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Topic 2 of 9

Gauss's Law & Electrostatics

Gauss's law — that the total electric flux through any closed surface equals the enclosed charge divided by ε₀ — is one of Maxwell's four equations. For high-symmetry geometries (spherical, cylindrical, planar), it gives the field immediately. For general distributions, Poisson's equation ∇²V = −ρ/ε₀ must be solved.

Key Concepts

Gauss's Law

\oint_S\vec E\cdot d\vec a=Q_{\rm enc}/\varepsilon_0

. Equivalent to Coulomb's law but more powerful for symmetric geometries. In differential form:

\nabla\cdot\vec E=\rho/\varepsilon_0

Gaussian Surfaces

Choose a closed surface that matches the symmetry: sphere (spherical charge), cylinder (line/cylindrical charge), pillbox (plane/sheet charge). Then

\vec E

is constant on the surface and parallel/perpendicular to it, making the integral trivial.

Poisson's & Laplace's Equations

\nabla^2 V=-\rho/\varepsilon_0

(Poisson's equation). Where

\rho=0

\nabla^2 V=0

(Laplace's equation). Boundary conditions uniquely determine

V

— this is the uniqueness theorem.

Shell Theorems

A spherically symmetric shell of charge: (1) outside, it looks like a point charge at the center; (2) inside the shell, the field is exactly zero.

Key Equations

Gauss's law (integral)

\oint_S \vec{E}\cdot d\vec{a} = \frac{Q_{\rm enc}}{\varepsilon_0}

Total electric flux = enclosed charge / ε₀.

Field of a sphere

E = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2}\quad (r > R)

Outside a uniformly charged sphere of total charge Q, radius R.

Poisson's equation

\nabla^2 V = -\frac{\rho}{\varepsilon_0}

Relates potential to charge density; reduces to Laplace where ρ = 0.

Worked Example

Electric Field of a Uniformly Charged Sphere

Problem

A solid sphere of radius $R=0.10$ m carries uniform charge density $\rho=8.85\times10^{-9}$ C/m³. Find $\vec E$ at $r=0.20$ m and at $r=0.05$ m.

Solution

Total charge: $Q=\rho\cdot(4\pi R^3/3)=8.85\times10^{-9}\times4.19\times10^{-3}=3.71\times10^{-11}$ C.

Outside ( $r=0.20$ m): $E=Q/(4\pi\varepsilon_0 r^2)=(3.71\times10^{-11})/(4\pi\times8.85\times10^{-12}\times0.04)=3.71\times10^{-11}/4.45\times10^{-12}\approx8.34$ N/C.

Inside ( $r=0.05$ m): $Q_{\rm enc}=\rho\cdot(4\pi r^3/3)=8.85\times10^{-9}\times5.24\times10^{-4}=4.64\times10^{-12}$ C.

E_{\rm in}=\frac{Q_{\rm enc}}{4\pi\varepsilon_0 r^2}=\frac{\rho r}{3\varepsilon_0}=\frac{8.85\times10^{-9}\times0.05}{3\times8.85\times10^{-12}}=\frac{4.43\times10^{-10}}{2.655\times10^{-11}}\approx16.7\text{ N/C}

Answer

E(0.20\text{ m})\approx8.34

N/C;

E(0.05\text{ m})\approx16.7

N/C.

Practice

Exercises

7 problems

1 of 7

A point charge $Q=2.0\times10^{-9}$ C. Find the electric flux through a sphere of radius $r=1.0$ m enclosing it (in N·m²/C). $\Phi=Q/\varepsilon_0$ .

Your answer

N·m²/C

2 of 7

Infinite line charge with linear density $\lambda=5.0\times10^{-9}$ C/m. Find $E$ at $r=0.10$ m (in N/C). $E=\lambda/(2\pi\varepsilon_0 r)$ .

Your answer

N/C

3 of 7

Infinite plane with surface charge density $\sigma=2.0\times10^{-9}$ C/m². Find $E$ (in N/C) on either side. $E=\sigma/(2\varepsilon_0)$ .

Unlock Exercise 3