$E = 500$ N/C, m², . Flux (N·m²/C)?

Concept Check 🔍

Practice

#	Surface orientation	$\Phi$
1	Perpendicular to $E$	$EA$
2	Parallel to $E$	$0$

Challenge Question 📋

Part 2: Gauss's Law Statement

⚡ Gauss's Law

Part 2 of 7 — Gauss's Law Statement

$\oint \vec{E} \cdot d\vec{A} = \frac{Q_{enc}}{epsilon_0}$

Part 3: Spherical Symmetry

⚡ Spherical Symmetry

Part 3 of 7 — Spherical Symmetry

For a spherically symmetric charge distribution:

Outside ( $r > R$ ): $E = \frac{kQ}{r^2}$ (as if all charge at center)

Part 4: Cylindrical Symmetry

⚡ Cylindrical Symmetry

Part 4 of 7 — Cylindrical Symmetry

For an infinite line charge with linear charge density $\lambda$ (C/m):

$E = \frac{\lambda}{2\pi\epsilon_0 r} = \frac{2k\lambda}{r}$

Part 5: Planar Symmetry

⚡ Planar Symmetry

Part 5 of 7 — Planar Symmetry

For an infinite plane of surface charge density $\sigma$ (C/m²):

$E = \frac{\sigma}{2epsilon_0}$

Part 6: Problem-Solving Workshop

⚡ Problem-Solving Workshop

Part 6 of 7 — Problem-Solving Workshop

Gauss's Law Strategy

Identify the symmetry (spherical, cylindrical, planar)
Choose a Gaussian surface matching the symmetry
Determine $Q_{enc}$ inside the surface
Evaluate $\oint \vec{E} \cdot d\vec{A}$ using symmetry

Part 7: Review & Applications

⚡ Review & Applications

Part 7 of 7 — Review & Applications

Key Results from Gauss's Law

Symmetry	Configuration	$E$
Spherical	Point/ $Q$ outside	$kQ/r^2$

SI unit: N·m²/C (or V·m)

$E = 500$ N/C, $A = 0.2$ m², $\theta = 60°$ . Flux (N·m²/C)?

The total electric flux through any closed surface equals the enclosed charge divided by $epsilon_0$ .

$epsilon_0 = 8.85 \times 10^{-12}$ C²/(N·m²)
The surface is called a Gaussian surface
It is most useful when the charge distribution has symmetry

Worked Example

A Gaussian surface encloses $Q = 5 \times 10^{-9}$ C. Find the total flux.

$\Phi = Q/epsilon_0 = 5 \times 10^{-9} / 8.85 \times 10^{-12} \approx 565$ N·m²/C ✅

Concept Check 🎯

Gauss's Law 🧮

A Gaussian surface encloses no charge. Net flux (N·m²/C)?
A Gaussian surface encloses $Q = 5$ nC. $\Phi \approx ?$ N·m²/C. (Use $epsilon_0 \approx 8.85 \times 10^{-12}$ . Round to nearest integer.)
Charges of $+3q$ and $-q$ are inside a Gaussian surface. The net enclosed charge is ___ $q$ .

Concept Check 🔍

Practice

#	Concept	Key Fact
1	Gauss's law	$\Phi = Q_{enc}/epsilon_0$
2	No enclosed charge	$\Phi = 0$
3	Symmetry types	Spherical, cylindrical, planar

Challenge Question 📋

Inside a uniformly charged sphere ( $r < R$ ): $E = \frac{kQr}{R^3} = \frac{\rho r}{3epsilon_0}$

Inside a conducting sphere: $E = 0$

Worked Example

A conducting sphere of radius 0.1 m has charge $Q = 10^{-8}$ C. Find $E$ at $r = 0.2$ m.

$E = kQ/r^2 = (9 \times 10^9)(10^{-8})/(0.04) = 2250$ N/C ✅

Inside the conductor: $E = 0$ .

Concept Check 🎯

Spherical Symmetry 🧮

Electric field inside a conducting sphere (N/C)?
Outside a sphere: $E = kQ/r^2$ . If $r$ doubles, $E$ decreases by a factor of ___
A sphere ( $Q = 10^{-8}$ C) at distance $r = 0.2$ m. $E$ (N/C)? ( $k = 9 \times 10^9$ )

Concept Check 🔍

Practice

#	Region	$E$
1	Outside sphere	$kQ/r^2$
2	Inside conductor	$0$
3	Inside uniform sphere	$kQr/R^3$

Challenge Question 📋

Use a cylindrical Gaussian surface coaxial with the charge distribution.

