A +5 μC point charge sits at the center of a spherical Gaussian surface of radius 2 m. What is the total electric flux?

Φ = Q/ε₀ = 5×10⁻⁶ / 8.854×10⁻¹² ≈ 5.65×10⁵ N·m²/C.

If you double the radius of the Gaussian surface (keeping the charge inside), how does the flux change?

The flux stays the same — Gauss's Law says Φ depends only on Q_enc, not the surface size.

A charge of -3 μC is outside a Gaussian surface. What is the net flux through the surface due to this charge?

Zero — charges outside the surface contribute zero net flux (field lines enter and exit).

Pro 🔒~30 min

Gauss's Law

Electric flux through Gaussian surfaces and enclosed charge relationships

How it works

Gauss's Law is one of Maxwell's four equations and relates the electric flux through a closed surface to the charge enclosed within it. The law states Φ_E = Q_enc/ε₀, where ε₀ = 8.854×10⁻¹² C²/(N·m²). For highly symmetric charge distributions (spherical, cylindrical, planar), Gauss's Law provides an elegant shortcut to calculate the electric field without integration. For a point charge Q, the field at distance r is E = kQ/r² (radial). A spherical Gaussian surface of radius r centered on the charge captures flux Φ = E·4πr² = Q/ε₀, independent of r — demonstrating that flux depends only on enclosed charge, not surface size. For an infinite line charge with linear density λ, a cylindrical Gaussian surface gives E = λ/(2πε₀r). For an infinite plane with surface density σ, E = σ/(2ε₀), uniform and independent of distance.

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Step-by-step

Select a charge distribution type and set the charge magnitude.
Adjust the Gaussian surface radius to enclose or exclude charges.
Observe the 3D electric field vectors (length proportional to field strength, color-coded by magnitude).
The flux panel shows the calculated Φ and verifies Gauss's Law.
Rotate the 3D view with mouse drag, zoom with scroll.

Key formulas

$\Phi_E = \oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enc}}}{\varepsilon_0}$ Gauss's Law: the total electric flux through a closed surface equals the enclosed charge divided by the permittivity of free space
$\vec{E} = \frac{1}{4\pi\varepsilon_0} \frac{Q}{r^2} \hat{r}$ Coulomb's field for a point charge: inversely proportional to r², radially outward for positive Q
$\Phi_E = E \cdot A = E \cdot 4\pi r^2$ For a spherical Gaussian surface with uniform radial field, flux equals E times the surface area

Frequently asked questions

A +5 μC point charge sits at the center of a spherical Gaussian surface of radius 2 m. What is the total electric flux?: Φ = Q/ε₀ = 5×10⁻⁶ / 8.854×10⁻¹² ≈ 5.65×10⁵ N·m²/C.
If you double the radius of the Gaussian surface (keeping the charge inside), how does the flux change?: The flux stays the same — Gauss's Law says Φ depends only on Q_enc, not the surface size.
A charge of -3 μC is outside a Gaussian surface. What is the net flux through the surface due to this charge?: Zero — charges outside the surface contribute zero net flux (field lines enter and exit).