Review / Chapter 1 — Electrostatics

Gauss's Law (Integral Form)

Chapter 1 — Electrostatics

Coulomb's law makes a bold prediction: the total electric flux through any closed surface depends only on the charge enclosed, not on where that charge sits inside, nor on the shape of the surface. That statement is Gauss's law — exactly equivalent to Coulomb's law, but dramatically more useful when there's symmetry. The differential form shows up in Chapter 2.

Statement

For any closed surface \(S\) bounding a volume \(V\),

\[\oint_S \vec E\cdot d\vec A = \frac{Q_{\text{enc}}}{\epsilon_0} = 4\pi k\,Q_{\text{enc}},\]

where \(Q_{\text{enc}} = \int_V \rho\, dV\) is the total charge inside \(S\). Charges outside the surface contribute zero net flux.

Why \(1/r^2\)?

Gauss's law plus symmetry forces the point-charge field to go like \(1/r^2\). For a sphere of radius \(r\) around a point charge, \(\Phi = E\cdot 4\pi r^2 = q/\epsilon_0\) gives \(E = kq/r^2\) immediately. In \(n\) spatial dimensions, the sphere surface scales as \(r^{n-1}\), so the Coulomb law would be \(E\propto 1/r^{n-1}\). We live in 3D, so inverse-square it is.

The Gaussian surface recipe

Gauss's law is always true, but it only gives you \(\vec E\) directly when you can find a surface where \(\vec E\cdot d\vec A\) is either zero or constant on each piece. That requires symmetry. The three textbook cases:

  1. Spherical symmetry → concentric sphere. (shell/ball)
  2. Cylindrical symmetry (infinite line/wire/cylinder) → coaxial cylinder + flat end caps. (infinite line)
  3. Planar symmetry (infinite plane/sheet) → pillbox straddling the plane. (infinite plane)

Workflow

Every Gauss's-law problem follows the same script:

  1. Identify the symmetry. Determine which direction \(\vec E\) must point and what it depends on.
  2. Choose a Gaussian surface that respects the symmetry — pieces where either \(\vec E\perp d\vec A\) (no contribution) or \(\vec E\parallel d\vec A\) with constant magnitude.
  3. Compute \(\oint \vec E\cdot d\vec A\) = \(E\) times area of the "active" portion.
  4. Compute \(Q_{\text{enc}}\) inside your surface.
  5. Solve for \(E\).
Sphere, cylinder, and pillbox Gaussian surfaces.
The three reliable Gaussian surfaces: sphere (spherical symmetry), cylinder (line/cylinder), pillbox (plane).

Practice Problems

Problem 1easy
A charge \(Q = 5\) nC is enclosed by a Gaussian surface of some complicated shape. What is the total flux through the surface?
Hint
Shape doesn't matter — only \(Q_{\text{enc}}\).
Solution

\(\Phi = Q/\epsilon_0 = 5\times 10^{-9}/8.854\times 10^{-12} \approx 565\) V·m.

Answer: \(\approx 565\) N·m\(^2\)/C.

Problem 2easy
Two charges \(+2\)nC and \(-3\)nC sit inside a closed surface, with a \(+5\)nC charge just outside. Find the flux through the surface.
Hint
Outside charges don't contribute.
Solution

\(Q_{\text{enc}} = +2 - 3 = -1\) nC. \(\Phi = -10^{-9}/\epsilon_0 \approx -113\) N·m\(^2\)/C.

Answer: \(\approx -113\) N·m\(^2\)/C.

Problem 3medium
Using Gauss's law, derive the electric field at distance \(r\) from a point charge \(q\).
Hint
Concentric sphere, use symmetry.
Solution

By spherical symmetry \(\vec E = E(r)\hat r\). Flux through a sphere of radius \(r\): \(E\cdot 4\pi r^2 = q/\epsilon_0\). Solve: \(E = q/(4\pi\epsilon_0 r^2) = kq/r^2\).

Answer: \(\vec E = (kq/r^2)\hat r\).

Problem 4medium
A uniform spherical volume charge has density \(\rho\) and radius \(R\). Find \(\vec E\) at an interior point \(r < R\).
Hint
Gaussian sphere of radius \(r\) encloses \(Q_{\text{enc}} = \rho\cdot\tfrac43\pi r^3\).
Solution

\(E\cdot 4\pi r^2 = \rho\cdot \tfrac43\pi r^3/\epsilon_0\), so \(E = \rho r/(3\epsilon_0)\), radially.

Answer: \(\vec E = \dfrac{\rho r}{3\epsilon_0}\hat r\), linear in \(r\) inside.