The flux through the end caps is zero (field is radial).

Worked Example

An infinite wire has $\lambda = 5 \times 10^{-9}$ C/m. Find $E$ at $r = 0.1$ m.

$E = 2k\lambda/r = 2(9 \times 10^9)(5 \times 10^{-9})/0.1 = 900$ N/C ✅

Concept Check 🎯

Cylindrical Symmetry 🧮

Line charge: $\lambda = 5$ nC/m, $r = 0.1$ m. $E$ (N/C)?
Same wire at $r = 0.2$ m. $E$ (N/C)?
$E$ for a line charge depends on $1/r^n$ . What is $n$ ?

Concept Check 🔍

Practice

#	Configuration	Field
1	Infinite line	$E = 2k\lambda/r$
2	Infinite cylinder (outside)	Same as line
3	Infinite cylinder (inside)	Depends on charge distribution

Challenge Question 📋

The field is uniform (independent of distance!)
Points away from a positive sheet on both sides
For a conductor's surface: $E = \sigma/epsilon_0$ (charge on one side only)

Worked Example

An infinite sheet has $\sigma = 4 \times 10^{-9}$ C/m². Find $E$ .

$E = \sigma/(2epsilon_0) = 4 \times 10^{-9}/(2 \times 8.85 \times 10^{-12}) \approx 226$ N/C ✅

Concept Check 🎯

Planar Symmetry 🧮

$\sigma = 4 \times 10^{-9}$ C/m². $E$ (N/C)? (round to nearest integer, $epsilon_0 \approx 8.85 \times 10^{-12}$ )
A conducting surface has $E = \sigma/epsilon_0$ . This is ___ times the field of a single sheet. (Give as integer.)
Two infinite sheets $+\sigma$ and $-\sigma$ : field outside (N/C)?

Concept Check 🔍

Practice

#	Configuration	Field
1	Single infinite sheet	$\sigma/(2epsilon_0)$
2	Conducting surface	$\sigma/epsilon_0$
3	Two parallel sheets	Superposition

Challenge Question 📋

Worked Example

A uniformly charged sphere ( $\rho = 2 \times 10^{-6}$ C/m³, $R = 0.1$ m). Find $E$ at $r = 0.05$ m.

$Q_{enc} = \rho \cdot \frac{4}{3}\pi r^3 = 2 \times 10^{-6} \cdot \frac{4}{3}\pi(0.05)^3$

$E(4\pi r^2) = Q_{enc}/epsilon_0$

$E = \rho r/(3epsilon_0) = (2 \times 10^{-6})(0.05)/(3 \times 8.85 \times 10^{-12}) \approx 3770$ N/C ✅

Concept Check 🎯

Problem-Solving Workshop 🧮

Name the three symmetry types: spherical, cylindrical, planar. How many are there?
For a line charge, the Gaussian surface is a cylinder. The flux through the curved surface has ___ end caps contributing zero flux. (How many end caps?)
For planar symmetry, the field passes through ___ pair(s) of flat faces of the pillbox.

Concept Check 🔍

Practice

#	Symmetry	Gaussian Surface
1	Sphere	Concentric sphere
2	Infinite wire	Coaxial cylinder
3	Infinite plane	Pillbox

Challenge Question 📋

Worked Example

Compare $E$ at $r = 1$ m from: (a) point charge $Q = 10^{-6}$ C, (b) line charge $\lambda = 10^{-6}$ C/m.

(a) $E = kQ/r^2 = 9 \times 10^3$ N/C

(b) $E = 2k\lambda/r = 18 \times 10^3$ N/C

The line charge field is stronger at this distance because it falls off as $1/r$ instead of $1/r^2$ . ✅

Concept Check 🎯

Review & Applications 🧮

$E$ for a point charge drops as $1/r^n$ . What is $n$ ?
$E$ for a line charge drops as $1/r^n$ . What is $n$ ?
$E$ for an infinite sheet drops as $1/r^n$ . What is $n$ ? (The field is constant.)

Concept Check 🔍

Practice

#	Topic	Formula
1	Point charge	$E = kQ/r^2$
2	Line charge	$E = 2k\lambda/r$
3	Sheet charge	$E = \sigma/(2epsilon_0)$

Challenge Question 📋