Gauss's Law

⚡ 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 $\hat{n}$
• SI unit: $N\cdot m^{2}/C$ (or V·m)

Worked Example

$E = 500$ N/C passes through a $0.2$ $m^{2}$ surface perpendicular to it. Find $\Phi$.

$\Phi = EA\cos 0° = 500(0.2)(1) = 100$ $N\cdot m^{2}/C$ ✅

Concept Check 🎯

Electric Flux 🧮

1. $E = 500$ N/C, $A = 0.2$ $m^{2}$, $\theta = 0°$. Flux $(N\cdot m^{2}/C)$?

2. $E = 500$ N/C, $A = 0.2$ $m^{2}$, $\theta = 90°$. Flux $(N\cdot m^{2}/C)$?

3. $E = 500$ N/C, $A = 0.2$ $m^{2}$, $\theta = 60°$. Flux $(N\cdot m^{2}/C)$?

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⚡ Gauss's Law

Part 2 of 7 — Gauss's Law Statement

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

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

• $epsilon_0 = 8.85 \times 10^{-12}$ $C^{2}/(N\cdot m^{2})$
• 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.

$\Phi = Q/epsilon_0 = 5 \times 10^{-9} / 8.85 \times 10^{-12} \approx 565$ $N\cdot m^{2}/C$ ✅

Concept Check 🎯

Gauss's Law 🧮

1. A Gaussian surface encloses no charge. Net flux $(N\cdot m^{2}/C)$?

2. A Gaussian surface encloses $Q = 5$ nC. $\Phi \approx ?$ $N\cdot m^{2}/C$. (Use $epsilon_0 \approx 8.85 \times 10^{-12}$. Round to nearest integer.)

3. Charges of $+3q$ and $-q$ are inside a Gaussian surface. The net enclosed charge is ___ $q$.

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⚡ 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)

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 🧮

1. Electric field inside a conducting sphere (N/C)?

2. Outside a sphere: $E = kQ/r^2$. If $r$ doubles, $E$ decreases by a factor of ___

3. A sphere ($Q = 10^{-8}$ C) at distance $r = 0.2$ m. $E$ (N/C)? ($k = 9 \times 10^9$)

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⚡ Cylindrical Symmetry

Part 4 of 7 — Cylindrical Symmetry

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

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

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 🧮

1. Line charge: $\lambda = 5$ nC/m, $r = 0.1$ m. $E$ (N/C)?

2. Same wire at $r = 0.2$ m. $E$ (N/C)?

3. $E$ for a line charge depends on $1/r^n$. What is $n$?

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⚡ Planar Symmetry

Part 5 of 7 — Planar Symmetry

For an infinite plane of surface charge density $\sigma$ $(C/m^{2})$:

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

• 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 $\sigma = 4 \times 10^{-9}$ $C/m^{2}$. 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 🧮

1. $\sigma = 4 \times 10^{-9}$ $C/m^{2}$. $E$ (N/C)? (round to nearest integer, $epsilon_0 \approx 8.85 \times 10^{-12}$)

2. A conducting surface has $E = \sigma/epsilon_0$. This is ___ times the field of a single sheet. (Give as integer.)

3. Two infinite sheets $+\sigma$ and $-\sigma$: field outside (N/C)?

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⚡ Problem-Solving Workshop

Part 6 of 7 — Problem-Solving Workshop

Gauss's Law Strategy

1. Identify the symmetry (spherical, cylindrical, planar)
2. Choose a Gaussian surface matching the symmetry
3. Determine $Q_{enc}$ inside the surface
4. Evaluate $\oint \vec{E} \cdot d\vec{A}$ using symmetry
5. Solve for $E$

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 🧮

1. Name the three symmetry types: spherical, cylindrical, planar. How many are there?

2. 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?)

3. For planar symmetry, the field passes through ___ pair(s) of flat faces of the pillbox.

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⚡ Review & Applications

Part 7 of 7 — Review & Applications

Key Results from Gauss's Law

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 🧮

1. $E$ for a point charge drops as $1/r^n$. What is $n$?

2. $E$ for a line charge drops as $1/r^n$. What is $n$?

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

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