Difficulty: Intermediate | Prerequisites: Electric field concepts, Coulomb's law, surface charge density, basic geometry of spheres and cylinders
Big picture: Gauss's law is the first of Maxwell's four equations, and it is the single most powerful tool in this course for finding electric fields when the geometry is sufficiently symmetric. It connects the total electric flux through a closed surface to the charge enclosed inside. This topic also covers how conductors behave in electrostatic equilibrium, including how charge distributes on shells. These ideas tie together the field concepts from earlier and set up the potential and capacitance topics that follow.
Gauss's law says the net electric flux through any closed surface equals the enclosed charge divided by ε₀. You choose a Gaussian surface that exploits symmetry so the field is constant over the surface and the integral becomes simple multiplication. Inside a conductor in equilibrium, the field is zero, and any excess charge sits on the surfaces.
Electric flux (Φ_E)
The "flow" of electric field through a surface, defined as Φ_E = ∮ E · dA. For a uniform field perpendicular to a flat surface, this simplifies to Φ_E = EA. Measured in N·m²/C (or equivalently V·m). In simple terms, flux counts how many field lines pass through a surface.
Gaussian surface
An imaginary closed surface you choose for applying Gauss's law. It is not a physical object. You pick its shape (sphere, cylinder, box) to match the symmetry of the charge distribution so the math simplifies.
Gauss's law
∮ E · dA = Q_enc / ε₀, where Q_enc is the total charge enclosed by the Gaussian surface. The integral is over the entire closed surface. In simple terms, the total flux out of any closed surface depends only on the net charge inside, regardless of where exactly that charge sits or what is happening outside.
Electric field line density
The number of field lines passing through a small area element. In regions where the field is stronger, the lines are more closely spaced. The density of lines through a patch on a Gaussian surface reflects the local field strength and direction at that patch.
Conductor in electrostatic equilibrium
A conductor where all charges have stopped moving. Three consequences follow: (1) the electric field inside the conducting material is zero, (2) any net charge resides on the surface(s), and (3) the field just outside is perpendicular to the surface.
Surface charge density on a conducting shell
When a conducting shell surrounds a charge Q, the inner surface of the shell acquires charge −Q (to make the field zero inside the conductor), and the outer surface acquires whatever is left to conserve total charge.
Consider two charges, q₁ = +3 μC at the origin and q₂ = −6 μC at (4 cm, 0), with a spherical Gaussian surface of radius 2 cm centred on q₂.
Two small patches da₁ and da₂ are on the sphere: da₁ faces toward q₁ (on the side nearer the origin) and da₂ is on the far side.
The field at da₁ is the superposition of q₁'s and q₂'s fields. Near da₁, q₁ is relatively close and its field adds to q₂'s. At da₂, q₁ is farther away and its contribution is weaker.
Because da₁ is closer to the additional charge q₁, the field is stronger there, so the line density through da₁ is greater: n₁ > n₂.
Gauss's law cares only about the enclosed charge.
The spherical Gaussian surface (radius 2 cm, centred on q₂) encloses only q₂ = −6 μC. The charge q₁ is outside (it is 4 cm away from q₂, and the sphere has radius 2 cm).
Net flux = Q_enc / ε₀ = (−6 × 10⁻⁶) / (8.85 × 10⁻¹²) = −6.78 × 10⁵ N·m²/C ≈ −6.8 × 10⁵ N·m²/C.
The negative sign indicates net flux is directed inward (toward the negative enclosed charge).
An insulating sphere of radius a = 6 cm carries total charge Q = +8 μC distributed uniformly throughout its volume.
Field at r = 4 cm (inside the sphere, r < a):
Draw a spherical Gaussian surface of radius r = 4 cm.
The enclosed charge is Q_enc = Q × (r/a)³ = 8 μC × (4/6)³ = 8 μC × (64/216) = 8 × 0.2963 μC ≈ 2.37 μC.
