The electric flux through a closed surface equals enclosed charge divided by $\epsilon_0$.

Electric Flux

$\Phi_E = \int \vec{E} \cdot d\vec{A}$

For uniform field and flat surface: $\Phi_E = \vec{E} \cdot \vec{A} = EA\cos\theta$

Applying Gauss's Law

Strategy:

1. Choose Gaussian surface with symmetry
2. Calculate flux on each part
3. Find enclosed charge
4. Solve for $E$

Works best for:

• Spherical symmetry
• Cylindrical symmetry
• Planar symmetry

Spherical Symmetry

Point Charge

Gaussian surface: sphere of radius $r$

$\oint E \, dA = E(4\pi r^2) = \frac{q}{\epsilon_0}$

$E = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2} = k\frac{q}{r^2}$

Uniform Spherical Shell

Shell of radius $R$, total charge $Q$:

Outside ($r > R$): $E(4\pi r^2) = \frac{Q}{\epsilon_0}$

$E = k\frac{Q}{r^2}$

Inside ($r < R$): $Q_{enc} = 0 \Rightarrow E = 0$

Uniform Solid Sphere

Sphere of radius $R$, uniform charge density $\rho$, total charge $Q$:

Outside ($r > R$): $E = k\frac{Q}{r^2}$

Inside ($r < R$): $Q_{enc} = \rho \cdot \frac{4}{3}\pi r^3 = Q\frac{r^3}{R^3}$

$E(4\pi r^2) = \frac{Q}{\epsilon_0}\frac{r^3}{R^3}$

$E = k\frac{Q}{R^3}r$

(Linear in $r$, maximum at surface)

Cylindrical Symmetry

Infinite Line Charge

Line with charge density $\lambda$:

Gaussian surface: cylinder of radius $r$, length $L$

$E(2\pi rL) = \frac{\lambda L}{\epsilon_0}$

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

Infinite Cylindrical Shell

Shell of radius $R$, charge per length $\lambda$:

Outside ($r > R$): $E = \lambda/(2\pi\epsilon_0 r)$

Inside ($r < R$): $E = 0$

Infinite Solid Cylinder

Cylinder of radius $R$, uniform volume charge density $\rho$:

Outside ($r > R$): $E = \lambda/(2\pi\epsilon_0 r)$ where $\lambda = \rho\pi R^2$

Inside ($r < R$): $Q_{enc} = \rho(\pi r^2 L)$

$E = \frac{\rho r}{2\epsilon_0}$

Planar Symmetry

Infinite Sheet

Surface charge density $\sigma$:

Gaussian surface: pillbox with area $A$

$2EA = \frac{\sigma A}{\epsilon_0}$

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

(Field is uniform, independent of distance!)

Two Parallel Sheets

Sheets with $+\sigma$ and $-\sigma$:

Between sheets: $E = \sigma/\epsilon_0$ (fields add)

Outside: $E = 0$ (fields cancel)

This is a capacitor.

Conductors in Electrostatic Equilibrium

1. $\vec{E} = 0$ inside conductor
2. Net charge resides on surface
3. $\vec{E}$ perpendicular to surface just outside
4. $E = \sigma/\epsilon_0$ just outside surface

Conducting Shell

Hollow conductor with charge $Q$:

• Inner surface: charge $-q$ (if $+q$ inside cavity)
• Outer surface: charge $Q + q$
• Field inside conductor: $E = 0$
• Field in cavity: depends on charge inside

Differential Form

Using divergence theorem:

$\nabla \cdot \vec{E} = \frac{\rho}{\epsilon_0}$

This is one of Maxwell's equations.

By Gauss's law: $E(2πrL) = 0$

$E = 0$

(b) Outside cylinder (r = 0.08 m > R):

Enclosed charge: $Q_{enc} = σ \cdot 2πRL$

By Gauss's law: $E(2πrL) = \frac{σ(2πRL)}{ε_0}$

$E = \frac{σR}{ε_0r}$

$E = \frac{(3.5 \times 10^{-6})(0.05)}{(8.85 \times 10^{-12})(0.08)}$

$E = \frac{1.75 \times 10^{-7}}{7.08 \times 10^{-13}} = 2.47 \times 10^5 \text{ N/C}$

(c) Charge per unit length:

$λ = \frac{Q}{L} = \frac{σ(2πRL)}{L} = 2πRσ$

$λ = 2π(0.05)(3.5 \times 10^{-6})$

$λ = 1.10 \times 10^{-6} \text{ C/m} = 1.10 \text{ μC/m}$

Alternative formula for outside: $E = \frac{λ}{2πε_0r} = \frac{2kλ}{r}$

$E = \frac{2(8.99 \times 10^9)(1.10 \times 10^{-6})}{0.08} = 2.47 \times 10^5 \text{ N/C}$ ✓

Answers: (a) E = 0 (inside cylindrical shell) (b) E = 2.47 × 10⁵ N/C (radially outward) (c) λ = 1.10 μC/m

$E = \frac{4.5 \times 10^{-7}}{8.85 \times 10^{-12}} = 5.08 \times 10^4 \text{ N/C}$

Direction: from positive to negative plate

(b) Field outside the plates:

On the left of both plates:

• Positive plate: field to the left
• Negative plate: field to the right

Fields cancel: $E_{outside} = \frac{σ}{2ε_0} - \frac{σ}{2ε_0} = 0$

Same reasoning for right side: E = 0

(c) Potential difference:

$ΔV = \int_0^d E \, dx = Ed$

$ΔV = (5.08 \times 10^4)(0.02) = 1.016 \times 10^3 \text{ V}$

$ΔV = 1.02 \text{ kV}$

Note: This configuration is a parallel-plate capacitor!

Capacitance: C = ε₀A/d $E = \frac{V}{d} = \frac{Q}{ε_0A} = \frac{σ}{ε_0}$

Answers: (a) E = 5.08 × 10⁴ N/C (between plates, uniform) (b) E = 0 (outside plates) (c) ΔV = 1.02 kV