Playground
Expand or shrink a Gaussian surface around a charged sphere. Flux stays constant while E changes — visualizing Gauss's law.
Variables
| Symbol | Name | SI | Dimension | Range |
|---|---|---|---|---|
| Electric fluxoutput Total electric flux through the closed surface | V·m | M·L³·T⁻³·I⁻¹ | 0 – 1000000 | |
| Enclosed charge Total charge enclosed by the Gaussian surface | C | I·T | -0.00001 – 0.00001 | |
| Vacuum permittivity Permittivity of free space | F/m | I²·T⁴·M⁻¹·L⁻³ | 8.854e-12 – 8.854e-12 |
Deep dive
Derivation
Start with Coulomb's law for a point charge. Integrate E·dA over a sphere of radius r centered on q: Φ = E(4πr²) = (q/4πε₀r²)(4πr²) = q/ε₀. Generalize via superposition: any enclosed charge contributes q/ε₀; external charges contribute zero net flux.
Experimental verification
Faraday's ice-pail experiment (1843) demonstrated that the induced charge on a hollow conductor equals the enclosed charge. Modern verification via high-precision Cavendish experiments confirming the inverse-square law exponent.
Common misconceptions
- Gauss's law is always true but not always useful — it only simplifies calculations when sufficient symmetry exists
- The flux depends only on enclosed charge, not on charges outside the surface
- E on the Gaussian surface may have contributions from external charges, but they cancel in the flux integral
Real-world applications
- Designing Faraday cages for electromagnetic shielding
- Calculating capacitor fields (parallel plate, cylindrical, spherical)
- Electrostatic shielding in sensitive electronics
- Understanding charge distribution on conductors
Worked examples
Field outside a charged sphere
Given:
- Q:
- 0.000002
- R:
- 0.1
- r:
- 0.2
Find: E at r = 0.2 m
Solution
E = Q / (4πε₀r²) = 8.9875×10⁹ × 2×10⁻⁶ / 0.04 = 4.49×10⁵ V/m
Field of an infinite line charge
Given:
- lambda:
- 5e-9
- r:
- 0.02
Find: E at r = 2 cm from the line
Solution
E = λ / (2πε₀r) = 5×10⁻⁹ / (2π × 8.854×10⁻¹² × 0.02) = 4494 V/m
Scenarios
What if…
- scenario:
- What if you double the Gaussian surface radius?
- answer:
- For a sphere, E drops by 4× (inverse-square). For a cylinder, E drops by 2× (inverse-linear). The flux stays the same — only the surface area changes.
- scenario:
- What if there's zero enclosed charge?
- answer:
- Net flux is zero, but E is not necessarily zero — external charges create fields that enter and exit the surface in equal amounts.
- scenario:
- What if the charge distribution lacks symmetry?
- answer:
- Gauss's law is still true (Φ = Q_enc/ε₀) but you can't extract E from the integral. Use Coulomb's law or numerical methods instead.
Limiting cases
- condition:
- Q_enc = 0
- result:
- Φ_E = 0
- explanation:
- No enclosed charge means zero net flux — field lines enter and exit in equal number.
- condition:
- Surface inside a conductor
- result:
- E = 0 everywhere on surface
- explanation:
- Charges reside on the conductor surface; any interior Gaussian surface encloses zero charge.
Context
Carl Friedrich Gauss · 1835
Gauss formulated this as a mathematical theorem relating surface integrals to volume integrals. Its power in electrostatics was recognized when Maxwell incorporated it as his first equation.
Hook
Why is the electric field zero inside a hollow metal sphere, even if it's highly charged?
A uniformly charged sphere of radius 0.1 m carries total charge +2 μC. Find the electric field at r = 0.2 m from the center.
Dimensions: [Φ_E] = [Q]/[ε₀] → C/(F/m) = C·m/F = V·m ✓
Validity: Universally valid in classical electrostatics. Most useful for symmetric charge distributions (spherical, cylindrical, planar). For arbitrary distributions, Coulomb's law or numerical methods are more practical.