By symmetry, E is radial and constant on the Gaussian surface: E × 4πr² = Q_enc / ε₀.
E = Q_enc / (4πε₀r²) = kQ_enc / r² = (8.99 × 10⁹)(2.37 × 10⁻⁶) / (0.04)² ≈ 1.33 × 10⁷ N/C.
This matches ≈ 1.3 × 10⁷ N/C.
A conducting spherical shell of radius b = 10 cm surrounds the insulating sphere. The shell is initially uncharged.
Inner surface charge: The field inside the conductor must be zero. A Gaussian surface just inside the conducting material (a < r < b) encloses the insulating sphere's charge Q = +8 μC. For E = 0 inside the conductor, the inner surface of the shell must carry charge −Q = −8 μC.
Surface charge density on the inner surface: σ_inner = Q_inner / (4πb²) = (−8 × 10⁻⁶) / (4π(0.10)²) = −8 × 10⁻⁶ / 0.1257 ≈ −6.4 × 10⁻⁵ C/m².
Outer surface charge: Since the shell is uncharged overall, the outer surface carries +8 μC.
An infinite line charge (λ = 1 μC/m, parallel to the z-axis, at x = −30 cm) is added to the sphere-and-shell system.
To find the field at point P = (−20 cm, 0), you superpose the contributions from (1) the line charge and (2) the sphere-and-shell system.
From the line charge: P is 10 cm to the right of the line. E_line = 2kλ/d = 2(8.99 × 10⁹)(1 × 10⁻⁶) / 0.10 = 1.798 × 10⁵ N/C, directed in the +x direction (away from the positive line charge).
From the sphere-and-shell system: P is at distance 20 cm from the origin. This is outside the shell (b = 10 cm), so the system looks like a point charge Q = +8 μC at the origin. E_sphere = kQ/r² = (8.99 × 10⁹)(8 × 10⁻⁶) / (0.20)² = 1.798 × 10⁶ N/C, directed in the −x direction (P is to the left of the origin, and the field from a positive charge at the origin points away from the origin, which is in the −x direction at P's location).
Net field at P: E_total = E_line(+x) + E_sphere(−x) = +1.798 × 10⁵ − 1.798 × 10⁶ ≈ −1.6 × 10⁶ N/C in the x-direction.
Gauss's law:
∮ E · dA = Q_enc / ε₀
Flux through a closed surface (simple case):
Φ_E = Q_enc / ε₀
Field inside a uniformly charged insulating sphere (r < a):
E = (Q / 4πε₀a³) r = kQr / a³
The enclosed charge scales as (r/a)³.
Field outside a spherically symmetric charge (r > a or r > b):
E = kQ / r²
Same as a point charge at the centre.
Field from an infinite line charge at distance d:
E = 2kλ / d = λ / (2πε₀d)
directed radially away from a positive line.
Surface charge density on a sphere of radius R carrying charge Q_surface:
σ = Q_surface / (4πR²)
Gauss's law is why a hollow metal enclosure (a Faraday cage) shields its interior from external electric fields. The charges on the conductor rearrange to cancel the external field inside. This is the principle behind shielded cables, microwave ovens, and the metal body of your car protecting you from lightning.
The insulating sphere result is used in modelling the electric field inside the Earth (treated as a uniformly charged volume) and in understanding how charge distributes in semiconductors and biological cells.
Students often think the field on a Gaussian surface depends on charges outside the surface. Gauss's law says the net flux depends only on enclosed charge. Charges outside can affect the local field at specific points on the surface, but their net contribution to the total flux is zero.
When applying Gauss's law inside a uniformly charged sphere, students sometimes use the total charge Q rather than the enclosed charge Q(r/a)³. Only the charge inside your Gaussian surface contributes to the flux through it.
Students frequently forget that a conducting shell's inner surface charge is determined by the enclosed charge alone, regardless of the shell's own total charge. If the shell is uncharged, its inner and outer surface charges sum to zero.
