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.

Enclosed charge: $Q_{enc} = ρ \cdot \frac{4}{3}πr^3$

$E = \frac{ρr^3}{3ε_0r^2} = \frac{ρr}{3ε_0}$

$E = \frac{(6.5 \times 10^{-6})(0.05)}{3(8.85 \times 10^{-12})} = \frac{3.25 \times 10^{-7}}{2.655 \times 10^{-11}}$

$E = 1.22 \times 10^4 \text{ N/C}$

Or using k = 1/(4πε₀): $E = \frac{4πkρr}{3} = \frac{4π(8.99 \times 10^9)(6.5 \times 10^{-6})(0.05)}{3}$

$E = 1.22 \times 10^4 \text{ N/C}$

(b) Outside sphere (r = 0.12 m > R):

Total charge enclosed: $Q = ρ \cdot \frac{4}{3}πR^3 = (6.5 \times 10^{-6}) \cdot \frac{4}{3}π(0.08)^3$

$Q = (6.5 \times 10^{-6})(2.144 \times 10^{-3}) = 1.39 \times 10^{-8} \text{ C}$

Outside, field is like a point charge: $E = \frac{kQ}{r^2} = \frac{(8.99 \times 10^9)(1.39 \times 10^{-8})}{(0.12)^2}$

$E = \frac{1.25 \times 10^2}{1.44 \times 10^{-2}} = 8.68 \times 10^3 \text{ N/C}$

(c) Total charge:

Already calculated: Q = 1.39 × 10⁻⁸ C = 13.9 nC

Verification: Field at surface using both formulas:

• From inside: E(R) = ρR/(3ε₀) = 1.96 × 10⁴ N/C
• From outside: E(R) = kQ/R² = 1.96 × 10⁴ N/C ✓

Answers: (a) E = 1.22 × 10⁴ N/C (b) E = 8.68 × 10³ N/C (c) Q = 13.9 nC