Formula & Calculator

Gauss's Law

Relates the net electric flux through a closed surface to the enclosed charge.

ElectromagneticsElectrostatics

Gauss's Law Calculator ∮E·dA = Qenc / ε₀

E · A = Q / ε₀
E = electric field (N/C)  ·  A = area (m²)  ·  Q = enclosed charge (C)  ·  ε₀ = permittivity (F/m)
⟹ Solve E, A, Q, ε₀
N/C
C
F/m
Please fix the errors above.
Solve for:
Presets:
Electric Field
E: A: Q: ε₀:
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E · A = Q / ε₀  ·  ε₀ ≈ 8.854 × 10⁻¹² F/m  ·  Valid for uniform field perpendicular to area.

Interpretation

Gauss's law: the total electric flux through a closed surface equals the enclosed charge divided by ε₀.
It is one of Maxwell's fundamental equations and relates charge to electric field.
Example: For a spherical surface containing charge Q, the net flux is Q/ε₀.

∮E·dA = Q_enc / ε₀
Gauss's Law

Variables

SymbolQuantityUnit
Φ_EElectric fluxN·m²/C
EElectric fieldN/C
dAInfinitesimal area vector
Q_encEnclosed chargeC
ε₀Vacuum permittivity (8.854×10⁻¹²)F/m

What it means

Gauss’s law states that the total electric flux through any closed surface is equal to the net charge enclosed divided by the permittivity of free space (ε₀). Mathematically, ∮ E·dA = Q_enc / ε₀. It is one of Maxwell’s four equations and is fundamental to electrostatics. The law is used to calculate electric fields for symmetric charge distributions, such as spherical, cylindrical, or planar symmetry. For example, the field of a point charge and the field inside a charged sphere can be derived using Gauss’s law. In electrical engineering, it is used in the design of capacitors and the analysis of conductors. The law also shows that the electric field inside a perfect conductor is zero. Example: For a sphere of radius R with total charge Q uniformly distributed, the electric field at distance r > R is E = Q/(4πε₀r²), which is the same as a point charge. At r < R, E = (Q r) / (4πε₀R³).

Worked example

Gauss's Law – Practical Example

Real‑World
Scenario: A point charge of 5 µC is placed at the centre of a spherical Gaussian surface of radius 0.1 m. Find the total electric flux through the surface.
ParameterValue
Qenc5 µC = 5×10⁻⁶ C
ε₀8.85×10⁻¹² F/m
FormulaΦ = Qenc / ε₀
1Substitute: Φ = (5×10⁻⁶) / (8.85×10⁻¹²)
2Calculate: Φ ≈ 5.65×10⁵ N·m²/C
Final Design Φ ≈ 5.65×10⁵ N·m²/C ✓ Total flux
Why: Gauss's law relates enclosed charge to electric flux – independent of the surface shape.

Common mistakes

  • Closed surface: The integral is over a closed surface – the flux is proportional to enclosed charge.
  • Electric field E: Must be normal to the surface for the dot product to simplify.
  • Permittivity ε₀: For media, replace with ε (absolute permittivity).
  • Symmetry: Gauss’s law is most useful when symmetry allows easy evaluation.
  • Sign: The flux is outward if charge positive; inward if negative.

Applications

Gauss's law relates the electric flux through a closed surface to the enclosed charge, forming one of Maxwell's equations. It is used to calculate electric fields for symmetric charge distributions, such as spheres, cylinders, and planes. Engineers apply it in the design of capacitors, insulation systems, and high‑voltage devices. It also underlies the concept of electric displacement in dielectric materials. By using Gauss's law, professionals can determine the field strength and potential distribution in complex geometries. This law is fundamental to electromagnetism and is essential for understanding the behaviour of electric fields.

  • Calculation of electric fields in capacitors and insulators
  • Design of high‑voltage equipment (bushings, cables)
  • Modelling of charge distributions in semiconductors
  • Understanding of electrostatic shielding and Faraday cages
  • Educational foundation of electromagnetic theory

Frequently Asked Questions

Q01What is Gauss's law and what does it state?
A01

Gauss's law states that the net electric flux through any closed surface is equal to the enclosed charge divided by ε₀: ∮E·dA = Q_enc / ε₀.

Q02What is electric flux?
A02

Electric flux is the product of the electric field and the area perpendicular to it, integrated over a surface.

Q03What is the significance of choosing a Gaussian surface?
A03

A Gaussian surface is chosen to exploit symmetry (spherical, cylindrical, planar) to simplify the integral.

Q04How is Gauss's law used to find electric fields?
A04

By choosing a symmetric Gaussian surface, the integral ∮E·dA simplifies to E × Area, allowing E to be determined.

Q05What is the electric field inside a conductor?
A05

Zero in electrostatic equilibrium because charges redistribute to cancel any internal field.

Q06How does Gauss's law apply to a point charge?
A06

For a spherical Gaussian surface centered on the charge, E = Q/(4πε₀r²), which matches Coulomb's law.

Q07What is the difference between Gauss's law and Coulomb's law?
A07

Gauss's law is a more general integral form; Coulomb's law is a special case for point charges.

Q08How is Gauss's law modified in dielectric materials?
A08

Replace ε₀ with ε = ε₀ ε_r, and the enclosed free charge is used; bound charges are handled by the permittivity.

Q09What are the applications of Gauss's law?
A09

Calculating electric fields of symmetric charge distributions, capacitance calculations, and electrostatics in materials.

Q10What are the common mistakes when using Gauss's law?
A10

Common errors include: 1) choosing the wrong Gaussian surface, 2) not considering symmetry, 3) forgetting the enclosed charge, 4) using the wrong permittivity, and 5) applying to non-static fields without modification.