For the superposition problem with the line charge and sphere, students sometimes forget that outside the conducting shell, the sphere-and-shell system behaves as a point charge at the centre. The shell does not "block" the field; it merely redistributes charge on its surfaces.
⚠️ The net flux question is a quick-points question if you remember: Φ = Q_enc / ε₀. Nothing outside the Gaussian surface matters for total flux. Do not waste time computing the field at every point on the surface.
⚠️ The field line density comparison (n₁ vs n₂) tests qualitative understanding. The field on a Gaussian surface is not uniform when the charge distribution inside is off-centre or when outside charges are present.
⚠️ Inside a uniformly charged insulating sphere, E grows linearly with r. Outside, it falls as 1/r². The transition at r = a is a common exam sketch question.
⚠️ The conducting shell's inner surface charge is always −Q_enc, regardless of the shell's own charge or anything outside. This is tested often.
⚠️ Superposition with mixed geometries (line charge + sphere) is a favourite exam problem. Treat each source independently, find its contribution at the point of interest, then add as vectors.
True or false: The net electric flux through a closed surface depends on charges both inside and outside the surface.
Fill in the blank: Inside a conductor in electrostatic equilibrium, the electric field is ________.
True or false: The electric field inside a uniformly charged insulating sphere increases linearly with distance from the centre.
Fill in the blank: A conducting shell that encloses a charge +Q will have a charge of ________ on its inner surface.
True or false: The field from an infinite line charge falls off as 1/r² with distance.
Q: A spherical Gaussian surface of radius 2 cm is centred on a charge q₂ = −6 μC. A second charge q₁ = +3 μC is located 4 cm away, outside the surface. What is the net flux through the surface?
A: Φ = Q_enc / ε₀ = (−6 × 10⁻⁶) / (8.85 × 10⁻¹²) ≈ −6.8 × 10⁵ N·m²/C.
Q: For the same configuration, is the electric field line density the same at all points on the Gaussian surface?
A: No. The nearby charge q₁ distorts the field, making it stronger on the side of the sphere closest to q₁. The line density through da₁ (nearer q₁) is greater than through da₂ (farther from q₁).
Q: An insulating sphere of radius 6 cm with total charge +8 μC. What is the magnitude of the electric field at r = 4 cm from the centre?
A: E = kQ_enc / r², where Q_enc = 8 μC × (4/6)³ ≈ 2.37 μC. So E ≈ (8.99 × 10⁹)(2.37 × 10⁻⁶) / (0.04)² ≈ 1.3 × 10⁷ N/C.
Q: What is the surface charge density on the inner surface of an uncharged conducting shell (radius 10 cm) surrounding the insulating sphere above?
A: The inner surface carries −8 μC. σ = −8 × 10⁻⁶ / (4π × 0.10²) ≈ −6.4 × 10⁻⁵ C/m².
Q: An infinite line charge (λ = +1 μC/m) at x = −30 cm and the sphere/shell system (net +8 μC centred at origin) are both present. What is the electric field at x = −20 cm?
A: Superpose: the line charge gives ≈ +1.8 × 10⁵ N/C (in +x), and the sphere/shell gives ≈ −1.8 × 10⁶ N/C (in −x, since P is to the left of the positive charge). Net ≈ −1.6 × 10⁶ N/C in the x-direction.
Gauss's law is the foundation for deriving the field in a parallel plate capacitor (the capacitors topic later in this midterm). The conducting shell result connects to shielding and to the idea that the potential is constant throughout a conductor in equilibrium, which is central to understanding circuits. The superposition approach here (line charge + sphere) previews how you will handle more complex configurations in later exams.
Gauss's law, electric flux, Gaussian surface, enclosed charge, ε₀, permittivity, conductor, electrostatic equilibrium, conducting shell, inner surface charge, insulating sphere, volume charge density, line charge, superposition, Faraday cage, shielding, PHYS 212, Physics 212 midterm, flux integral, field line density