Quick answer: Electric flux is Φ = ∮E · dA, the field passing through a surface. Also, Gauss’s law states that the net flux through a closed surface equals the enclosed charge over ε₀. Thus, with enough symmetry, it gives the field of spheres, lines and sheets quickly.
Level: AP Physics C (Mechanics and E&M, calculus-based) · Curriculum: AP PC:E&M Unit 8
What is electric flux?
Electric flux measures how much electric field passes through a surface. For example, picture field lines as streams of water and a surface as a net held in the flow. Also, the more lines that cross the net, the larger the flux, and a net held edge-on catches none.
The electric flux through a surface is Φ = ∫ E · dA, in N m²/C. For a uniform field through a flat area A, Φ = E A cosθ, where θ is the angle between E and the normal to the surface. On a closed surface, the normal points outward.
What does Gauss’s law say?
Gauss’s law: the net flux through any closed surface equals the enclosed charge divided by ε₀. ∮ E · dA = Q_enc / ε₀, where ε₀ = 8.85 × 10⁻¹² C²/(N m²) and k = 1/(4πε₀) = 8.99 × 10⁹ N m²/C².
In addition, the law holds for every closed surface and every charge arrangement. Charges outside the surface contribute to E on the surface, but their field lines enter and leave, so they add nothing to the net flux. Gauss’s law is most useful when symmetry lets E be constant over the surface and perpendicular to it, so that the integral becomes E times an area.
How do you find the field of a continuous charge distribution?
Two methods exist. In addition, the direct method divides the charge into pieces dq, each giving dE = k dq / r², and integrates the components. For example, for a ring of charge Q and radius R, the perpendicular parts cancel on the axis, leaving E = kQx / (x² + R²)^(3/2). Moreover, the second method uses Gauss’s law and symmetry.
Next, for a long line of charge, choose a coaxial cylinder of radius r and length L. Therefore, flux leaves only through the curved side: E (2πrL) = λL/ε₀, which gives E = λ / (2πε₀r).
| Charge distribution | Gaussian surface | Result for E |
|---|---|---|
| Point charge or sphere (shell or solid) | Concentric sphere, area 4πr² | E = kQ_enc/r² |
| Long line of charge λ | Coaxial cylinder, side area 2πrL | E = λ/(2πε₀r) |
| Large flat sheet σ | Pillbox through the sheet, 2 end faces | E = σ/(2ε₀) |
| Surface of a conductor | Small pillbox, one face outside | E = σ/ε₀ |
A long wire carries a uniform charge of λ = 2.0 μC/m. At r = 0.50 m the field is E = λ / (2πε₀r) = 2.0 × 10⁻⁶ / (2π × 8.85 × 10⁻¹² × 0.50) = 7.2 × 10⁴ N/C, directed radially away from the wire. Check: 2kλ/r = 2 × 8.99 × 10⁹ × 2.0 × 10⁻⁶ / 0.50 = 7.2 × 10⁴ N/C.
What is the field of a charged sphere?
Gauss’s law notes: field inside and outside a charged sphere
For a uniformly charged solid insulating sphere of radius R, use a concentric spherical surface of radius r. Outside (r ≥ R), the surface encloses all of Q, so E = kQ/r². Inside the sphere (r less than R) the enclosed charge is Q (r/R)³, so E (4πr²) = Q r³/(R³ε₀), giving E = kQ r / R³. Similarly, the field peaks at the surface.
A solid insulating sphere has Q = +6.0 nC and R = 0.10 m. First, at r = 0.05 m, E = kQr/R³ = 8.99 × 10⁹ × 6.0 × 10⁻⁹ × 0.05 / (0.10)³ = 2.7 × 10³ N/C. at r = 0.20 m, E = kQ/r² = 8.99 × 10⁹ × 6.0 × 10⁻⁹ / 0.040 = 1.3 × 10³ N/C. Finally, at the surface the field is 5.4 × 10³ N/C.
Gauss’s law notes: conductors in electrostatic equilibrium
However, conductors behave differently. In electrostatic equilibrium the field inside a conductor is zero, all net charge sits on the outer surface, and the field just outside is σ/ε₀ perpendicular to the surface.
Also, do not say that the field at a point on a Gaussian surface comes only from the enclosed charge. The flux depends only on the enclosed charge, but all charges, inside and outside, produce the field E at each point.
Before using Gauss’s law, ask: can I draw a surface where E has the same size everywhere on it and is perpendicular or parallel to it? If not, the law is true but will not help, and you should integrate dE directly.
- Flux: Φ = ∮E · dA = E A cosθ for a uniform field (N m²/C).
- Gauss’s law: ∮E · dA = Q_enc / ε₀, with ε₀ = 8.85 × 10⁻¹² C²/(N m²).
- Line of charge: E = λ/(2πε₀r); infinite sheet: E = σ/(2ε₀); conductor surface: E = σ/ε₀.
- Solid insulating sphere: E = kQr/R³ inside, E = kQ/r² outside.
- Inside a conductor in equilibrium E = 0 and the net charge is on its surface.
Flux, Gauss's law and continuous charge: frequently asked questions
What are the units of electric flux?
We measure electric flux in N m²/C, which is the same as V m. Also, the unit comes from multiplying a field in N/C by an area in m². Flux through a closed surface equals Q/ε₀ and so tells you the enclosed charge.
Why is the flux zero for a charge outside the surface?
Every field line from an outside charge that enters the closed surface must leave it again. The inward flux, which is negative, equals the outward flux, so the net flux is zero. In addition, the charge still affects E at points on the surface.
When can you use Gauss’s law to find E?
You can use it when symmetry makes E constant in size and either perpendicular or parallel to each part of a chosen surface. For example, spherical, cylindrical and planar symmetry satisfy this. However, without symmetry, integrate dE = k dq/r² directly.
Why is the field inside a conductor zero?
Free charges move until the net force on each is zero. Moreover, if E were non-zero inside, charges would keep moving. In equilibrium the field vanishes in the bulk, so any extra charge sits on the surface.
How does the field of a sheet differ from that of a point charge?
A large uniform sheet gives E = σ/(2ε₀), which does not depend on distance. Similarly, a point charge’s field falls as 1/r². The sheet field is constant because the area of field lines is constant, unlike the spreading area of a sphere.
